使用 while 循环的 C++ 可能的硬币组合

Question

我在编程课上遇到了一个挑战，我们必须使用一个 void 函数来计算可能的硬币组合，给定的变化值从 1 到 99 美分。

到目前为止，我的代码如下所示:

    #include <iostream>

using namespace std;

void computeCoins(int coinValue, int& num, int& amount_left);

int main()
{
    //Define varibles then get user input for change
    int leftOver, quarters=0, dimes=0, nickels=0, pennies=0, coins=0, originalAmount;
    do
    {
    cout << "Enter change amount: ";
    cin >> originalAmount;
    leftOver = originalAmount;
    } while ((leftOver > 99) || (leftOver < 1)); //Larger than 99 or smaller than 1? If yes, then try again.

    //Quarters
    computeCoins(25, coins, leftOver);
    quarters = coins;

    //Dimes
    computeCoins(10, coins, leftOver);
    dimes = coins;

    //Nickels
    computeCoins(5, coins, leftOver);
    nickels = coins;
    pennies = leftOver;

    cout << originalAmount << " cent/s could be given as " << quarters << " quarter/s, " << dimes << " dime/s, " << nickels << " nickel/s, " << " and " << pennies << " pennies.";
    cout << endl;
    system("PAUSE");
    return 0;
}

void computeCoins(int coinValue, int& num, int& amount_left)
{
    //Using the specified coin value, find how many can go into it then subtract it
    while (amount_left % coinValue == 0)
    {
        // Still dividable by the coin value
        num += 1;
        amount_left -= coinValue;
    }
}

现在我的问题是，当我运行该程序时，它会为 25 美分、10 美分和 5 分硬币返回一个非常大的负值。我肯定这与我的循环条件的设置方式有关，有人知道为什么会这样吗？

Answer 1

两个问题:一是未定义硬币初始值。两个 amount_left % coinValue == 0 部分 - 我想你的意思是 amount_left >= coinValue

虽然没有必要一直在那个函数里迭代

void computeCoins(int coinValue, int& num, int& amount_left)
{
    // add as many coinValues as possible.    
    num += amount_left / coinValue;
    // the modulus must be what is left.
    amount_left = amount_left % coinValue;
}

请注意(除其他事项外)，您最好使用 unsigned ints 来表示数量。