Question

因此，我必须制作一个“硬币找零制作器”程序，用户在其中输入价格和支付的金额，而输出则必须是他们的四分之一硬币，硬币，小费和便士硬币。由于某种原因，我的程序无法提供正确的输出。如果我输入价格为40美分，输入我支付的金额为50美分，则表示所需的零钱为10美分，但2角钱，所以这给了我一毛钱。如果有人可以帮助我，将不胜感激。

int main()
{

    int numberOfQuarters =0;
    int numberOfDimes =0;
    int numberOfNickels =0;
    int numberOfPennies =0;

    int price;
    int paid;
    int change;



    printf("Please enter the price of your item in cents: ");
    scanf("%d", &price);

    printf("Please enter the amount of money you gave in cents: ");
    scanf("%d", &paid);

    change = (paid - price);
    printf("Change required: %d", change);

    while(change >= 25)
    {
        numberOfQuarters++;
        if(numberOfQuarters >0 )
        {
            change = (change - (numberOfQuarters * 25));
        }
        printf("\nNumber of Quarters:%d", numberOfQuarters);
    }

    while((change >= 10) && (change < 25))
    {
        numberOfDimes++;
        if(numberOfDimes > 1)
        {
            change = (change - (numberOfDimes * 10));
        }
    }
    while((change >= 5) && (change < 10))
    {
        numberOfNickels++;
        if(numberOfNickels >0)
        {
            change = (change - (numberOfNickels * 5));
        }
    }

    printf("\nThe number of quarters: %d", numberOfQuarters);
    printf("\nThe number of dimes: %d", numberOfDimes);
    printf("\nThe number of nickels: %d", numberOfNickels);
    printf("\nThe number of pennies: %d", change);

}

Answer 1

一个简单的逻辑错误。

将if(numberOfDimes > 1)更改为if(numberOfDimes > 0)。

更好的是，您不需要块内的if语句。如果程序到达while循环内部，则if语句将始终为true。

Answer 2

while(change >= 25)
{
    numberOfQuarters++;
    if(numberOfQuarters >0 )
    {
        change = (change - (numberOfQuarters * 25));
    }
    printf("\nNumber of Quarters:%d", numberOfQuarters);
}

此结构是错误的，因为在此代码中，第二季度的价值将增加两倍，而第三季度的价值将增加第三倍。您不需要乘以numberOfQuarters，也不需要检查numberOfQuarters是否为正，因为从零开始递增（如果不发生溢出），它必须为正。

此部分应如下所示：

while(change >= 25)
{
    numberOfQuarters++; /* add one more quarter */
    change = change - 25; /* subtract the value of one quarter because one is added */
    printf("\nNumber of Quarters:%d", numberOfQuarters);
}

角钱和镍价的计算也应采用这种方式。

使用两个一角硬币的原因是，由于某种原因，您倾向于在只有一个一角硬币的情况下不降低一角硬币的值。

Answer 3

因此，既然已经解决了主要问题，那么我不妨提出另一种实现方式：

#include <stdio.h>

int main()
{

    int numberOfQuarters =0;
    int numberOfDimes =0;
    int numberOfNickels =0;
    int numberOfPennies =0;

    int price;
    int paid;
    int change;

    printf("Please enter the price of your item in cents: ");
    scanf("%d", &price);

    printf("Please enter the amount of money you gave in cents: ");
    scanf("%d", &paid);

    change = (paid - price);
    printf("Change required: %d", change);

    numberOfQuarters += change / 25;
    change %= 25;
    numberOfDimes += change / 10;
    change %= 10;
    numberOfNickels += change / 5;
    change %= 5;
    numberOfPennies = change;

    printf("\nThe number of quarters: %d", numberOfQuarters);
    printf("\nThe number of dimes: %d", numberOfDimes);
    printf("\nThe number of nickels: %d", numberOfNickels);
    printf("\nThe number of pennies: %d\n", change);

}

这样，您可以避免出现任何循环，而只需使用除法和提醒即可。

C语言中的简单硬币更改制作程序

3 个答案: