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.
代码语言：javascript复制#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;
}
