基础算法
快速排序
#include <iostream>
using namespace std;
const int N = 100010;
int q[N];
void quick_sort(int q[], int l, int r){
if (l >= r) return;
int i = l - 1, j = r + 1, x = q[l + r >> 1];
while (i < j){
do i ++ ; while (q[i] < x);
do j -- ; while (q[j] > x);
if (i < j) swap(q[i], q[j]);
}
quick_sort(q, l, j);
quick_sort(q, j + 1, r);
}
int main(){
int n;
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);
quick_sort(q, 0, n - 1);
for (int i = 0; i < n; i ++ ) printf("%d ", q[i]);
return 0;
}
第 k k k个数
#include <iostream>
using namespace std;
const int N = 100010;
int q[N];
int quick_sort(int q[], int l, int r, int k){
if (l >= r) return q[l];
int i = l - 1, j = r + 1, x = q[l + r >> 1];
while (i < j){
do i ++ ; while (q[i] < x);
do j -- ; while (q[j] > x);
if (i < j) swap(q[i], q[j]);
}
if (j - l + 1 >= k) return quick_sort(q, l, j, k);
else return quick_sort(q, j + 1, r, k - (j - l + 1));
}
int main(){
int n, k;
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);
cout << quick_sort(q, 0, n - 1, k) << endl;
return 0;
}
归并排序
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int a[N], tmp[N];
void merge_sort(int q[], int l, int r){
if (l >= r) return;
int mid = l + r >> 1;
merge_sort(q, l, mid), merge_sort(q, mid + 1, r);
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r)
if (q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
else tmp[k ++ ] = q[j ++ ];
while (i <= mid) tmp[k ++ ] = q[i ++ ];
while (j <= r) tmp[k ++ ] = q[j ++ ];
for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
}
int main(){
int n;
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
merge_sort(a, 0, n - 1);
for (int i = 0; i < n; i ++ ) printf("%d ", a[i]);
return 0;
}
逆序对数
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int a[N], tmp[N];
LL merge_sort(int q[], int l, int r){
if (l >= r) return 0;
int mid = l + r >> 1;
LL res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r)
if (q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
else{
res += mid - i + 1;
tmp[k ++ ] = q[j ++ ];
}
while (i <= mid) tmp[k ++ ] = q[i ++ ];
while (j <= r) tmp[k ++ ] = q[j ++ ];
for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
return res;
}
int main(){
int n;
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
cout << merge_sort(a, 0, n - 1) << endl;
return 0;
}
数的范围
#include <iostream>
using namespace std;
const int N = 100010;
int n, m;
int q[N];
int main(){
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);
while (m -- ){
int x;
scanf("%d", &x);
int l = 0, r = n - 1;
while (l < r){
int mid = l + r >> 1;
if (q[mid] >= x) r = mid;
else l = mid + 1;
}
if (q[l] != x) cout << "-1 -1" << endl;
else{
cout << l << ' ';
int l = 0, r = n - 1;
while (l < r){
int mid = l + r + 1 >> 1;
if (q[mid] <= x) l = mid;
else r = mid - 1;
}
cout << l << endl;
}
}
return 0;
}
三次方根
#include <iostream>
using namespace std;
int main(){
double x;
cin >> x;
double l = -100, r = 100;
while (r - l > 1e-8){
double mid = (l + r) / 2;
if (mid * mid * mid >= x) r = mid;
else l = mid;
}
printf("%.6lf\n", l);
return 0;
}
高精加
#include <iostream>
#include <vector>
using namespace std;
vector<int> add(vector<int> &A, vector<int> &B){
if (A.size() < B.size()) return add(B, A);
vector<int> C;
int t = 0;
for (int i = 0; i < A.size(); i ++ ){
t += A[i];
if (i < B.size()) t += B[i];
C.push_back(t % 10);
t /= 10;
}
if (t) C.push_back(t);
return C;
}
int main(){
string a, b;
vector<int> A, B;
cin >> a >> b;
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
auto C = add(A, B);
for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
cout << endl;
return 0;
}
高精减
#include <iostream>
#include <vector>
using namespace std;
bool cmp(vector<int> &A, vector<int> &B){
if (A.size() != B.size()) return A.size() > B.size();
for (int i = A.size() - 1; i >= 0; i -- )
if (A[i] != B[i])
return A[i] > B[i];
return true;
}
vector<int> sub(vector<int> &A, vector<int> &B){
vector<int> C;
for (int i = 0, t = 0; i < A.size(); i ++ ){
t = A[i] - t;
if (i < B.size()) t -= B[i];
C.push_back((t + 10) % 10);
if (t < 0) t = 1;
else t = 0;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main(){
string a, b;
vector<int> A, B;
cin >> a >> b;
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
vector<int> C;
if (cmp(A, B)) C = sub(A, B);
else C = sub(B, A), cout << '-';
for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
cout << endl;
return 0;
}
高精乘
#include <iostream>
#include <vector>
using namespace std;
vector<int> mul(vector<int> &A, int b){
vector<int> C;
int t = 0;
for (int i = 0; i < A.size() || t; i ++ ){
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main(){
string a;
int b;
cin >> a >> b;
vector<int> A;
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
auto C = mul(A, b);
for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);
return 0;
}
高精除
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> div(vector<int> &A, int b, int &r){
vector<int> C;
r = 0;
for (int i = A.size() - 1; i >= 0; i -- ){
r = r * 10 + A[i];
C.push_back(r / b);
r %= b;
}
reverse(C.begin(), C.end());
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main(){
string a;
vector<int> A;
int B;
cin >> a >> B;
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
int r;
auto C = div(A, B, r);
for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
cout << endl << r << endl;
return 0;
}
前缀和
#include <iostream>
using namespace std;
const int N = 100010;
int n, m;
int a[N], s[N];
int main(){
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
for (int i = 1; i <= n; i ++ ) s[i] = s[i - 1] + a[i]; // 前缀和的初始化
while (m -- ){
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", s[r] - s[l - 1]); // 区间和的计算
}
return 0;
}
矩阵前缀和
#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int s[N][N];
int main(){
scanf("%d%d%d", &n, &m, &q);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
scanf("%d", &s[i][j]);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
while (q -- ){
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
}
return 0;
}
差分
#include <iostream>
using namespace std;
const int N = 100010;
int n, m;
int a[N], b[N];
void insert(int l, int r, int c){
b[l] += c;
b[r + 1] -= c;
}
int main(){
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
for (int i = 1; i <= n; i ++ ) insert(i, i, a[i]);
while (m -- ){
int l, r, c;
scanf("%d%d%d", &l, &r, &c);
insert(l, r, c);
}
for (int i = 1; i <= n; i ++ ) b[i] += b[i - 1];
for (int i = 1; i <= n; i ++ ) printf("%d ", b[i]);
return 0;
}
差分矩阵
#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int a[N][N], b[N][N];
void insert(int x1, int y1, int x2, int y2, int c){
b[x1][y1] += c;
b[x2 + 1][y1] -= c;
b[x1][y2 + 1] -= c;
b[x2 + 1][y2 + 1] += c;
}
int main(){
scanf("%d%d%d", &n, &m, &q);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
scanf("%d", &a[i][j]);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
insert(i, j, i, j, a[i][j]);
while (q -- ){
int x1, y1, x2, y2, c;
cin >> x1 >> y1 >> x2 >> y2 >> c;
insert(x1, y1, x2, y2, c);
}
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];
for (int i = 1; i <= n; i ++ ){
for (int j = 1; j <= m; j ++ ) printf("%d ", b[i][j]);
puts("");
}
return 0;
}
最长连续不重复子序列
#include <iostream>
using namespace std;
const int N = 100010;
int n;
int q[N], s[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);
int res = 0;
for (int i = 0, j = 0; i < n; i ++ ){
s[q[i]] ++ ;
while (j < i && s[q[i]] > 1) s[q[j ++ ]] -- ;
res = max(res, i - j + 1);
}
cout << res << endl;
return 0;
}
数组元素的目标和
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int n, m, x;
int a[N], b[N];
int main(){
scanf("%d%d%d", &n, &m, &x);
for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
for (int i = 0; i < m; i ++ ) scanf("%d", &b[i]);
for (int i = 0, j = m - 1; i < n; i ++ ){
while (j >= 0 && a[i] + b[j] > x) j -- ;
if (j >= 0 && a[i] + b[j] == x) cout << i << ' ' << j << endl;
}
return 0;
}
二进制中1的个数
#include <iostream>
using namespace std;
int main(){
int n;
scanf("%d", &n);
while (n -- ){
int x, s = 0;
scanf("%d", &x);
for (int i = x; i; i -= i & -i) s ++ ;
printf("%d ", s);
}
return 0;
}
区间和离散化
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 300010;
int n, m;
int a[N], s[N];
vector<int> alls;
vector<PII> add, query;
int find(int x){
int l = 0, r = alls.size() - 1;
while (l < r){
int mid = l + r >> 1;
if (alls[mid] >= x) r = mid;
else l = mid + 1;
}
return r + 1;
}
int main(){
cin >> n >> m;
for (int i = 0; i < n; i ++ ){
int x, c;
cin >> x >> c;
add.push_back({x, c});
alls.push_back(x);
}
for (int i = 0; i < m; i ++ ){
int l, r;
cin >> l >> r;
query.push_back({l, r});
alls.push_back(l);
alls.push_back(r);
}
// 去重
sort(alls.begin(), alls.end());
alls.erase(unique(alls.begin(), alls.end()), alls.end());
// 处理插入
for (auto item : add){
int x = find(item.first);
a[x] += item.second;
}
// 预处理前缀和
for (int i = 1; i <= alls.size(); i ++ ) s[i] = s[i - 1] + a[i];
// 处理询问
for (auto item : query){
int l = find(item.first), r = find(item.second);
cout << s[r] - s[l - 1] << endl;
}
return 0;
}
区间合并
