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Lucky Numbers

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29 Discussions

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  • Agarwalsanket11
     + 0 comments

    Quite easy using recursion: import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

    public class Solution {

    public static boolean lucky_check(long num){
    if(num<4 && num!=0){
        return false;
    }
    if(num%7==0 || num%4==0){
        return true;
    }
    else{

        boolean check1 =  lucky_check(num-7);
        boolean check2 =  lucky_check(num-4);

        if(check1 || check2){
            return true;
        }
        else{
            return false;
        }

    }


}
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int q = in.nextInt();
    for(int a0 = 0; a0 < q; a0++){
        long n = in.nextLong();
        if(lucky_check(n)){
            System.out.println("Yes");
        }
        else{
             System.out.println("No");
        }
    }
}

    }

  • jameshemmaplardh
     + 0 comments

    Use modulus. Psudo code: if n % 4, 7, 11 != 0: print("No") else: print("yes")

  • agyaeytiwari
     + 0 comments
    #include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    int q;
    cin >> q;
    for(int a0 = 0; a0 < q; a0++){
        long n,x=0,i=0;
        cin >> n;
        while(n>=7*i)
            {
            if(((n-7*i)%4)==0)
                {
                x=1;
                cout<<"Yes"<<endl;
                break;
            }
            i++;
        }
        if(x==0)
            cout<<"No"<<endl;
    }
    return 0;
}

  • 1729_pulkit
     + 0 comments

    what's the issue with this code :

    def isdivi(num):
    if num <0 :
        return False 
    if num == 0:
        return True
    
    return (isdivi(n-7)  or isdivi(n-4))

q = int(raw_input().strip())
for a0 in xrange(q):
    n = long(raw_input().strip())
    if (isdivi(n)):
        print "Yes"
    else :
        print "No"

  • architgarg1515
     + 0 comments

    We have

    7(x-4)+4(y+7)=7x+4y

    So if (x, y) is a solution, then (x-4,y+7) is also a solution. Hence if there is a solution then there is one with x<4. That's why you only need to test x=0..3 which runs in constant time.

    This can be extended to any equation of the form ax+by=n, you only need to test x=0..b-1.