本文主要是介绍uva 11609 Teams，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
原题：
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
3
1
2
3
Sample Output
Case #1: 1
Case #2: 4
Case #3: 12
#include <bits/stdc++.h>
using namespace std;
const long mod=1000000007;
long quickmod(long a,long b)
{long ans=1;while(b){if(b&1){ans=(ans*a)%mod;b--;}b/=2;a=a*a%mod;}return ans;
}
int main()
{ios::sync_with_stdio(false);long n,t,k=1;cin>>t;while(t--){cin>>n;long ans=(n%mod)*(quickmod(2,n-1)%mod)%mod;cout<<"Case #"<<k++<<": "<<ans<<endl;}return 0;
}解答：
非常简答的数学题，公式能够很容易的就列出来等于
0×C(0,n)+1×C(1,n)+2×C(2,n)+….+n×C(n,n)
如果记得这个公式的最后结果可以直接写出来等于n×2^(n-1)
如果不记得，可以直接把组合数展开，然后把n提出来后能够得到结果
或者利用K×C(K,n)=n×C(K-1,n-1)的公式替换。
这篇关于uva 11609 Teams的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