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
void merge(vector<PII> &segs){
vector<PII> res;
sort(segs.begin(), segs.end());
int st = -2e9, ed = -2e9;
for (auto seg : segs)
if (ed < seg.first){
if (st != -2e9) res.push_back({st, ed});
st = seg.first, ed = seg.second;
}
else ed = max(ed, seg.second);
if (st != -2e9) res.push_back({st, ed});
segs = res;
}
int main(){
int n;
scanf("%d", &n);
vector<PII> segs;
for (int i = 0; i < n; i ++ ){
int l, r;
scanf("%d%d", &l, &r);
segs.push_back({l, r});
}
merge(segs);
cout << segs.size() << endl;
return 0;
}
数据结构
单调栈
给定一个长度为 N N N的整数数列，输出每个数左边第一个比它小的数，如果不存在则输出 −1。
#include<iostream>
#include<stack>
using namespace std;
stack<int> stk;
const int N = 100005;
int arr[N];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++)
cin >> arr[i];
for (int i = 0; i < n; i++) {
while (!stk.empty()&&arr[i] <= stk.top())
stk.pop();
if (stk.empty())
cout << -1 << " ";
else cout << stk.top() << " ";
stk.push(arr[i]);
}
return 0;
}
滑动窗口
你的任务是确定滑动窗口位于每个位置时，窗口中的最大值和最小值。
#include<iostream>
#include<deque>
using namespace std;
deque<int> dq;
const int N = 1000005;
int arr[N];
int main() {
int n, k;
cin >> n >> k;
for (int i = 0; i < n; i++)
cin >> arr[i];
for (int i = 0; i < n; i++) {
if (!dq.empty() && i - k + 1 > dq.front())
dq.pop_front();
while (!dq.empty() && arr[dq.back()] >= arr[i])
dq.pop_back();
dq.push_back(i);
if (i >= k - 1)
cout << arr[dq.front()]<<" ";
}
dq.clear();
cout << endl;
for (int i = 0; i < n; i++) {
if (!dq.empty() && i - k + 1 > dq.front())
dq.pop_front();
while (!dq.empty() && arr[dq.back()] <= arr[i])
dq.pop_back();
dq.push_back(i);
if (i >= k - 1)
cout << arr[dq.front()] << " ";
}
return 0;
}
KMP
#include <iostream>
using namespace std;
const int N = 100010, M = 1000010;
int n, m;
int ne[N];
char s[M], p[N];
int main(){
cin >> n >> p + 1 >> m >> s + 1;
for (int i = 2, j = 0; i <= n; i ++ ){
while (j && p[i] != p[j + 1]) j = ne[j];
if (p[i] == p[j + 1]) j ++ ;
ne[i] = j;
}
for (int i = 1, j = 0; i <= m; i ++ ){
while (j && s[i] != p[j + 1]) j = ne[j];
if (s[i] == p[j + 1]) j ++ ;
if (j == n){
printf("%d ", i - n);
j = ne[j];
}
}
return 0;
}
/*下标从0开始
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1000010;
int n, m;
char s[N], p[N];
int ne[N];
int main(){
cin >> m >> p >> n >> s;
ne[0] = -1;
for (int i = 1, j = -1; i < m; i ++ ){
while (j >= 0 && p[j + 1] != p[i]) j = ne[j];
if (p[j + 1] == p[i]) j ++ ;
ne[i] = j;
}
for (int i = 0, j = -1; i < n; i ++ ){
while (j != -1 && s[i] != p[j + 1]) j = ne[j];
if (s[i] == p[j + 1]) j ++ ;
if (j == m - 1){
cout << i - j << ' ';
j = ne[j];
}
}
return 0;
}
Trie
#include <iostream>
using namespace std;
const int N = 100010;
int son[N][26], cnt[N], idx;
char str[N];
void insert(char *str){
int p = 0;
for (int i = 0; str[i]; i ++ ){
int u = str[i] - 'a';
if (!son[p][u]) son[p][u] = ++ idx;
p = son[p][u];
}
cnt[p] ++ ;
}
int query(char *str){
int p = 0;
for (int i = 0; str[i]; i ++ ){
int u = str[i] - 'a';
if (!son[p][u]) return 0;
p = son[p][u];
}
return cnt[p];
}
int main(){
int n;
scanf("%d", &n);
while (n -- ){
char op[2];
scanf("%s%s", op, str);
if (*op == 'I') insert(str);
else printf("%d\n", query(str));
}
return 0;
}
最大异或对
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010, M = 3100010;
int n;
int a[N], son[M][2], idx;
void insert(int x){
int p = 0;
for (int i = 30; i >= 0; i -- ){
int &s = son[p][x >> i & 1];
if (!s) s = ++ idx;
p = s;
}
}
int search(int x){
int p = 0, res = 0;
for (int i = 30; i >= 0; i -- ){
int s = x >> i & 1;
if (son[p][!s]){
res += 1 << i;
p = son[p][!s];
}
else p = son[p][s];
}
return res;
}
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ){
scanf("%d", &a[i]);
insert(a[i]);
}
int res = 0;
for (int i = 0; i < n; i ++ ) res = max(res, search(a[i]));
printf("%d\n", res);
return 0;
}
并查集
#include <iostream>
using namespace std;
const int N = 100010;
int p[N];
int find(int x){
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
int main(){
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++ ) p[i] = i;
while (m -- ){
char op[2];
int a, b;
scanf("%s%d%d", op, &a, &b);
if (*op == 'M') p[find(a)] = find(b);
else{
if (find(a) == find(b)) puts("Yes");
else puts("No");
}
}
return 0;
}
连通块中点的数量
#include <iostream>
using namespace std;
const int N = 100010;
int n, m;
int p[N], cnt[N];
int find(int x){
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
int main(){
cin >> n >> m;
for (int i = 1; i <= n; i ++ ){
p[i] = i;
cnt[i] = 1;
}
while (m -- ){
string op;
int a, b;
cin >> op;
if (op == "C"){
cin >> a >> b;
a = find(a), b = find(b);
if (a != b){
p[a] = b;
cnt[b] += cnt[a];
}
}
else if (op == "Q1"){
cin >> a >> b;
if (find(a) == find(b)) puts("Yes");
else puts("No");
}
else{
cin >> a;
cout << cnt[find(a)] << endl;
}
}
return 0;
}
堆排序
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int n, m;
int h[N], cnt;
void down(int u){
int t = u;
if (u * 2 <= cnt && h[u * 2] < h[t]) t = u * 2;
if (u * 2 + 1 <= cnt && h[u * 2 + 1] < h[t]) t = u * 2 + 1;
if (u != t){
swap(h[u], h[t]);
down(t);
}
}
int main(){
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++ ) scanf("%d", &h[i]);
cnt = n;
for (int i = n / 2; i; i -- ) down(i);
while (m -- ){
printf("%d ", h[1]);
h[1] = h[cnt -- ];
down(1);
}
puts("");
return 0;
}
模拟堆
维护一个集合，初始时集合为空，支持如下几种操作：
I x，插入一个数 x x x；
PM，输出当前集合中的最小值；
DM，删除当前集合中的最小值（数据保证此时的最小值唯一）；
D k，删除第 k k k 个插入的数；
C k x，修改第 k k k 个插入的数，将其变为 x x x；
现在要进行 N 次操作，对于所有第 2 个操作，输出当前集合的最小值。
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e5+10;
int h[N]; //堆
int ph[N]; //存放第k个插入点的下标
int hp[N]; //存放堆中点的插入次序
int cur_size; //size 记录的是堆当前的数据多少
//这个交换过程其实有那么些绕 但关键是理解 如果hp[u]=k 则ph[k]=u 的映射关系
//之所以要进行这样的操作是因为 经过一系列操作 堆中的元素并不会保持原有的插入顺序
//从而我们需要对应到原先第K个堆中元素
//如果理解这个原理 那么就能明白其实三步交换的顺序是可以互换
//h,hp,ph之间两两存在映射关系 所以交换顺序的不同对结果并不会产生影响
void heap_swap(int u,int v){
swap(h[u],h[v]);
swap(hp[u],hp[v]);
swap(ph[hp[u]],ph[hp[v]]);
}
void down(int u){
int t=u;
if(u*2<=cur_size&&h[t]>h[u*2]) t=u*2;
if(u*2+1<=cur_size&&h[t]>h[u*2+1]) t=u*2+1;
if(u!=t){
heap_swap(u,t);
down(t);
}
}
void up(int u){
if(u/2>0&&h[u]<h[u/2]) {
heap_swap(u,u/2);
up(u>>1);
}
}
int main(){
int n;
cin>>n;
int m=0; //m用来记录插入的数的个数
//注意m的意义与cur_size是不同的 cur_size是记录堆中当前数据的多少
//对应上文 m即是hp中应该存的值
while(n--){
string op;
int k,x;
cin>>op;
if(op=="I"){
cin>>x;
m++;
h[++cur_size]=x;
ph[m]=cur_size;
hp[cur_size]=m;
//down(size);
up(cur_size);
}
else if(op=="PM") cout<<h[1]<<endl;
else if(op=="DM"){
heap_swap(1,cur_size);
cur_size--;
down(1);
}
else if(op=="D"){
cin>>k;
int u=ph[k];//这里一定要用u=ph[k]保存第k个插入点的下标
heap_swap(u,cur_size);//因为在此处heap_swap操作后ph[k]的值已经发生
cur_size--;//如果在up,down操作中仍然使用ph[k]作为参数就会发生错误
up(u);
down(u);
}
else if(op=="C"){
cin>>k>>x;
h[ph[k]]=x;//此处由于未涉及heap_swap操作且下面的up、down操作只会发生一个所以
down(ph[k]);//所以可直接传入ph[k]作为参数
up(ph[k]);
}
}
return 0;
}
字符串哈希
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long ULL;
const int N = 100010, P = 131;
int n, m;
char str[N];
ULL h[N], p[N];
ULL get(int l, int r){
return h[r] - h[l - 1] * p[r - l + 1];
}
int main(){
scanf("%d%d", &n, &m);
scanf("%s", str + 1);
p[0] = 1;
for (int i = 1; i <= n; i ++ ){
h[i] = h[i - 1] * P + str[i];
p[i] = p[i - 1] * P;
}
while (m -- ){
int l1, r1, l2, r2;
scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
if (get(l1, r1) == get(l2, r2)) puts("Yes");
else puts("No");
}
return 0;
}
搜索与图论
全排列
#include <iostream>
using namespace std;
const int N = 10;
int n;
int path[N];
void dfs(int u, int state){
if (u == n){
for (int i = 0; i < n; i ++ ) printf("%d ", path[i]);
puts("");
return;
}
for (int i = 0; i < n; i ++ )
if (!(state >> i & 1)){
path[u] = i + 1;
dfs(u + 1, state + (1 << i));
}
}
int main(){
scanf("%d", &n);
dfs(0, 0);
return 0;
}
组合型枚举
#include<iostream>
#include<vector>
using namespace std;
bool chosen[30];
int n, m;
vector<int> nums;
void dfs(int pos) {
if (nums.size() > m || nums.size() + n - pos + 1 < m)
return;
if (nums.size() == m) {
for (int i = 0; i < nums.size(); i++)
cout << nums[i]<<" ";
cout << endl;
return;
}
nums.push_back(pos);
dfs(pos + 1);
nums.pop_back();
dfs(pos + 1);
}
int main() {
cin >> n >> m;
dfs(1);
return 0;
}
八皇后
#include <iostream>
using namespace std;
const int N = 20;
int n;
char g[N][N];
bool col[N], dg[N], udg[N];
void dfs(int u){
if (u == n){
for (int i = 0; i < n; i ++ ) puts(g[i]);
puts("");
return;
}
for (int i = 0; i < n; i ++ )
if (!col[i] && !dg[u + i] && !udg[n - u + i]){
g[u][i] = 'Q';
col[i] = dg[u + i] = udg[n - u + i] = true;
dfs(u + 1);
col[i] = dg[u + i] = udg[n - u + i] = false;
g[u][i] = '.';
}
}
int main(){
cin >> n;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n; j ++ )
g[i][j] = '.';
dfs(0);
return 0;
}
走迷宫
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 110;
int n, m;
int g[N][N], d[N][N];
int bfs(){
queue<PII> q;
memset(d, -1, sizeof d);
d[0][0] = 0;
q.push({0, 0});
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
while (q.size()){
auto t = q.front();
q.pop();
for (int i = 0; i < 4; i ++ ){
int x = t.first + dx[i], y = t.second + dy[i];
if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1){
