POJ2249--一道简单的排列组合题

Binomial Showdown

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14435		Accepted: 4403

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 2³¹. 
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

题目大意就是从n个数中取k个数的情况种数，就是求C（n，k）；

刚开始时用递推，RE了几次，最后改成数组来，但有些细节没注意到，WA了几次，总的说来，这是一道比较简单的组合数学的基本功是的运用

排列组合的基本公式：

Pascal公式

和一些恒等式

要解答出这道题主要运用的就是恒等式（1）

我的提交情况

参考代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
__int64 a[ 100000000];
int main()
{

    __int64 m , n ;
    while ( scanf("%I64d%I64d",&m,&n), m || n )
    {
        a[0]=1;
        if ( n > m / 2 )
            n = m - n ;
        
        for ( int i = 1; i <= n ; i ++ )
            a[i] = a[i-1] * ( m - i + 1 ) / i ;
        printf("%I64d\n", a[n] );
    }
    return 0;
}