# C Program to Accept an Amount and Display the Number of Currency Notes This C program will accept an amount, and will convert and display that amount as currency notes… That is currency denomination.

For example:
If amount is: 123
We can write amount 123 as :

=> Number of notes of 100       = 1
Number of notes of  20        = 2
Number of notes of  2          = 1
Number of notes of  1          = 1
Total number of currency notes = 1 + 2 + 1 +1 = 5

Example 2:

If amount is: 512
We can write amount 143 as :

=> Number of notes of 100       = 5
Number of notes of  20        = 0
Number of notes of  10        = 1
Number of notes of  2          = 1

Total number of currency notes = 5 + 0 + 1 +1 = 7

`#include <stdio.h>`

#include <conio.h>

#include <stdio.h>

#include <conio.h>

void main()

{

int amt,n1,n2,n3,n4,n5,n6,n7,n8,n9,total;

printf(“n Enter the amount :”);

scanf(“%d”,&amt);

n1=amt/1000;

amt=amt%1000;

n2=amt/500;

amt=amt%500;

n3=amt/100;

amt=amt%100;

n4=amt/50;

amt=amt%50;

n5=amt/20;

amt=amt%20;

n6=amt/10;

amt=amt%10;

n7=amt/5;

amt=amt%5;

n8=amt/2;

n9=amt%2;

total=n1+n2+n3+n4+n5+n6+n7+n8+n9;

printf(” n Number of notes of 1000 = %d “,n1);

printf(” n Number of notes of 500   = %d “,n2);

printf(” n Number of notes of 100   = %d “,n3);

printf(” n Number of notes of 50     = %d “,n4);

printf(” n Number of notes of 20     = %d “,n5);

printf(” n Number of notes of 10     = %d “,n6);

printf(” n Number of notes of 5       = %d “,n7);

printf(” n Number of notes of 2       = %d “,n8);

printf(” n Number of notes of 1       = %d “,n9);

printf(” n Total number of currency notes = %d “,total);

getch( );

}

Output

Enter the amount : 1865
Number of notes of 1000 = 1
Number of notes of 500   = 1
Number of notes of 100 = 3
Number of notes of 50 = 1
Number of notes of 20 = 0
Number of notes of 10 = 1
Number of notes of 5 = 1
Number of notes of 2 = 0
Number of notes of 1 = 0
Total number of currency notes = 8 You can convert this C program  to C++ program by changing input – output keywords…