題目：

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.InputThe 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.OutputFor 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

思路：

先把素數打表，然後把組合數中各個素數的次數求出來，然後全部乘起來就不會溢位

程式碼：

#include<iostream>
using namespace std;

int list[50000], p[5133];//5132個素數
void getp()//在p陣列中存所有不超過50000的素數
{
	p[0] = 2;
	int key = 0;
	for (int i = 0; i < 50000; i++)list[i] = i % 2;
	for (int i = 3; i < 50000; i += 2)if (list[i])
	{
		p[++key] = i;
		for (int j = i + i; j < 50000; j += i)list[j] = 0;
	}
}
 


int degree(int m, int p)//求m！中素數p的次數
{
	if (m)return degree(m / p, p) + m / p;
	return 0;
}

int main()
{
	getp();
	int n, k, mi;
	while (cin >> n >> k)
	{
		if (n == 0)return 0;
		int ans = 1;
		for (int i = 0; i < 5133; i++)
		{
			mi = degree(n, p[i]) - degree(k, p[i]) - degree(n - k, p[i]);
			while (mi--)ans *= p[i];
		}	
		cout << ans << endl;
	}
	return 0;
}