传送门：点击打开链接
题意：将一个矩阵A先转置，然后再与原A矩阵相乘，求最后得到的矩阵中的数字之和
思路：刚开始没看懂题目，，，随便写几个矩阵A，然后倒置后相乘找规律，一下子就能发现规律，最后的答案等于每一行数字之和的平方之和
#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;const int MX = 1e5 + 5;
const int INF = 0x3f3f3f3f;int cnt[MX];
int main() {int n, m;//FIN;while(~scanf("%d%d", &n, &m)) {memset(cnt, 0, sizeof(cnt));for(int i = 1; i <= m; i++) {int u, v;scanf("%d%d", &u, &v);if(u != v) {cnt[u]++; cnt[v]++;} else cnt[u]++;}LL ans = 0;for(int i = 1; i <= n; i++) {ans += (LL)cnt[i] * cnt[i];}printf("%lld\n", ans);}return 0;
}