编程c语言 存款利率计算器

Question

设capital是最初的存款总额（即本金），rate是整存整取的存款年利率，n 是储蓄的年份，deposit是第n年年底账号里的存款总额。已知如下两种本利之和的计算方式：
按复利方式计息的本利之和计算公式为：deposit = capital * (1 + rate) n
按普通计息方式计算本利之和的公式为：deposit = capital * (1 + rate * n)
已知银行整存整取不同期限存款的年息利率分别为：
存期1年，利率为 0.0225
存期2年，利率为 0.0243
存期3年，利率为 0.0270
存期5年，利率为 0.0288
存期8年，利率为 0.0300
若输入其他年份，则输出"Error year!"

编程从键盘输入存钱的本金和存款期限，然后再输入按何种方式计息，最后再计算并输出到期时能从银行得到的本利之和，要求结果保留到小数点后4位。

程序的运行结果示例1：
Input capital, year:10000,2↙
Compound interest (Y/N)?Y↙
rate = 0.0243, deposit = 10491.9049

程序的运行结果示例2：
Input capital, year:10000,2↙
Compound interest (Y/N)?n↙
rate = 0.0243, deposit = 10486.0000

程序的运行结果示例3：
Input capital, year:1000,4↙
Compound interest (Y/N)?y↙
Error year!

输入提示信息："Input capital, year:"
输入提示信息："Compound interest (Y/N)?"
存期输入错误的提示信息： "Error year!\n"
本金及存款期限的输入格式: "%lf,%d"
是否选择复利计算的输入格式： " %c" (注意：%c的前面有一个空格。输入的字符大小写皆可，即Y或y，N或n皆可)
输出格式："rate = %.4f, deposit = %.4f\n"
为避免出现格式错误，请直接拷贝粘贴题目中给的格式字符串和提示信息到你的程序中。
时间限制：500ms内存限制：32000kb

#include
#include
int main()
{
int year;
double rate, deposit, capital;
char x, y, Y, n, N;
printf("Input capital, year:");
printf("Compound interest (Y/N)?");
scanf("%lf,%d", &capital, &year);
switch (year)
{
case 1:
rate = 0.0225; break;
case 2:
rate = 0.0243; break;
case 3:
rate = 0.0270; break;
case 5:
rate = 0.0288; break;
case 8:
rate = 0.0300; break;
default:printf("Error year!\n");
}
switch (x)
{
case'y':
deposit = capital * pow((1 + rate), year);
printf("deposit = %.4f\n", deposit);
break;
case'Y':
deposit = capital * pow((1 + rate), year);
printf("deposit = %.4f\n", deposit);
break;
case'N':
deposit = capital * (1 + rate * year);
printf("deposit = %.4f\n", deposit);
break;
case'n':
deposit = capital * (1 + rate * year);
printf("deposit = %.4f\n", deposit);
break;
}
printf("rate = %.4f, deposit = %.4f\n", rate, deposit);
return 0;
}

Answer 1

#include<stdio.h>
#include<math.h>
int main()
{
    char Ci;
    int year;
    double capital, rate;
    printf("Input capital, year:");
    scanf("%lf,%d", &capital, &year);
    printf("Compound interest (Y/N)?");
    scanf(" %c",&Ci);
    if(Ci=='Y'||Ci=='y')
    {
        switch(year)
        {
            case 1 : rate = 0.0225;
            printf("rate = 0.0225, deposit = %.4lf\n", capital * pow(1+rate,year));break;
            case 2 : rate = 0.0243;
            printf("rate = 0.0243, deposit = %.4lf\n", capital * pow(1+rate,year));break;
            case 3 :rate = 0.0270;
            printf("rate = 0.0270, deposit = %.4lf\n", capital * pow(1+rate,year));break;
            case 5 :rate = 0.0288;
            printf("rate = 0.0288, deposit = %.4lf\n", capital * pow(1+rate,year));break;
            case 8 :rate = 0.0300;
            printf("rate = 0.0300, deposit = %.4lf\n", capital * pow(1+rate,year));break;
            default : printf("Error year!\n");
        }
    }
     else
    {
        switch(year)
        {
            case 1 : rate = 0.0225;
            printf("rate = 0.0225, deposit = %.4lf\n", capital * (1+rate*year));break;
            case 2 : rate = 0.0243;
            printf("rate = 0.0243, deposit = %.4lf\n", capital * (1+rate*year));break;
            case 3 :rate = 0.0270;
            printf("rate = 0.0270, deposit = %.4lf\n", capital * (1+rate*year));break;
            case 5 :rate = 0.0288;
            printf("rate = 0.0288, deposit = %.4lf\n", capital * (1+rate*year));break;
            case 8 :rate = 0.0300;
            printf("rate = 0.0300, deposit = %.4lf\n", capital * (1+rate*year));break;
            default : printf("Error year!\n");
        }
    }

return 0;

}