Q:

# C program to count total number of notes in entered amount

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