Clarke and puzzle

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 951    Accepted Submission(s): 349

Problem DescriptionClarke 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.

 InputThe 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. OutputFor each testcase, for each operation 1, print Yes if a can win this game, otherwise print No. Sample Input1
1 2 3
1 2
1 1 1 1 2
2 1 2 1
1 1 1 1 2 Sample OutputYes
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.

 SourceBestCoder Round #56 (div.2) 　　　　对于操作1而言就是一个裸的nim博弈，答案就是 {c[x1][y1]^……^c[x2][y2]}　　 操作二就是改变了格子中的一个数。        用二维BIT来维护矩阵内的异或和即可。 　　(最近总把下标写错，，把x1,y1写成了i,j检查半天， (误

 #include<bits/stdc++.h>
 using namespace std;
 #define ULL unsigned long long
 #define LL long long
 int c[][];
 int C[][];
 int N,M;
 inline int lowbit(int x){return x&-x;}
 void change(int x,int y,int d){
     for(int i=x;i<=N;i+=lowbit(i))
         for(int j=y;j<=M;j+=lowbit(j))
             C[i][j]^=d;
 }
 int ask(int x,int y){
     int r=;
     for(int i=x;i;i-=lowbit(i))
         for(int j=y;j;j-=lowbit(j))
             r^=C[i][j];
     return r;
 }
 int main(){
     int t,n,m,i,q,j,k,x1,x2,y1,y2;
     scanf("%d",&t);
     while(t--){
         memset(C,,sizeof(C));
         scanf("%d%d%d",&n,&m,&q);
         N=n,M=m;
         for(i=;i<=n;++i){
             for(j=;j<=m;++j){
                 scanf("%d",&c[i][j]);
                 change(i,j,c[i][j]);
             }
         }         int opt,d;
         while(q--){
             scanf("%d",&opt);
             if(opt==){
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 int ans=(ask(x2,y2)^ask(x1-,y1-)^ask(x1-,y2)^ask(x2,y1-));
                 ans?puts("Yes"):puts("No");
             }
             else{
                 scanf("%d%d%d",&x1,&y1,&d);
                 change(x1,y1,(d^c[x1][y1]));
                 c[x1][y1]=d;
             }
         }
     }
     return ;
 }