Question

输入一个正整数 n,输出 2/1+3/2+5/3+8/5+……前n项之和,保留2位小数

Answer 1

// 试试看

#include "stdio.h"

int main()
{
double x = 1, y = 1, z = 2;
double sum = 0;
int n;
int i = 0;

scanf("%d", &n);
while(i++ < n){
sum += z / y;
x = y;
y = z;
z = x + y;
}
printf("%.2f\n", sum);
return 0;
}

Answer 2

#include<stdio.h>
void main()
{ int fz,fm,k,n,i;
double sum;
sum=0; fz=2; fm=1;
scanf("%d",&n);
for ( i=0;i<n;i++ ) { sum+=((double)fz/fm); k=fz; fz+=fm; fm=k; }
printf("%.2lf\n",sum+0.005);
}