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 題解:這個PDF講的挺好-> 淺談資訊學競賽中的“0 ” 和 “1” ——二進制思想在資訊學競賽中的應用
代碼:
1 #include <stdio.h>
2 #include <string.h>
3 #include <math.h>
4 #include <algorithm>
5 #include <iostream>
6 #include <ctype.h>
7 #include <iomanip>
8 #include <queue>
9 #include <stdlib.h>
10 using namespace std;
11
12 int a[1111][1111];
13 int n,m;
14 char s[10];
15
16 int lowbit(int x)
17 {
18 return x & (-x);
19 }
20
21 int get(int x,int y)
22 {
23 int sum=0;
24 for(int i=x; i>0; i-=lowbit(i)){
25 for(int j=y; j>0; j-=lowbit(j)){
26 sum+=a[i][j];
27 }
28 }
29 return sum;
30 }
31
32 void add(int x,int y)
33 {
34 for(int i=x; i<=n; i+=lowbit(i)){
35 for(int j=y; j<=n; j+=lowbit(j)){
36 a[i][j]++;
37 }
38 }
39 }
40
41 int main()
42 {
43 int t;
44 scanf("%d",&t);
45 while(t--){
46 scanf("%d%d",&n,&m);
47 memset(a,0,sizeof(a));
48 while(m--){
49 scanf("%s",s);
50 if(s[0]=='C'){
51 int x1,y1,x2,y2;
52 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
53 add(x2+1,y2+1);
54 add(x1,y1);
55 add(x1,y2+1);
56 add(x2+1,y1);
57 }
58 else{
59 int x,y;
60 scanf("%d%d",&x,&y);
61 printf("%d\n",get(x,y)%2);
62 }
63 }
64 if(t)
65 printf("\n");
66 }
67 return 0;
68 }