View Code of Problem 48

#include <stdio.h>
int main()
{
	int n,a[20]={1,2};
	double p,sum=0;//必须为double
	int i;
	scanf("%d",&n);
    for(i=0;i<n;i++)
	{
		p=(double)a[i+1]/a[i];
		sum=sum+p;
		a[i+2]=a[i]+a[i+1];
	}
    printf("%.6f",sum);

	return 0;
}