7-1 编写程序求m!/(n!*(m-n)!)的值(这是题目)
条件是n<=m<=20
#include <stdio.h>
int jc(int n)
{
if(n==0)
return 1;
return n*jc(n-1);
}
int main()
{
int m,n;
scanf("%d %d",&m,&n);
printf("%d",jc(m)/(jc(m-n)*jc(n)));
return 0;
}
7-1 编写程序求m!/(n!*(m-n)!)的值(这是题目)
条件是n<=m<=20
#include <stdio.h>
int jc(int n)
{
if(n==0)
return 1;
return n*jc(n-1);
}
int main()
{
int m,n;
scanf("%d %d",&m,&n);
printf("%d",jc(m)/(jc(m-n)*jc(n)));
return 0;
}