C Program to count minimum number of denomination value for given amount

#include<stdio.h>
int main()
{
	int amount, note2000,note500,note200,note100,note50,note20,note10,note5,note2,note1;
	note2000 = note500 = note200 = note100 = note50 = note20 = note10 = note5 = note2 = note1 = 0;
	
	printf("Enter Amount: ");
	scanf("%d",&amount);
	
	if(amount >= 2000)
	{
		note2000 = amount/2000;
		amount -= note2000 * 2000;
	}
	if(amount >= 500)
	{
		note500 = amount/500;
		amount -= note500 * 500;
	}
	if(amount >= 200)
	{
		note200 = amount/200;
		amount -= note200 * 200;
	}
	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/1;
		amount -= note1 * 1;
	}
	
	/* Print required notes */
	printf("Total number of notes : \n");
    printf("2000 = %d\n", note2000);
	printf("500 = %d\n", note500);
	printf("200 = %d\n", note200);
    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: 6338
Total number of notes :
2000 = 3
500 = 0
200 = 1
100 = 1
50 = 0
20 = 1
10 = 1
5 = 1
2 = 1
1 = 1