"First player is starting from $(x,y)$. A cell is marked $W$ if a player starting from that cell (or who has his turn when stone is on that cell) wins. A cell is marked $L$ if he loses if he has to make a move from that cell.

Now, assume we already found out which of previous cells (or cells which we might visit in our moves) are $W$ or $L$. If all the cells to which I can move to (from current cell) are $W$, then that means that my opponent will win (because a cell is $W$ if player who gets to make a turn when stone is on that cell wins). This means, I will lose no matter what, as irrespective of how I move, there is a winning strategy possible for my opponent. What if all are not $W$, i.e. I can move to a cell labelled $L$? It means my opponent cannot win no matter what if stone is at that cell when he gets his turn. Now, obviously, I will move the stone to that $L$ labelled cell to win as I move optimally."

//Applicable only after Row 1. Not to be used to find states of Row 1 or Row 2.
1. If($A_{i-1,j}==L$) then $A_{ij}=W$
2. Else if ($A_{i,j-1}==L$) then $A_{ij}=W$
3. Else $\implies$ both $A_{i-1,j}==W$ and $A_{i,j-1}==W$ $\implies A_{ij}=L$

How to find for Row 1 and Column 1 then?-

Row 1-

1. If($A_{0,j}==W$) then $A_{ij}=W$//I can win by moving to cell $(0,i)$

2. Else if ($A_{1,j-1}==L$) then $A_{ij}=W$//I can force my opponent to lose by moving to cell $(1,j-1)$**

3. Else $A_{1j}=L$

Use this to derive the same for Column 1.

Say $A_{ij}=L$. Now, what does this imply for $A_{i+1,j}$ and $A_{i,j+1}$? We can see that, from both these cells, the starting player can move to $A_{ij}$ and force his opponent into a losing move!! This means that if $A_{ij}=L \implies A_{i+1,j} =A_{i,j+1}=W$

Now, if $A_{ij}=L \implies A_{i+1,j} =A_{i,j+1}=W$, then this means that, if I am at cell $A_{i+1,j+1}$, then no matter where I move, opponent will have a winning move. This implies that $A_{i+1,j+1}=L$ as we cannot force the opponent to lose and the game cannot tie. (i.e. the opponent will be able to force us to lose).

No!! The base point of above proof was that, just $1$ $L$ (among cells where we can move) is enough to give us the chance to make opponent lose. But $1$ $W$ among them doesnt mean that we lose. All of them must be $W$ to enforce that!

The significance is that, while we can say that if $A_{ij}=L \implies A_{i+1,j+1}=L$, we cannot imply the same for $W$. Hence, the observation becomes "A state of $W$ may or may not change diagonally, but the state of $L$ will definitely NOT change.

Assume that we have three consecutive $1's$ at cells $(i,j-1)$ $(i,j)$ and $(i,j+1)$. From the $W$'s at $(i,j+1)$, go back diagonally. The state at $(i-1,j)$ must also be $W$! Why? Because had it been $L$, then state at $(i,j+1)$ would had been $L$ as well as $L$ forces its diagonally next elements to be $L$ as well.

Now, this is the corner case. Why? Because say, there do exist $3$ consecutive $W$ in the matrix. What effect does it have on the states of next row? Say, we talk of cell $(i+1,j+1)$. We know that $(i,j)$ is a $W$, but so is $(i,j+1)$ and $(i+1,j)$, i.e. both the cells where we can move are giving opponent a winning strategy. This means that $(i+1,j+1)$ is actually $L$!!.

This proof is significant in $2$ ways. First it tells that, if I have $W$ diagonally, they can convert to $L$ and remain constant thereafter. The second is that it proves that I will not see $3$ consecutive $1$'s in any row except row 1, because row 0 is not determined by our rules and is given as an input from user.

void solve() {
    scanf("%s", s);
    n = strlen(s);
    for (int i = 1; i <= n; i++)
        a[0][i] = (int)(s[i - 1] - '0');//Base case, assigning input to 0'th row.
    scanf("%s", s);
    m = strlen(s);
    for (int i = 1; i <= m; i++)
        a[i][0] = (int)(s[i - 1] - '0');
    for (int x = 1; x <= m; x++) {
        for (int y = 1; y <= n && y < (int)a[x].size(); y++)
            a[x][y] = 1 ^ (a[x - 1][y] & a[x][y - 1]);//Calulating the states. Refer to formula 
                                                               //we discussed
    }
    int q;
    scanf("%d", &q);
    while(q--) {
        int x, y;
        scanf("%d%d", &x, &y);
        int d = min(x, y) - 3;//Number of steps to reach at least 3rd row or column
        if (d > 0) {//d<= means we are at <=3rd row/column already (eg- Row 2 etc.) 
            x -= d;//We already have ans of above part. Else, we move diagonally backwards
            y -= d;//Until we reach third row.
        }
        printf("%d", a[x][y]);//print the character instead of involving strings at all.
    }
    printf("\n");
}

