Nimble Game
Problem Statement :
Two people are playing Nimble! The rules of the game are: The game is played on a line of squares, indexed from to . Each square (where ) contains coins. For example: The players move in alternating turns. During each move, the current player must remove exactly coin from square and move it to square if and only if . The game ends when all coins are in square and nobody can make a move. The first player to have no available move loses the game. Given the value of 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. 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: An integer, , denoting the number of squares. space-separated integers, , where each describes the number of coins at square .
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int i,j,t,n,c,result;
scanf("%d", &t);
for (i=1; i<=t; i++) {
scanf("%d", &n);
result = 0;
if (n==0) {
printf("Second\n");
continue;
}
scanf("%d", &c);
if (n==1) {
printf("Second\n");
continue;
}
for(j=1; j<n; j++) {
scanf("%d", &c);
if (c%2) {
result ^= j;
}
}
if (result) {
printf("First\n");
} else {
printf("Second\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;
scanf("%ld",&nTest);
while (nTest--)
{
ll res=0,x;
scanf("%ld",&n);
for (long i=0; i<n; ++i)
{
scanf("%lld",&x);
if (x&1LL && i>0) res^=i;
}
if (n==1) puts("Second");
else 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 {
//Day 2: Nimble Game
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
int[] s = new int[100];
int i,j,n;
int nimsum;
for(int tt =0;tt<t;tt++){
n = in.nextInt();
if( n==1 ) {
n=in.nextInt();//consume it
System.out.println("Second");
} else {//n>1
for(i = 0;i<n;i++){
s[i]=in.nextInt();
}
nimsum = 0;
for(i = 1;i<n;i++){
nimsum^=(s[i]%2==1)?i:0;
}
if (nimsum > 0) System.out.println("First");
else System.out.println("Second");
}
}
}
}
Solution in Python :
In Python3 :
test = int(input())
for _ in range(test):
n = int(input())
ar = list(map(int,input().strip().split()))
#print(n,ar)
result = []
for i in range(len(ar)):
result.append(ar[i]%2)
#print(result)
xor = 0
for i,n in enumerate(result):
xor = xor ^(i*n)
if xor==0:
print('Second')
else:
print('First')
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