# 题目

## Description

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

## Input

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.

## Output

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

## Sample Output

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.

# 代码

```/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！

*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

unsigned long long gcd(unsigned long long a, unsigned long long b) {
return b == 0 ? a : gcd(b, a%b);
}

bool Do() {
int n, m;
if (scanf("%d%d", &n, &m), n == 0 && m == 0)
return false;

unsigned long long ans = 1;
int a = max(m, n - m);
int b = min(m, n - m);
unsigned long long t = 1;
for (int i = n, j = 2; i > a; i--, j++) {
ans *= i;
if (j <= b || t > 1) {
if (j <= b)
t *= j;
if (t > 1) {
unsigned long long q = gcd(ans, t);
ans /= q;
t /= q;
}
}

}

printf("%d things taken %d at a time is %llu exactly.\n", n, m, ans);

return true;
}

int main() {
while (Do());
return 0;
}
```