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c - 开始程序有问题

转载 作者:塔克拉玛干 更新时间:2023-11-03 04:33:59 25 4
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此程序旨在获取用户输入(金额)并以硬币形式输出输入。该程序尝试使用尽可能少的硬币。该程序在大多数情况下都有效。例如,当我输入 2.30 的值时,我得到的输出仅为 9 个季度。我需要帮助解决这个编码问题。

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

int main()
{

int coins, quarter, money3, money2, pennies, nickels, dimes, quarters;

int penny, nickel, dime, money1;

float money;

quarter = 25;
dime = 10;
nickel = 5;
penny = 1;

printf("How much money would you like in change?\n");
scanf("%f", & money);

coins = (money * 100) + .5;

quarters = coins/quarter;
money1 = coins % quarter;
dimes = money1/dime;
money2 = coins % dime;
nickels = money2/nickel;
money3 = coins % nickel;
pennies = money3/penny;

printf("The number of quarters is %d\n", quarters);
printf("The number of dimes is %d\n", dimes);
printf("The number of nickels are %d\n", nickels);
printf("The number of pennies is %d\n", pennies);
printf("%d\n", money1);
printf("%d\n", coins);
}

最佳答案

您应该使用调试器单步执行程序并在变量运行时检查分配给变量的值,或者添加更多的 printf 以在分配/更改变量时输出变量的值。您遇到的问题基本上是一个简单的错字。我建议您再试一次尝试解决它,如果您自己没有弄清楚,请参阅以下 ideone 现场演示:http://ideone.com/BsIgg0

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

int main()

{
int coins, quarter, money3, money2, pennies, nickels, dimes, quarters;
int penny, nickel, dime, money1;
float money;

quarter = 25;
dime = 10;
nickel = 5;
penny = 1;

printf("How much money would you like in change?\n");
scanf("%f", & money);

coins = (money * 100) + .5;

printf("coins = %d\n", coins);

quarters = coins/quarter;
money1 = coins % quarter;
printf("money1 = %d\n", money1);
dimes = money1/dime;
money2 = coins % dime;
printf("money2 = %d\n", money2);
nickels = money2/nickel;
money3 = coins % nickel;
printf("money3 = %d\n", money3);
pennies = money3/penny;

printf("The number of quarters is %d\n", quarters);
printf("The number of dimes is %d\n", dimes);
printf("The number of nickels are %d\n", nickels);
printf("The number of pennies is %d\n", pennies);
printf("%d\n", money1);
printf("%d\n", coins);
}

关于c - 开始程序有问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20790036/

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