Binomial Showdown（组合计数模板）

最新推荐文章于 2019-01-18 22:43:20 发布

林下的码路

最新推荐文章于 2019-01-18 22:43:20 发布

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分类专栏： ACM 组合数学文章标签： ACM 算法优化

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本文介绍了如何计算组合数，并提供了使用C++实现的代码示例。通过输入两个整数n和k，程序将计算从n个元素中选择k个元素的组合数，输出结果始终不超过2^31。

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Link：http://poj.org/problem?id=2249

Binomial Showdown

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18459		Accepted: 5635

Description

In how many ways can you choose k elements out of n elements, not taking order into account?
Write a program to compute this number.

Input

The input will contain one or more test cases.
Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).
Input is terminated by two zeroes for n and k.

Output

For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 2 ³¹.
Warning: Don't underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.

Sample Input

Sample Output

6
252
13983816

Source

Ulm Local 1997

题意&解题思路：计算组合数c(n,m)

AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#define LL long long
#define MAXN 1000100
using namespace std;
LL n,m;
LL Cnm(LL n,LL m)//计算组合数C(n,m) 
{
	if(m>n/2)//根据组合公式，可以减少枚举量 
	m=n-m;
	LL a=1,b=1;
	for(int i=1;i<=m;i++)//顺序进行m次运算 
	{
		a*=n+1-i;//计算前i项运算结果的分子a和分母 b 
		b*=i;
		if(a%b==0)
		{
			a/=b;
			b=1;
		}
	}
	return a/b;
}

int main()
{
	while(cin>>n>>m)
	{
		if(n==0&&m==0)
			break;
		cout<<Cnm(n,m)<<endl;
	}
	return 0;
 }