输入一个正整数 n,输出 2⼀1+3⼀2+5⼀3+8⼀5+……前n项之和,保留2位小数

2025-02-25 14:05:30
推荐回答(2个)
回答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;
}

回答2:

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