组队 UVa11609

1.题目描述：点击打开链接

2.解题思路：首先选择一个人当队长，有n种选法；对于每一个队长，剩下的可以有0,1,2,...n-1个人，一共有2^(n-1)种情况。答案就是n*2^(n-1)。

3.代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<long long, long long> PL;
#define me(s) memset(s,0,sizeof(s))
#define For(i,n) for(int i=0;i<(n);i++)

const int MOD = 1000000007;

ll pow_mod(ll a, ll n)
{
	ll res = 1;
	while (n>0)
	{
		if (n & 1)res = res*a%MOD;
		a = a*a%MOD;
		n >>= 1;
	}
	return res;
}
int main()
{
	//freopen("t.txt", "r", stdin);
	int T;
	scanf("%d", &T);
	int rnd = 0;
	while (T--)
	{
		int n;
		scanf("%d", &n);
		ll ans = pow_mod(2, n - 1);
		ans = ans*n%MOD;
		printf("Case #%d: %lld\n", ++rnd, ans);
	}
	return 0;
}