11609 - Teams UVA

最新推荐文章于 2024-07-27 09:15:12 发布

Just_Try

最新推荐文章于 2024-07-27 09:15:12 发布

阅读量352

点赞数

CC 4.0 BY-SA版权

分类专栏：基本数学问题 UVA

本文链接：https://blog.youkuaiyun.com/abc123456789cbaabc/article/details/8691098

UVA 同时被 2 个专栏收录

20 篇文章

订阅专栏

基本数学问题

7 篇文章

订阅专栏

本文介绍了一种在大规模数据集上进行团队选择可能性计算的方法，使用快速幂取模技术解决了大数运算的问题，并提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Teams

Input: Standard Input

Output: Standard Output

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 ≤ 10^9), 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 Output for Sample Input

Case #1: 1

Case #2: 4

Case #3: 12

本题因为数据很大所以要用到快速幂取模！！

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include <math.h>
using namespace std;
#define LL long long
LL p(LL a,LL b,LL c)//当a不太大时（操作过程中不会数据溢出）
{
    if(b==0)
    return 1;
    LL x = p(a,b/2,c);
    long long ans = (long long) x*x%c;
    if(b%2==1)
    ans  = ans*a%c;
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int i = 0;i<t;++i)
    {
        LL n;
        scanf("%lld",&n);
        printf("Case #%d: %lld\n",i+1,(p(2,n-1,1000000007)*(n%1000000007))%1000000007);//*(n%1000000007)%100000007);
    }
    return 0;
}