Introduction to Nim Game
Problem Statement :
Nim is the most famous two-player algorithm game. The basic rules for this game are as follows: The game starts with a number of piles of stones. The number of stones in each pile may not be equal. The players alternately pick up or more stones from pile The player to remove the last stone wins. For example, there are piles of stones having stones in them. Play may proceed as follows: Player Takes Leaving pile=[3,2,4] 1 2 from pile[1] pile=[3,4] 2 2 from pile[1] pile=[3,2] 1 1 from pile[0] pile=[2,2] 2 1 from pile[0] pile=[1,2] 1 1 from pile[1] pile=[1,1] 2 1 from pile[0] pile=[0,1] 1 1 from pile[1] WIN Given the value of and the number of stones in each pile, determine the game's winner if both players play optimally. Function Desctription Complete the nimGame function in the editor below. It should return a string, either First or Second. nimGame has the following parameter(s): pile: an integer array that represents the number of stones in each pile Input Format The first line contains an integer, , denoting the number of games they play. Each of the next pairs of lines is as follows: The first line contains an integer , the number of piles. The next line contains space-separated integers , the number of stones in each pile. Output Format For each game, 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>
int main()
{
int t,n,fuck,ans,i;
scanf("%d",&t);
while(t--)
{
ans=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&fuck);
ans=ans^fuck;
}
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;
scanf("%ld",&nTest);
while (nTest--)
{
scanf("%ld%ld",&n,&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 s=new Scanner(System.in);
int no=s.nextInt();
int t=0;
while(t<no)
{
int n=s.nextInt();
int i=0;
int x=0;
for(i=0;i<n;i++)
{
int in=s.nextInt();
x=x^in;
}
if(x>0)
{
System.out.println("First");
}
else
{
System.out.println("Second");
}
t++;
}
}
}
Solution in Python :
In Python3 :
test = int(input())
for _ in range(test):
pile = int(input())
ar = list(map(int,input().strip().split()))
xor = 0
for n in ar:
xor = xor^n
if xor == 0:
print('Second')
else:
print('First')
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