用函数求1+1/2+1/3+……+1/n之和,要求函数原型为“double fnsum(int n)”。
#include<stdio.h>
double fnsum(int n)
{
int i;
double sum = 0;
for(i = 1; i <= n; i++)
sum += (double)1.0/(double)i;
return sum;
}
int main()
{
double sum;
int n = 5;
sum = fun(n);
printf("sum = %lf\n",sum);
return 0;
}