D. Iahub and Xors
time limit per test 1 second
memory limit per test 256 megabytes
Iahub does not like background stories, so he'll tell you exactly what this problem asks you for.
You are given a matrix a with n rows and n columns. Initially, all values of the matrix are zeros. Both rows and columns are 1-based, that is rows are numbered 1, 2, ..., n and columns are numbered 1, 2, ..., n. Let's denote an element on the i-th row and j-th column as ai, j.
We will call a submatrix (x0, y0, x1, y1) such elements ai, j for which two inequalities hold: x0 ≤ i ≤ x1, y0 ≤ j ≤ y1.
Write a program to perform two following operations:
- Query(x0, y0, x1, y1): print the xor sum of the elements of the submatrix (x0, y0, x1, y1).
- Update(x0, y0, x1, y1, v): each element from submatrix (x0, y0, x1, y1) gets xor-ed by value v.
Input
The first line contains two integers: n (1 ≤ n ≤ 1000) and m (1 ≤ m ≤ 105). The number m represents the number of operations you need to perform. Each of the next m lines contains five or six integers, depending on operation type.
If the i-th operation from the input is a query, the first number from i-th line will be 1. It will be followed by four integers x0, y0, x1, y1. If the i-th operation is an update, the first number from the i-th line will be 2. It will be followed by five integers x0, y0, x1, y1, v.
It is guaranteed that for each update operation, the following inequality holds: 0 ≤ v < 262. It is guaranteed that for each operation, the following inequalities hold: 1 ≤ x0 ≤ x1 ≤ n, 1 ≤ y0 ≤ y1 ≤ n.
Output
For each query operation, output on a new line the result.
Examples
input
Copy
3 5 2 1 1 2 2 1 2 1 3 2 3 2 2 3 1 3 3 3 1 2 2 3 3 1 2 2 3 2
output
Copy
3 2
Note
After the first 3 operations, the matrix will look like this:
1 1 2 1 1 2 3 3 3
The fourth operation asks us to compute 1 xor 2 xor 3 xor 3 = 3.
The fifth operation asks us to compute 1 xor 3 = 2.
题目链接：http://codeforces.com/contest/341/problem/D
题目大意：给一个n*n的矩阵，初始值全为0，有两种操作，1表示查询以(x0, y0)和(x1, y1)为对顶点的矩阵中所有值的异或和，2表示将以(x0, y0)和(x1, y1)为对顶点的矩阵中的所有值都异或v
题目分析：先考虑一维的情况，让点i维护[1,i]的前缀异或和，对于(l, r)这段区间
1）更新：通过后缀异或和的方式，更新l和r+1两个点 =》update(l, v)；update(r+1, v)；
2）查询：利用前缀异或和(a^a=0) =》query(l-1)^query(r)
需要特别注意的一点是，异或操作与加法不同，a+a=2a而a^a=0，因此对于区间(l, r)，更新时真正影响的其实只有l,l+2,l+4...这些，因此需要分奇偶讨论，所以需要两个树状数组来维护。
将其拓展到二维即可
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1e3 + 5;
int n, m;
ll xorsum[2][2][MAX][MAX];
int lowbit(int x) {
return x & (-x);
}
void update(int x, int y, ll val) {
for (int i = x; i <= n; i += lowbit(i)) {
for (int j = y; j <= n; j += lowbit(j)) {
xorsum[x & 1][y & 1][i][j] ^= val;
}
}
}
ll query(int x, int y) {
ll ans = 0;
for (int i = x; i > 0; i -= lowbit(i)) {
for (int j = y; j > 0; j -= lowbit(j)) {
ans ^= xorsum[x & 1][y & 1][i][j];
}
}
return ans;
}
int main() {
int tp, x0, y0, x1, y1;
ll v;
scanf("%d %d", &n, &m);
while (m--) {
scanf("%d", &tp);
if (tp == 1) {
scanf("%d %d %d %d", &x0, &y0, &x1, &y1);
ll ans = 0;
ans ^= query(x0 - 1, y0 - 1);
ans ^= query(x0 - 1, y1);
ans ^= query(x1, y0 - 1);
ans ^= query(x1, y1);
printf("%I64d\n", ans);
} else {
scanf("%d %d %d %d %I64d", &x0, &y0, &x1, &y1, &v);
update(x0, y0, v);
update(x0, y1 + 1, v);
update(x1 + 1, y0, v);
update(x1 + 1, y1 + 1, v);
}
}
}