关于PTA的问题，如何解决？

问题遇到的现象和发生背景

问题相关代码，请勿粘贴截图

函数接口定义：

double m_tax(double salary,int month);//计算每月累计应纳税额

函数返回每月累计应纳税额;
裁判测试程序样例：

#include<stdio.h>

double m_tax(double salary,int month);

int main()
{
    double money,tax,sum=0;
    int i;
    for(i=1;i<=12;i++)
    {
        scanf("%lf",&money);
        sum+=money;
        tax=m_tax(money,i);
        printf("the sum of %d months tax is %.2f\n",i,tax);
    }
    return 0;
}
/* 请在这里填写答案 */

我想要达到的结果

double sum=0;
double m_tax(double salary,int month)
{
    double s,tax;
    sum+=salary;
    if(sum<=month*5000)
        tax=0;
    else
    {
        s=sum-month*5000;
        if(s<=36000)
            tax=s*0.03;
        else if(s<=144000)
            tax=s*0.1-2520;
        else if(s<=300000)
            tax=s*0.2-16910;
        else if(s<=420000)
            tax=s*0.25-31920;
        else if(s<=660000)
            tax=s*0.30-52920;
        else if(s<=960000)
            tax=s*0.35-85920;
        else
            tax=sum*0.45-181920;
    }
    return tax;
}

double m_tax(double salary,int month)
{
    int count_salary = salary - month * 5000;
    int tax = 0;

    if( count_salary > 0)
    {        
        if(count_salary<36000)
        {
            return count_salary * 0.03;
        }
        else if( count_salary>36000 && count_salary < 144000)
        {
            return 36000 * 0.03 + (count_salary - 36000)*0.1;
        }
        else if( count_salary > 144000 && count_salary < 300000)
        {
            return 36000 * 0.03 + (144000 - 36000)*0.1 + count_salary - 144000 * 0.2;
        }
        else if( count_salary > 300000 && count_salary < 420000)
        {
            return 36000 * 0.03 + (144000 - 36000)*0.1 + (300000 - 144000) * 0.2 + (count_salary - 300000) * 0.25;
        }
        else if( count_salary > 420000 && count_salary < 660000)
        {
            return 36000 * 0.03 + (144000 - 36000)*0.1 + (300000 - 144000) * 0.2 + (420000 - 300000) * 0.25 + (count_salary - 420000) * 0.3;            
        }
        else if( count_salary > 660000 && count_salary < 960000)
        {
            return 36000 * 0.03 + (144000 - 36000)*0.1 + (300000 - 144000) * 0.2 + (420000 - 300000) * 0.25 + (660000 - 420000) * 0.3 + (count_salary - 660000) * 0.35;            
        }
        else if( count_salary > 960000)
        {
            return 36000 * 0.03 + (144000 - 36000)*0.1 + (300000 - 144000) * 0.2 + (420000 - 300000) * 0.25 + (660000 - 420000) * 0.3 + (960000 - 660000) * 0.35 + (count_salary - 960000) * 0.45;            
        }
    }
    else
    {
        return 0;
    }
}


int main()
{

    double money,tax,sum=0;
    int i;
    for(i=1;i<=12;i++)
    {
        scanf("%lf",&money);
        sum+=money;
        tax=m_tax(sum,i);//这里需要传sum 
        printf("the sum of %d months tax is %.2f\n",i,tax);
    }
    return 0;
}

#include<stdio.h>

double m_tax(double salary,int month)
{
    if(salary<=5000*month)
        return 0.0;
    else
    {
        if((salary-5000*month)<36000)
        {
            return (salary-5000*month)*0.03;
        }
        else if((salary-5000*month)>=36000&&(salary-5000*month)<144000)
        {
            return (salary-5000*month)*0.1;
        }
        else if((salary-5000*month)>=144000&&(salary-5000*month)<300000)
        {
            return (salary-5000*month)*0.2;
        }
        else if((salary-5000*month)>=300000&&(salary-5000*month)<420000)
        {
            return (salary-5000*month)*0.25;
        }
        else if((salary-5000*month)>=420000&&(salary-5000*month)<660000)
        {
            return (salary-month*5000)*0.3;
        }
        else if((salary-5000*month)>=660000&&(salary-5000*month)<960000)
        {
            return (salary-month*5000)*0.35;
        }
        else
        {
            return (salary-month*5000)*0.45;
        }
    }
}

int main()
{
    double money[12],tax[12],sum=0;
    for(int i=0; i<12; i++)
    {
        scanf("%lf",&money[i]);
        sum+=money[i];
        tax[i]=m_tax(sum,i+1);
    }
    for(int i=1; i<=12; i++)
    {

        printf("the sum of %d months tax is %.2f\n",i,tax[i-1]);
    }

    return 0;
}