Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 33682 | Accepted: 12194 |
Description
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 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
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string> using namespace std;
const int N=1e3+;
int s[N][N];
int n;
int lowbit(int x)
{
return x&(-x);
}
void updata(int x,int y,int z)
{
for(int i=x;i<=n;i+=lowbit(i)){
for(int j=y;j<=n;j+=lowbit(j)){
s[i][j]+=z;
}
}
} int sum(int x,int y)
{
int res=;
for(int i=x;i>;i-=lowbit(i)){
for(int j=y;j>;j-=lowbit(j)){
res+=s[i][j];
}
}
return res;
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
memset(s,,sizeof(s));
int m;
scanf("%d %d",&n,&m);
while(m--){
char t[];
scanf("%s",t);
if(t[]=='C'){
int x1,y1,x2,y2;
scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
updata(x1,y1,);
updata(x2+,y2+,);
updata(x2+,y1,-);
updata(x1,y2+,-);
}
else{
int x,y;
scanf("%d %d",&x,&y);
printf("%d\n",sum(x,y)%);
}
}
if(T) printf("\n");
} return ;
}