POJ 2249 Binomial Showdown 求组合数C(n,k)-优快云博客

本文介绍了一个程序，用于计算从n个元素中选择k个元素的方法数量，不考虑顺序。该程序通过一个公式计算组合数，并提供了样例输入输出来验证正确性。

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Binomial Showdown

Time Limit:1000MS

Memory Limit:65536K

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

/* Author : yan
 * Question : POJ 2249 Binomial Showdown
 * Date && Time : Thursday, February 03 2011 06:05 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
double comb(int n,int k)
{
	double ans = 1;
	int i;
	for(i=1;i<=k;i++)
	{
		ans*=(double)(n-i+1)/(double)i;
	}
	return ans;
}
int main()
{
	freopen("input","r",stdin);
	int n,m;
	while(scanf("%d %d",&n,&m) && n)
	{
		m=m<(n-m)?m:(n-m);
		printf("%.0f/n",comb(n,m));
	}
	return 0;
}