Bestcoders 56 Clarke and puzzle

Clarke and puzzle

 

 Accepts: 129

 

 Submissions: 322

 Time Limit: 4000/2000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke split into two personality $a$ and $b$, they are playing a game.
There is a $n\ast m$ matrix, each grid of this matrix has a number c_{i, j}c​i,j​​.
$a$ wants to beat $b$ every time, so $a$ ask you for a help.
There are $q$ operations, each of them is belonging to one of the following two types:

1. They play the game on a (x_1, y_1)-(x_2, y_2)(x​1​​,y​1​​)−(x​2​​,y​2​​) sub matrix. They take turns operating. On any turn, the player can choose a grid which has a positive integer from the sub matrix and decrease it by a positive integer which less than or equal this grid's number. The player who can't operate is loser. $a$ always operate first, he wants to know if he can win this game.
2. Change c_{i, j}c​i,j​​ to $b$.

Input

The first line contains a integer $T(1\le T\le 5)$, the number of test cases.
For each test case:
The first line contains three integers n, m, q(1 \le n, m \le 500, 1 \le q \le 2*10^5)n,m,q(1≤n,m≤500,1≤q≤2∗10​5​​)
Then $n\ast m$ matrix follow, the $i$ row $j$ column is a integer c_{i, j}(0 \le c_{i, j} \le 10^9)c​i,j​​(0≤c​i,j​​≤10​9​​)
Then $q$ lines follow, the first number is $opt$.
if $opt=1$, then $4$ integers x_1, y_1, x_1, y_2(1 \le x_1 \le x_2 \le n, 1 \le y_1 \le y_2 \le m)x​1​​,y​1​​,x​1​​,y​2​​(1≤x​1​​≤x​2​​≤n,1≤y​1​​≤y​2​​≤m) follow, represent operation $1$.
if $opt=2$, then $3$ integers $i,j,b$ follow, represent operation $2$.

Output

For each testcase, for each operation $1$, print $Yes$ if $a$ can win this game, otherwise print $No$.

Sample Input

1
1 2 3
1 2
1 1 1 1 2
2 1 2 1
1 1 1 1 2

Sample Output

Yes
No

Hint:
The first enquiry: $a$ can decrease grid $(1,2)$'s number by $1$. No matter what $b$ operate next, there is always one grid with number $1$ remaining . So, $a$ wins.
The second enquiry: No matter what $a$ operate, there is always one grid with number $1$ remaining. So, $b$ wins.

#include 
#include 
#include 

using namespace std;
#define N 500 + 5

int n, m, q;
int c[N][N];
int sum[N][N];

inline int lowbit(int x)
{
    return x & (-x);
}

void update(int x, int y, int data)
{
    for(int i = x; i <= n; i += lowbit(i))
        for(int j = y; j <= m; j += lowbit(j))
        sum[i][j] ^= data;
}

int get_sum(int x, int y)
{
    int res = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        for(int j = y; j > 0; j -= lowbit(j))
        res ^= sum[i][j];
    return res;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        memset(sum, 0, sizeof sum);
        scanf("%d%d%d", &n, &m, &q);

        for(int i = 1; i <= n ; i++)
            for(int j = 1; j <= m; j++)
            {
                scanf("%d", &c[i][j]);
                update(i, j, c[i][j]);
            }

        int x1, x2, y1, y2, op;
        while(q--)
        {
            scanf("%d", &op);
            if(op == 1)
            {
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                int ans = 0;
                ans = get_sum(x2, y2) ^ get_sum(x1 - 1, y1 - 1) ^ get_sum(x2, y1 - 1) ^ get_sum(x1 - 1, y2);
                (ans == 0) ? printf("No\n") : printf("Yes\n");
            }
            else
            {
                scanf("%d%d%d", &x1, &y1, &x2);
                update(x1, y1, c[x1][y1] ^ x2);
                c[x1][y1] = x2;
            }
        }
    }
    return 0;
}

/*

1
3 3 10
1 1 1
1 1 1
1 1 1



*/