根据用户输入计算尽可能少的硬币

Question

相当新的编码员正在学习创建一个程序，该程序将从 0 到 99 之间的任何给定美分输入中输出尽可能少的硬币。我不需要为此代码添加任何限制，这就是为什么你不看不到 0 到 99 之间的任何限制。

这是我所拥有的，但我似乎无法弄清楚如何正确使用模数来结转之前计算的剩余数字。

非常感谢您提供的任何建议！我想我已经很接近了，但是我在 % 模数的心理壁垒上。我知道我可以使用 long 算术计算前一个数字，但我想弄清楚如何使用 % 修饰符使其更短。

#include <iostream>
using namespace std;

int main()
{
int cents; 
const int quarter = 25; 
const int dime = 10;
const int nickel = 5;
const int penny = 1;

// Get the amount of cents
cout << "Please enter an amount in cents less than a dollar." << endl;
cin >> cents;

// Calculate how many of each coin you can get from cent input
int Q = cents % quarter;
int D = Q % dime;
int N = D % nickel;
int P = N % penny;

// Display the coins you can get from cent input
cout << "Your change will be:" << endl;
cout << "Q: " << Q << endl;
cout << "D: " << D << endl;
cout << "N: " << N << endl;
cout << "P: " << P << endl;

return 0;
}

Answer 1

% returns除法的余数，不是四舍五入的数字。所以它分配给 Q 除法的剩余部分。建议你先算出X型硬币的个数，再把余数留到下一次计算。像

int Q = cents / quarter;
int D = (cents%quarter) / dime;

以此类推

Answer 2

每种硬币类型的总数：

int Q = cents / quarter;
int D = cents / dime;
int N = cents / nickel;
int P = cents / penny;

最少的硬币数量

int Q = cents / quarter; cents %= quarter; 
int D = cents / dime; cents %= dime; 
int N = cents / nickel; cents %= nickel; 
int P = cents;

Answer 3

我不太明白你想要什么，但据我了解，这就是我相信你正在尝试做的事情

int cointypes[4]={25,10,5,1};
int coinnumber[4];
int temp=cents;
for(int i=0;i<=3;i++)
{
 coinnumber[i]=temp/cointypes[i];
 temp=temp%cointypes[i];

}
cout << "Your change will be:" << endl;
cout << "Q: " << coinnumber[0] << endl;
cout << "D: " <<coinnumber[1] << endl;
cout << "N: " << coinnumber[2]<< endl;
cout << "P: " << coinnumber[3]<< endl;

Answer 4

这是另一个使用 std::div 和 C++17 结构化绑定的解决方案：

#include <iostream>
#include <string>
#include <cstdlib>
#include <utility>
#include <vector>

int main()
{
    std::vector<std::pair<std::string, int>> coins {
        {"quarter", 25},
        {"dime", 10},
        {"nickel", 5},
        {"penny", 1} };

    std::cout << "Please enter an amount in cents less than a dollar.\n";
    int cents;
    std::cin >> cents;

    std::cout << "Your change will be:\n";
    for(const auto& [coin_name, value] : coins) {
        auto[coin_count, remainder] = std::div(cents, value);
        cents = remainder;
        std::cout << coin_name << "\t: " << coin_count << '\n';
    }
}

根据用户输入计算尽可能少的硬币

Calculating the fewest possible coins possible from user input

c++

modulus