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 aa and bb,
they are playing a game.

There is a n*mn∗m matrix,
each grid of this matrix has a number c_{i,
j}c​i,j​​.

aa wants
to beat bb every
time, so aa ask
you for a help.

There are qq operations,
each of them is belonging to one of the following two types:

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. aa always
operate first, he wants to know if he can win this game.
Change c_{i,
j}c​i,j​​ to bb.

Input

The first line contains a integer T(1
\le T \le 5)T(1≤T≤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*mn∗m matrix
follow, the ii row jj 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 qq lines
follow, the first number is optopt.

if opt=1opt=1,
then 44 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 11.

if opt=2opt=2,
then 33 integers i,
j, bi,j,b follow,
represent operation 22.

Output

For each testcase, for each operation 11,
print YesYes if aa can
win this game, otherwise print NoNo.

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: aa can decrease grid (1, 2)(1,2)'s number by 11. No matter what bb operate next, there is always one grid with number 11 remaining . So, aa wins.
The second enquiry: No matter what aa operate, there is always one grid with number 11 remaining. So, bb wins.

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
#define N 500 + 5

int n, m, q;
int c

;
int sum

;

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

*/