Count the total number of notes in the given amount with C language

#include <stdio.h>

int main()
{
    int amount;
    int note500, note100, note50, note20, note10, note5, note2, note1;

    note500 = note100 = note50 = note20 = note10 = note5 = note2 = note1 = 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;
}

Output:-

Enter amount : 797
Total number of notes =
500 = 1
100 = 2
50 = 1
20 = 2
10 = 0
5 = 1
2 = 1
1 = 0