如题:
代码如下:
#include <stdlib.h>
#include <conio.h>
#include <stdio.h>
double fun (int n)
{
int a,b,c, k;
double s;
s=0.0;a=2;b=1;
for(k=1;k<=n;k++)
{
s=s+(double)a/b;
c=a;a=a+b;b=c;
}
return s;
}
void main()
{
int n=5;
system("CLS");
printf("\nThe value of function is:%lf\n",fun(n));
}
关键点:
s=s+(double)a/b;
c=a;a=a+b;b=c;