**更新：这是由Loc Tran解决的！谢谢LocTran !! **
我正在编写一个程序（严格使用C）来计算美国面额的最小变化。在我的程序达到一角硬币（第一个面额不能被整数整除之后，程序似乎不想继续使用镍币和硬币。
我试图使用if语句，如果返回零则排除dimesDue值，但似乎无法弄明白。如果你注意到，我还必须根据从总变化金额中扣除的先前面额，为每个面额制作一个新的变更变量。我宁愿简化这个并在每次计算后指定新值，但不能。
// Amount Tendered and Purchase amount converted to pennies
amountDue = 2117;
amountGiven = 10000;
// Creating a new change amount for each denomination, based on each previous computation
change = amountGiven - amountDue;
change10s = change % (20 * 100);
change5s = change % (10 * 100);
change1s = change % (5 * 100);
changeQs = change % (1 * 100);
changeDs = change % 25;
changeNs = change % 10;
changePs = change % 1;
// Using each new change amount to calculate amount of denomination
twentiesDue = (change / 20) / 100;
tensDue = (change10s / 10) / 100;
fivesDue = (change5s / 5) / 100;
onesDue = (change1s / 1) / 100;
quartersDue = (changeQs / 25);
dimesDue = (changeDs / 10);
nickelsDue = (changeNs / 5);
penniesDue = (changePs / 1);
printf("Amount Due: $21.17\nAmount Tendered: $100\n\n");
printf("Change Due:\n(by denomination)\n");
printf("Twenties: %d\n", twentiesDue);
printf("Tens: %d\n", tensDue);
printf("Fives: %d\n", fivesDue);
printf("Ones: %d\n", onesDue);
printf("Quarters: %d\n", quartersDue);
printf("Dimes: %d\n", dimesDue);
printf("Nickels: %d\n", nickelsDue);
printf("Pennies: %d\n", penniesDue);
程序达到一角硬币（第一个面额不等于整数）并且不会继续计算镍和硬币的数量。因为剩下的季度变化后的剩余部分是8美分，这不能被10整除。但是我无法弄清楚如何使用if语句指定忽略它！
因此，结果是一旦程序达到一角硬币，之后所有变量都计算为零。但应该有一个镍和三个便士！这是我运行时的结果：
到期金额：21.17美元招标金额：100美元
变更到期日:(按面额计算）二十年代：3十分：1个法孚：1分：3分：3分：0镍：0便士：0
虽然这个解决方案有效，但我无法使用它。必须使用模数运算符。
使用货币值更改后计算剩余金额时出现问题。获取数字中的所有数字是类似的过程。例如，在获得1之后有198，剩下的是198-1 * 100 = 98，然后我们得到9，剩下的是98-9 * 10 = 8，然后得到8然后我们完成了。你用错误的公式计算。这是我的计算方法，以及在数字中获取数字的相同方式，以及整个解决方案：
#include <stdlib.h>
#include <stdio.h>
int main()
{
// Amount Tendered and Purchase amount converted to pennies
int amountDue = 2117;
int amountGiven = 10000;
// Using each new change amount to calculate amount of denomination
// getting 20dollars notes
int change = amountGiven - amountDue;
int twentiesDue = (change / 20) / 100;
// get the remaining after change 20dollar notes, and
// then divide 10*100 for getting 10dollars notes
int change10s = change % (20*100);
int tensDue = (change10s / 10) / 100;
// get the remaining after change 10dollar notes, and
// then divide 5*100 for getting 5dollars notes
int change5s = change10s % (10*100);
int fivesDue = (change5s / 5) / 100;
// get the remaining after change 5dollar notes, and
// then divide 1*100 for getting 1dollars notes
int change1s = change5s % (5*100);
int onesDue = (change1s / 1) / 100;
// get the remaining after change 1dollar notes, and
// then divide 25 for getting quarter coins
int changeQs = change1s % (1*100);
int quartersDue = (changeQs / 25);
// get the remaining after change quarter coins, and
// then divide 10 for getting dime coins
int changeDs = changeQs % 25;
int dimesDue = (changeDs / 10);
// get the remaining after change dime coins, and
// then divide 5 for getting nickel coins
int changeNs = changeDs % 10;
int nickelsDue = (changeNs / 5);
// get the remaining after change nickel coins, and
// then divide 1 for getting 1cent coins
int changePs = changeNs % 5;
int penniesDue = (changePs / 1);
printf("Amount Due: $21.17\nAmount Tendered: $100\n\n");
printf("Change Due:\n(by denomination)\n");
printf("Twenties: %d\n", twentiesDue);
printf("Tens: %d\n", tensDue);
printf("Fives: %d\n", fivesDue);
printf("Ones: %d\n", onesDue);
printf("Quarters: %d\n", quartersDue);
printf("Dimes: %d\n", dimesDue);
printf("Nickels: %d\n", nickelsDue);
printf("Pennies: %d\n", penniesDue);
return 0;
}