https://codeforces.com/gym/105229
A

#include <bits/stdc++.h>
using namespace std;

int main() {
        int n,m;
        cin>>n>>m;
    for (int a = 0; a <= n; a++) {
        for (int b = 0; b <= m; b++) {
            int ans = 0;
            for (int c = 0; c <= n; c++) {
                for (int d = 0; d <= m; d++) {
                    if (a == c && b == d) continue;
                    // 正方形面积不为零，所以对角线上的两个点不能重合
                    if ((b + c - a - d) % 2 != 0) continue;
                    int num = (b + c - a - d) / 2;
                    int x1 = a + num, y1 = d + num;
                    int x2 = c - num, y2 = b - num;
                    if (x1 >= 0 && x1 <= n && y1 >= 0 && y1 <= m && x2 >= 0 && x2 <= n && y2 >= 0 && y2 <= m) ans++;
                }
            }
            printf("%d ", ans);
        }
        puts("");
    }
    return 0;
}

M

#include <bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int a[N][N];
vector<pair<int,int>>v1,v2;
int main()
{
        int n;
        cin>>n;
        int t=sqrt(n);
        if(n-t*t>t)t++;
        int cnt=0;
        for(int i=1;i<=t+1;i++)
        {
        for(int j=1;j<=t;j++)
        {
        a[i][j]=++cnt;  
        if(cnt>n)
        a[i][j]=0;
        }
        }
        for(int i=1;i<=t+1;i++)
        {
                for(int j=1;j<=t;j++)
                {
                if(a[i+1][j])v1.push_back({a[i][j],a[i+1][j]});
                if(a[i][j+1])v2.push_back({a[i][j],a[i][j+1]});
                }
        }
         int size = min(v1.size(), v2.size());
    printf("%d\n", size);
    for (int i = 0; i < size; i++) printf("%d %d\n", v1[i].first, v1[i].second);
    for (int i = 0; i < size; i++) printf("%d %d\n", v2[i].first, v2[i].second);
}

E

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
void solve()
{
    int n;
    cin >> n;
    string s;
    cin >> s;
    array<int,26> cnt{};
    for(auto &c : s) {
        c = tolower(c);
        cnt[c - 'a'] ++;
    }
    cout << min({cnt['s'-'a'],cnt['h'-'a']/2,cnt['a'-'a']/2,cnt['n'-'a'],cnt['g'-'a'],cnt['i'-'a']}) << endl;
}
signed main()
{   
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T=1;
    //cin>>T;
    while(T--)
    solve();
    return 0;
}

J

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
vector<int> pri;
bool not_prime[N];

void pre(int n) {
  for (int i = 2; i <= n; ++i) 
  {
    if (!not_prime[i])
    {
      pri.push_back(i);
    }
    for (int pri_j : pri) 
    {
      if (i * pri_j > n) break;
      not_prime[i * pri_j] = true;
      if (i % pri_j == 0) {
        // i % pri_j == 0
        // 换言之，i 之前被 pri_j 筛过了
        // 由于 pri 里面质数是从小到大的，所以 i 乘上其他的质数的结果一定会被
        // pri_j 的倍数筛掉，就不需要在这里先筛一次，所以这里直接 break
        // 掉就好了
        break;
      }
    }
  }
}
void solve()
{
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for(int i = 1 ; i <= n ; i ++)
    {
        cin >> a[i];
    }
    for(int k = 0 ; k <= n ; k ++)
    {
        for(int i = 1 ; i <= n ; i ++)
        {
            int s = 0;
            for(int j = i ; j - i <= k && j <= n; j ++)
            {
                s += a[j];
            }
            if(not_prime[s]) {
                cout << k << endl;
                return ;
            }
        }
    }
    cout << -1 << endl;
}


signed main()
{   
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    cin >> t;
    pre(N);
    //cout<<not_prime[9]<<endl;
    while(t--)
        solve();
    return 0;
}