求本金，试过了，调试了还是不行
题目内容：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。
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"
#include <stdio.h>
#include <math.h>
int main()
{
int year;
float deposit,capital;
float rate;
char Y,y,N,n;
printf("Input capital, year:");
scanf("%lf,%d",&capital,&n);
switch (n)
{
printf("Compound interest (Y/N)?");
if(scanf(" %c",&y)||scanf(" %c",&y))
{
switch(n)
{
case 1: rate=0.0225;
deposit = capital * pow((1 + rate),year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 2: rate=0.0243;
deposit = capital * pow((1 + rate),year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 3: rate=0.0270;
deposit = capital * pow((1 + rate),year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 5: rate=0.0288;
deposit = capital * pow((1 + rate),year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 8: rate=0.0300;
deposit = capital * pow((1 + rate),year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
default:
printf("Error year!\n");
}
}
if(scanf(" %c",&N)||scanf(" %c",&n))
{
switch(n)
{
case 1:rate=0.0225;
deposit = capital * (1 + rate * year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 2: rate=0.0243;
deposit = capital * (1 + rate * year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 3: rate=0.0270;
deposit = capital * (1 + rate * year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 5: rate=0.0288;
deposit = capital * (1 + rate * year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
case 8: rate=0.0300;
deposit = capital * (1 + rate * year);
printf("rate = %.4f, deposit = %.4f\n",rate,deposit);
break;
default:
printf("Error year!\n");
}
}
}
}
