Article directory
Baidu Pinecone Elite Class – OJ Competition (4th time)
1, before K decimal
**Title: **Brother Xiaoma now has two sequences a, b, (1≤i, j≤n) of length n in his hand, and the numbers ai + bjcan be constructed through n * nit, and the first k of the constructed numbers are small number.
/**
输入格式:第一行2个数n, k;
第二行n个数，表示数列a;
第三行n个数，表示数列b。
输出格式:从小到大输出前k小的数。
样例1 输入：3 3
2 6 6
1 4 8
输出：3 6 7
备注
其中:1≤n ≤1e4，1≤k≤n*n≤1e4， 1≤ai，bj ≤1e6
*/
#include<bits/stdc++.h>
using namespace std;
const int N = 1e8 + 5;
int a[N];
int b[N];
struct Node{
int ida, idb, num;
bool operator > (const Node &a) const{
return num > a.num;
}
} node;
priority_queue<Node, vector<Node>, greater<Node>> q;//小根堆
int main(){
int n, k;
cin >> n >> k;
for(int i = 0; i < n; i++){
cin >> a[i];
}
for(int i = 0; i < n; i++){
cin >> b[i];
}
sort(a, a + n);//将数列a从小到大排序
sort(b, b + n);//将数列b从小到大排序
//将a数列的值与b的第一个相加入堆
for(int i = 0; i < n; i++){
q.push({i, 0, a[i] + b[0]});
}
while(k--){
node = q.top();//此时堆的最前面是最小的
q.pop();//将其从堆中去除
cout << node.num << " ";
//再将b的下一个与a的第一个相加
node.num = a[node.ida] + b[++node.idb];
q.push(node);//入堆得到最小堆
}
return 0;
}
2. The first k decimals (advanced)
**Title:**Give you n known series of length m, you can select one element from each series at a time, a total of n numbers, put the sum of these n numbers into the α array, Exhaustively enumerate all possible options and put them into the α array.
Brother Xiaoma, please calculate the first k decimals in the sequence a?
/**
输入格式:输入n，m，k ;
接下来n行，每一行m个数，空格分隔，一行表示一个数列，共n行。
输出格式:从小到大输出前k 小的数。
样例1 输入：2 2 2
1 2
1 2
输出：2 3
备注
其中:1<n ≤200,1 ≤k≤m≤200
*/
#include<bits/stdc++.h>
using namespace std;
const int N = 4e5 + 5;
int rec[400][400], n, m, k;
vector<int> a;
vector<int> u, v;
int main(){
cin >> n >> m >> k;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
scanf("%d", &rec[i][j]);
}
}
for (int i = 1; i <= m; i++){
u.push_back(rec[1][i]);
}
for (int i = 2; i <= n; i++) {
sort(rec[i] + 1, rec[i] + 1 + m);
for (int j = 1; j <= k; j++){
v.push_back(rec[i][j]);
}
for (int x: u) {
for (int y: v) {
a.push_back(x + y);
}
}
sort(a.begin(), a.end());
while (u.size() != 0){
u.pop_back();
}
while (v.size() != 0){
v.pop_back();
}
for (int i = 0; i < k; i++){
u.push_back(a[i]);
}
while (a.size() != 0){
a.pop_back();
}
}
for (int i = 0; i < k; i++) {
printf("%d ", u[i]);
}
return 0;
}
Three, hacker code brother
**Title: **Brother Xiaoma is a hacker. He recently won the lottery. Since he has not redeemed it, Brother Xiaoma is very worried that the lottery ticket will be stolen (Brother Xiaoma is too cautious). He wants to set up a new one for his safe. password, by the way he wants you to test the encoding.
There are two encoding methods:
- For a certain character in a known string, all become another character. For example, all the A appearing in it are replaced with B;
- For the current string, shuffle the character order.
First give you an encrypted string and a string before encryption, and judge whether the string before encryption can get the string after encryption. All characters in the string are capital letters.
