**Poker Nim**

### Problem Statement :

Poker Nim is another -player game that's a simple variation on a Nim game. The rules of the games are as follows: The game starts with piles of chips indexed from to . Each pile (where ) has chips. The players move in alternating turns. During each move, the current player must perform either of the following actions: Remove one or more chips from a single pile. Add one or more chips to a single pile. At least chip must be added or removed during each turn. To ensure that the game ends in finite time, a player cannot add chips to any pile more than times. The player who removes the last chip wins the game. Given the values of , , and the numbers of chips in each of the piles, determine whether the person who wins the game is the first or second person to move. Assume both players move optimally. Input Format The first line contains an integer, , denoting the number of test cases. Each of the subsequent lines defines a test case. Each test case is described over the following two lines: Two space-separated integers, (the number of piles) and (the maximum number of times an individual player can add chips to some pile ), respectively. space-separated integers, , where each describes the number of chips at pile . Output Format For each test case, print the name of the winner on a new line (i.e., either First or Second ).

### Solution :

` ````
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int tc,n,k,i,j;
long int a[100],ans;
scanf("%d",&tc);
for(i=0;i<tc;i++)
{
scanf("%d %d",&n,&k);
for(j=0;j<n;j++)
{
scanf("%ld",&a[j]);
}
ans=0;
//while(l)
// {
for(j=0;j<n;j++)
{
ans=ans^a[j];
}
//}
if(ans==0)
printf("Second\n");
else
printf("First\n");
}
return 0;
}
```

` ````
Solution in C++ :
In C ++ :
#include <bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
int main()
{
long nTest,n,res,x,k;
scanf("%ld",&nTest);
while (nTest--)
{
scanf("%ld%ld%ld",&n,&k,&res);
for (long i=1; i<n; ++i) scanf("%ld",&x),res^=x;
puts((!res)?"Second":"First");
}
}
```

` ````
Solution in Java :
In Java :
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();
int[] s = new int[100];
int i,j,k,n;
int nimsum;
for(int tt =0;tt<t;tt++){
n = in.nextInt();
k = in.nextInt();
for(i = 0;i<n;i++){
s[i]=in.nextInt();
}
nimsum = s[0];
for(i = 1;i<n;i++){
nimsum^=s[i];
}
if (nimsum > 0) System.out.println("First");
else System.out.println("Second");
}
}
}
```

` ````
Solution in Python :
In Python3 :
def nim(array):
ans = 0
while array:
ans ^= array.pop()
return ans
def main():
for _ in range(int(input())):
input()
print(['Second', 'First'][bool(nim(list(map(int, input().split()))))])
if __name__ == '__main__':
main()
```

## View More Similar Problems

## Insert a Node at the head of a Linked List

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

View Solution →## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →## Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →## Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →