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C Program to find the number of denominations for a given amount

Last updated on September 23, 2020

The following program asks the user to enter an amount and prints number of notes (of denominations 500, 100, 50, 20, 10, 1) to be distributed. For example, if the user enters $451, then 4 note of 100, 1 note of 50 and 1 note of 1 is required.

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/***************************************************************
 Program to find the number of denominations for a given amount
 ***************************************************************/

#include <stdio.h>
#define SIZE 6

int main() 
{    
    int amount, notes;   

    // currency denominations
    int denominations[SIZE] = {500, 100, 50, 20, 10, 1};

    printf("Enter amount: ");
    scanf("%d", &amount);

    printf("\n");

    for (int i = 0; i < SIZE; i++) 
    {
        notes = amount / denominations[i];

        if (notes)
        {
            amount = amount % denominations[i]; // remaining money

            printf("%d * %d = %d \n", notes, denominations[i], 
                notes * denominations[i]);
        }                
    }

    return 0;
}

Try it now

Expected Output: 1st run:

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Enter amount: 642

1 * 500 = 500 
1 * 100 = 100 
2 * 20 = 40 
2 * 1 = 2

2nd run:

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Enter amount: 9241

18 * 500 = 9000 
2 * 100 = 200 
2 * 20 = 40 
1 * 1 = 1

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