/**
输入格式:第一行加密后的字符串;
第二行加密前的字符串;
有多组数据输入。
输出格式:输出YES或NO。
样例1 输入：JWPUDJSTVP
VICTORIOUS
MAMA
ROME
HAHA
HEHE
AAA
AAA
NEERCISTHEBEST
SECRETMESSAGES
输出：YES
NO
YES
YES
NO
备注
所有字符串长度不超过100。
*/
#include<bits/stdc++.h>
using namespace std;
const int LETTER = 26;
// 判断两个字符串是否有可能是经过指定加密算法处理前后的两串
bool isSamePossiblely(string str1, string str2){
// 长度不同就一定不是
int strlen = str1.length();
if(strlen != str2.length()){
return false;
}
// 将数组全部初始化为0
int ary1[LETTER]={0}, ary2[LETTER]={0};
// 先统计各字符串中不同字符的出现次数
for(int i = 0; i < strlen; i++){
ary1[str1[i]-'A']++;
ary2[str2[i]-'A']++;
}
// 消除字符间的顺序差异
sort(ary1, ary1 + LETTER);
sort(ary2, ary2 + LETTER);
// 判断是否可能是变换前后的两字符串
for(int i = 0; i < LETTER; i++){
if(ary1[i] != ary2[i]){
return false;
}
}
return true;
}
int main(){
string str1, str2;
while(cin >> str1 >> str2) {
if(isSamePossiblely(str1, str2)){
cout << "YES" << endl;
}else{
cout << "NO" << endl;
}
}
return 0;
}
4. To classify words
**Title:** There are the following word classification principles: two words can be classified into one category if and only if the number of letters that make up the two words is equal.
The existing n words are all composed of capital letters, and the length of each word is less than or equal to 100. Ask words to be divided into categories.
/**
输入格式:第一行输入单词个数n ;以下n行每行一个单词。
输出格式:—个整数表示类数。
样例1 输入：3
AABAC
CBAAA
AAABB
输出：2
备注
对于70%的数据满足n≤100 ;
对于100%的数据满足n≤10000。
*/
#include<bits/stdc++.h>
using namespace std;
int n, ans;
string s;
map<string, int> mp;
int main(){
cin >> n;
for(int i = 1; i <= n; i++){
cin >> s;
//给输入的单词排序
sort(s.begin(), s.end());
//如果value值存在则返回1，不存在则返回0
if(!mp[s]){
ans++;
mp[s] = 1;
}
}
cout << ans;
return 0;
}
5. Addition of unary polynomials
**Title:**Brother Xiaoma gives two unary polynomials LA and LB, please add these two unary polynomials to get a new unary polynomial Lc.
For example:
- THE =
1 - 10x^6 + 2x^8 + 7x^14 - LB =
- x^4 + 10x^6 - 3x^10 + 8x^14 + 4x^18 - LC = LA + LB =
1- x^4 + 2x^8 - 3x^10+15x^14 +4x^18
/**
输入格式:第一行两个整数n和m，分别表示LA和LB的项数;接下来n行，每行输入LA的每一项的信息，两个整数分别表示该项的系数coef和次数earpn，输入保证次数递增;接下来m行，每行输入LB 的每一项的信息，两个整数分别表示该项的系数coef和次数earpn，输入保证次数递增。
输出格式:输出k行，k为Lc的项数，每行输出Lc的每一项的信息，两个整数分别表示该项的系数coef和次数expn，输出按次数递增。
样例1 输入：4 5
1 0
-10 6
2 8
7 14
-1 4
10 6
-3 10
8 14
4 18
输出：1 0
-1 4
2 8
-3 10
15 14
4 18
备注
对于100%的数据:0≤n, m ≤ 1000000，一1000 ≤coef ≤ 1000，一1000000000 ≤expn ≤ 1000000000 。
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N =2e6 + 5;
struct NODE {
ll nex, coef, expn;//下一个节点 系数 指数
}node[N];
int n, m, head, tail, pos;
ll coefA[N], expnA[N], coefB[N], expnB[N];
void insert(int curr,ll val1,ll val2){
node[++pos].coef = val1;//给节点赋值
node[pos].expn = val2;
node[pos].nex = node[curr].nex;//将节点插入到curr的后面
node[curr].nex = pos;
if (!node[pos].nex){//到链表尾部为空
tail = pos;//将当前节点赋值为尾节点
}
}
int main(){
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++){
scanf("%lld%lld", &coefA[i], &expnA[i]);
}
for (int i = 1; i <= m; i++){
scanf("%lld%lld", &coefB[i], &expnB[i]);
}
int l = 1, r = 1;
while (l <= n && r <= m) {
if (expnA[l] == expnB[r]) {//A和B的指数相等
insert(tail, coefA[l] + coefB[r], expnA[l]);
l++;
r++;
}else{
