Uri 1021 solution in C | Banknotes and Coins

/*Before seeing the code solution make sure that you have tried enough. Don’t copy- paste the whole code. Find out the logic. If you face any trouble, just inform me in comment.*/

#include <stdio.h>

int main() {
    int n100, n50, n20, n10, n5, n2;
    int m1, m50, m25, m10, m05, m01;
    double n;

    scanf("%lf", &n);
    int notas = n;
    int moedas = (n - notas) * 100;

    if((moedas * 1000) % 10 == 9){
        moedas++;
    }


    n100 = notas/100;
    notas = notas%100;
    n50 = notas/50;
    notas = notas%50;
    n20 = notas/20;
    notas = notas%20;
    n10 = notas/10;
    notas = notas%10;
    n5 = notas/5;
    notas = notas%5;
    n2 = notas/2;
    notas = notas%2;

    m1 = notas/1;
    notas = notas%1;
    m50 = moedas/50;
    moedas = moedas%50;
    m25 = moedas/25;
    moedas = moedas%25;
    m10 = moedas/10;
    moedas = moedas%10;
    m05 = moedas/5;
    moedas = moedas%5;
    m01 = moedas/1;

    printf("NOTAS:\n");

    printf("%d nota(s) de R$ 100.00\n", n100);
    printf("%d nota(s) de R$ 50.00\n", n50);
    printf("%d nota(s) de R$ 20.00\n", n20);
    printf("%d nota(s) de R$ 10.00\n", n10);
    printf("%d nota(s) de R$ 5.00\n", n5);
    printf("%d nota(s) de R$ 2.00\n", n2);

    printf("MOEDAS:\n");

    printf("%d moeda(s) de R$ 1.00\n", m1);
    printf("%d moeda(s) de R$ 0.50\n", m50);
    printf("%d moeda(s) de R$ 0.25\n", m25);
    printf("%d moeda(s) de R$ 0.10\n", m10);
    printf("%d moeda(s) de R$ 0.05\n", m05);
    printf("%d moeda(s) de R$ 0.01\n", m01);

    return 0;

}

 //Happy_Coding