HDU 5465 Clarke and puzzle

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∗m  matrix, each grid of this matrix has a number  ci,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  (x1,y1)−(x2,y2)  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  ci,j  to  b . 

 

Input

The first line contains a integer  T(1≤T≤5) , the number of test cases. 
For each test case: 
The first line contains three integers  n,m,q(1≤n,m≤500,1≤q≤2∗105)  
Then  n∗m  matrix follow, the  i  row  j  column is a integer  ci,j(0≤ci,j≤109)  
Then  q  lines follow, the first number is  opt . 
if  opt=1 , then  4  integers  x1,y1,x1,y2(1≤x1≤x2≤n,1≤y1≤y2≤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.
   
   
   
   

 

nim取子游戏+前缀和

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn = 505;
int T, n, m, q, cs, x, y, X, Y, z, a[maxn][maxn], f[maxn][maxn];

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d%d", &n, &m, &q);
		for (int i = 1; i <= n; i++)
		{
			f[i][0] = 0;
			for (int j = 1; j <= m; j++)
			{
				scanf("%d", &a[i][j]);
				f[i][j] = a[i][j] ^ f[i][j - 1];
			}
		}
		while (q--)
		{
			scanf("%d", &cs);
			if (cs == 1)
			{
				scanf("%d%d%d%d", &x, &y, &X, &Y);
				int flag = 0;
				for (int i = x; i <= X; i++) flag ^= f[i][y - 1] ^ f[i][Y];
				if (flag) printf("Yes\n"); else printf("No\n");
			}
			else
			{
				scanf("%d%d%d", &x, &y, &z);
				a[x][y] = z;
				for (int i = y; i <= m; i++) f[x][i] = a[x][i] ^ f[x][i - 1];
			}
		}
	}
	return 0;
}