if (expnA[l] < expnB[r]) {//A的指数小，插入A
insert(tail, coefA[l], expnA[l]);
l++;
}else{//B的指数小，插入B
insert(tail, coefB[r], expnB[r]);
r++;
}
}
}
while (l <= n){//如B插完，则将剩余的A插入
insert(tail, coefA[l], expnA[l]);
l++;
}
while (r <= m){//如A插完，则将剩余的B插入
insert(tail, coefB[r], expnB[r]);
r++;
}
for (int i = node[head].nex; i != 0; i = node[i].nex){
if (node[i].coef != 0){
printf("%lld %lld\n", node[i].coef, node[i].expn);
}
}
return 0;
}
6. Queue Arrangement
**Title:** In a school, the teacher wants to arrange N students in the class in a row. The students are numbered from 1 to N. He adopts the following method:
- Arrange student No. 1 into the queue first, and he is the only one in the queue at this time;
- Students numbered 2-N are listed one by one, and students numbered i are listed in the following way: the teacher designates the student numbered i to stand with a student numbered 1~(i-1) (that is, the student who has been listed before) to the left or right of the
- Remove M (M<N) students from the queue, and the order of other students remains unchanged.
After all the students have been arranged in line according to the above method, the teacher wants to know the numbers of all the students from left to right.
/**
输入格式:第1行为一个正整数N，表示了有N个同学;
第2―N行，第i行包含两个整数k,p，其中k为小于i的正整数, p为0或者1。若p为0，则表示将i号同学插入到k号同学的左边，p为1则表示插入到右边;
第N＋1行为一个正整数M，表示去掉的同学数目;
接下来M行，每行一个正整数x，表示将x号同学从队列中移去，如果x号同学已经不在队列中则忽略这—条指令。
输出格式:1行，包含最多N个由空格隔开的正整数，表示了队列从左到右所有同学的编号。
样例1 输入:4
1 0
2 1
1 0
2
3
3
输出：2 4 1
备注
样例解释:
将同学2插入至同学1左边，此时队列为:21;
将同学3插入至同学2右边，此时队列为:231;
将同学4插入至同学1左边，此时队列为:2341;
将同学3从队列中移出，此时队列为:241;
同学3已经不在队列中，忽略最后一条指令
最终队列:241
数据范围:N，M ≤100000 。
*/
#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
struct line{
int pr, ne;//前一个，当前节点
bool mk;//判断是否已经删除
}a[N];
int f[N];
int n , m;
int ans = 0;
int main(){
cin >> n;
int i , j;
int k , p;
a[0].ne = 1;//头节点
a[0].pr = 0;//链头为null
a[1].mk = 1;//未删除
a[1].pr = 0;
for(i = 2 ; i <= n ; i++){
cin >> k >> p;
a[i].mk = 1;
if(p == 0){//插到左边
a[a[k].pr].ne = i;
a[i].pr = a[k].pr;
a[k].pr = i;
a[i].ne = k;
}else{//插到右边
a[a[k].ne].pr = i;
a[i].pr = k;
a[i].ne = a[k].ne;
a[k].ne = i;
}
}
cin >> m;
int x;
for(i = 1 ; i <= m ; i++){
cin >> x;
if(!a[x].mk){//当mk=0时表示已经删除
continue;
}
a[x].mk = 0;
a[a[x].pr].ne = a[x].ne;
a[a[x].ne].pr = a[x].pr;
}
printf("%d",a[0].ne);
int now = a[0].ne;
now = a[now].ne;
while(now){
printf(" %d",now);
now = a[now].ne;
}
cout << endl;
return 0;
}
7. Water supply pipeline
**Title:** It was originally planned to build many sewer pipes between n cities. The managers found that these pipes would form a loop, and the sewers only needed to connect the cities, so the loop would increase the construction cost. So I hope you can help find redundant pipelines for optimization. Of course, there is a difference between a pipeline and a pipeline. For example, sij is used to represent the pipeline management fee from i to j, and the smaller the sij, the lower the pipeline management fee. Can some piping be eliminated to minimize total management costs. Find the lowest management cost (there is no pipeline that becomes a loop with itself).
