UVA

In a galaxy far far awaythere is an ancient game played among the planets. The specialty of the game isthat there is no limitation on the number of players in each team, as long asthere is a captain in the team. (The game is totally strategic, so sometimesless player increases the chance to win). So the coaches who have a total of Nplayers to play, selects K (1 ≤ K ≤ N) players and makeone of them as the captain for each phase of the game. Your task is simple,just find in how many ways a coach can select a team from his N players.Remember that, teams with same players but having different captain areconsidered as different team.

 

Input

Thefirst line of input contains the number of test cases T ≤ 500.Then each of the next T lines contains the value of N (1 ≤ N ≤10^9), the number of players the coach has.

Output

 

Foreach line of input output the case number, then the number of ways teams can beselected. You should output the result modulo 1000000007.

Forexact formatting, see the sample input and output.

 

Sample Input                                                                               Outputfor Sample Input

3 1 2 3

Case #1: 1 Case #2: 4 Case #3: 12

ProblemSetter: Towhidul Islam Talukdar

SpecialThanks: Md. Arifuzzaman Arif

题意：有n个人。选一个或者多个人參加比赛。当中一名当队长，有多少种方案？假设參赛者全然同样，但队长不同，算作不同方案。

思路：非常easy得到sum=∑i=1nC[n][i]∗i ,      当中C[n][i]    表示组合数， 那么依据排列数公式我们得到C[n][i]∗i=n∗C[n−1][i−1]

那么结果就变成了sum=n∗∑i=1nC[n−1][i−1]        =   n∗2n−1    高速幂取模

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
typedef long long ll;
using namespace std;
const ll mod = 1000000007;

ll pow_mod(ll x, ll y) {
	ll ans = 1;
	while (y > 0) {
		if (y & 1)
			ans = ans * x % mod;
		y >>= 1;
		x = x * x % mod;
	}
	return ans;
}

ll n;

int main() {
	int t, cas = 1;
	scanf("%d", &t);
	while (t--) {
		scanf("%lld", &n);
		printf("Case #%d: %lld
", cas++, n*pow_mod(2ll, n-1) % mod);
	}
	return 0;
}