code works strangely in greedy without modulo

Question

I have been trying to help my friend with greedy without modulo (mine I did with modulo). We used the if conditions and for loops. The code works but counts coins strangely. When I give an input of 0.61 the result is 3, instead of 4. input 0.40 - result 2, when it should be 3.

I have changed the if conditions and for loops and put while loops instead. The program works perfectly. I just can't understand what is wrong with the first code. Why is machine skipping some if conditions?

Here's my code with if conditions:

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

int main(void)

{
printf ("Hi! How much is owed?\n");

int money;
float mon;
int count = 0;
int count1 = 0;
int count2 = 0;
int count3 = 0;
int count4 = 0;


do
{
    mon = get_float();
}
while (mon < 0);

money = round(mon * 100);

// minus 25c
if (money >= 25)
{
    for (int i = 0; money > 0; i++)
    {
        money = money - 25;
        count1 = count1 + 1;
    }
}

// minus 10c
if (money >= 10)
{
    for (int i = 0; money > 0; i++)
    {
        money = money - 10;
        count2 = count2 + 1;
    }
}

// minus 5c
if (money >= 5)
{
    for (int i = 0; money > 0; i++)
    {
        money = money - 5;
        count3 = count3 + 1;
    }
}
// for ones
    if (money >= 1)
{
    for (int i = 0; money > 0; i++)
    {
        money = money - 1;
        count4 = count4 + 1;
    }
}

count = count1 + count2 + count3 + count4;

printf ("%d\n", count);
return 0;
}

Answer 1

In your for loops, int i = 0 and i++ can be removed, since you don't use i's value other than in that i++. You're left with what is essentially a while loop.

You let your loop run as long as money is greater than 0, which means you would subtract a quarter even if money is only one cent. Replace both the if and the for with a single while, using the correct comparison of the if.