369 - Combinations

Time limit: 3.000 seconds

Combinations

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:

Compute the EXACT value of:

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 and Output

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.

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.

===========================================

#include <stdio.h>
#include <stdlib.h>

int gcd(int,int);
void combi(const int,const int);

int main()
{
　freopen("Input.txt","r",stdin);
　freopen("Output.txt","w",stdout);
　int N,M;
　while(scanf("%d%d",&N,&M)==2)
　　if(N|M)
　　　combi(N,M);
　return 0;
}

/* calculate the combinition number */
void combi(const int N,const int M)
{
　int A[M],B[M];
　int i,j;
　for(i=0;i<M;i++)
　{
　　A[i]=N-i;
　　B[i]=i+1;
　}
　for(j=1;j<M;j++)
　　for(i=0;i<M;i++)
　　{
　　　/* find the GCD of each pair when B[] is not 1 */
　　　int GCD=gcd(A[i],B[j]);
　　　if(GCD>1)
　　　{
　　　　A[i]/=GCD;
　　　　B[j]/=GCD;
　　　}
　　　if(B[j]==1)
　　　　break;
　　}
　/* when all elements in B[] are 1 then multiply every element in A[] */
　long ans=1;
　for(i=0;i<M;i++)
　　ans*=A[i];
　printf("%d things taken %d at a time is %d exactly.\n",N,M,ans);
}

/* return (x,y) by recursive */
int gcd(int x,int y)
{
　return (!y)?x:gcd(y,x%y);
}