Sherlock and The Beast

                            Solution in C :

In  C :







#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char **argv)
{
  int T, N, i, j, c;
  fscanf(stdin, "%d", &T);

  for (i=0; i<T; ++i) {
	if (fscanf(stdin, "%d", &N)==EOF) return 1;
	if (N%3==0) {
		for (j=0; j<N/3; ++j) printf("555");
	}
	else {
		c=0;
		while (N>=0 && N%3!=0) {
			++c;
			N-=5;
		}
		if (N<0) printf("-1");
		else {
			for (j=0; j<N/3; ++j) printf("555");
			for (j=0; j<c; ++j) printf("33333");
		}
	}
	printf("\n");
  }
  return 0;
}
                        

                        Solution in C++ :

In  C++  :






#include <string>
#include <vector>
#include <map>
#include <list>
#include <iterator>
#include <set>
#include <queue>
#include <iostream>
#include <sstream>
#include <stack>
#include <deque>
#include <cmath>
#include <memory.h>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <utility> 
using namespace std;
 
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define RFOR(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define REP(i, N) FOR(i, 0, N)
#define RREP(i, N) RFOR(i, N, 0)
 
#define ALL(V) V.begin(), V.end()
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define Pi 3.14159265358979

typedef long long Int;
typedef unsigned long long UInt;
typedef vector <int> VI;
typedef pair <int, int> PII;



int main()
{
		
	int T;
	cin>>T;
	REP(tests,T)
	{
		int n;
		cin>>n;
		
		int res = -1;
		
		for (int i = 0; i <= n; ++i)
		{
			if (i % 5 == 0)
			{
				int j = n - i;
				
				if (j % 3 == 0)
				{
					res = i;
					break;
				}
			}
		}
		
		if (res == -1)
		{
			cout << -1 << endl;
			continue;
		}
		
		REP(i,n-res)
			putchar('5');
		REP(i,res)
			putchar('3');
		puts("");
	}

	
	return 0;
}
                    

                        Solution in Java :

In  Java :







import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Solution {
	static BufferedReader in = new BufferedReader(new InputStreamReader(
			System.in));
	static StringBuilder out = new StringBuilder();
	

	public static void main(String[] args) throws NumberFormatException, IOException {
		int numCases = Integer.parseInt(in.readLine());
		for(int t = 0; t < numCases; t ++)
		{
			int k = Integer.parseInt(in.readLine());
			int numFives = (k / 3) * 3;
			int numThrees = 0;
			if(k-numFives == 2)
			{
				numFives -= 3;
				numThrees += 5;
			}
			else if(k-numFives == 1)
			{
				numFives -= 9;
				numThrees += 10;
			}
			
			if(numFives >= 0)
			{
				for(int i = 0; i < numFives; i ++)
				{
					out.append(5);
				}
				for(int i = 0; i < numThrees; i ++)
				{
					out.append(3);
				}
				out.append("\n");
			}
			else
			{
				out.append("-1\n");
			}
		}

		System.out.print(out);
	}
}
                    

                        Solution in Python : 
                            
In  Python3 :







T = int(input())
for i in range(T):
    k = int(input())
    n5 = k
    n3 = 0
    done = False
    while n5 >= 0:
        if n5%3 == 0 and n3%5 == 0:
            for j in range(n5):
                print('5',end='')
            for j in range(n3):
                print('3',end='')
            print()
            done = True
            break
        n5 -= 1
        n3 += 1
    if not done:
        print(-1)