hdu 5465 Clarke and puzzle(前缀和，异或，nim博弈)

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:
. 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.
. Change ci,j to b. 

Input

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

 

Output

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

Sample Input

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.

 

Source

BestCoder Round #56 (div.2)

 

题目要求二维的nim游戏，考虑到nim的结论是xor和为0则必败、否则必胜，那么我们只需要维护子矩阵的xor和。由于xor有前缀和性质，所以我们可以用一个二维bit来维护(1, 1)-(a, b)的矩阵的xor和，然后由sum(x2,y2) xor sum(x2,y1−1) xor sum(x1−1,y2) xor sum(x1−1,y1−1)sum(x2, y2) \ xor \ sum(x2, y1-1) \ xor \ sum(x1-1, y2) \ xor \ sum(x1-1, y1-1)sum(x2,y2) xor sum(x2,y1−1) xor sum(x1−1,y2) xor sum(x1−1,y1−1)来得到答案即可。单点修改在bit上是很容易的。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 int dirx[]={,,-,};
 int diry[]={-,,,};
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 506
 #define inf 1e12
 int n,m,q;
 int mp[N][N];
 int a[N][N];
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         scanf("%d%d%d",&n,&m,&q);
         for(int i=;i<=n;i++){
             for(int j=;j<=m;j++){
                 scanf("%d",&mp[i][j]);
                 a[i][j]=a[i][j-]^mp[i][j];
             }
         }
         for(int i=;i<q;i++){
             int opt;
             scanf("%d",&opt);
             int x,y,z;
             if(opt==){
                 scanf("%d%d%d",&x,&y,&z);
                 mp[x][y]=z;
                 for(int j=y;j<=m;j++){
                     a[x][j]=a[x][j-]^mp[x][j];
                 }
             }
             else{
                 int ans=;
                 int x1,y1,x2,y2;
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 for(int j=x1;j<=x2;j++){
                     ans=ans^(a[j][y2]^a[j][y1-]);
                 }
                 if(ans==){
                     printf("No\n");
                 }
                 else{
                     printf("Yes\n");
                 }
             }
         }
     }
     return ;
 }