51nod 1632 B君的连通

最新推荐文章于 2022-05-03 18:35:35 发布

原创最新推荐文章于 2022-05-03 18:35:35 发布 · 362 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

12 篇文章

订阅专栏

思路：得出结果 ans = 1*C(0, n-1) + 2*C(1, n-1)+.....+n*C(n-1,n-1) = 2 ^(n-1) + (n-1) * 2^(n - 2)，再快速幂取模就行了

#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long ll;
const int maxn = 1e5 + 10;
const ll mod = 1e9 + 7;
using namespace std;

ll po(ll a, ll n, ll m) {
    ll res = 1;
    while(n) {
        if(n & 1) res = res * a % m;
        a = a * a % m;
        n >>= 1;
    }
    return res;
}

int main() {
    ll n, a, b;
    while(scanf("%lld", &n) != EOF) {
        for(int i = 0; i < n - 1; i++) {
            scanf("%lld %lld", &a, &b);
        }
        ll r1 = po(2, n - 1, mod);
        ll r2 = po(2, n - 2, mod);
        ll r3 = ((n - 1) * r2) % mod;
        printf("%lld\n", (r1 + r3) % mod);
    }
    return 0;
}