**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')
```

## View More Similar Problems

## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

View Solution →## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

View Solution →## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

View Solution →## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =

View Solution →## Kindergarten Adventures

Meera teaches a class of n students, and every day in her classroom is an adventure. Today is drawing day! The students are sitting around a round table, and they are numbered from 1 to n in the clockwise direction. This means that the students are numbered 1, 2, 3, . . . , n-1, n, and students 1 and n are sitting next to each other. After letting the students draw for a certain period of ti

View Solution →