UVa11609

Posted 2020-08-07 Yan_Bin

11609 Teams
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 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 are considered as different team.
Input
The first 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
3123
Sample Output
Case #1: 1
Case #2: 4
Case #3: 12

题意：
       有n个人，选不少于一个人参加比赛，其中一人当队长，有多少种选择方案。

分析：
       不论谁当队长，情况都是等价的。假设第一个人当队长，剩下的n-1个人可选可不选，有2^(n-1)种选法，所以答案就是n*2^(n-1)。

 1 #include <cstdio>
 2 #include <cmath>
 3 #define LL long long
 4 const LL MOD = 1000000007;
 5 LL pow_mod(LL a,LL p,LL n){
 6     if(p == 0) return 1;
 7     LL ans = pow_mod(a,p / 2,n);
 8     ans = ans * ans % n;
 9     if(p % 2 == 1) ans = ans * a % n;
10     return ans;
11 }
12 int main(){
13     int T; scanf("%d",&T); int kase = 0;
14     while(T--){
15         int n; scanf("%d",&n);
16         printf("Case #%d: %lld\\n",++kase,n * pow_mod(2,n - 1,MOD) % MOD);
17     }
18     return 0;
19 }

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