d[x][y] = d[t.first][t.second] + 1;
q.push({x, y});
}
}
}
return d[n - 1][m - 1];
}
int main(){
cin >> n >> m;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < m; j ++ )
cin >> g[i][j];
cout << bfs() << endl;
return 0;
}
树的重心
重心定义：重心是指树中的一个结点，如果将这个点删除后，剩余各个连通块中点数的最大值最小，那么这个节点被称为树的重心。
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
const int N = 100005;
vector<int> g[N];
int n;
int st[N];
int ans = N;
int dfs(int u) {
st[u] = true;
int sum = 1, res = 0;
for (auto node : g[u]) {
if (!st[node]) {
int s = dfs(node);
res = max(res, s);
sum += s;
}
}
res = max(res, n - sum);
ans = min(ans, res);
return sum;
}
int main() {
cin >> n;
for(int i=0;i<n-1;i++) {
int a, b;
cin >> a >> b;
g[a].push_back(b);
g[b].push_back(a);
}
dfs(1);
cout << ans;
return 0;
}
拓扑排序
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
const int N = 100010;
int n, m;
vector<int> graph[N];
vector<int> path;
int ind[N];//入度
bool toposort() {
int indnode = 0;
queue<int> q;
for (int i = 1; i <= n; i++) {
if (ind[i] == 0) {
indnode = i;
q.push(i);
}
}
if (indnode == 0)
return false;
while (!q.empty()) {
int cur = q.front();
q.pop();
path.push_back(cur);
for (auto node : graph[cur]) {
ind[node]--;
if (ind[node] == 0)
q.push(node);
}
}
for (int i = 1; i <= n; i++)
if (ind[i] > 0)
return false;
return true;
}
int main() {
cin >> n >> m;
while (m--) {
int x, y;
cin >> x >> y;
graph[x].push_back(y);
ind[y]++;
}
if (toposort()) {
for (auto node : path)
cout << node << " ";
}
else
cout << -1;
return 0;
}
八数码
#include<iostream>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef pair<int, vector<int> > PIV;
priority_queue<PIV, vector<PIV>,greater<PIV> > q;
map<vector<int>, string> path;
map<vector<int>, int> dist;
vector<int> endst = { 1,2,3,4,5,6,7,8,0 };
int pred(vector<int> state,int steps) {
int res = 0;
for (int i = 0; i < 9; i++)//这里1~8对应的下标为0~7
if (state[i] != 0) {
int t = state[i] - 1;//对应下标
res += abs(i / 3 - t / 3) + abs(i % 3 - t % 3);//曼哈顿距离
}
return res+steps;//返回总曼哈顿距离
}
int mfind(vector<int> state) {
for (int i = 0; i < 9; i++)
if (state[i] == 0)
return i;
return -1;
}
void bfs() {
while (!q.empty()) {
PIV cur = q.top();
q.pop();
if (cur.second ==endst) {
cout << path[cur.second];
return;
}
int pos = mfind(cur.second);
if (pos / 3 == 0) {
vector<int> newc = cur.second;
swap(newc[pos], newc[pos + 3]);
if (path.count(newc) == 0||dist[newc]>dist[cur.second]+1) {
path[newc] = path[cur.second] + "d";
dist[newc] = dist[cur.second] + 1;
q.push({pred(newc,dist[newc]),newc });
}
}
if (pos / 3 == 1) {
vector<int> newc = cur.second;
swap(newc[pos], newc[pos + 3]);
if (path.count(newc) == 0 || dist[newc] > dist[cur.second] + 1) {
path[newc] = path[cur.second] + "d";
dist[newc] = dist[cur.second] + 1;
q.push({ pred(newc,dist[newc]),newc });
}
newc = cur.second;
swap(newc[pos], newc[pos - 3]);
if (path.count(newc) == 0 || dist[newc] > dist[cur.second] + 1) {
path[newc] = path[cur.second] + "u";
dist[newc] = dist[cur.second] + 1;
q.push({ pred(newc,dist[newc]),newc });
}
}
if (pos / 3 == 2) {
vector<int> newc = cur.second;
swap(newc[pos], newc[pos - 3]);
if (path.count(newc) == 0 || dist[newc] > dist[cur.second] + 1) {
path[newc] = path[cur.second] + "u";
dist[newc] = dist[cur.second] + 1;
q.push({ pred(newc,dist[newc]),newc });
}
}
if (pos % 3 == 0) {
vector<int> newc = cur.second;
swap(newc[pos], newc[pos + 1]);
if (path.count(newc) == 0 || dist[newc] > dist[cur.second] + 1) {
path[newc] = path[cur.second] + "r";
dist[newc] = dist[cur.second] + 1;
q.push({ pred(newc,dist[newc]),newc });
}
}
if (pos % 3 == 1) {
vector<int> newc = cur.second;
swap(newc[pos], newc[pos + 1]);
if (path.count(newc) == 0 || dist[newc] > dist[cur.second] + 1) {
path[newc] = path[cur.second] + "r";
dist[newc] = dist[cur.second] + 1;
q.push({ pred(newc,dist[newc]),newc });
}
newc = cur.second;
swap(newc[pos], newc[pos - 1]);
if (path.count(newc) == 0 || dist[newc] > dist[cur.second] + 1) {
path[newc] = path[cur.second] + "l";
dist[newc] = dist[cur.second] + 1;
q.push({ pred(newc,dist[newc]),newc });
}
}
if (pos % 3 == 2) {
vector<int> newc = cur.second;
swap(newc[pos], newc[pos - 1]);
if (path.count(newc) == 0 || dist[newc] > dist[cur.second] + 1) {
path[newc] = path[cur.second] + "l";
dist[newc] = dist[cur.second] + 1;
q.push({ pred(newc,dist[newc]),newc });
}
}
}
}
int nixu(vector<int> state) {
int res = 0;
for (int i = 0; i < 9; i++) {
for (int j = i; j < 9; j++)
if (state[j] < state[i] && state[j] != 0)
res++;
}
return res;
}
int main() {
vector<char> ch(9);
for (int i = 0; i < 9; i++)
cin >> ch[i];
vector<int> init(9);
for (int i = 0; i < 9; i++) {
if (ch[i] == 'x')
init[i] = 0;
else init[i] = ch[i] - '0';
}
if (nixu(init) % 2 == 0) {
q.push({pred(init,0),init });
dist[init] = 0;
path[init] = "";
bfs();
return 0;
}
else {
cout << "unsolvable";
return 0;
}
}
解数独
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 9, M = 1 << N;
int ones[M], map[M];
int row[N], col[N], cell[3][3];
char str[100];
void init(){
for (int i = 0; i < N; i ++ )
row[i] = col[i] = (1 << N) - 1;
for (int i = 0; i < 3; i ++ )
for (int j = 0; j < 3; j ++ )
cell[i][j] = (1 << N) - 1;
}
void draw(int x, int y, int t, bool is_set){
if (is_set) str[x * N + y] = '1' + t;
else str[x * N + y] = '.';
int v = 1 << t;
if (!is_set) v = -v;
row[x] -= v;
col[y] -= v;
cell[x / 3][y / 3] -= v;
}
int lowbit(int x){
return x & -x;
}
int get(int x, int y){
return row[x] & col[y] & cell[x / 3][y / 3];
}
bool dfs(int cnt){
if (!cnt) return true;
int minv = 10;
int x, y;
for (int i = 0; i < N; i ++ )
for (int j = 0; j < N; j ++ )
if (str[i * N + j] == '.'){
int state = get(i, j);
if (ones[state] < minv){
minv = ones[state];
x = i, y = j;
}
}
int state = get(x, y);
for (int i = state; i; i -= lowbit(i)){
int t = map[lowbit(i)];
draw(x, y, t, true);
if (dfs(cnt - 1)) return true;
draw(x, y, t, false);
}
return false;
}
int main(){
for (int i = 0; i < N; i ++ ) map[1 << i] = i;
for (int i = 0; i < 1 << N; i ++ )
for (int j = 0; j < N; j ++ )
ones[i] += i >> j & 1;
while (cin >> str, str[0] != 'e'){
init();
int cnt = 0;
for (int i = 0, k = 0; i < N; i ++ )
for (int j = 0; j < N; j ++, k ++ )
if (str[k] != '.')
{
int t = str[k] - '1';
draw(i, j, t, true);
}
else cnt ++ ;
dfs(cnt);
puts(str);
}
return 0;
}
堆优化Dijkstra
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
int t,c,ts,te;
int ans=0;
typedef pair<int,int> PII;
priority_queue<PII,vector<PII>,greater<PII> > q;
int st[2505];
int dist[2505];
vector<PII> g[2505];
void dijkstra(){
memset(dist,0x3f3f3f3f,sizeof dist);
dist[ts]=0;
q.push({0,ts});
while(!q.empty()){
auto [d,cur]=q.top();
q.pop();
if (st[cur])
continue;
st[cur]=1;
for(auto node:g[cur]){
auto [d,next]=node;
if (d+dist[cur]<dist[next]){
dist[next]=d+dist[cur];
q.push({dist[next],next});
}
}
}
}
int main(){
cin>>t>>c>>ts>>te;
for(int i=1;i<=c;i++){
int rs,re,ci;
cin>>rs>>re>>ci;
g[rs].push_back({ci,re});
g[re].push_back({ci,rs});
}
dijkstra();
cout<<dist[te];
}
SPFA
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
int n, m;
const int N = 100005;
int dist[N];
typedef pair<int, int> PII;//first表示指向点，second表示距离
vector<PII> graph[N];
void spfa() {
dist[1] = 0;
queue<int> q;
q.push(1);
while (!q.empty()) {
int curnode = q.front();
q.pop();
for (auto node : graph[curnode]) {
if (dist[node.first] > dist[curnode] + node.second) {
dist[node.first] = dist[curnode] + node.second;
q.push(node.first);
}
}
}
}
int main() {
cin >> n >> m;
while (m--) {
int x, y, z;
cin >> x >> y >> z;
graph[x].push_back({ y,z });
}
memset(dist, 0x3f3f3f3f, sizeof dist);
spfa();
if (dist[n] > 0x3f3f3f3f / 2)
cout << "impossible";
else
cout << dist[n];
return 0;
}
SPFA判负环
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
int n, m;
const int N = 100005;
int cnt[N];
int dist[N];
typedef pair<int, int> PII;//first表示指向点，second表示距离
vector<PII> graph[N];
bool spfa() {
queue<int> q;
for(int i=1;i<=n;i++)
q.push(i);
while (!q.empty()) {
int curnode = q.front();
q.pop();
for (auto node : graph[curnode]) {
if (dist[node.first] > dist[curnode] + node.second) {
dist[node.first] = dist[curnode] + node.second;
cnt[node.first]=cnt[curnode]+1;
if (cnt[node.first] >= n)
return true;
q.push(node.first);
}
}
}
return false;
}
int main() {
cin >> n >> m;
while (m--) {
int x, y, z;
cin >> x >> y >> z;
graph[x].push_back({ y,z });
}
if (spfa())
cout << "Yes";
else
cout << "No";
return 0;
}
Bellman-Ford
#include<iostream>
#include<cstring>
using namespace std;
int n, m, k;
struct Edge {
int x, y, z;
};
Edge edges[10005];
int dist[505];
int backup[505];
int main() {
cin >> n >> m >> k;
memset(dist, 0x3f3f3f3f, sizeof dist);
memset(backup, 0x3f3f3f3f, sizeof backup);
for(int i=0;i<m;i++){
int x, y, z;
cin >> x >> y >> z;
edges[i] = { x,y,z };
}
dist[1] = 0;
while (k--) {
for (int i = 1; i <= n; i++)
backup[i] = dist[i];
for (int i = 0; i < m; i++)
dist[edges[i].y] = min( dist[edges[i].y],backup[edges[i].x] + edges[i].z );
}
if (dist[n] >= 0x3f3f3f3f / 2)
cout << "impossible";
else
cout << dist[n];
return 0;
}
Prim
#include<cstring>
