For anyone that may be a little confused by the challenge, and failing a test case here or there... think of it this way. The best choice for the player is to choose the index that benefits him the most, but is not as simple as finding the largest un-used number left in the player's array. It might be a smarter strategy to block the other player's largest number. Example:

Player1nums : 100 200 500 Player2nums : 900 200 50

Player1 should pick index 0 because it blocks player2 from collecting on that 900.

Is there a way to assess the benefit of picking a given i value? Find that way of assessing the benefit and you are on your way to solving this.

this is not good they should have explained it without leaving any room for confusion eg:- 2 20 2 1 10000

c++ solution

#include <bits/stdc++.h>

using namespace std;

struct p2 {
    int i , poss ;
    bool operator < ( p2& other ){
        return i < other.i ;
    }
};

int main()
{
    int Testcases ;
    scanf("%d",&Testcases);
    for( int t = 0 ; t < Testcases ; t++ ){
        int Number = 0 ;
        scanf("%d",&Number);
        int A[Number+1] , B[Number+1] ;
        for( int r = 0 ; r < Number ; r++ ) scanf( "%d" ,&A[ r ] );
        for( int r = 0 ; r < Number ; r++ ) scanf( "%d" ,&B[ r ] );
        p2 Converted[Number+1] = { 0 } ;
        int P1 , P2 ;
        P1 = P2 = 0 ;
        for( int c = 0 ; c < Number ; c++ ) {
            Converted[ c ].i = A[ c ] + B[ c ] ;
            Converted[ c ].poss = c ;
        }
        
        sort( Converted , Converted+Number );
        
        bool playing = true ;
        
        for( int p = Number - 1 ; p >= 0 ; p-- ){
            if( playing ){
                P1 += A[ Converted[ p ].poss ] ;
            }else{
                P2 += B[ Converted[ p ].poss ] ;
            }
            playing = !playing ;
        }
        if( P1 > P2 )printf("First\n");
        else if( P1 < P2 )printf("Second\n");
        else printf("Tie\n");
    }
    return 0;
}

1

Another ambiguous English.

It should be given as below:

Each value of i can be chosen only once, either by Player 1 or Player 2, not both. The game always ends after N moves.

2

One of the 10 tests in Test Case #4 looks goofy.

PYTHON:

.

for _ in xrange(input()):
    n,a,b = input(),map(int, raw_input().split()),map(int, raw_input().split())
    s = [a[x] + b[x] for x in xrange(n)]
    add = c = 0
    for i in range(n):
        index = s.index(max(s))
        if c == 0: add += a[index]
        if c == 1: add -= b[index]
        s.pop(index)
        a.pop(index)
        b.pop(index)
        c = 1 - c
    if add > 0: print "First"
    elif add < 0: print "Second"
    else: print "Tie"