uva 11609

可以想到 答案为 1*C(1,n)+2*C(2,n)+3*C(3,n)+....+n*C(n,n);

由公式 k*C(k,n) = n*C(k-1,n-1)

所以最终答案 n*2^(n-1)

用到快速幂取余

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 1010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;

ll mymod(int n)
{
    if(n == 0) return 1;
    ll x = mymod(n/2);
    ll ans = x*x%MOD;
    if(n%2) ans = ans*2%MOD;
    return ans;
}

int main()
{
    int n, t, ca = 1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        ll ans = mymod(n-1);
        printf("Case #%d: %lld\n", ca++, ans*n%MOD);
    }
    return 0;
}