----------------解决方案--------------------------------------------------------
#include "stdio.h"
double fun(int m,int n)
{
int i;
double t=1;
for(i=1;i<=n;i++)
t*=m;
return t;
}
main()
{
int m=3,n=4;
double s=0;
s=fun(m,n);
printf("s=%f\n",s);
}
----------------解决方案--------------------------------------------------------
不会做
----------------解决方案--------------------------------------------------------
/*返回两数加乘方*/
/*返回两个数a 和 b,和m的n次方*/#include "stdio.h"
max_min(int *x,int *y)
{
*x=*x+5;
*y=*y+5;
}
main()
{
int a=2,b=3,m,n;
int i,s;
printf("Please input m and n :\n");
scanf("%d %d",&m,&n);
s=m;
for(i=1;i<n;i++)
m=m*s;
max_min(&a,&b);
printf("a=%d b=%d m=%d\n",a,b,m);
}
----------------解决方案--------------------------------------------------------