#include<iostream>
using namespace std;
const int N = 505;
int graph[N][N];
int dist[N];
int s[N];
int n, m;
int res = 0;
void prim() {
s[1] = true;
dist[1] = 0;
for (int i = 1; i <= n; i++)
dist[i] = graph[1][i];
for (int i = 1; i <= n; i++) {
int curnode = 0;
for (int j = 1; j <= n; j++) {
if (!s[j] && (curnode==0||dist[j] < dist[curnode]))
curnode = j;
}
if (curnode&&dist[curnode] == 0x3f3f3f3f) {
res = 0x3f3f3f3f;
return;
}
if (curnode) {
s[curnode] = true;
res += dist[curnode];
for (int j = 1; j <= n; j++)
if (!s[j] && dist[j] > graph[curnode][j])
dist[j] = graph[curnode][j];
}
}
}
int main() {
cin >> n >> m;
memset(graph, 0x3f3f3f3f, sizeof graph);
memset(dist, 0x3f3f3f3f, sizeof dist);
while (m--) {
int x, y, z;
cin >> x >> y >> z;
if (z < graph[x][y])
graph[x][y] =graph[y][x]= z;
}
for (int i = 0; i <= n; i++)
graph[i][i] = 0;
prim();
if (res < 0x3f3f3f3f)
cout << res;
else
cout << "impossible";
return 0;
}
Kruskal
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1000005;
struct edge {
int u, v, w;
bool operator <(const edge& b) const {
return w < b.w;
}
};
int p[N];
edge edges[N];
int find(int a) {
if (a!= p[a])
p[a] = find(p[a]);
else return p[a];
}
int main() {
int n, m;
cin >> n >> m;
for (int i = 0; i < m; i++) {
int u, v, w;
cin >> u >> v >> w;
if (u!=v)
edges[i] = { u,v,w };
}
for (int i = 0; i <=n; i++)
p[i] = i;
sort(edges, edges + m);
long long res = 0;
int cnt = 0;
for (int i = 0; i < m; i++) {
int u = edges[i].u;
int v = edges[i].v;
int w = edges[i].w;
u = find(u);
v = find(v);
if (u != v) {
res += w;
cnt++;
p[find(u)] =v;
}
}
if (cnt < n - 1)
cout << "impossible";
else
cout << res;
return 0;
}
染色法判定二分图
#include<iostream>
#include<vector>
using namespace std;
const int N = 100005;
int color[N];
int res = true;
vector<int> G[N];
void dfs(int i, int clr) {
color[i] = clr;
for (auto node : G[i]) {
if(!color[node])
dfs(node, 3 - clr);
else if (color[node] == color[i]) {
res = false;
return;
}
}
}
int main() {
int n, m;
cin >> n >> m;
while (m--){
int u, v;
cin >> u >> v;
if (u != v) {
G[u].push_back(v);
G[v].push_back(u);
}
}
for (int i = 1; i <= n; i++) {
if(!color[i])
dfs(i, 1);
}
if (res)
cout << "Yes";
else
cout << "No";
return 0;
}
匈牙利算法
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
const int N = 1005;
vector<int> graph[N];
int match[N];
int st[N];
bool find(int x) {
for (auto node : graph[x]) {
if (!st[node]) {
st[node] = true;
if (!match[node] || find(match[node])) {
match[node]=x;
return true;
}
}
}
return false;
}
int main() {
int n1, n2, m;
cin >> n1 >> n2 >> m;
while (m--) {
int u, v;
cin >> u >> v;
graph[u].push_back(v);
}
int res = 0;
for (int i = 1; i <= n1; i++) {
memset(st, false, sizeof st);
if (find(i))
res++;
}
cout << res << endl;
return 0;
}
数论
线性筛
#include <iostream>
#include <algorithm>
using namespace std;
const int N= 1000010;
int primes[N], cnt;
bool st[N];
void get_primes(int n){
for (int i = 2; i <= n; i ++ ){
if (!st[i]) primes[cnt ++ ] = i;
for (int j = 0; primes[j] <= n / i; j ++ ){
st[primes[j] * i] = true;
if (i % primes[j] == 0) break;
}
}
}
int main(){
int n;
cin >> n;
get_primes(n);
cout << cnt << endl;
return 0;
}
分解质因数
#include <iostream>
#include <algorithm>
using namespace std;
void divide(int x){
for (int i = 2; i <= x / i; i ++ )
if (x % i == 0){
int s = 0;
while (x % i == 0) x /= i, s ++ ;
cout << i << ' ' << s << endl;
}
if (x > 1) cout << x << ' ' << 1 << endl;
cout << endl;
}
int main(){
int n;
cin >> n;
while (n -- ){
int x;
cin >> x;
divide(x);
}
return 0;
}
试除法求约数
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> get_divisors(int x){
vector<int> res;
for (int i = 1; i <= x / i; i ++ )
if (x % i == 0){
res.push_back(i);
if (i != x / i) res.push_back(x / i);
}
sort(res.begin(), res.end());
return res;
}
int main(){
int n;
cin >> n;
while (n -- ){
int x;
cin >> x;
auto res = get_divisors(x);
for (auto x : res) cout << x << ' ';
cout << endl;
}
return 0;
}
约数个数
#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 110, mod = 1e9 + 7;
int main(){
int n;
cin >> n;
unordered_map<int, int> primes;
while (n -- ){
int x;
cin >> x;
for (int i = 2; i <= x / i; i ++ )
while (x % i == 0){
x /= i;
primes[i] ++ ;
}
if (x > 1) primes[x] ++ ;
}
LL res = 1;
for (auto p : primes) res = res * (p.second + 1) % mod;
cout << res << endl;
return 0;
}
约数之和
#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 110, mod = 1e9 + 7;
int main(){
int n;
cin >> n;
unordered_map<int, int> primes;
while (n -- ){
int x;
cin >> x;
for (int i = 2; i <= x / i; i ++ )
while (x % i == 0){
x /= i;
primes[i] ++ ;
}
if (x > 1) primes[x] ++ ;
}
LL res = 1;
for (auto p : primes){
LL a = p.first, b = p.second;
LL t = 1;
while (b -- ) t = (t * a + 1) % mod;
res = res * t % mod;
}
cout << res << endl;
return 0;
}
gcd
#include <iostream>
#include <algorithm>
using namespace std;
int gcd(int a, int b){
return b ? gcd(b, a % b) : a;
}
int main(){
int n;
cin >> n;
while (n -- ){
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", gcd(a, b));
}
return 0;
}
欧拉函数
#include <iostream>
using namespace std;
int phi(int x){
int res = x;
for (int i = 2; i <= x / i; i ++ )
if (x % i == 0){
res = res / i * (i - 1);
while (x % i == 0) x /= i;
}
if (x > 1) res = res / x * (x - 1);
return res;
}
int main(){
int n;
cin >> n;
while (n -- ){
int x;
cin >> x;
cout << phi(x) << endl;
}
return 0;
}
筛法欧拉函数
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 1000010;
int primes[N], cnt;
int euler[N];
bool st[N];
void get_eulers(int n){
euler[1] = 1;
for (int i = 2; i <= n; i ++ ){
if (!st[i]){
primes[cnt ++ ] = i;
euler[i] = i - 1;
}
for (int j = 0; primes[j] <= n / i; j ++ ){
int t = primes[j] * i;
st[t] = true;
if (i % primes[j] == 0){
euler[t] = euler[i] * primes[j];
break;
}
euler[t] = euler[i] * (primes[j] - 1);
}
}
}
int main(){
int n;
cin >> n;
get_eulers(n);
LL res = 0;
for (int i = 1; i <= n; i ++ ) res += euler[i];
cout << res << endl;
return 0;
}
快速幂
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL qmi(int a, int b, int p){
LL res = 1 % p;
while (b){
if (b & 1) res = res * a % p;
a = a * (LL)a % p;
b >>= 1;
}
return res;
}
int main(){
int n;
scanf("%d", &n);
while (n -- ){
int a, b, p;
scanf("%d%d%d", &a, &b, &p);
printf("%lld\n", qmi(a, b, p));
}
return 0;
}
快速幂求逆元
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL qmi(int a, int b, int p){
LL res = 1;
while (b){
if (b & 1) res = res * a % p;
a = a * (LL)a % p;
b >>= 1;
}
return res;
}
int main(){
int n;
scanf("%d", &n);
while (n -- ){
int a, p;
scanf("%d%d", &a, &p);
if (a % p == 0) puts("impossible");
else printf("%lld\n", qmi(a, p - 2, p));
}
return 0;
}
扩展欧几里得算法
给定 n n n对正整数 a i , b i a_i,b_i ai,bi，对于每对数，求出一组 x i , y i x_i,y_i xi,yi，使其满足 a i × x i + b i × y i = g c d ( a i , b i ) a_i×x_i+b_i×y_i=gcd(a_i,b_i) ai×xi+bi×yi=gcd(ai,bi)。
#include <iostream>
#include <algorithm>
using namespace std;
int exgcd(int a, int b, int &x, int &y){
if (!b){
x = 1, y = 0;
return a;
}
int d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
int main(){
int n;
scanf("%d", &n);
while (n -- ){
int a, b;
scanf("%d%d", &a, &b);
int x, y;
exgcd(a, b, x, y);
printf("%d %d\n", x, y);
}
return 0;
}
线性同余方程
给定 n n n组数据 a i , b i , m i a_i,b_i,m_i ai,bi,mi，对于每组数求出一个 x i x_i xi，使其满足 a i × x i ≡ b i ( m o d m i ) a_i×x_i≡b_i(\mod m_i) ai×xi≡bi(modmi)，如果无解则输出 impossible。
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
int exgcd(int a, int b, int &x, int &y){
if (!b){
x = 1, y = 0;
return a;
}
int d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
int main(){
int n;
scanf("%d", &n);
while (n -- ){
int a, b, m;
scanf("%d%d%d", &a, &b, &m);
int x, y;
int d = exgcd(a, m, x, y);
if (b % d) puts("impossible");
else printf("%d\n", (LL)b / d * x % m);
}
return 0;
}
扩展中国剩余定理
表达整数的奇怪方式。给定 2 n 2n 2n个整数 a 1 , a 2 , … , a n a_1,a_2,…,a_n a1,a2,…,an和 m 1 , m 2 , … , m n m_1,m_2,…,m_n m1,m2,…,mn，求一个最小的非负整数 x x x，满足 ∀ i ∈ [ 1 , n ] , x ≡ m i ( m o d a i ) ∀i∈[1,n],x≡m_i(\mod a_i) ∀i∈[1,n],x≡mi(modai)。
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long LL;
int n;
LL exgcd(LL a, LL b, LL &x, LL &y){
if(b == 0){
x = 1, y = 0;
return a;
}
LL d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
LL inline mod(LL a, LL b){
return ((a % b) + b) % b;
}
int main(){
scanf("%d", &n);
LL a1, m1;
scanf("%lld%lld", &a1, &m1);
for(int i = 1; i < n; i++){
LL a2, m2, k1, k2;
scanf("%lld%lld", &a2, &m2);
LL d = exgcd(a1, -a2, k1, k2);
if((m2 - m1) % d){ puts("-1"); return 0; }
k1 = mod(k1 * (m2 - m1) / d, abs(a2 / d));
m1 = k1 * a1 + m1;
a1 = abs(a1 / d * a2);
}
printf("%lld\n", m1);
return 0;
}
同余方程
关于 x x x的方程 a x ≡ 1 ( m o d b ) ax≡1(modb) ax≡1(modb)的最小正整数解
#include<iostream>
typedef long long LL;
using namespace std;
LL exgcd(LL a,LL b,LL&x,LL&y){
if(!b){
x=1;
y=0;
return a;
}
int d=exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
int main(){
LL a,b,x,y;
cin>>a>>b;
exgcd(a,b,x,y);
cout<<(x%b+b)%b;
return 0;
}
斐波那契(矩阵快速幂)
#include<iostream>
typedef long long LL;
using namespace std;
LL n,m;
LL M[2][2]={
{0,1},
{1,1}
};
LL res[2]={1,0};
void mulrm(){
