POJ 1306 Combinations

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.

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;

int main()
{ int n,m; while(~scanf("%d%d",&n,&m)&&(n||m)){ int s[200]; bool vis[200]; memset(vis,0,sizeof(vis)); int x,y; if(n-m<m){ x=n-m;/**x为小的那个数**/ y=m; } else{ x=m; y=n-m; } int sum=1; for(int i=y+1;i<=n;i++){ sum*=i; for(int j=1;j<=x;j++){ if(sum%j==0&&vis[j]==0){ vis[j]=1; sum=sum/j; } } } printf("%d things taken %d at a time is %d exactly.\n",n,m,sum); } return 0;
}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/49427953