Question

请帮帮我！我的c ++程序似乎有错误。代码显示关于面额的错误输出，给出输入比索的整数量。编写一个程序，显示1000,500,100,50,20和10比索的账单数量。输出以及所有面额被删除后的剩余金额。

这是我的程序代码。

#include <iostream>

using namespace std;

int main()
{
    int a;
    cout << "Enter the amount: ";
    cin  >> a;

    cout << "No. of 1000 peso bills:  "  << a/1000;
    cout << "\nNo. of 500  peso bills:  "  << a%1000/500;
    cout << "\nNo. of 100  peso bills:  "  << a%500/100;
    cout << "\nNo. of 50   peso bills:  "  << a%100/50;
    cout << "\nNo. of 20   peso bills:  "  << a%50/20;
    cout << "\nNo. of 10   peso bills:  "  << a%20/10;
    cout << "\n\nThe rest of the amount: " << a%10;
}

输出显示：

Enter the amount: 34757
No. of 1000 peso bills:  34
No. of 500  peso bills:  1
No. of 100  peso bills:  2
No. of 50   peso bills:  1
No. of 20   peso bills:  0
No. of 10   peso bills:  1

The rest of the amount: 7

Process returned 0 (0x0)   execution time : 2.156 s
Press any key to continue.

10比索账单的数量必须是0而不是1，我怎么能想更正呢？提前谢谢。

Answer 1

问题在于数学。由于50不是20的倍数，您需要修复10比索账单的计算：

(a%50-(a%50)/20*20)/10;

这当然只是针对这一具体案例的解决方案。一般来说，如果你有其他账单，事情会更复杂。在你的情况下，你很幸运，因为大多数账单都是所有小账单的完全倍数，其中50/20对是唯一的例外。

一个更通用和可读的解决方案，受到另一个答案的启发：

#include <iostream>
using namespace std;

int main()
{
    int a;
    cout << "Enter the amount: ";
    cin  >> a;

    cout << "No. of 1000 peso bills:  "  << a/1000;
    cout << "\nNo. of 500  peso bills:  "  << (a%=1000)/500;
    cout << "\nNo. of 100  peso bills:  "  << (a%=500)/100;
    cout << "\nNo. of 50   peso bills:  "  << (a%=100)/50;
    cout << "\nNo. of 20   peso bills:  "  << (a%=50)/20;
    cout << "\nNo. of 10   peso bills:  "  << (a%=20)/10;
    cout << "\n\nThe rest of the amount: " << a%10;
}

Answer 2

解决问题的一种方法是从总数中减去已用较大账单表示的金额，如下所示：

cout << "No. of 1000 peso bills:  "  << a/1000;
a %= 1000;
cout << "\nNo. of 500  peso bills:  "  << a%1000/500;
a %= 500;
cout << "\nNo. of 100  peso bills:  "  << a%500/100;
a %= 100;
cout << "\nNo. of 50   peso bills:  "  << a%100/50;
a %= 50;
cout << "\nNo. of 20   peso bills:  "  << a%50/20;
a %= 20;
cout << "\nNo. of 10   peso bills:  "  << a%20/10;
a %= 10;
cout << "\n\nThe rest of the amount: " << a;

这将模拟进行更改的过程，方法是在您获得较小的账单时，从您已经提供较大账单的金额中删除。

Demo on ideone

使用c ++模运算符计算票据的面额

2 个答案: