UVA 11609 Teams 组合数学+快速幂

In a galaxy far far away there is an ancient game played among the planets. The specialty of the game
is that there is no limitation on the number of players in each team, as long as there is a captain in
the team. (The game is totally strategic, so sometimes less player increases the chance to win). So the
coaches who have a total of N players to play, selects K (1 ≤ K ≤ N) players and make one of them
as the captain for each phase of the game. Your task is simple, just ?nd in how many ways a coach
can select a team from his N players. Remember that, teams with same players but having di?erent
captain are considered as di?erent team.
Input
The ?rst 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 ≤ 109
), the number of players the coach has.
Output
For each line of input output the case number, then the number of ways teams can be selected. You
should output the result modulo 1000000007.
For exact formatting, see the sample input and output.
Sample Input
3
1
2
3
Sample Output
Case #1: 1
Case #2: 4
Case #3: 12

题意：给你一个n，n个人，标号为1~n，现在选若干人组成一队，并且选出一个队长，问说可以选多少种队伍，队长，人数，成员不同均算不同的队伍。

题解：我们枚举选择k个人(1<=k<=n)

答案就是： 1*c(n,1)+2*c(n,2)+.......+n*c(n,n);

　　　接着提出n

　　　　化为：    n*(c(n-1,0)+c(n-1,1)+c(n-1,2)+......+c(n-1,n-1));

　　答案就是：n*（2^(n-1)）;快速幂求解

//meek///#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <sstream>
#include <vector>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;

const int N = 110;
const int inf = 99999999;
const int MOD= 1000000007;

ll quick_pow(ll x,ll p) {
    if(!p) return 1;
    ll ans = quick_pow(x,p>>1);
    ans = ans*ans%MOD;
    if(p & 1) ans = ans*x%MOD;
    return ans;
}
int main() {
    int T,cas=1;
    ll n;
    scanf("%d",&T);
    while(T--) {
        scanf("%lld",&n);
        printf("Case #%d: %lld\n",cas++,n*quick_pow(2,n-1)%MOD);
    }
    return 0;
}

DAIMA