c - 循环和函数 :Retirment lab for class

Question

我无法弄清楚我的代码有什么问题。有两个问题是年份不会从 0 1 2 等它变成 1 1 或 0 0，还有第二个问题是赢得利息像我预期的那样加起来。我不知道它出了什么问题，所以我希望能有第二双眼睛看到它，所以提前谢谢。这是源代码和我输出的一部分。

#include <stdio.h>
#define TARGET_VALUE 2000000
#define RATE .10




int main(void) 
{

int year;
double balance, target, finalbalance, deposit, rate, interest;

printf("Enter starting balance ($): ");
scanf("%lf", &balance);
printf("Enter amount deposited every year ($): ");
scanf("%lf", &deposit);
printf("Enter estimated annual interest rate (\%%): ");
scanf("%lf", &rate);
printf("Enter target balance ($): ");
scanf("%lf", &target);

year = 0;
interest = 0;

finalbalance = balance + deposit + interest;

printf("\nYear    Deposit    Interest    Balance");
printf("\n----    -------    --------    -------");
printf("\n%d   %.2lf    %.2lf    %.2lf", year, deposit, interest,
      finalbalance);
do {
    finalbalance = finalbalance + deposit + interest;
    printf("\n%d    %.2lf    %.2lf    %.2lf", year, deposit, interest,
    finalbalance);
    year += 1;
    interest = (finalbalance + deposit) * RATE;
} while (finalbalance <= target);

printf("\nIn year %d, balance %.2lf reaches target %.2lf", year, balance, target);
}



Enter starting balance ($):250000
Enter amount deposited every year($):20000
Enter estimated annual interest rate(%):10
Enter target balance ($): 2000000

year    deposit    interest    balance
0       20000.0     0.00        270000.00
0       20000.0     0.00        290000.00
1       20000.0     31000.00    341000.00
2       20000.0      36100.00    397100.00

这是我的老师给出的预期输出，但我的代码没有匹配。

Enter starting balance ($): 250000.0
Enter amount deposited every year ($): 20000.0
Enter estimated annual interest rate (%): 10.0
Enter target balance ($): 2000000.0
 Year Deposit Interest Balance
 ---- ------- -------- -------
 0 250000.00 0.00 250000.00
 1 20000.00 27000.00 297000.00
 2 20000.00 31700.00 348700.00
 3 20000.00 36870.00 405570.00
 4 20000.00 42557.00 468127.00
 . . .
 . . .
In year ??, balance ????? reaches target 2000000.00 

Answer 1

切换循环中发生的一些事情。打印前递增年份。这些宏是不必要的，您的版本忽略了输入利率。通常最好将 \n 换行符放在 printf() 格式字符串的末尾。将输入利率除以 100。balance 和 finalbalance 是多余的。在第 0 年，存款是初始余额。最重要的是，您可以通过为 printf() 指定宽度来使一切变得漂亮。

#include <stdio.h>

int main(void) 
{

    int year;
    double balance, target, deposit, rate, interest;

    printf("Enter starting balance ($): ");
    scanf("%lf", &balance);
    printf("Enter amount deposited every year ($): ");
    scanf("%lf", &deposit);
    printf("Enter estimated annual interest rate (%%): ");
    scanf("%lf", &rate);
    rate = rate * 0.01;
    printf("Enter target balance ($): ");
    scanf("%lf", &target);

    year = 0;
    interest = 0;

    printf("Year      Deposit      Interest       Balance\n");
    printf("----    ---------    ----------    ----------\n");

    static const char fmt_str[] = "%4d   %10.2lf    %10.2lf    %10.2lf\n";
    printf(fmt_str, year, balance, interest, balance);

    // we haven't checked if target was reached with the initial deposit so use
    // while instead of do...while
    while (balance < target)
    {
        year += 1;
        interest = (balance + deposit) * rate;
        balance = balance + deposit + interest;
        printf(fmt_str, year, balance, interest, balance);
    }

    printf("In year %3d, balance %.2lf reaches target %.2lf\n", year, balance, target);
    return 0;

}

输出:

Enter starting balance ($): 250000.0
Enter amount deposited every year ($): 20000.0
Enter estimated annual interest rate (%): 10.0
Enter target balance ($): 2000000.0
Year      Deposit      Interest       Balance
----    ---------    ----------    ----------
   0    250000.00          0.00     250000.00
   1    297000.00      27000.00     297000.00
   2    348700.00      31700.00     348700.00
   3    405570.00      36870.00     405570.00
   4    468127.00      42557.00     468127.00
   5    536939.70      48812.70     536939.70
   6    612633.67      55693.97     612633.67
   7    695897.04      63263.37     695897.04
   8    787486.74      71589.70     787486.74
   9    888235.41      80748.67     888235.41
  10    999058.96      90823.54     999058.96
  11   1120964.85     101905.90    1120964.85
  12   1255061.34     114096.49    1255061.34
  13   1402567.47     127506.13    1402567.47
  14   1564824.22     142256.75    1564824.22
  15   1743306.64     158482.42    1743306.64
  16   1939637.30     176330.66    1939637.30
  17   2155601.03     195963.73    2155601.03
In year  17, balance 2155601.03 reaches target 2000000.00