LL ans[2]={0};
for(LL i=0;i<2;i++)
for(LL j=0;j<2;j++)
ans[i]+=res[j]*M[i][j]%m;
for(LL i=0;i<2;i++)
res[i]=ans[i]%m;
}
void mulmm(){
LL ans[2][2]={0};
for(LL i=0;i<2;i++){
for(LL j=0;j<2;j++){
for(LL k=0;k<2;k++)
ans[i][j]+=M[i][k]*M[k][j]%m;
}
}
for(LL i=0;i<2;i++)
for(LL j=0;j<2;j++)
M[i][j]=ans[i][j]%m;
}
void qpow(LL n){
while(n){
if (n&1)
mulrm();
mulmm();
n>>=1;
}
}
int main(){
cin>>n>>m;
qpow(n+2);
cout<<res[1]-1;
return 0;
}
高斯消元
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 110;
const double eps = 1e-8;
int n;
double a[N][N];
int gauss(){ // 高斯消元，答案存于a[i][n]中，0 <= i < n
int c, r;
for (c = 0, r = 0; c < n; c ++ ){
int t = r;
for (int i = r; i < n; i ++ ) // 找绝对值最大的行
if (fabs(a[i][c]) > fabs(a[t][c]))
t = i;
if (fabs(a[t][c]) < eps) continue;
for (int i = c; i <= n; i ++ ) swap(a[t][i], a[r][i]); // 将绝对值最大的行换到最顶端
for (int i = n; i >= c; i -- ) a[r][i] /= a[r][c]; // 将当前行的首位变成1
for (int i = r + 1; i < n; i ++ ) // 用当前行将下面所有的列消成0
if (fabs(a[i][c]) > eps)
for (int j = n; j >= c; j -- )
a[i][j] -= a[r][j] * a[i][c];
r ++ ;
}
if (r < n){
for (int i = r; i < n; i ++ )
if (fabs(a[i][n]) > eps)
return 2; // 无解
return 1; // 有无穷多组解
}
for (int i = n - 1; i >= 0; i -- )
for (int j = i + 1; j < n; j ++ )
a[i][n] -= a[i][j] * a[j][n];
return 0; // 有唯一解
}
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n + 1; j ++ )
scanf("%lf", &a[i][j]);
int t = gauss();
if (t == 2) puts("No solution");
else if (t == 1) puts("Infinite group solutions");
else{
for (int i = 0; i < n; i ++ ){
if (fabs(a[i][n]) < eps) a[i][n] = 0; // 去掉输出 -0.00 的情况
printf("%.2lf\n", a[i][n]);
}
}
return 0;
}
组合数
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 2010, mod = 1e9 + 7;
int c[N][N];
void init(){
for (int i = 0; i < N; i ++ )
for (int j = 0; j <= i; j ++ )
if (!j) c[i][j] = 1;
else c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
}
int main(){
int n;
init();
scanf("%d", &n);
while (n -- ){
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", c[a][b]);
}
return 0;
}
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100010, mod = 1e9 + 7;
int fact[N], infact[N];
int qmi(int a, int k, int p){
int res = 1;
while (k){
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
int main(){
fact[0] = infact[0] = 1;
for (int i = 1; i < N; i ++ ){
fact[i] = (LL)fact[i - 1] * i % mod;
infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
}
int n;
scanf("%d", &n);
while (n -- ){
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", (LL)fact[a] * infact[b] % mod * infact[a - b] % mod);
}
return 0;
}
C a b = C a p b p C a m o d p b m o d p ( m o d p ) C_a^b=C_{\frac ap}^{\frac bp}C_{a \mod p}^{b\mod p}(\mod p) Cab=CpapbCamodpbmodp(modp)
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
int qmi(int a, int k, int p){
int res = 1;
while (k){
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
int C(int a, int b, int p){
if (b > a) return 0;
int res = 1;
for (int i = 1, j = a; i <= b; i ++, j -- ){
res = (LL)res * j % p;
res = (LL)res * qmi(i, p - 2, p) % p;
}
return res;
}
int lucas(LL a, LL b, int p){
if (a < p && b < p) return C(a, b, p);
return (LL)C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}
int main(){
int n;
cin >> n;
while (n -- ){
LL a, b;
int p;
cin >> a >> b >> p;
cout << lucas(a, b, p) << endl;
}
return 0;
}
卡特兰数
C 2 n n n + 1 C 1 = 1 , C n = C n − 1 ∗ 4 n − 2 n + 1 \frac {C_{2n}^n}{n+1}\\ C_1=1,C_n=C_{n-1}*\frac{4n-2}{n+1} n+1C2nnC1=1,Cn=Cn−1∗n+14n−2
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100010, mod = 1e9 + 7;
int qmi(int a, int k, int p){
int res = 1;
while (k){
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
int main(){
int n;
cin >> n;
int a = n * 2, b = n;
int res = 1;
for (int i = a; i > a - b; i -- ) res = (LL)res * i % mod;
for (int i = 1; i <= b; i ++ ) res = (LL)res * qmi(i, mod - 2, mod) % mod;
res = (LL)res * qmi(n + 1, mod - 2, mod) % mod;
cout << res << endl;
return 0;
}
能被整除的数
给定一个整数 n n n和 m m m个不同的质数 p 1 , p 2 , … , p m p_1,p_2,…,p_m p1,p2,…,pm。
请你求出 1 ∼ n 1∼n 1∼n中能被 p 1 , p 2 , … , p m p_1,p_2,…,p_m p1,p2,…,pm中的至少一个数整除的整数有多少个。
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 20;
int p[N];
int main(){
int n, m;
cin >> n >> m;
for (int i = 0; i < m; i ++ ) cin >> p[i];
int res = 0;
for (int i = 1; i < 1 << m; i ++ ){
int t = 1, s = 0;
for (int j = 0; j < m; j ++ )
if (i >> j & 1){
if ((LL)t * p[j] > n){
t = -1;
break;
}
t *= p[j];
s ++ ;
}
if (t != -1){
if (s % 2) res += n / t;
else res -= n / t;
}
}
cout << res << endl;
return 0;
}
龟速乘法
#include<iostream>
using namespace std;
typedef long long LL;
LL a,b,p;
LL qadd(LL a,LL b,LL p){
LL res=0;
while(b){
if (b&1)
res=(res+a)%p;
a=(a+a)%p;
b>>=1;
}
return res;
}
int main(){
cin>>a>>b>>p;
cout<<qadd(a,b,p);
return 0;
}
DP
01背包
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int v[N], w[N];
int f[N];
int main(){
cin >> n >> m;
for (int i = 1; i <= n; i ++ ) cin >> v[i] >> w[i];
for (int i = 1; i <= n; i ++ )
for (int j = m; j >= v[i]; j -- )
f[j] = max(f[j], f[j - v[i]] + w[i]);
cout << f[m] << endl;
return 0;
}
完全背包
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int v[N], w[N];
int f[N];
int main(){
cin >> n >> m;
for (int i = 1; i <= n; i ++ ) cin >> v[i] >> w[i];
for (int i = 1; i <= n; i ++ )
for (int j = v[i]; j <= m; j ++ )
f[j] = max(f[j], f[j - v[i]] + w[i]);
cout << f[m] << endl;
return 0;
}
多重背包
#include<iostream>
#include<algorithm>
using namespace std;
int dp[10005];
int v[1005];
int w[1005];
int s[1005];
int vf[10005], wf[10005];
int N, V;
int main() {
cin >> N >> V;
for (int i = 1; i <= N; i++)
cin >> v[i] >> w[i]>>s[i];
int idx = 0;
for (int i = 1; i <= N; i++) {
int cur = 1;
int res = s[i];
while (res>=cur) {
res = res - cur;
idx++;
vf[idx] = cur * v[i];
wf[idx] = cur * w[i];
cur = cur * 2;
}
if (res>0) {
idx++;
vf[idx] = res * v[i];
wf[idx] = res * w[i];
}
}
for(int i=1;i<=idx;i++)
for (int j = V; j >= vf[i]; j--) {
dp[j] = max(dp[j], dp[j - vf[i]] + wf[i]);
}
cout << dp[V];
return 0;
}
分组背包问题
有 N N N组物品和一个容量是 V V V的背包。
每组物品有若干个，同一组内的物品最多只能选一个。
每件物品的体积是 v i j v_{ij} vij，价值是 w i j w_{ij} wij，其中 i i i是组号, j j j是组内编号。
求解将哪些物品装入背包，可使物品总体积不超过背包容量，且总价值最大。
输出最大价值。
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 110;
int n, m;
int v[N][N], w[N][N], s[N];
int f[N];
int main(){
cin >> n >> m;
for (int i = 1; i <= n; i ++ ){
cin >> s[i];
for (int j = 0; j < s[i]; j ++ )
cin >> v[i][j] >> w[i][j];
}
for (int i = 1; i <= n; i ++ )
for (int j = m; j >= 0; j -- )
for (int k = 0; k < s[i]; k ++ )
if (v[i][k] <= j)
f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
cout << f[m] << endl;
return 0;
}
数字三角形
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 510, INF = 1e9;
int n;
int a[N][N];
int f[N][N];
int main(){
scanf("%d", &n);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= i; j ++ )
scanf("%d", &a[i][j]);
for (int i = 0; i <= n; i ++ )
for (int j = 0; j <= i + 1; j ++ )
f[i][j] = -INF;
f[1][1] = a[1][1];
for (int i = 2; i <= n; i ++ )
for (int j = 1; j <= i; j ++ )
f[i][j] = max(f[i - 1][j - 1] + a[i][j], f[i - 1][j] + a[i][j]);
int res = -INF;
for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]);
printf("%d\n", res);
return 0;
}
石子合并
设有 N N N堆石子排成一排，其编号为 1 ， 2 ， 3 ， … ， N 1，2，3，…，N 1，2，3，…，N。
每堆石子有一定的质量，可以用一个整数来描述，现在要将这 N N N堆石子合并成为一堆。
每次只能合并相邻的两堆，合并的代价为这两堆石子的质量之和，合并后与这两堆石子相邻的石子将和新堆相邻，合并时由于选择的顺序不同，合并的总代价也不相同。
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 310;
int n;
int s[N];
int f[N][N];
int main(){
scanf("%d", &n);
for (int i = 1; i <= n; i ++ ) scanf("%d", &s[i]);
for (int i = 1; i <= n; i ++ ) s[i] += s[i - 1];
for (int len = 2; len <= n; len ++ )
for (int i = 1; i + len - 1 <= n; i ++ ){
int l = i, r = i + len - 1;
f[l][r] = 1e8;
for (int k = l; k < r; k ++ )
f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r] + s[r] - s[l - 1]);
}
printf("%d\n", f[1][n]);
return 0;
}
lis
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int n;
int a[N];
int q[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
int len = 0;
for (int i = 0; i < n; i ++ ){
int l = 0, r = len;
while (l < r){
int mid = l + r + 1 >> 1;
if (q[mid] < a[i]) l = mid;
else r = mid - 1;
}
len = max(len, r + 1);
q[r + 1] = a[i];
}
printf("%d\n", len);
return 0;
}
整数划分
一个正整数 n n n可以表示成若干个正整数之和，形如 n = n 1 + n 2 + … + n k n=n_1+n_2+…+n_k n=n1+n2+…+nk，其中 n 1 ≥ n 2 ≥ … ≥ n k , k ≥ 1 n_1≥n_2≥…≥n_k,k≥1 n1≥n2≥…≥nk,k≥1。
我们将这样的一种表示称为正整数 n n n的一种划分。
现在给定一个正整数 n n n，请你求出 n n n共有多少种不同的划分方法。
完全背包解法
状态表示：
f[i][j]表示只从1~i中选，且总和等于j的方案数
状态转移方程:
f[i][j] = f[i - 1][j] + f[i][j - i];
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010, mod = 1e9 + 7;
int n;
int f[N];
int main(){
cin >> n;
f[0] = 1;
for (int i = 1; i <= n; i ++ )
for (int j = i; j <= n; j ++ )
f[j] = (f[j] + f[j - i]) % mod;
cout << f[n] << endl;
return 0;
}
其他算法
状态表示：
f[i][j]表示总和为i，总个数为j的方案数
状态转移方程：
f[i][j] = f[i - 1][j - 1] + f[i - j][j];
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010, mod = 1e9 + 7;
int n;
int f[N][N];
int main(){
cin >> n;
f[1][1] = 1;
for (int i = 2; i <= n; i ++ )
for (int j = 1; j <= i; j ++ )
