Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!

You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000

The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read.

The output from this program should be in the form:
N things taken M at a time is C exactly.

100  6
20  5
18  6
0  0

100 things taken 6 at a time is 1192052400 exactly.
20 things taken 5 at a time is 15504 exactly.
18 things taken 6 at a time is 18564 exactly.

     暴力枚举，利用C(n,m)=  C(n-1,m-1)+C(n-1,m) 。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int c[105][105];
int main()
{
	int n, m;
	c[1][1] = 1;
	c[1][0] = 1;
	for(int i = 1;i <= 100;i ++)
		c[i][0] = 1;
	while(scanf("%d%d", &n, &m))
	{
		if(n == 0 && m == 0)
			break;
		for(int i = 1;i <= n;i ++)
			for(int j = 1;j <= m;j ++)
			{
				c[i+1][j] = c[i][j] + c[i][j-1];
			}
		printf("%d things taken %d at a time is %d exactly.\n", n, m, c[n][m]);
	}
	return 0;
}