Write a C program to accept amount as integer and display total number of notes of Rs. 500, 100, 50, 20, 10, 5, 2, 1. How to count the minimum number of notes required for the given amount in C programming. Program to find minimum number of notes required for the given denomination. Logic to find minimum number of denomination for a given amount in C program.

**Example**

**Input**

Input amount: 575

**Output**

Total number of notes: 500: 1 100: 0 50: 1 20: 1 10: 0 5: 1 2: 0 1: 0

## Required knowledge

## Logic to count minimum number of denomination for given amount

There any many optimum algorithms to solve the given problem. For this exercise to make things simple I have used brute force algorithm. Below is the step by step descriptive logic to find minimum number of denomination.

- Input amount in some variable say amt.
- If the amount is greater than 500 then, divide amount by 500 to get maximum 500 notes required. Store the division result in some variable say note500.
- Subtract the resultant amount of 500 notes to get final amount. Perform amt = amt - (note500 * 500).
- Repeat step 2-3, for finding minimum number of notes of 200, 100, 50 and so on.

## Program to count minimum number of denomination for given amount

```
/**
* C program to count minimum number of notes in an amount
*/
#include <stdio.h>
int main()
{
int amount;
int note500, note100, note50, note20, note10, note5, note2, note1;
// Initialize all notes to 0
note500 = note100 = note50 = note20 = note10 = note5 = note2 = note1 = 0;
/* Read amount from user */
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;
}
```

Output

Enter amount: 4673 Total number of notes = 500 = 9 100 = 1 50 = 1 20 = 1 10 = 0 5 = 0 2 = 1 1 = 1

Happy coding 😉

<pre><code> ----Your Source Code---- </code></pre>