/**
输入格式:第一行有两个数n和k表示城市数量和管道数量;
接下来k行，每行都有三个数i，j，c表示城市i和城市j之间的管道成本为c
输出格式:一个正整数，最低管理成本。
样例1 输入：3 3
1 2 3
1 3 5
2 3 8
输出：8
备注
其中:1 <n, k <100
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 7;
struct NODE{
int i, j, c;
bool operator<(const NODE &t)const {
return c < t.c;//升序
}
}p[N];
int fa[N], n, k, ans;
void init(){//数据初始化
for (int i = 1; i < N; i++){
fa[i] = i;
}
}
int find(int x){//查找
return x == fa[x] ? x: (fa[x] = find(fa[x]));
}
void mer(int x, int y) {//合并
x = find(x);
y = find(y);
if (x != y){
fa[x] = y;
}
}
int main(){
init();
cin >> n >> k;
for (int i = 1; i <= k; i++){
cin >> p[i].i >> p[i].j >> p[i].c;
}
sort(p + 1, p + 1 + k);//将管道按照成本升序
for (int i = 1; i <= k; i++){
if (find(p[i].i) != find(p[i].j)){
ans += p[i].c;
mer(p[i].i, p[i].j);
}
}
cout << ans;
return 0;
}
Eight, quick row deformation
**Title:**There are n numbers, ask how to make the elements in the array into ascending order by exchanging two numbers of adjacent items each time.
eg : 9 8 7 6 5，
into ascending order:
5 6 7 8 9。
Requires 10 neighbor exchanges.
/**
输入格式:第一行输入一个整数n ，表示序列长度;
第二行输入n个数。
输出格式:输出一个整数，表示最少的交换次数。
样例1 输入：5
9 8 7 6 5
输出：10
备注
其中:1<n ≤200000
*/
#include <bits/stdc++.h>
using namespace std;
const int N=2e5 +7;
int a[N], b[N], c[N], n;
long long ans;
//树状数组模板
int lowbit(int x){
return x &-x;
}
void add(int i, int x){
for (; i <= n; i += lowbit(i))
c[i] += x;
}
int sum(int i){
int ans = 0;
for (; i > 0; i -= lowbit(i))
ans += c[i];
return ans;
}
bool cmp(const int x, const int y){
if (b[x] == b[y]){
return x > y;
}
return b[x] > b[y];
}
int main(){
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> b[i];
a[i] = i;
}
sort(a + 1, a + n + 1, cmp);
for (int i = 1; i <= n; i++){
add(a[i], 1);
ans += sum(a[i] - 1);
}
cout << ans <<endl;
return 0;
}
Nine, reverse order
**Title:**Given an arrangement p of length n, find an element so that the records arranged after taking this element out of the arrangement are the most. A record is an element ai such that for every positive integer 1≤j <i, ai > aj.
