贪婪硬币计数中的整数溢出 [英] Integer overflow in greedy coin counting

问题描述

#include <stdio.h>
#include <cs50.h>
#include <math.h>

int main (void) { 

    printf ("Enter amount: ");
    float amount = GetFloat();
    int coins = 0;

    while (amount != 0) {

        if (fmod(amount, 0.25) == 0) {
            amount = amount - 0.25;
            coins += 1;
        }

        else if (fmod(amount, 0.10) == 0) {
            amount = amount - 0.10;
            coins += 1;
        }

        else if (fmod(amount, 0.05) == 0) {
            amount = amount - 0.05;
            coins += 1;
        }

        else {
            amount = amount - 0.01;
            coins += 1;
        }
    }

    printf ("Coins : %d\n", coins);
}

我正在尝试实现一种小型贪心算法，用户可以在其中输入一定数量的货币（例如：9.25），我们输出的货币兑换量最少（仅25美分，10美分，5美分和1美分）。

I'm trying to implement a small greedy algorithm, in which a user inputs an amount of money ( Ex: 9.25 ) and we output the least amount of coins that it takes for us to exchange it in change( 25 cents, 10 cents, 5 cents and 1 cent only).

此算法适用于int数量（例如10或20）以及仅要求程序使用25美分硬币的数量。

This algorithm works with int amounts like 10 or 20 and with amounts that only requires the program to use the 25 cents coins.

如果我尝试像9.10或9.01这样的金额，出现运行时错误，有符号整数溢出。我理解这意味着什么，但我不明白硬币的价值怎么突然变得如此之高。

If I try an amount like 9.10 or 9.01, I get a runtime error, signed integer overflow. I understand what it means, but I don't understand how can the value of coins go so high all of a sudden.

推荐答案

正如Danial Tran所说，在执行逻辑运算时最好使用int。请阅读为什么不使用Double或Float表示货币？也可以避免while循环。

As Danial Tran said it is better to use int when you do logical operations. Please read Why not use Double or Float to represent currency? Also you can avoid while loop.

#include <stdio.h>
#include <cs50.h>
#include <math.h>

int main (void) { 

    printf ("Enter amount: ");
    //float amount = GetFloat(); //Please refer to chqrlie's comments below.
    double amount  = 0.0;
    scanf("%lf",&amount); //I don't know GetFloat() equivalent for double. So using scanf().

    long long int amountInt = (long long int) (amount * 100.0);

    int coins = 0;

    if (25 <= amountInt) {
        coins += (amountInt/25);
        amountInt = amountInt % 25;
    }

    if (10 <= amountInt) {
        coins += (amountInt/10);
        amountInt = amountInt % 10;
    }

    if (5 <= amountInt) {
        coins += (amountInt/5);
        amountInt = amountInt % 5;
    }

    if (1 <= amountInt) {
        coins += amountInt;
    }

    printf ("Coins : %d\n", coins);
}

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