有n个人，其中有m对朋友，第i对朋友的友谊值是f[i], 有k次旅行，每次旅行随机、独立地选两人，如果两个人是朋友，那么快乐值增加f，并且下一次旅行这两个人f值加一；否则快乐值和f都不变。求最大快乐值

题目

思路：

#include 
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e12, maxm = 4e4 + 5, mod = 1e9 + 7, N = 1e6;
int a[maxn], b[maxn];
// bool vis[maxn];
int n, m;
string s;
int f[maxn];
int fac[maxn], inv[maxn];

int C(int n, int m){
	if(n < m) return 0;
	return fac[n] * inv[m] % mod * inv[n - m] % mod;
}

int qpow(int a, int b){
	int res = 1;
	while(b){
		if(b & 1) res = res * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return res;
}

void solve(){
    int res = 0;
    int k;
    int x;
    int q;
    // cin >> n;
    cin >> n >> m >> k;
	int sum = 0;
	for(int i = 1; i <= m; i++){
		cin >> a[i] >> b[i] >> f[i];
		sum = (sum + f[i]) % mod;
	}
	int d = n * (n - 1) / 2 % mod;
	res = sum * k % mod * qpow(d, mod - 2) % mod;
	for(int x = 0; x <= k; x++){
		int tmp = C(k, x) * qpow(d - 1, k - x) % mod;
		int t = qpow(d, k);
		tmp = (tmp * qpow(t, mod - 2)) % mod * ((x * (x - 1) / 2) % mod) % mod;
		res = (res + m * tmp % mod) % mod;
	}
	cout << res << '\n';
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    fac[0] = 1;
    for(int i = 1; i <= N; i++){
        fac[i] = fac[i - 1] * i % mod;
    }
    inv[N] = qpow(fac[N], mod - 2);
    for(int i = N - 1; i >= 0; i--){
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}