Compute the optimal spare money - With only coins of 2, bills of 5 and 10

rendu-monnaie.py

def impossible(n):
    return n == 1 or n == 3


def spare(m):
    if impossible(m):
        return ((0, 0, 0))

    # Compute the bills of 10
    pn10 = int(m / 10)
    n10 = pn10 if not impossible(m - (pn10 * 10)) else pn10 - 1

    # Compute the bills of 5
    m = m - (n10 * 10)
    pn5 = int(m / 5)
    n5 = pn5 if not impossible(m - (pn5 * 5)) else pn5 - 1

    # Compute the coins of 2
    m = m - (n5 * 5)
    pn2 = int(m / 2)

    return (n10, n5, pn2)


if __name__ == '__main__':
    m = int(input())
    n10, n5, n2 = spare(m)
    print("Bill of 10: %d\nBill of 5: %d\nCoins of 2: %d" % (n10, n5, n2))


""" A quick implementation in java

public class Change {

   private long coins2 = 0;
   private long bill5 = 0;
   private long bill10 = 0;

   private static boolean impossible(long n) {
       return n == 1 || n == 3;
   }

   private static Change change(long m) {
       Change change = new Change();

       if(impossible(m))
           return change;

       // Compute the bills of 10
       long pn10 = m / 10;
       long n10 = !impossible(m - (pn10 * 10))? pn10 : pn10 - 1;

       // Compute the bills of 5
       m = m - (n10 * 10);
       long pn5 = m / 5;
       long n5 = !impossible(m - (pn10 * 5))? pn5 : pn5 - 1;

       // Compute the coins of 2
       m = m - (n5 * 5);
       long pn2 = m / 2;

       change.bill10 = n10;
       change.bill5 = n5;
       change.coins2 = pn2;

       return change;
   }


   public static void main(String[] args) {
       Change change = Change.change(17L);
       System.out.printf("Bill of 10: %d\nBill of 5: %d\nCoins of 2: %d" , change.bill10, change.bill5, change.coins2);
   }
}
"""