寫的話就是排列組合。。。但能化簡。。。ΣC(n,i)*C(i,1) 化簡為n*2^(n-1) ;

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define
 
 MOD 1000000007
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1000000, INF = 0x7fffffff;
LL down[maxn], up[maxn];

LL qpow(LL a, LL b)
{
    LL res 
 
= 1;
    while(b)
    {
        if(b & 1) res = res * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return res;
}
//
//void init()
//{
//    up[0] = 1;
//    down[0] = 1;
//    for(int i=1; i<maxn; i++)
//    {
//        up[i] = up[i-1] * i % MOD;
//        down[i] = qpow(up[i], MOD - 2);
//    }
 
//}
//
//LL C(LL n, LL m)
//{
//    return up[n] * down[m] % MOD * down[n-m] % MOD;
//}

int main()
{
    int T, kase = 0;
  //  init();
    cin>> T;
    while(T--)
    {
        LL n, res = 0, m;
        cin>> n;

        printf("Case #%d: %lld\n",++kase, n * qpow(2, n-1) % MOD);

   //     cout<< C(n, m) <<endl;



    }


    return 0;
}

Teams UVA - 11609（快速冪板題）