Code:
#include<stdio.h>
#include<conio.h>
int main()
{
    long amount;
    long note5000, note1000, note500, note100, note50, note20, note10, coin5, coin2, coin1;
    clrscr();



    note5000 = note1000 = note500 = note100 = note50 = note20 = note10 = coin5 = coin2 = coin1 = 0;




    printf("Enter the amount: ");
    scanf("%ld", &amount);




    if(amount >= 5000)
    {
    note5000 = amount/5000;
    amount -= note5000 * 5000;
    }
    if(amount >= 1000)
    {
    note1000 = amount/1000;
    amount -= note1000 * 1000;
    }
    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)
    {
    coin5 = amount/5;
    amount -=coin5 * 5;
    }
    if(amount >= 2)
    {
    coin2 = amount/2;
    amount -=coin2 * 2;
    }
    if(amount >= 1)
    {
    coin1 = amount;
    }


    printf("Total number of notes = \n");


    textcolor(GREEN);
    cprintf("5000 = %ld\n", note5000);
    textcolor(MAGENTA);
    cprintf("1000 = %ld\n", note1000);
    textcolor(MAGENTA);
    cprintf("500  = %ld\n", note500);
    textcolor(MAGENTA);
    cprintf("100  = %ld\n", note100);
    textcolor(MAGENTA);
    cprintf("50   = %ld\n", note50);
    textcolor(MAGENTA);
    cprintf("20   = %ld\n", note20);
    textcolor(MAGENTA);
    cprintf("10   = %ld\n", note10);
    textcolor(MAGENTA);
    cprintf("5    = %ld\n", coin5);
    textcolor(MAGENTA);
    cprintf("2    = %ld\n", coin2);
    textcolor(MAGENTA);
    cprintf("1    = %ld\n", coin1);


    getch();


    return 0;
}


Above is my code,I want this program to ask amount and then check the amount and ask in which denominations the money is required depending on the amount and I want to add font style please help me.