http://poj.org/problem?id=2249
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.
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.
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.
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
4 2 10 5 49 6 0 0
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;
}
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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