f[i][j] = (f[i - 1][j - 1] + f[i - j][j]) % mod;
int res = 0;
for (int i = 1; i <= n; i ++ ) res = (res + f[n][i]) % mod;
cout << res << endl;
return 0;
}
1050. 鸣人的影分身
在火影忍者的世界里，令敌人捉摸不透是非常关键的。
我们的主角漩涡鸣人所拥有的一个招数——多重影分身之术——就是一个很好的例子。
影分身是由鸣人身体的查克拉能量制造的，使用的查克拉越多，制造出的影分身越强。
针对不同的作战情况，鸣人可以选择制造出各种强度的影分身，有的用来佯攻，有的用来发起致命一击。
那么问题来了，假设鸣人的查克拉能量为 M M M，他影分身的个数最多为 N N N，那么制造影分身时有多少种不同的分配方法？
注意：
影分身可以分配 0 0 0点能量。
分配方案不考虑顺序，例如： M = 7 , N = 3 M=7,N=3 M=7,N=3，那么 (2,2,3) 和 (2,3,2) 被视为同一种方案。
#include<cstring>
#include<iostream>
using namespace std;
int f[15][15];
int main(){
int t;
cin>>t;
while(t--){
int m,n;
cin>>m>>n;
memset(f,0,sizeof f);
f[0][0]=1;
//把m划分成n个数
for(int i=0;i<=m;i++){
for(int j=1;j<=n;j++){
f[i][j]+=f[i][j-1];
if (i>=j)
f[i][j]+=f[i-j][j];
}
}
cout<<f[m][n]<<endl;
}
return 0;
}
NOIP2001数字划分
将整数 n n n分成 k k k份，且每份不能为空，任意两个方案不能相同(不考虑顺序)。
#include<iostream>
using namespace std;
int dp[205][10];
int main(){
int n,k;
cin>>n>>k;
dp[0][0]=0;
for(int i=0;i<=n;i++)
dp[i][1]=1;
for(int i=2;i<=n;i++){
for(int j=1;j<=k;j++){
dp[i][j]=dp[i-1][j-1];
if(i>j)
dp[i][j]+=dp[i-j][j];
}
}
cout<<dp[n][k];
}
滑雪
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 310;
int n, m;
int g[N][N];
int f[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
int dp(int x, int y){
int &v = f[x][y];
if (v != -1) return v;
v = 1;
for (int i = 0; i < 4; i ++ ){
int a = x + dx[i], b = y + dy[i];
if (a >= 1 && a <= n && b >= 1 && b <= m && g[x][y] > g[a][b])
v = max(v, dp(a, b) + 1);
}
return v;
}
int main(){
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
scanf("%d", &g[i][j]);
memset(f, -1, sizeof f);
int res = 0;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
res = max(res, dp(i, j));
printf("%d\n", res);
return 0;
}
没有上司的舞会
Ural 大学有 N N N名职员，编号为 1 ∼ N 1∼N 1∼N。
他们的关系就像一棵以校长为根的树，父节点就是子节点的直接上司。
每个职员有一个快乐指数，用整数 H i H_i Hi给出，其中 1 ≤ i ≤ N 1≤i≤N 1≤i≤N。
现在要召开一场周年庆宴会，不过，没有职员愿意和直接上司一起参会。
在满足这个条件的前提下，主办方希望邀请一部分职员参会，使得所有参会职员的快乐指数总和最大，求这个最大值。
#include<iostream>
#include<vector>
using namespace std;
const int N=6005;
int f[N][2];
vector<int> G[N];
bool st[N];
int H[N];
void dfs(int root){
f[root][1]=H[root];//选自己
for(auto node:G[root]){
dfs(node);
f[root][0]+=max(f[node][0],f[node][1]);//不选自己
f[root][1]+=f[node][0];//选自己
}
}
int main(){
int n;
cin>>n;
for(int i=1;i<=n;i++)
cin>>H[i];
for(int i=0;i<n-1;i++){
int u,v;
cin>>u>>v;
G[v].push_back(u);
st[u]=1;
}
int root=0;
for(int i=1;i<=n;i++){
if (st[i]==0){
root=i;
break;
}
}
dfs(root);
cout<<max(f[root][0],f[root][1]);
return 0;
}
最短Hamilton路径
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 20, M = 1 << N;
int n;
int w[N][N];
int f[M][N];
int main(){
cin >> n;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n; j ++ )
cin >> w[i][j];
memset(f, 0x3f, sizeof f);
f[1][0] = 0;
for (int i = 0; i < 1 << n; i ++ )
for (int j = 0; j < n; j ++ )
if (i >> j & 1)
for (int k = 0; k < n; k ++ )
if (i >> k & 1)
f[i][j] = min(f[i][j], f[i - (1 << j)][k] + w[k][j]);
cout << f[(1 << n) - 1][n - 1];
return 0;
}
蒙德里安的梦想
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 12, M = 1 << N;
int n, m;
LL f[N][M];
vector<int> state[M];
bool st[M];
int main(){
while (cin >> n >> m, n || m){
for (int i = 0; i < 1 << n; i ++ ){
int cnt = 0;
bool is_valid = true;
for (int j = 0; j < n; j ++ )
if (i >> j & 1){
if (cnt & 1){
is_valid = false;
break;
}
cnt = 0;
}
else cnt ++ ;
if (cnt & 1) is_valid = false;
st[i] = is_valid;
}
for (int i = 0; i < 1 << n; i ++ ){
state[i].clear();
for (int j = 0; j < 1 << n; j ++ )
if ((i & j) == 0 && st[i | j])
state[i].push_back(j);
}
memset(f, 0, sizeof f);
f[0][0] = 1;
for (int i = 1; i <= m; i ++ )
for (int j = 0; j < 1 << n; j ++ )
for (auto k : state[j])
f[i][j] += f[i - 1][k];
cout << f[m][0] << endl;
}
return 0;
}
计数问题
给定两个整数 a a a和 b b b，求 a a a和 b b b之间的所有数字中 0 ∼ 9 0∼9 0∼9的出现次数。
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 10;
/*
001~abc-1, 999
abc
1. num[i] < x, 0
2. num[i] == x, 0~efg
3. num[i] > x, 0~999
*/
int get(vector<int> num, int l, int r){
int res = 0;
for (int i = l; i >= r; i -- ) res = res * 10 + num[i];
return res;
}
int power10(int x){
int res = 1;
while (x -- ) res *= 10;
return res;
}
int count(int n, int x){
if (!n) return 0;
vector<int> num;
while (n){
num.push_back(n % 10);
n /= 10;
}
n = num.size();
int res = 0;
for (int i = n - 1 - !x; i >= 0; i -- ){
if (i < n - 1){
res += get(num, n - 1, i + 1) * power10(i);
if (!x) res -= power10(i);
}
if (num[i] == x) res += get(num, i - 1, 0) + 1;
else if (num[i] > x) res += power10(i);
}
return res;
}
int main(){
int a, b;
while (cin >> a >> b , a){
if (a > b) swap(a, b);
for (int i = 0; i <= 9; i ++ )
cout << count(b, i) - count(a - 1, i) << ' ';
cout << endl;
}
return 0;
}
Nim游戏
给定 n n n堆石子，两位玩家轮流操作，每次操作可以从任意一堆石子中拿走任意数量的石子（可以拿完，但不能不拿），最后无法进行操作的人视为失败。
问如果两人都采用最优策略，先手是否必胜。
#include <iostream>
#include <cstdio>
using namespace std;
/*
先手必胜状态：先手操作完，可以走到某一个必败状态
先手必败状态：先手操作完，走不到任何一个必败状态
先手必败状态：a1 ^ a2 ^ a3 ^ ... ^an = 0
先手必胜状态：a1 ^ a2 ^ a3 ^ ... ^an ≠ 0
*/
int main(){
int n;
scanf("%d", &n);
int res = 0;
for(int i = 0; i < n; i++) {
int x;
scanf("%d", &x);
res ^= x;
}
if(res == 0) puts("No");
else puts("Yes");
}
贪心
区间选点
给定 N N N个闭区间 [ a i , b i ] [a_i,b_i] [ai,bi]，请你在数轴上选择尽量少的点，使得每个区间内至少包含一个选出的点。
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int n;
struct Range{
int l, r;
bool operator< (const Range &W)const{
return r < W.r;
}
}range[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d%d", &range[i].l, &range[i].r);
sort(range, range + n);
int res = 0, ed = -2e9;
for (int i = 0; i < n; i ++ )
if (range[i].l > ed){
res ++ ;
ed = range[i].r;
}
printf("%d\n", res);
return 0;
}
最大不相交区间数量
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int n;
struct Range{
int l, r;
bool operator< (const Range &W)const{
return r < W.r;
}
}range[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d%d", &range[i].l, &range[i].r);
sort(range, range + n);
int res = 0, ed = -2e9;
for (int i = 0; i < n; i ++ )
if (ed < range[i].l){
res ++ ;
ed = range[i].r;
}
printf("%d\n", res);
return 0;
}
区间分组
给定 N N N个闭区间 [ a i , b i ] [a_i,b_i] [ai,bi]，请你将这些区间分成若干组，使得每组内部的区间两两之间（包括端点）没有交集，并使得组数尽可能小。
输出最小组数。
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 100010;
int n;
struct Range{
int l, r;
bool operator< (const Range &W)const{
return l < W.l;
}
}range[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ){
int l, r;
scanf("%d%d", &l, &r);
range[i] = {l, r};
}
sort(range, range + n);
priority_queue<int, vector<int>, greater<int>> heap;
for (int i = 0; i < n; i ++ ){
auto r = range[i];
if (heap.empty() || heap.top() >= r.l) heap.push(r.r);
else{
heap.pop();
heap.push(r.r);
}
}
printf("%d\n", heap.size());
return 0;
}
区间覆盖
给定 N N N个闭区间 [ a i , b i ] [a_i,b_i] [ai,bi]以及一个线段区间 [s,t]，请你选择尽量少的区间，将指定线段区间完全覆盖。
输出最少区间数，如果无法完全覆盖则输出 −1。
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int n;
struct Range{
int l, r;
bool operator< (const Range &W)const{
return l < W.l;
}
}range[N];
int main(){
int st, ed;
scanf("%d%d", &st, &ed);
scanf("%d", &n);
for (int i = 0; i < n; i ++ ){
int l, r;
scanf("%d%d", &l, &r);
range[i] = {l, r};
}
sort(range, range + n);
int res = 0;
bool success = false;
for (int i = 0; i < n; i ++ ){
int j = i, r = -2e9;
while (j < n && range[j].l <= st){
r = max(r, range[j].r);
j ++ ;
}
if (r < st){
res = -1;
break;
}
res ++ ;
if (r >= ed){
success = true;
break;
}
st = r;
i = j - 1;
}
if (!success) res = -1;
printf("%d\n", res);
return 0;
}
合并果子
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
int main(){
int n;
scanf("%d", &n);
priority_queue<int, vector<int>, greater<int>> heap;
while (n -- ){
int x;
scanf("%d", &x);
heap.push(x);
}
int res = 0;
while (heap.size() > 1){
int a = heap.top(); heap.pop();
int b = heap.top(); heap.pop();
res += a + b;
heap.push(a + b);
}
printf("%d\n", res);
return 0;
}
排队打水
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100010;
int n;
int t[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &t[i]);
sort(t, t + n);
reverse(t, t + n);
LL res = 0;
for (int i = 0; i < n; i ++ ) res += t[i] * i;
printf("%lld\n", res);
return 0;
}
货仓选址
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int n;
int q[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);
sort(q, q + n);
int res = 0;
for (int i = 0; i < n; i ++ ) res += abs(q[i] - q[n / 2]);
printf("%d\n", res);
return 0;
}
耍杂技的牛
农民约翰的 N N N头奶牛（编号为 1.. N 1..N 1..N）计划逃跑并加入马戏团，为此它们决定练习表演杂技。
奶牛们不是非常有创意，只提出了一个杂技表演：
叠罗汉，表演时，奶牛们站在彼此的身上，形成一个高高的垂直堆叠。
奶牛们正在试图找到自己在这个堆叠中应该所处的位置顺序。
这 N N N头奶牛中的每一头都有着自己的重量 W i W_i Wi以及自己的强壮程度 S i S_i Si。
一头牛支撑不住的可能性取决于它头上所有牛的总重量（不包括它自己）减去它的身体强壮程度的值，现在称该数值为风险值，风险值越大，这只牛撑不住的可能性越高。
您的任务是确定奶牛的排序，使得所有奶牛的风险值中的最大值尽可能的小。
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 50010;
int n;
PII cow[N];
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i ++ ){
int s, w;
scanf("%d%d", &w, &s);
cow[i] = {w + s, w};
}
sort(cow, cow + n);
int res = -2e9, sum = 0;
for (int i = 0; i < n; i ++ ){
int s = cow[i].first - cow[i].second, w = cow[i].second;
res = max(res, sum - s);
sum += w;
}
printf("%d\n", res);
return 0;
}
判断子序列
