Discussion on Sherlock and The Beast Challenge

mcotse 4 years ago+ 3 comments

Loops are very unnecessary in this problem, here is the more elegant way

a=int(raw_input(''))
for i in range(0,a):
    n=(int(raw_input('')))
    if n%3==0 and n!=0:
        print n*str(5)
    elif n%3==1 and n>=10:
        print (n-10)*str(5)+10*str(3)
    elif n%3==2 and n>=5:
        print (n-5)*str(5)+5*str(3)
    else:
        print '-1'

AntFitch 4 years ago+ 0 comments

Nice!!!

liewsanmin 4 years ago+ 1 comment

what's the java version of this?

sxffff 3 years ago+ 0 comments

import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    for(int a0 = 0; a0 < t; a0++){
        int n = in.nextInt();
        StringBuffer sb = new StringBuffer();
        if (n%3==0) {
            System.out.println(append(n,"5"));
        } else if (n%3==1) {
            if (n<10) {
                System.out.println("-1");   
            } else {
                System.out.println(append(n-10,"5") + (append(10,"3")));
            }
        } else if (n%3==2) {
            if (n<5) {
                System.out.println("-1");   
            } else {
                System.out.println(append(n-5,"5") +(append(5,"3")));
            }
        }

    }
}
private static String append(int numberOfTimes , String value){
    StringBuffer sb = new StringBuffer();
    for (int i = 0; i < numberOfTimes; i++) {
        sb.append(value);           
    }
    return sb.toString();
}

}

ktleary 3 years ago+ 0 comments

A not as pretty JS version:

 if (n%3==0 && n!=0) {  
     console.log( Array(n+1).join("5"));
 } else if (n%3==1 && n>=10) {
     console.log(Array((n-10)+1).join("5")+Array(10 + 1).join("3"));
} else if (n%3==2 && n>=5){
    console.log(Array((n-5)+1).join("5") + Array(5+1).join("3"));
} else {
    console.log('-1');
}