麻烦各位帮我看看这段代码的问题。

胡漠杨 @ 2020-03-14 21:06:04

#include<stdio.h>
float sum(float (1/n)); 
int main() 
{
    int n,k;
    scanf ("%d",&k);
    for (n=1;;n++)
    {
        if (k<sum(1/n))
        {
            break;
        }
        printf ("%d",n);
    }
    return 0;
}

float sum(float (1/n))
{
    int n;
    if (n==1)
    {
        return 1;
    }
    else
    {
        return sum(1/(n-1))+(1/n); 
    }
}

by UnyieldingTrilobite @ 2020-03-14 21:06:36

这个函数······看不懂

by 胡漠杨 @ 2020-03-14 21:08:40

@return20071007 我想实现当n=1时返回1，当n≠1时返回1/（n-1) + 1/n。

by CSP_Sept @ 2020-03-14 21:08:52

float sum(float (1/n)); 

是什么意思

by Marser @ 2020-03-14 21:09:09

能不能好好学一下函数啊

by Marser @ 2020-03-14 21:09:21

形参哪有这么定义的啊

by Marser @ 2020-03-14 21:09:55

还有float有什么用啊，不用double精度怎么够啊

by UnyieldingTrilobite @ 2020-03-14 21:10:25

用循环好了······

伪代码：

I have some integers k,n;
I have one decimal s;
Read in k;
While the value of s is less than the value of k or the value of s equals the value of k do{
    Please let the value of n be the value of n plus 1;
    Then Let the value of s be the value of s plus 1.0 divided by n;
}
Print out n;
Print out '\n';

by CSP_Sept @ 2020-03-14 21:10:43

double sum(double n)

才对

by UnyieldingTrilobite @ 2020-03-14 21:10:52

怎么感觉今天Marser神仙到处D人

by Marser @ 2020-03-14 21:11:03

@return20071007 真实自然语言编程