# HackerRank Kitty and Katty problem solution

In this HackerRank Kitty and Katty problem solution, we have given the value of N plastic blocks. Do we need to find and print the name of the winner? Assume both plays optimally.

## Problem solution in Python.

```#!/bin/python3

import math
import os
import random
import re
import sys

if __name__ == '__main__':
T = int(input())

for T_itr in range(T):
n = int(input())
if(n==1):
print("Kitty")
elif(n%2==0):
print("Kitty")
else:
print("Katty")```

## Problem solution in Java.

```import java.io.*;
import java.util.*;

public class KittyAndKatty {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;

void solve() throws IOException {
int t = nextInt();
while (t-- > 0) {
int n = nextInt();
out.println((n > 1 && n % 2 == 1) ? "Katty" : "Kitty");
}
}

KittyAndKatty() throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
solve();
out.close();
}

public static void main(String[] args) throws IOException {
new KittyAndKatty();
}

String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
eof = true;
return null;
}
}
return st.nextToken();
}

String nextString() {
try {
return br.readLine();
} catch (IOException e) {
eof = true;
return null;
}
}

int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}

long nextLong() throws IOException {
return Long.parseLong(nextToken());
}

double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
}```

## Problem solution in C++.

```#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 T;
cin >> T;
for(int a0 = 0; a0 < T; a0++){
int n;
cin >> n;
if(n==1)
cout<<"Kitty"<<endl;
else if(n%2==0)
cout<<"Kitty"<<endl;
else
cout<<"Katty"<<endl;
}
return 0;
}```

## Problem solution in C.

```#include<stdio.h>
#include<string.h>
main()
{
int u;
scanf("%d",&u);
while(u!=0)
{
int n;
scanf("%d",&n);

if(n==1){
printf("Kitty\n");
}
else if(n%2==0){
printf("Kitty\n");
}
else{
printf("Katty\n");

}

u--;
}

}```