A - Odd Divisor
判断一个数是否能被大于1的奇数整除
首先排除所有的奇数，在偶数中通常能拆成2*n，如果n为奇数，那么就可以被奇数整除。如果要求没有奇数的因子，那么这个数一定是2n ，所以用log2函数来判断一下就可以了。
#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
const int N = 1e5 + 10;
const int M = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
typedef long long ll;
typedef pair<int, int> PII;
int T, n, m;
int a[N], b[N], c[N];
void solve() {
cin >> T;
while (T--) {
ll x; cin >> x;
if (int(log2(x)) == log2(x))
cout << "NO" << endl;
else
cout << "YES" << endl;
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
solve();
}
B - New Year’s Number
给出一个数，问能否由任意个2020和2021相加而成。
数据很小，直接暴力枚举就行
#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
const int N = 1e5 + 10;
const int M = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
typedef long long ll;
typedef pair<int, int> PII;
int T, n, m;
int a[N], b[N], c[N];
void solve() {
cin >> T;
while (T--) {
int x; cin >> x;
int a1 = x / 2020;
int b1 = x / 2021;
int f = 0;
for (int j = 0; j <= a1; j++)
for (int k = 0; k <= b1; k++) {
if ((j * 2020 + k * 2021) == x) {
f = 1;
}
}
if (f) cout << "YES" << endl;
else cout << "NO" << endl;
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
solve();
}
C - Ball in Berland
找有多少对二元组可以满足题中所给得要求
假设a为当前同类男生集合，b为当前同类女生集合
因此满足要求的组数是k-a∪b，因为a∩b=1，所以ans+=k-a-b+1
#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
const int N = 2e5 + 10;
const int M = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
const int MOD = 1e9 + 7;
typedef long long ll;
typedef pair<int, int> PII;
int T, n, m, k;
int a[N], b[N];
struct node {
int x;
int y;
}p[N];
void solve() {
cin >> T;
while (T--) {
cin >> n >> m >> k;
for (int i = 1; i <= n; i++) a[i] = 0;
for (int i = 1; i <= m; i++) b[i] = 0;
for (int i = 1; i <= k; i++) {
cin >> p[i].x; a[p[i].x]++;}
for (int i = 1; i <= k; i++) {
cin >> p[i].y; b[p[i].y]++;}
ll ans = 0;
for (int i = 1; i <= k; i++)
ans += (k - a[p[i].x]) - (b[p[i].y]-1);
cout << ans / 2 << endl;
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
solve();
}
D. Cleaning the Phone
有n个物品有不同的价值，花费分别为1和2
问最少花费多少价值可以到达m
同样是贪心的做法，按照性价比排序，当最后一个取的价值为2的物品时会有三种情况出现。
1.可被后面花费为1的物品代替
2.前面可以剔除一件买过的商品
3.什么都不做
#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
const int N = 2e5 + 10;
const int M = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
const int MOD = 1e9 + 7;
typedef long long ll;
typedef pair<int, int> PII;
int T, n, m, k;
int a[N], b[N];
struct node {
int x;
int y;
double v;
}p[N];
bool cmp(node x, node y) {
return x.v > y.v;
}
void solve() {
cin >> T;
while (T--) {
cin >> n >> m;
ll sum = 0;
for (int i = 1; i <= n; i++) cin >> p[i].x, sum += p[i].x;
for (int i = 1; i <= n; i++) cin >> p[i].y;
if (sum < m) {
cout << -1 << endl;
continue;
}
for (int i = 1; i <= n; i++) p[i].v = p[i].x * 1.0 / p[i].y;
sort(p+1, p+n+1, cmp);
// for (int i = 1; i <= n; i++) cout << p[i].x << " ";
// cout << endl;
// for (int i = 1; i <= n; i++) cout << p[i].y << " ";
int ans = 0, count = 0;
int f = 0;
for (int i = 1; i <= n; i++) {
ans += p[i].x;
count += p[i].y;
if (ans >= m) {
if (p[i].y == 2) {
for (int j = i + 1; j <= n; j++) {
if (p[j].y == 1 && ans - p[i].x + p[j].x >= m) {
count--;
f = 1;
break;
}
}
if (f) break;
int Maxy = 0;
for (int j = 1; j < i; j++) {
if (ans - p[j].x >= m) {
Maxy = max(Maxy, p[j].y);
f = 1;
}
}
count -= Maxy;
}
break;
}
}
cout << count << endl;
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
solve();
}
E - Advertising Agency
有n个数，一次取k个，问取到最大值的方案有几个
从大到小取，唯一产生分歧的就是第k大的数不唯一
因此我们算一下有多少种取法也就是在取第k大之前取了几个，还需要取几个，相同的右几个，就是简单的组合数问题，逆元处理。
#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
const int N = 2e5 + 10;
const int M = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
const int MOD = 1e9 + 7;
typedef long long ll;
typedef pair<int, int> PII;
int T, n, m, k;
int a[N], b[N], c[N];
int vis[N];
vector<int> boy[N], girl[N];
struct node {
int x, y;
}p[N];
ll ksm(ll a, ll b) {
ll res = 1, base = a % MOD;
while (b) {
if (b & 1) res = res * base % MOD;
base = base * base % MOD;
b >>= 1;
}
return res % MOD;
}
ll comb(ll a, ll b) {
ll ans = 1;
for (ll i = a; i >= a - b + 1; i--) ans = ans * i % MOD;
for (ll i = b; i >= 1; i--) ans = ans * ksm(i, MOD-2) % MOD;
return ans % MOD;
}
void solve() {
cin >> T;
while (T--) {
cin >> n >> k;
for (int i = 0; i <= n; i++) vis[i] = 0;
for (int i = 1; i <= n; i++) cin >> a[i], vis[a[i]]++;
sort(a+1, a+n+1);
reverse(a+1, a+n+1);
ll now = a[k], count = 0;
for (int i = 1; i <= n; i++) {
if (now == a[i]) count++;
}
ll need = 1;
for (int i = k; i >= 1; i--) {
if (a[i] != a[i-1]) break;
need++;
}
ll ans = comb(count, need);
cout << ans << endl;
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
solve();
}