给定一个长度为 n n n的整数序列 a 1 , a 2 , … , a n a_1,a_2,…,a_n a1,a2,…,an以及一个长度为 m m m的整数序列 b 1 , b 2 , … , b m b_1,b_2,…,b_m b1,b2,…,bm。
请你判断 a a a序列是否为 b b b序列的子序列。
子序列指序列的一部分项按原有次序排列而得的序列，例如序列{
a 1 , a 3 , a 5 a_1,a_3,a_5 a1,a3,a5} 是序列 {
a 1 , a 2 , a 3 , a 4 , a 5 a_1,a_2,a_3,a_4,a_5 a1,a2,a3,a4,a5} 的一个子序列。
#include <iostream>
#include <cstring>
using namespace std;
const int N = 100010;
int n, m;
int a[N], b[N];
int main(){
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
for (int i = 0; i < m; i ++ ) scanf("%d", &b[i]);
int i = 0, j = 0;
while (i < n && j < m){
if (a[i] == b[j]) i ++ ;
j ++ ;
}
if (i == n) puts("Yes");
else puts("No");
return 0;
}
表达式求值
#include <iostream>
#include <cstring>
#include <algorithm>
#include <stack>
#include <unordered_map>
using namespace std;
stack<int> num;
stack<char> op;
void eval(){
auto b = num.top(); num.pop();
auto a = num.top(); num.pop();
auto c = op.top(); op.pop();
int x;
if (c == '+') x = a + b;
else if (c == '-') x = a - b;
else if (c == '*') x = a * b;
else x = a / b;
num.push(x);
}
int main(){
unordered_map<char, int> pr{
{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}};
string str;
cin >> str;
for (int i = 0; i < str.size(); i ++ ){
auto c = str[i];
if (isdigit(c)){
int x = 0, j = i;
while (j < str.size() && isdigit(str[j]))
x = x * 10 + str[j ++ ] - '0';
i = j - 1;
num.push(x);
}
else if (c == '(') op.push(c);
else if (c == ')'){
while (op.top() != '(') eval();
op.pop();
}
else{
while (op.size() && op.top() != '(' && pr[op.top()] >= pr[c]) eval();
op.push(c);
}
}
while (op.size()) eval();
cout << num.top() << endl;
return 0;
}
上机课
进制转换
将一个长度最多为 30 位数字的十进制非负整数转换为二进制数输出。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> div(vector<int> A, int b){
vector<int> C;
for (int i = A.size() - 1, r = 0; i >= 0; i -- ){
r = r * 10 + A[i];
C.push_back(r / b);
r %= b;
}
reverse(C.begin(), C.end());
while (C.size() && C.back() == 0) C.pop_back();
return C;
}
int main(){
string s;
while (cin >> s){
vector<int> A;
for (int i = 0; i < s.size(); i ++ )
A.push_back(s[s.size() - i - 1] - '0');
string res;
if (s == "0") res = "0";
else{
while (A.size()){
res += to_string(A[0] % 2);
A = div(A, 2);
}
}
reverse(res.begin(), res.end());
cout << res << endl;
}
return 0;
}
3374. 进制转换2
将 M 进制的数 X 转换为 N 进制的数输出。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
int main(){
int a, b;
string s;
cin >> a >> b >> s;
vector<int> A;
for (int i = 0; i < s.size(); i ++ ){
char c = s[s.size() - 1 - i];
if (c >= 'A') A.push_back(c - 'A' + 10);
else A.push_back(c - '0');
}
string res;
if (s == "0") res = "0";
else{
while (A.size()){
int r = 0;
for (int i = A.size() - 1; i >= 0; i -- ){
A[i] += r * a;
r = A[i] % b;
A[i] /= b;
}
while (A.size() && A.back() == 0) A.pop_back();
if (r < 10) res += to_string(r);
else res += r - 10 + 'a';
}
reverse(res.begin(), res.end());
}
cout << res << endl;
return 0;
}
重排链表
一个包含 n n n 个元素的线性链表 L = ( a 1 , a 2 , … , a n − 2 , a n − 1 , a n ) L=(a_1,a_2,…,a_{n−2},a_{n−1},a_n) L=(a1,a2,…,an−2,an−1,an)。
现在要对其中的结点进行重新排序，得到一个新链表 L ′ = ( a 1 , a n , a 2 , a n − 1 , a 3 , a n − 2 … ) L′=(a_1,a_n,a_2,a_{n−1},a_3,a_{n−2}…) L′=(a1,an,a2,an−1,a3,an−2…)
class Solution {
public:
void rearrangedList(ListNode* head) {
if (!head->next) return;
int n = 0;
for (auto p = head; p; p = p->next) n ++ ;
int left = (n + 1) / 2; // 前半段的节点数
auto a = head;
for (int i = 0; i < left - 1; i ++ ) a = a->next;
auto b = a->next, c = b->next;
a->next = b->next = NULL;
while (c) {
auto p = c->next;
c->next = b;
b = c, c = p;
}
for (auto p = head, q = b; q;) {
auto o = q->next;
q->next = p->next;
p->next = q;
p = p->next->next;
q = o;
}
}
};
打印日期
给出年份 y y y 和一年中的第 d d d 天，算出第 d d d 天是几月几号。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int months[13] = {
0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};
int is_leap(int year){ // 闰年返回1，平年返回0
if (year % 4 == 0 && year % 100 || year % 400 == 0)
return 1;
return 0;
}
int get_days(int y, int m){ // y年m月有多少天
if (m == 2) return months[m] + is_leap(y);
return months[m];
}
int main(){
int y, s;
while (cin >> y >> s){
int m = 1, d = 1;
s -- ;
while (s -- ){
if ( ++ d > get_days(y, m)){
d = 1;
if ( ++ m > 12){
m = 1;
y ++ ;
}
}
}
printf("%04d-%02d-%02d\n", y, m, d);
}
return 0;
}
日期累加
设计一个程序能计算一个日期加上若干天后是什么日期。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int months[13] = {
0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};
int is_leap(int year){
if (year % 4 == 0 && year % 100 || year % 400 == 0)
return 1;
return 0;
}
int get_days(int y, int m){
if (m == 2) return months[m] + is_leap(y);
return months[m];
}
int get_year_days(int y, int m){
if (m <= 2) return 365 + is_leap(y);
return 365 + is_leap(y + 1);
}
int main(){
int T;
cin >> T;
while (T -- ){
int y, m, d, a;
cin >> y >> m >> d >> a;
if (m == 2 && d == 29) a --, m = 3, d = 1;
while (a > get_year_days(y, m)){
a -= get_year_days(y, m);
y ++ ;
}
while (a -- ){
if ( ++ d > get_days(y, m)){
d = 1;
if ( ++ m > 12){
m = 1;
y ++ ;
}
}
}
printf("%04d-%02d-%02d\n", y, m, d);
}
return 0;
}
二叉树的带权路径长度
二叉树的带权路径长度(WPL)是二叉树中所有叶结点的带权路径长度之和，也就是每个叶结点的深度与权值之积的总和。
class Solution {
public:
int dfs(TreeNode* root, int depth) {
if (!root) return 0;
if (!root->left && !root->right) return root->val * depth;
return dfs(root->left, depth + 1) + dfs(root->right, depth + 1);
}
int pathSum(TreeNode* root) {
return dfs(root, 0);
}
};
二叉排序树
你需要写一种数据结构，来维护一些数，其中需要提供以下操作：
插入数值 x。
删除数值 x。
输出数值 x 的前驱(前驱定义为现有所有数中小于 x 的最大的数)。
输出数值 x 的后继(后继定义为现有所有数中大于 x 的最小的数)。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 1e8;
struct TreeNode{
int val;
TreeNode *left, *right;
TreeNode(int _val): val(_val), left(NULL), right(NULL) {}
}*root;
void insert(TreeNode* &root, int x){
if (!root) root = new TreeNode(x);
else if (x < root->val) insert(root->left, x);
else insert(root->right, x);
}
void remove(TreeNode* &root, int x){
if (!root) return;
if (x < root->val) remove(root->left, x);
else if (x > root->val) remove(root->right, x);
else{
if (!root->left && !root->right) root = NULL;
else if (!root->left) root = root->right;
else if (!root->right) root = root->left;
else{
auto p = root->left;
while (p->right) p = p->right;
root->val = p->val;
remove(root->left, p->val);
}
}
}
int get_pre(TreeNode* root, int x){
if (!root) return -INF;
if (root->val >= x) return get_pre(root->left, x);
return max(root->val, get_pre(root->right, x));
}
int get_suc(TreeNode* root, int x){
if (!root) return INF;
if (root->val <= x) return get_suc(root->right, x);
return min(root->val, get_suc(root->left, x));
}
int main(){
int n;
cin >> n;
while (n -- ){
int t, x;
cin >> t >> x;
if (t == 1) insert(root, x);
else if (t == 2) remove(root, x);
else if (t == 3) cout << get_pre(root, x) << endl;
else cout << get_suc(root, x) << endl;
}
return 0;
}
表达式树
请设计一个算法，将给定的表达式树(二叉树)转换为等价的中缀表达式(通过括号反映操作符的计算次序)并输出。
class Solution {
public:
string dfs(TreeNode* root) {
if (!root) return "";
if (!root->left && !root->right) return root->val;
return '(' + dfs(root->left) + root->val + dfs(root->right) + ')';
}
string expressionTree(TreeNode* root) {
return dfs(root->left) + root->val + dfs(root->right);
}
};
未出现过的最小正整数
给定一个长度为 n 的整数数组，请你找出未在数组中出现过的最小正整数。
class Solution {
public:
int findMissMin(vector<int>& nums) {
int n = nums.size();
vector<bool> hash(n + 1);
for (int x: nums)
if (x >= 1 && x <= n)
hash[x] = true;
for (int i = 1; i <= n; i ++ )
if (!hash[i])
return i;
return n + 1;
}
};
三元组的最小距离
定义三元组$ (a,b,c) （ （ （a , , ,b , , ,c 均 为 整 数 ） 的 距 离 均为整数）的距离 均为整数）的距离D=|a−b|+|b−c|+|c−a| 。 给 定 3 个 非 空 整 数 集 合 。 给定 3 个非空整数集合 。给定3个非空整数集合S_1,S_2,S_3 ， 按 升 序 分 别 存 储 在 3 个 数 组 中 。 请 设 计 一 个 尽 可 能 高 效 的 算 法 ， 计 算 并 输 出 所 有 可 能 的 三 元 组 ，按升序分别存储在 3 个数组中。 请设计一个尽可能高效的算法，计算并输出所有可能的三元组 ，按升序分别存储在3个数组中。请设计一个尽可能高效的算法，计算并输出所有可能的三元组 (a,b,c)（a∈S_1,b∈S_2,c∈S_3）$中的最小距离。
例如 S 1 = − 1 , 0 , 9 , S 2 = − 25 , − 10 , 10 , 11 , S 3 = 2 , 9 , 17 , 30 , 41 S_1={−1,0,9},S_2={−25,−10,10,11},S_3={2,9,17,30,41} S1=−1,0,9,S2=−25,−10,10,11,S3=2,9,17,30,41则最小距离为 2，相应的三元组为 (9,10,9)。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100010;
int l, m, n;
int a[N], b[N], c[N];
int main(){
scanf("%d%d%d", &l, &m, &n);
for (int i = 0; i < l; i ++ ) scanf("%d", &a[i]);
for (int i = 0; i < m; i ++ ) scanf("%d", &b[i]);
for (int i = 0; i < n; i ++ ) scanf("%d", &c[i]);
LL res = 1e18;
for (int i = 0, j = 0, k = 0; i < l && j < m && k < n;){
int x = a[i], y = b[j], z = c[k];
res = min(res, (LL)max(max(x, y), z) - min(min(x, y), z));
if (x <= y && x <= z) i ++ ;
else if (y <= x && y <= z) j ++ ;
else k ++ ;
}
printf("%lld\n", res * 2);
return 0;
}
众数
给定一个整数序列，其中包含 n 个非负整数，其中的每个整数都恰好有 m 位，从最低位到最高位，依次编号为 1∼m 位。
现在，请你统计该序列各个位的众数。
第 i 位的众数是指，在给定的 n 个整数的第 i 位上，出现次数最多的最小数字。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int s[6][10];
int main(){
int n, m;
scanf("%d%d", &n, &m);
while (n -- ){
int x;
scanf("%d", &x);
for (int i = 0; i < m; i ++ ){
s[i][x % 10] ++ ;
x /= 10;
}
}
for (int i = 0; i < m; i ++ ){
int k = 0;
for (int j = 0; j < 10; j ++ )
if (s[i][k] < s[i][j])
k = j;
printf("%d\n", k);
}
return 0;
}
玛雅人的密码