/**
输入格式:第一行为n ;第二行为排列p 。
输出格式:一个整数即笞菜。
样例1 输入：5
5 1 2 3 4
输出：5
备注
其中:1 ≤n, a[i]≤1e5
当答案不唯一时，输出较小的数。如:
(1) 3 4 5 1 2
(2) 3 4 5 2 1
上面两种情况，均输出1。
*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e5 + 7;
int n;
int c[N],chg [N];
int lowbit(int x){
return x & -x;
}
void add(int x){
for (;x < N;x += lowbit(x)){
c[x]++;
}
}
int sum(int x){
int ret = 0;
for (;x> 0;x -= lowbit(x)){
ret += c[x];
}
return ret;
}
int main(){
cin >> n;
int x=0,maxx =0;
for (int i=1;i<=n;i++){
cin >>x;
maxx = max(maxx,x);
int cnt = sum(x);
if (cnt ==i-1){
chg [maxx]--;
}
else if (cnt ==i-2){
chg [maxx]++;
}
add (x);
}
int ans = 1;
for (int i = 1;i <= n;i++){
if (chg[i] > chg[ans]){
ans = i;
}
}
cout <<ans <<endl;
return 0;
}
10. Segment tree
**Title:**Line segment tree template question. Knowing a number sequence, you need to perform the following two operations: add k to each number in a certain interval, and find the sum of each number in a certain interval.
/**
输入格式:第一行包含两个整数n,m ( 5 ≤n, m ≤10%),分别表示该数列数字的个数和操作的总个数;
第二行包含n个用空格分隔的整数(0一10°)，其中第i个数字表示数列第i项的初始值;
接下来m行每行包含3或4个整数，表示一个操作，具体如下:1 x y k :将区间[x, y]内每个数加上k。
2 x y: 输出区间[x, y]内的数的和。其中: 1≤≤y ≤n,0 ≤k ≤105。
输出格式:输出包含若干行整数，即为所有操作2的结果。
样例1 输入：5 5
1 5 4 2 3
2 2 4
1 2 3 2
2 3 4
1 1 5 1
2 1 4
输出：11
8
20
备注
线段树模板
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll N = 1e5 + 5;
struct NODE{
ll l, r, val, lz;
}tree[4 * N];
ll a[N];
void build(ll p, ll l, ll r){
tree[p].l = l;
tree[p].r = r;
if(l == r){
tree[p].val = a[l];
return;
}
ll mid = l + ((r - l) >> 1);
build(p * 2, l, mid);
build(p * 2 + 1, mid + 1, r);
tree[p].val = tree[p * 2].val + tree[p * 2 + 1].val;
}
void lazy(ll p, ll v){
ll s = tree[p].l;
ll t = tree[p].r;
tree[p].val += (t - s + 1) * v;
tree[p].lz += v;
}
void pushdown(ll p){
lazy(2 * p, tree[p].lz);
lazy(2 * p + 1, tree[p].lz);
tree[p].lz = 0;
}
void update(ll l, ll r, ll c, ll p){
ll s = tree[p].l;
ll t = tree[p].r;
if(l <= s && t <= r){
return lazy(p, c);
}
if(tree[p].lz && s != t){
pushdown(p);
}
ll mid = s + ((t - s) >> 1);
if(l <= mid){
update(l, r, c, p * 2);
}
if(r > mid){
update(l, r, c, p * 2 + 1);
}
tree[p].val = tree[p * 2].val + tree[p * 2 + 1].val;
}
ll query(ll l, ll r, ll p){
ll s = tree[p].l;
ll t = tree[p].r;
if(l <= s && t <= r){
return tree[p].val;
}
if(tree[p].lz){
pushdown(p);
}
ll mid = s + ((t - s) >> 1);
ll sum = 0;
if(l <= mid){
sum = query(l, r, p * 2);
}
if(r > mid){
sum += query(l, r, p * 2 + 1);
}
return sum;
}
int main(){
ll n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++){
cin >> a[i];
}
build(1, 1, n);
while(m--){
ll op, x, y, k;
cin >> op;
if(op == 1){
cin >> x >> y >> k;
update(x, y, k, 1);
}else{
cin >> x >> y;
cout << query(x, y, 1) << endl;
}
}
return 0;
}
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