O - Combinations （组合数学）

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. 

Sample Input

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.
解题思路：和上题一样，n很小，最大只有100，直接暴力求解，类型全开long long，水过！
AC代码：

 #include<iostream>
 using namespace std;
 typedef long long LL;
 LL n,k,m,ans;
 int main(){
     while(cin>>n>>k&&(n+k)){
         m=k;//记录原来的取法数量
         if(n-k<k)k=n-k;//取最小的取法数量
         ans=;
         for(LL i=;i<=k;++i)ans=ans*(n-i+)/i;
         cout<<n<<" things taken "<<m<<" at a time is "<<ans<<" exactly."<<endl;
     }
     return ;
 }