 # C Programming Code Examples

## C > If Else and Switch Case Code Examples

### count total number of notes in given amount

``` count total number of notes in given amount Write a C program to input amount from user and print minimum number of notes (Rs. 500, 100, 50, 20, 10, 5, 2, 1) required for the amount. How to 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. There any many optimal algorithms to solve the given problem. For this exercise to make things simple I have used greedy algorithm. Input amount from user. Store it in some variable say amt. If 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 = amt / 500;. After division, subtract the resultant amount of 500 notes from original amount. Perform amt = amt - (note500 * 500). Repeat above step, for each note 200, 100, 50, 20, 10, 5, 2 and 1. #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; /* Input 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; } /* Print required notes */ 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; } ```