int main()
{
//  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);

    a.resize(N);
    for (int i = 0; i < N; i++) {
        if (i < 4)
            a[i].resize(N);//For first 4 rows, find states of all columns. After that, keep 
                                   //track of only first 3-4 columns
        else
            a[i].resize(4);
    }

    int t;
    scanf("%d", &t);
    while(t--) solve();

    return 0;
}

//teja349
#include <bits/stdc++.h>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <utility>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <iomanip> 
//setbase - cout << setbase (16); cout << 100 << endl; Prints 64
//setfill -   cout << setfill ('x') << setw (5); cout << 77 << endl; prints xxx77
//setprecision - cout << setprecision (14) << f << endl; Prints x.xxxx
//cout.precision(x)  cout<<fixed<<val;  // prints x digits after decimal in val

using namespace std;
#define f(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) f(i,0,n)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define vi vector< int >
#define vl vector< ll >
#define ss second
#define ff first
#define ll long long
#define pii pair< int,int >
#define pll pair< ll,ll >
#define sz(a) a.size()
#define inf (1000*1000*1000+5)
#define all(a) a.begin(),a.end()
#define tri pair<int,pii>
#define vii vector<pii>
#define vll vector<pll>
#define viii vector<tri>
#define mod (1000*1000*1000+7)
#define pqueue priority_queue< int >
#define pdqueue priority_queue< int,vi ,greater< int > >
vector<vii> vec(123456);
char ans[123456];
int main(){
    std::ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        string c;
        string r;
        cin>>r>>c;
        int i,j;
        int n=c.length();
        int m=r.length();
        rep(i,n+10){
            vec[i].clear();
        }
        int q,pos;
        cin>>q;
        int a,b;//note down queries.
        rep(i,q){
            cin>>a>>b;
            vec[a].pb(mp(b,i));
        }
        string s="";
        fd(i,r.length()-1,0)
            s+=r[i];
        r=s;
        int len=r.length();
        r+='0';
        f(i,1,n+1){
            // simulating the problem statement for first 10 rows. Becuase there is no good pattern in first three rows
            if(i<10){
                r[len]=c[i-1];
                fd(j,len-1,0){
                    if(r[j+1]=='1' && r[j]=='1')
                        r[j]='0';
                    else
                        r[j]='1';
                }
            }
            else{
                // correspondingly updating pattern for each row from previous row based on the element.
                if(r[len-1]=='0'){
                    r[len]='1';

                }
                else{
                    if(c[i-1]=='1'){
                        r[len]='0';
                    }
                    else{
                        r[len]='1';
                        if(len-2>=0 && r[len-2]=='1')
                            r[len-1]='0';
                    }
                }
                len++;
                r+='0';
            }
            // answering queries in that row.
            rep(j,vec[i].size()){
                pos=vec[i][j].ss;
                b=vec[i][j].ff;
                ans[pos]=r[len-b];
            }

        }
        rep(i,q){
            cout<<ans[i];
        }
        cout<<endl;

    }
    return 0;   
}

$$"Just \space do \space what \space the \space problem \space says!"$$
$$-Oleksandr Kulkov$$

Assume $x=y=n$ and considered diagonal $[(0,n), (1,n-1),...,(n,0)]$. If you start in $(n, n)$, you will always go through this diagonal and reach it in exactly $n$ steps.

Now consider mapping $(x,y) \implies (x+y,x-y)$. You can see that each step now is actually decreasing first coordinate by $1$ and increasing or decreasing second coordinate by $1$.

Thus we start in $(2n, 0)$ and will end up in $(n, k)$ when we step on. Now we can see that either first or second player has victorious strategy which ends up in $-2 \le k \le 2$.

Assume first player also makes last move. Then if $(n,-1)$ is winning, he will start by stepping in $(2n-1,-1)$ and will return to it on each his step. Same for $(n,1)$. If they both are losing, second player has winning strategy.

You may see that if first players wins or not starting in $(2n, 0)$ is actually same as for $(n+1,0)$, or in initial coordinates $(\lfloor n/2\rfloor +1,\lfloor n/2 \rfloor +1).$

Now if second player makes a step on diagonal, first player will win iff $(n,-2)$ and $(n,0)$ are winning OR $(n,0)$ and $(n,2)$ are winning. In first case he starts with $(2n-1,-1)$ and with $(2n-1,1)$ in second. You may see that now winning of first player is identical to (n+2,0), or in initial coordinates, well, also $(n/2+1,n/2+1)$. This will reduce you size almost twice, unless $n=2$, for which you should calculate answers manually.

To generalize, assume x<y and consider diagonal $[(y-x,x), (y-x+1,x-1),...,(y,0)]$.