C++典型贪心算法--找硬币问题

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最新推荐文章于 2025-05-19 19:25:44 发布

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分类专栏：技术交流 C/C++ 文章标签： c++ 算法 input output each date

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本文介绍了一个简单的找零程序实现方案，该程序通过输入所需的找零金额及可用的硬币数量来计算最少硬币数目的找零组合。文章通过示例展示了如何使用C++语言进行逻辑处理，包括对输入数据的读取、找零逻辑的设计以及最终输出结果。

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/************************************************************************/
/*FileName:Making Change
/*Author:PenglueR
/*Date:2009/07/23
/*Comment:
Problem description
In the old days before everything was electronic, you could pay for stuff with cash and (maybe) get change back. The number of coins you get could vary. For example, 10 cents in change can be done with 1 dime, or 2 nickels, or 1 nickel and 5 pennies, or 10 pennies. The rule of thumb was for the cashier to give the least amount of coins from the selection of coins in the register. Now since the cashiers may not be quick enough to give you the coins, you're to write a program that produces the correct amount of change in the fewest coins

Input
Each line of input will have 5 integers on it. The first number is the amount of change to make. The second number is the number of pennies available, followed by the number of nickels, then the number of dimes and finally the number of quarters.

Output
For each input, print a line that has the number of pennies, the number of nickels, the number of dimes and the number of quarters needed to make the given change using the least number of coins. If the requested amount of change can't be made from the given amount of coins, then print the message "Not enough change".

Sample Input
89 3 0 1 2
89 10 10 10 10
89 0 0 0 4
45 45 0 0 0

Sample Output
Not enough change
4 0 1 3
Not enough change
45 0 0 0
/************************************************************************/

#include <iostream>
using namespace std;
int main()
{
    int money,penny,nickel,dime,quarter,p,n,d,q;
    while (cin>>money>>penny>>nickel>>dime>>quarter||(getchar() != EOF))
    {
        q = money/25;
        if (q <= quarter)
            money -= q*25;
        else
        {
            q = quarter;
            money -= q*25;
        }
        d = money/10;
        if (d <= dime)
            money -= d*10;
        else
        {
            d = dime;
            money -= d*10;
        }
        n = money/5;
        if (n <= nickel)
            money -= n*5;
        else
        {
            n = nickel;
            money -= n*5;
        }
        p = money;
        if (money <=  penny)
            cout<<p<<" "<<n<<" "<<d<<" "<<q<<endl;
        else
            cout<<"Not enough change"<<endl;
    }
    return 0;
}