A:
要使s,t相等只要互相删除对方没有的字母即可，即找到a-z字母拥有最少的
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;string s1, s2;
int a1[30], a2[30];
void solve(){cin >> s1 >> s2;for(int i = 0; i < s1.size(); i++){int x = s1[i]-'a';a1[x]++;}for(int i = 0; i < s2.size(); i++){int x = s2[i]-'a';a2[x]++;}int sum = 0;for(int i = 0; i <26; i++) sum+=min(a1[i], a2[i]);cout << sum;
}
signed main (){ios::sync_with_stdio(false);cin.tie(nullptr); int T;
// cin >> T;T = 1;while (T--) {solve();}return 0;
}
B:
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;
const int mod = 1e9+7;int k;
void solve(){cin >> k;int ret = 7, num=1;for(int i = 1; i <= 1e7; i++){if(ret%k==0){cout<< num ;return ;}ret=ret*10+7; ret%=k;num++;}cout << "-1" ;
}
signed main (){ios::sync_with_stdio(false);cin.tie(nullptr); int T;
// cin >> T;T = 1;while (T--) {solve();}return 0;
}
C:
对于任意两个数求和，即为每个数与之后的前缀和的积的和，注意MOD使用
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;
const int mod = 1e9+7;int n, a[200010], b[200010];
void solve(){cin >> n;int sum = 0;for(int i = 1; i <= n; i++){cin >> a[i];b[i] = a[i]+b[i-1];}for(int i = 1; i < n; i++){sum = (sum + ( (a[i]%mod) * ((b[n]-b[i])%mod) ) %mod ) %mod;}cout << sum%mod;
}
signed main (){ios::sync_with_stdio(false);cin.tie(nullptr); int T;
// cin >> T;T = 1;while (T--) {solve();}return 0;
}
D:
求出每个与i相乘不会大于n的数即可，其值为1到n的i的倍数，即为n/i
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;
const int mod = 1e9+7;int n;
void solve(){cin >> n;int num = 0;for(int i = 1; i < n; i++){num+=(n-1)/i;}cout << num;
}
signed main (){ios::sync_with_stdio(false);cin.tie(nullptr); int T;
// cin >> T;T = 1;while (T--) {solve();}return 0;
}
E:
数据范围为1e12，枚举到1到1e6判断该数字是否成立，即将其接一个上去与n比较即可
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;
const int mod = 1e9+7;int n;
int se(string s){s+=s;int ret = 0;for(int i = 0; i < s.size(); i++){ret = ret*10+(s[i]-'0');}return ret;
}
void solve(){cin >> n;int num = 0;for(int i = 1; i <= 1e7; i++){if(n>=se(to_string(i))) num++;else break;}cout << num;
}
signed main (){ios::sync_with_stdio(false);cin.tie(nullptr); int T;
// cin >> T;T = 1;while (T--) {solve();}return 0;
}
F:
dfs实现，对于一块颜色相同的相连区域，只需要搜索一遍即可，而只要走到颜色相同且已经走过的位置即为成环，注意不能往回走
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;int n, m, a[60][60];
int s1[4] = {-1,0,0,1};
int s2[4] = {0,-1,1,0};
char c[60][60];
int p = 0;
void dfs(int x, int y, int lx, int ly){if(a[x][y]){p = 1;return ;}a[x][y] = 1;for(int i = 0; i < 4; i++){int nx = x+s1[i];int ny = y+s2[i];if(nx==lx && ny==ly) continue;if(nx>=1 && nx<=n && ny>=1 && ny<=m && c[nx][ny]==c[x][y]){dfs(nx, ny, x, y);}if(p) break;}
}
void solve(){cin >> n >> m;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)cin >> c[i][j];for(int i = 1; i <= n;i++){for(int j = 1; j <= m; j++){if(p) break;if(a[i][j]==0){dfs(i,j,i,j);}if(p) break;}if(p) break;}if(p) cout << "Yes" ;else cout << "No";
}
signed main (){ios::sync_with_stdio(false);cin.tie(nullptr);int T;
// cin >> T;T = 1;while (T--) {solve();}return 0;
}
