CS50第1周Cash作业逻辑疑问与代码测试异常求助

我搞不懂CS50第1周Cash作业的测试逻辑——按贪心算法优先用最大面额硬币的话，73美分找零应该是2个quarter、2个dime和3个penny，28美分是1个quarter加3个penny，但测试要求73美分的calculate_dimes返回7，28美分的calculate_nickels返回5，这到底是怎么回事？

我的代码 #

#include <cs50.h>
#include <stdio.h>

int get_cents(void);
int calculate_quarters(int cents);
int calculate_dimes(int cents);
int calculate_nickels(int cents);
int calculate_pennies(int cents);

int main(void)
{
    // Ask how many cents the customer is owed
    int cents = get_cents();

    // Calculate the number of quarters to give the customer
    int quarters = calculate_quarters(cents);
    cents = cents - quarters * 25;

    // Calculate the number of dimes to give the customer
    int dimes = calculate_dimes(cents);
    cents = cents - dimes * 10;

    // Calculate the number of nickels to give the customer
    int nickels = calculate_nickels(cents);
    cents = cents - nickels * 5;

    // Calculate the number of pennies to give the customer
    int pennies = calculate_pennies(cents);
    cents = cents - pennies * 1;

    // Sum coins
    int coins = quarters + dimes + nickels + pennies;

    // Print total number of coins to give the customer
    printf("%i\n", coins);
    printf("quarters: %i\ndimes: %i\nnickels:%i\npennies:%i\n ", quarters, dimes, nickels, pennies);
}
int quarters, dimes, nickels, pennies;
int get_cents(void)
{
    int cents;
    do
    {
        cents = get_int("Change owed: ");
    }
    while (cents < 0);
    return  cents;
}

int calculate_quarters(int cents)
{
    // TODO
    quarters = cents / 25;
    return quarters;
}

int calculate_dimes(int cents)
{
    // TODO
    dimes = (cents % 25) / 10;
    return dimes;
}

int calculate_nickels(int cents)
{
    // TODO
    nickels = (cents % 25 % 10) / 5;
    return nickels;
}

int calculate_pennies(int cents)
{
    // TODO
    pennies = cents % 25 % 10;
    return pennies;
}

CS50测试结果 #

:) cash.c exists
:) cash.c compiles
:) get_cents returns integer number of cents
:) get_cents rejects negative input
:) get_cents rejects a non-numeric input of "foo" 
:) calculate_quarters returns 2 when input is 50
:) calculate_quarters returns 1 when input is 42
:) calculate_dimes returns 1 when input is 10
:) calculate_dimes returns 1 when input is 15
:( calculate_dimes returns 7 when input is 73
    expected "7", not "2"
:) calculate_nickels returns 1 when input is 5
:( calculate_nickels returns 5 when input is 28
    expected "5", not "0"
:) calculate_pennies returns 4 when input is 4
:) input of 41 cents yields output of 4 coins
:) input of 160 cents yields output of 7 coins

我的程序输出 #

cash/ $ ./cash
Change owed: 73
7
quarters: 2
dimes: 2
nickels:0
pennies:3

 cash/ $ ./cash
Change owed: 28
4
quarters: 1
dimes: 0
nickels:0
pennies:3

问题原因与修正方案 #

核心错误是你误解了测试用例的调用逻辑：测试是单独调用每个计算函数的，不是像main函数那样先扣减大面额后传值。比如：

• 测试calculate_dimes(73)时，输入的是完整的73美分，不是扣完quarter剩下的23美分，所以正确结果是73 / 10 = 7
• 测试calculate_nickels(28)时，输入的是完整的28美分，不是扣完quarter剩下的3美分，所以正确结果是28 /5 =5

你的代码还有两个问题：

1. 没必要用全局变量存储硬币数量，容易引发意外问题
2. 每个计算函数里错误地对输入做了%25这类取余操作——这是main函数的职责，单个计算函数只需要接收当前剩余美分数，返回该面额最多能用的数量（即cents / 面额值）

修正后的计算函数如下：

int calculate_quarters(int cents)
{
    return cents / 25;
}

int calculate_dimes(int cents)
{
    return cents / 10;
}

int calculate_nickels(int cents)
{
    return cents / 5;
}

int calculate_pennies(int cents)
{
    return cents;
}

main函数里的逻辑是对的：先算quarters，扣减后传剩余值给dimes，以此类推——测试只是单独验证每个函数的独立计算能力，和main里的串联逻辑不冲突。

内容的提问来源于stack exchange，提问作者user21322673