Clarke and puzzle

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 673    Accepted Submission(s): 223

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.

Source

BestCoder Round #56 (div.2)

题目大意：给你t组测试数据。每组测试数据中有一组n，m，q，分别表示有一个n*m的矩阵，有q组询问。每组询问中，操作1表示查询（x1，y1） ---（x2，y2）在矩阵中做Nim游戏先手是否会赢。操作2表示要把矩阵中某个值修改为c。

知识补充：对于一维树状数组见的比较多。二维还是见的少（弱）。那么二维跟一维的区别是：一般意义上二维用来求的是矩阵的和。对于C[x][y]，这里的定义就是从（1,1）---（x,y）矩阵的元素和。我们可以将每行抽象成一个点，这样其实就是抽象成了求一维的情况。其实一维跟二维实现的写法差别很容易想到。对于二维BIT，我们得到C数组，需要n^2log(n)^2的时间复杂度。

求和：

int sum(int x,int y){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i)){
        for(int j=y;j>0;j-=lowbit(j)){
            ret+=C[i][j];
        }
    }
    return ret;
}

修改：

void modify(int x,int y,int val){
    for(int i=x;i<=n;i+=lowbit(i)){
        for(int j=y;j<=m;j+=lowbit(j)){
            C[i][j] += val;
        }
    }
}

解题思路：

#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
const int maxn=550;
int a[maxn][maxn];
int C[maxn][maxn];
int n,m;
int lowbit(int x){
    return x & (-x);
}
void modify(int x,int y,int val){
    for(int i=x;i<=n;i+=lowbit(i)){
        for(int j=y;j<=m;j+=lowbit(j)){
            C[i][j] ^= val;
        }
    }
}
int sum(int x,int y){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i)){
        for(int j=y;j>0;j-=lowbit(j)){
            ret ^=C[i][j];
        }
    }
    return ret;
}
int main(){
    int t,q;
    scanf("%d",&t);
    while(t--){
        memset(a,0,sizeof(a));
        memset(C,0,sizeof(C));
        scanf("%d%d%d",&n,&m,&q);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                modify(i,j,a[i][j]);
            }
        }
        int typ,x1,x2,y1,y2,c;
        for(int i=0;i<q;i++){
            scanf("%d",&typ);
            if(typ==1){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                int ans1,ans2,ans3,ans4;
                ans1=sum(x2,y2);
                ans2=sum(x2,y1-1);
                ans3=sum(x1-1,y2);
                ans4=sum(x1-1,y1-1);
                int ans=ans1^ans2^ans3^ans4;
                printf("%s\n",ans==0?"No":"Yes");
            }else{
                scanf("%d%d%d",&x1,&y1,&c);
                modify(x1,y1,c^a[x1][y1]);
                a[x1][y1]=c;
            }
        }
    }
    return 0;
}