玛雅人有一种密码，如果字符串中出现连续的 2012 四个数字就能解开密码。
给定一个长度为 N 的字符串，该字符串中只含有 0,1,2 三种数字。
可以对该字符串进行移位操作，每次操作可选取相邻的两个数字交换彼此位置。
请问这个字符串要移位几次才能解开密码。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <unordered_map>
using namespace std;
int n;
int bfs(string start){
queue<string> q;
unordered_map<string, int> dist;
dist[start] = 0;
q.push(start);
while (q.size()){
auto t = q.front();
q.pop();
for (int i = 0; i < n; i ++ )
if (t.substr(i, 4) == "2012")
return dist[t];
for (int i = 1; i < n; i ++ ){
string r = t;
swap(r[i], r[i - 1]);
if (!dist.count(r)){
dist[r] = dist[t] + 1;
q.push(r);
}
}
}
return -1;
}
int main(){
string start;
cin >> n >> start;
cout << bfs(start) << endl;
return 0;
}
等差数列
有一个特殊的 n n n行 m m m列的矩阵 Aij，每个元素都是正整数，每一行和每一列都是独立的等差数列。
在某一次故障中，这个矩阵的某些元素的真实值丢失了，被重置为 0。
现在需要你想办法恢复这些元素，并且按照行号和列号从小到大的顺序（行号为第一关键字，列号为第二关键字，从小到大）输出能够恢复的元素。
#include<bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
const int N = 1010;
int n, m, a[N][N], r[N][4], c[N][4];
queue<int> q;
//给坐标[i, j]赋值为x
//同时判断第i 行是否有两个点，第j 列是否有两个点
void add(int i, int j, int x){
a[i][j] = x;
//如果第i 行一个点都没有，给定第一个点的值到r[i][0],列数为r[i][1]
//其次如果第i 行没第二个点，给定第二个点的值到r[i][2],列数为r[i][3]，并且加入到bfs中
//如果还有第三个点其实就不进行操作
if(!r[i][0]) r[i][0] = x, r[i][1] = j;
else if(!r[i][2]) r[i][2] = x, r[i][3] = j, q.push(i * 2 + 1);//如果是行扩展，加入奇数
//对列进行同等操作
if(!c[j][0]) c[j][0] = x, c[j][1] = i;
else if(!c[j][2]) c[j][2] = x, c[j][3] = i, q.push(j * 2);//如果是列扩展，加入偶数
}
int main(){
cin >> n >> m;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++){
int x;
scanf("%d",&x);
if(x) add(i, j, x);//给[i, j] 赋值 x
}
vector<pii> ans;//frist 里面存[x, y]坐标，sceond里面存值
while(q.size()){
int now = q.front() / 2, f = q.front() % 2; q.pop();//取出行号或者列号
if(f){//如果是行扩展
int suf = (r[now][2] - r[now][0]) / (r[now][3] - r[now][1]);//得到等差值
int start = r[now][0] - (r[now][1] * suf);//求得第一个点的值
for(int i = 0; i < m; i++){
if(!a[now][i]){//如果a[now][i]是0，将a[now][i]按照等差公式赋值
add(now, i, start + i * suf);//赋值
ans.push_back({now * 2000 + i, start + i * suf});//加入[坐标，值]答案数组中
}
}
}
else{//列扩展，以下同理
int suf = (c[now][2] - c[now][0]) / (c[now][3] - c[now][1]);
int start = c[now][0] - (c[now][1] * suf);
for(int i = 0; i < n; i++){
if(!a[i][now]){
add(i, now, start + i * suf);
ans.push_back({i * 2000 + now, start + i * suf});
}
}
}
}
sort(ans.begin(), ans.end());//按照坐标排序
for(auto [xy, val] : ans){
printf("%d %d %d\n",xy/2000+1,xy%2000+1,val);//输出答案
}
return 0;
}
3441. 重复者
给定一个仅包含一种字符和空格的模板，将之不断重复扩大。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
int n;
vector<string> p;
vector<string> g(int k){
if (k == 1) return p;
auto s = g(k - 1);
int m = s.size();
vector<string> res(n * m);
for (int i = 0; i < n * m; i ++ )
res[i] = string(n * m, ' ');
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n; j ++ )
if (p[i][j] != ' ')
for (int x = 0; x < m; x ++ )
for (int y = 0; y < m; y ++ )
res[i * m + x][j * m + y] = s[x][y];
return res;
}
int main(){
while (cin >> n, n){
p.clear();
getchar(); // 读掉n后的回车
for (int i = 0; i < n; i ++ ){
string line;
getline(cin, line);
p.push_back(line);
}
int k;
cin >> k;
auto res = g(k);
for (auto& s: res) cout << s << endl;
}
return 0;
}
3382. 整数拆分
一个整数总可以拆分为 2 的幂的和
用 f(n) 表示 n 的不同拆分的种数，例如 f(7)=6。
要求编写程序，读入 n，输出 f(n)mod1e9
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1000010, MOD = 1e9;
int n;
int f[N];
int main(){
scanf("%d", &n);
f[0] = 1;
for (int i = 1; i <= n; i *= 2)
for (int j = i; j <= n; j ++ )
f[j] = (f[j] + f[j - i]) % MOD;
cout << f[n] << endl;
return 0;
}
位操作练习
给出两个不大于 65535 的非负整数，判断其中一个的 16 位二进制表示形式，是否能由另一个的 16 位二进制表示形式经过循环左移若干位而得到。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int main(){
int a, b;
while (cin >> a >> b){
string x, y;
for (int i = 15; i >= 0; i -- ){
x += to_string(a >> i & 1);
y += to_string(b >> i & 1);
}
y += y;
if (y.find(x) != -1) puts("YES");
else puts("NO");
}
return 0;
}
最小面积子矩阵
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 110, INF = 100000;
int n, m, k;
int s[N][N];
int main(){
cin >> n >> m >> k;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ ){
cin >> s[i][j];
s[i][j] += s[i - 1][j];
}
int res = INF;
for (int x = 1; x <= n; x ++ )
for (int y = x; y <= n; y ++ ){
for (int i = 1, j = 1, sum = 0; i <= m; i ++ ){
sum += s[y][i] - s[x - 1][i];
while (sum - (s[y][j] - s[x - 1][j]) >= k){
sum -= s[y][j] - s[x - 1][j];
j ++ ;
}
if (sum >= k) res = min(res, (y - x + 1) * (i - j + 1));
}
}
if (res == INF) res = -1;
cout << res << endl;
return 0;
}
矩阵幂
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 15;
int n, m;
int w[N][N];
void mul(int c[][N], int a[][N], int b[][N]){
static int tmp[N][N];
memset(tmp, 0, sizeof tmp);
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n; j ++ )
for (int k = 0; k < n; k ++ )
tmp[i][j] += a[i][k] * b[k][j];
memcpy(c, tmp, sizeof tmp);
}
int main(){
cin >> n >> m;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n; j ++ )
cin >> w[i][j];
int res[N][N] = {0};
for (int i = 0; i < n; i ++ ) res[i][i] = 1;
while (m -- ) mul(res, res, w);
for (int i = 0; i < n; i ++ ){
for (int j = 0; j < n; j ++ )
cout << res[i][j] << ' ';
cout << endl;
}
return 0;
}
C翻转
给定一个 5×5 的矩阵，对其进行翻转操作。
操作类型共四种，具体形式如下：
1 2 x y，表示将以第 x 行第 y 列的元素为左上角，边长为 2 的子矩阵顺时针翻转 90 度。
1 3 x y，表示将以第 x 行第 y 列的元素为左上角，边长为 3 的子矩阵顺时针翻转 90 度。
2 2 x y，表示将以第 x 行第 y 列的元素为左上角，边长为 2 的子矩阵逆时针翻转 90 度。
2 3 x y，表示将以第 x 行第 y 列的元素为左上角，边长为 3 的子矩阵逆时针翻转 90 度。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 5;
int n;
int g[N][N];
void rotate(int x, int y, int m){
int w[N][N];
memcpy(w, g, sizeof g);
for (int i = 0; i < m; i ++ )
for (int j = 0, k = m - 1; j < m; j ++, k -- )
w[i][j] = g[x + k][y + i];
for (int i = 0; i < m; i ++ )
for (int j = 0; j < m; j ++ )
g[x + i][y + j] = w[i][j];
}
int main(){
for (int i = 0; i < 5; i ++ )
for (int j = 0; j < 5; j ++ )
cin >> g[i][j];
int a, b, x, y;
cin >> a >> b >> x >> y;
x --, y -- ;
if (a == 1) rotate(x, y, b);
else{
for (int i = 0; i < 3; i ++ )
rotate(x, y, b);
}
for (int i = 0; i < 5; i ++ ){
for (int j = 0; j < 5; j ++ )
cout << g[i][j] << ' ';
cout << endl;
}
return 0;
}
最长平衡串
给定一个只含 01 的字符串，找出它的最长平衡子串的长度。
如果一个 01 字符串包含的 0 和 1 的个数相同，则称其为平衡串。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
using namespace std;
const int N = 1000010;
int n;
char str[N];
int main(){
scanf("%s", str + 1);
unordered_map<int, int> hash;
n = strlen(str + 1);
int res = 0;
hash[0] = 0;
for (int i = 1, s = 0; i <= n; i ++ ){
if (str[i] == '0') s ++ ;
else s -- ;
if (hash.count(s)) res = max(res, i - hash[s]);
else hash[s] = i;
}
printf("%d\n", res);
return 0;
}
Python
差
读取四个整数 A,B,C,D，并计算 (A×B−C×D) 的值。
a = int(input())
b = int(input())
c = int(input())
d = int(input())
print("DIFERENCA = %d" % (a * b - c * d))
倍数
a, b = map(int, input().split(' '))
if a % b == 0 or b % a == 0:
print("Sao Multiplos")
else:
print("Nao sao Multiplos")
零食
x, y = map(int, input().split(' '))
prices = [0, 4, 4.5, 5, 2, 1.5]
print("Total: R$ %.2lf" % (prices[x] * y))
字符串长度
print(len(input()))
递增序列
while True:
x = int(input())
if x == 0:
break
for i in range(1, x + 1):
print(i, end=' ')
print()
质数
import math
n = int(input())
for i in range(n):
x = int(input())
for j in range(2, int(math.sqrt(x)) + 1):
if x % j == 0:
print(x, "is not prime")
break
else:
print(x, "is prime")
数组的右上
t = input()
s, c = 0, 0
for i in range(12):
d = list(map(float, input().split(' ')))
for j in range(12):
if j > i:
s += d[j]
c += 1
if t == "M":
s /= c
print("%.1f" % (s))
蛇形矩阵
n, m = map(int, input().split())
res = [[0 for j in range(m)] for i in range(n)]
dx, dy = [-1, 0, 1, 0], [0, 1, 0, -1]
x, y, d = 0, 0, 1
for i in range(1, n * m + 1):
res[x][y] = i
a, b = x + dx[d], y + dy[d]
if a < 0 or a >= n or b < 0 or b >= m or res[a][b]:
d = (d + 1) % 4
a, b = x + dx[d], y + dy[d]
x, y = a, b
for i in range(n):
for j in range(m):
print(res[i][j], end = ' ')
print()
排列
n = int(input())
path = [0 for i in range(n)]
used = [False for i in range(n)]
def dfs(u):
if u == n:
for i in range(n):
print(path[i] + 1, end=' ')
print()
else:
for i in range(n):
if not used[i]:
path[u] = i
used[i] = True
dfs(u + 1)
used[i] = False
path[u] = 0
dfs(0)
a+b
#输入无行数限制
while True:
try:
a, b = map(int, input().split())
print(a + b)
except:
break
#先输入行数
n = int(input())
while n:
a, b = map(int, input().split())
print(a + b)
n -= 1
#输入00结束
while True:
a, b = map(int, input().split())
if a != 0 and b != 0: print(a + b)
else: break
行内数组求和
#给定数组长度,输入0停止
while True:
arr = list(map(int, input().split()))
if arr[0] == 0: break
else: print(sum(arr[1:]))
#给定总数组数量
n = int(input())
while n:
print(sum(list(map(int, input().split()))[1:]))
n -= 1
#不给定总数组数量
while True:
try: # 对于没有行提示的，使用try except就很简单
print(sum(list(map(int, input().split()))[1:]))
except:
break
# 不给行数和每行数组长度
while True:
try:
print(sum(list(map(int, input().split()))))
except:
break
字符串排序
#给定字符数量
n = int(input())
arr = input().split()
arr.sort()
print(' '.join(arr))
#不给定字符数量
while True:
try:
arr = input().split()
arr.sort()
print(" ".join(arr))
except:
break
#逗号分割
while True:
try:
arr = input().split(",")
arr.sort()
print(",".join(arr))
except:
break