C Program to Count Total Notes in a Given Amount

How to write a C Program to Count Total Notes in a Given Amount with an example using For Loop, Functions, and If Statement.

C Program to Count Total Notes in a Given Amount Example 1

This program helps the user to enter the amount of cash, and then it’s going to find the total number of denominators using For Loop

/* C Program to Count Total Notes in a Given Amount */
#include<stdio.h>
 
int main()
{
 int a[8] = {500, 100, 50, 20, 10, 5, 2, 1};
 int Amount, i, temp;
 
 printf("\n Please Enter the Amount of Cash  =  ");
 scanf("%d", &Amount);
 
 temp = Amount;
  
 for(i = 0; i < 8; i++)
 {
 	printf("\n %d Notes is = %d", a[i], temp / a[i]);
 	temp = temp % a[i];
 }

 return 0;
}

For Loop First Iteration : for(i = 0; 0 < 8; 0++)
a[i]  = a[0] = 500
temp / a[i] = 698 / 500  = 1
Next, temp = 698 % 500 = 198

Now temp = 198, and i value will be incremented to 1

For Loop Second Iteration : for(i = 1; 1 < 8; 1++)
a[i]  = a[1] = 100
temp / a[i] = 198 / 100  = 1
Next, temp = 198 % 100 = 98

Now temp = 98, and i value will be incremented to 1

Third Iteration : for(i = 2; 2 < 8; 2++)
a[i]  = a[2] = 50
temp / a[i] = 98 / 50  = 1
Next, temp = 98 % 50 = 48

Now temp = 48, and i value will be 3

Fourth Iteration : for(i = 3; 3 < 8; 3++)
a[i]  = a[3] = 20
temp / a[i] = 48 / 20  = 2
Next, temp = 48 % 20 = 8

Now temp = 8, and i value will be 4

Fifth Iteration : for(i = 4; 4 < 8; 4++)
a[i]  = a[4] = 10
temp / a[i] = 8 / 10  = 0
Next, temp = 8 % 10 = 8

Now temp = 8, and i value will be 5

Do the same for remaining C Programming iterations

C Program to Count Total Notes in a Given Amount Example 2

It is the same program that we used in the first example, but this time we separated the logic using Functions.

/* C Program to Count Total Notes in a Given Amount */
#include<stdio.h>

void Total_Notes(int Amount);
 
int main()
{
 	int Amount;
 
 	printf("\n Please Enter the Amount of Cash  =  ");
 	scanf("%d", &Amount);
 
	Total_Notes(Amount);

 	return 0;
}

void Total_Notes(int Amount)
{
	int a[8] = {500, 100, 50, 20, 10, 5, 2, 1};
	int i, temp;
	
	temp = Amount;
  
 	for(i = 0; i < 8; i ++)
 	{
 		printf("\n %d Notes is = %d", a[i], temp / a[i]);
 		temp = temp % a[i];
 	}
}

Program to Count Total Notes in a Given Amount Example 3

I know this is a horrible idea, but it is good to know that you can achieve this using the If statement. Here, for every If statement, the amount will be reduced.

/* C Program to Count Total Notes in a Given Amount */

#include <stdio.h>
 
int main()
{
	int Amount;
	int Note500, Note100, Note50, Note20, Note10, Note5, Note2, Coin1;
	Note500 = Note100 = Note50 = Note20 = Note10 = Note5 = Note2 = Coin1 = 0;
  	
	printf("\n Please Enter the Amount of Cash  = ");
  	scanf("%d", &Amount);
  
  	if (Amount > 500)
  	{
  		Note500 = Amount / 500;
  		Amount = Amount - (Note500 * 500);	
  	} 
  	if (Amount >= 100)
  	{
  		Note100 = Amount / 100;
  		Amount = Amount - (Note100 * 100);	
  	}
  	if (Amount >= 50)
  	{
  		Note50 = Amount / 50;
  		Amount = Amount - (Note50 * 50);	
  	}
  	if (Amount >= 20)
  	{
  		Note20 = Amount / 20;
  		Amount = Amount - (Note20 * 20); 	
  	}
	if (Amount >= 10)
  	{
  		Note10 = Amount / 10;
  		Amount = Amount - (Note10 * 10); 	
  	} 
  	if (Amount >= 5)
  	{
  		Note5 = Amount / 5;
  		Amount = Amount - (Note5 * 5); 	
  	}
   	if (Amount >= 2)
  	{
  		Note2 = Amount / 2;
  		Amount = Amount - (Note2 * 2); 	
  	} 	 	      	
  	if (Amount >= 1)
  	{
	   	Coin1 = Amount;
	}
	printf("\n Total Number of Notes presenet in the Cash that you entered are \n");
	printf("\n 500 Notes  =  %d", Note500); 
	printf("\n 100 Notes  =  %d", Note100); 
	printf("\n 50 Notes  =  %d", Note50); 
	printf("\n 20 Notes  =  %d", Note20); 
	printf("\n 10 Notes  =  %d", Note10); 
	printf("\n 5 Notes  =  %d", Note5); 
	printf("\n 2 Notes  =  %d", Note2); 
	printf("\n 1 Coin  =  %d", Coin1); 
	
  	return 0;
}