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 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

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：

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 __int64 a[ 100000000];
 9 int main()
10 {
11 
12     __int64 m , n ;
13     while ( scanf("%I64d%I64d",&m,&n), m || n )
14     {
15         a[0]=1;
16         if ( n > m / 2 )
17             n = m - n ;
18         
19         for ( int i = 1; i <= n ; i ++ )
20             a[i] = a[i-1] * ( m - i + 1 ) / i ;
21         printf("%I64d\n", a[n] );
22     }
23     return 0;
24 }