C Exercises: Convert a given amount to possible number of notes and coins

C Basic Declarations and Expressions: Exercise-100 with Solution

Write a C program to convert a currency value (floating point with two decimal places) to possible number of notes and coins.

Possible Notes: 100, 50, 20, 10, 5, 2, 1
Possible Coins: 0.50, 0.25, 0.10, 0.05 and 0.01

Sample Solution:

C Code:

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

int main() {
    double amt;

    unsigned int int_amt, frac_amt;
    printf("Input the currency value (floating point with two decimal places):\n");

    scanf("%lf", &amt);

    int_amt = (int) amt;
    amt -= int_amt;

    frac_amt = round((amt * 100));

    printf("\nCurrency Notes:");
    printf("\n100 number of Note(s): %d", int_amt / 100);

    int_amt -= (int_amt / 100) * 100;

    if (int_amt > 50) {
        printf("\n50 number of Note(s): 1");
        int_amt -= 50;
    } 

    if (int_amt/20 > 0)   
    printf("\n20 number of Note(s): %d", int_amt / 20);

    int_amt -= (int_amt / 20) * 20;
    if (int_amt/10 > 0)   
    printf("\n10 number of Note(s): %d", int_amt / 10);

    int_amt -= (int_amt / 10) * 10;
    if (int_amt/5 > 0)   
    printf("\n5 number of Note(s): %d", int_amt / 5);
		
	
    int_amt -= (int_amt / 5) * 5;
    
    if (int_amt > 0)   
    printf("\n2 number of Note(s): %d", int_amt / 2);
    
    
    int_amt -= (int_amt / 2) * 2;
   
    if (int_amt > 0)           
    printf("\n1 number of Note(s): %d", int_amt);
    
    
    printf("\n\nCurrency Coins:");

    if (frac_amt > 50) {
        printf("\n.50 number of Coin(s): 1");        
        frac_amt -= 50;
    } 

    if (frac_amt/25 > 0)   
    printf("\n.25 number of Coin(s): %d", frac_amt / 25);
       
    frac_amt -= (frac_amt / 25) * 25;
    if (frac_amt/10 > 0)  
    printf("\n.10 number of Coin(s): %d", frac_amt / 10);

    frac_amt -= (frac_amt / 10) * 10;
    if (frac_amt/5 > 0) 
    printf("\n.05 number of Coin(s): %d", frac_amt / 5);
    frac_amt -= (frac_amt / 5) * 5;
    if (frac_amt > 0) 
    printf("\n.01 number of Coin(s): %d", frac_amt);
    return 0;
}

Sample Output:

Input the currency value (floating point with two decimal places):
10387.75

Currency Notes:
100 number of Note(s): 103
50 number of Note(s): 1
20 number of Note(s): 1
10 number of Note(s): 1
5 number of Note(s): 1
2 number of Note(s): 1

Currency Coins:
.50 number of Coin(s): 1
.25 number of Coin(s): 1

Flowchart:

C programming Code Editor:

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