Combinations(组合数)

Combinations

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9095 Accepted: 4199

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  620  518  60  0

Sample Output

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

#include <stdio.h>int main(){    int a,b,c,i,temp_a,temp_b;    double s1,s2;    while(scanf("%d%d",&a,&b)!=EOF)    {        if (a==0&&b==0)        {            break;        }        temp_a=a;        temp_b=b;        if(a-b<b)        {            b=a-b;        }        s1=s2=1;        for (i=1;i<=b;i++)        {            s1*=i;//s1为b的阶乘            s2*=a;//s2为a的阶乘除以a-b的阶乘            a--;        }        printf("%d things taken %d at a time is %.0lf exactly.\n",temp_a,temp_b,(s2/s1));    }    return 0;}

#include <stdio.h>  #include <stdlib.h>    int main()  {      int n,m,j;      double sum,i;        while(scanf("%d%d",&n,&m)!=EOF&&(n||m))      {if(n<5||n>100||m<5||m>100||n<m)         break;          sum=1.0;          j=m;          for(i=n; i>n-m; i--)          {              sum*=(i/j);              j--;          }              printf("%d things taken %d at a time is %0.lf exactly.\n",n,m,sum);      }          return 0;  }