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C program to count total number of notes in entered amount

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C program to count total number of notes in entered amount

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I have used DEV-C++ compiler for debugging purpose. But you can use any C programming language compiler as per your availability.

#include <stdio.h>
 
int main()
{
    int amount;
        
    int note1, note2, note5, note10, note20, note50, note100, note500;
    
    note1 = note2 = note5 = note10 = note20 = note50 = note100 = note500 = 0;     
    
    printf("Enter amount: ");
    scanf("%d", &amount); 
 
    if(amount >= 500)
    {
        note500 = amount/500;
        amount -= note500 * 500;
    }
    if(amount >= 100)
    {
        note100 = amount/100;
        amount -= note100 * 100;
    }
    if(amount >= 50)
    {
        note50 = amount/50;
        amount -= note50 * 50;
    }
    if(amount >= 20)
    {
        note20 = amount/20;
        amount -= note20 * 20;
    }
    if(amount >= 10)
    {
        note10 = amount/10;
        amount -= note10 * 10;
    }
    if(amount >= 5)
    {
        note5 = amount/5;
        amount -= note5 * 5;
    }
    if(amount >= 2)
    {
        note2 = amount /2;
        amount -= note2 * 2;
    }
    if(amount >= 1)
    {
        note1 = amount;
    }
 
    printf("Total number of notes = \n");
    printf("500 = %d\n", note500);
    printf("100 = %d\n", note100);
    printf("50 = %d\n", note50);
    printf("20 = %d\n", note20);
    printf("10 = %d\n", note10);
    printf("5 = %d\n", note5);
    printf("2 = %d\n", note2);
    printf("1 = %d\n", note1);
 
    return 0;
}

Result:

Enter amount: 685

Total number of notes = 

500 = 1

100 = 1

50 = 1

20 = 1

10 = 1

5 = 1

2 = 0

1 = 0

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