HackerRank Nimble Game problem solution

In this HackerRank Nimble Game problem solution, we have given the value of N and the number of coins in each square, determine whether the person who wins the game is the first or second person to move. Assume both players move optimally.

HackerRank Nimble Game problem solution in Python.

t = int(input().strip())
for _ in range(t):
    n = int(input().strip())
    piles = list(map(int, input().strip().split(' ')))
    xor_sum = 0
    for n,pile in enumerate(piles):
        if pile % 2 == 1:
            xor_sum = xor_sum ^ n
    if xor_sum == 0:
        print('Second')
    else:
        print('First')

Nimble Game problem solution in Java.

import java.util.Scanner;

public class NimbleGame {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            long s = 0;
            for (int i = 0; i < n; i++) {
                long x = in.nextLong();
                if (x % 2 > 0) {
                    s ^= i;
                }
            }
            System.out.println(s == 0 ? "Second" : "First");
        }
    }

}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

void testcase(){
    int N; 
    cin >> N; 
    vector<long> vals(N); 
    long nimval = 0; 
    for(int i=0; i<N; i++){
        cin >> vals[i]; 
        if (vals[i]%2==1) nimval ^=i; 
    }
  
    if (nimval==0) cout << "Second"<< endl; 
    else cout << "First" << endl; 
}

int main() {
    int T; 
    cin >> T; 
    while(T--) testcase(); 
    return 0;
}

Problem solution in C.

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

int main() {
    int t,i,j,n,ans;
    long int a[200];
    scanf("%d",&t);
    for(i=0;i<t;i++){
        scanf("%d",&n);
        for(j=0;j<n;j++){
            scanf("%ld",&a[j]);
        }
        ans = 0;
        for(j=1;j<n;j++){
            if(a[j]%2!=0){
                ans = ans^j;
            }
        }
        if(ans ==0){
            printf("Secondn");
        }
        else{
            printf("Firstn");
        }
    }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}