rip scan sea int write and log cheng sin

題目鏈接：

解題思路：首先考慮一維的情況，更改某個區間[l, r]時相當於l出加一，r + 1除減一，這樣就能保證每次計算的時候兩者總是能抵消的。當然也可以在l與r + 1除都加一，而每次求和之後mod 2。

接下來考慮二維，更改區間[x1, y1], [x2, y2]的時候，只需在(x1, y1), (x1, y2 + 1), (x2 + 1, y1), (x2 + 1, y2 + 1)四個點除加一即可（好吧其實我只推出了一維，二維是看別人博客的）。然後求和就交給二維樹狀數組了。

比較有意思的是，當一個函數這樣寫的時候

1 void add(int i, int j, int val){
 
2     for(i; i <= n; i += lowbit(i)){
3         for(j; j <= n; j += lowbit(j)){
4             bit[i][j] += val;
5         }
6     } 
7 }

得到的結果竟然是錯誤的，不知道是什麽bug。

代碼：

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <functional>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cstring>
 6
 
 #include <cassert>
 7 #include <cstdio>
 8 #include <cctype>
 9 #include <vector>
10 #include <string>
11 #include <queue>
12 #include <stack>
13 #include <cmath>
14 #include <map>
15 #include <set>
16 using namespace std;
17 #define rep(i,a,n) for (int i=a;i<n;i++)
18
 
 #define per(i,a,n) for (int i=n-1;i>=a;i--)
19 #define pb push_back
20 #define mp make_pair
21 #define all(x) (x).begin(),(x).end()
22 #define fi first
23 #define se second
24 #define SZ(x) ((int)(x).size())
25 typedef vector<int> VI;
26 typedef long long ll;
27 typedef pair<int, int> PII;
28 const ll mod = 1000000007;
29 ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }return res; }
30 // head
31 const int inf = 0x3f3f3f3f;
32 const int maxn = 1010;
33 
34 int bit[maxn][maxn], n;
35 
36 int lowbit(int x){
37     return x & (-x);
38 }
39 int sum(int x, int y){
40     int ans = 0;
41     for(int i = x; i > 0; i -= lowbit(i)){
42         for(int j = y ; j > 0; j -= lowbit(j)){
43             ans += bit[i][j];
44         }
45     }
46     return ans;
47 }
48 void add(int x, int y, int val){
49     for(int i = x; i <= n; i += lowbit(i)){
50         for(int j = y; j <= n; j += lowbit(j)){
51             bit[i][j] += val;
52         }
53     } 
54 }
55 
56 int main(){
57     int T;
58     scanf("%d", &T);
59     while(T--){
60         memset(bit, 0, sizeof(bit));
61         int m;
62         scanf("%d %d", &n, &m);
63         while(m--){
64             char ch;
65             scanf(" %c", &ch);
66             if(ch == ‘C‘){
67                 int x1, y1, x2, y2;
68                 scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
69                 
70                 add(x1, y1, 1); 
71                 add(x2 + 1, y1, 1);
72                 add(x1, y2 + 1, 1);
73                 add(x2 + 1, y2 + 1, 1);
74                 
75             }
76             else{
77                 int x1, y1;
78                 scanf("%d %d", &x1, &y1);
79                 if(sum(x1, y1) % 2 == 0) puts("0");
80                 else puts("1");
81             }
82         }
83         if(T)
84             puts("");
85     }
86 }

題目：

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 28786		Accepted: 10494

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a ‘0‘ then change it into ‘1‘ otherwise change it into ‘0‘). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.

Output

For each querying output one line, which has an integer representing A[x, y].

There is a blank line between every two continuous test cases.

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

Source

POJ Monthly,Lou Tiancheng

POJ 2155 Matrix