There is a board with

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本文介绍了一个简单的程序设计案例，该程序可在100x100的网格上绘制黑白方块并计算指定区域内黑色方块的数量。通过执行不同指令（如画白色方块、画黑色方块及测试区域内的黑色方块数量），程序实现了基本的绘图功能。

There is a board with 100 * 100 grids as shown below. The left-top gird is denoted as (1, 1) and the right-bottom grid is (100, 100).

We may apply three commands to the board:

WHITE x, y, L // Paint a white square on the board,
// the square is defined by left-top grid (x, y)
// and right-bottom grid (x+L-1, y+L-1)
BLACK x, y, L // Paint a black square on the board,
// the square is defined by left-top grid (x, y)
// and right-bottom grid (x+L-1, y+L-1)
TEST x, y, L // Ask for the number of black grids
// in the square (x, y)- (x+L-1, y+L-1)

In the beginning, all the grids on the board are white. We apply a series of commands to the board. Your task is to write a program to give the numbers of black grids within a required region when a TEST command is applied.
输入
The first line of the input is an integer t (1 <= t <= 100), representing the number of commands. In each of the following lines, there is a command. Assume all the commands are legal which means that they won’t try to paint/test the grids outside the board.
输出
For each TEST command, print a line with the number of black grids in the required region.
样例输入
5
BLACK 1 1 2
BLACK 2 2 2
TEST 1 1 3
WHITE 2 1 1
TEST 1 1 3
样例输出
7
6

#include<stdio.h>
#include<string.h>
char arr[1050][1000],m;
int brr[1050][40],ans[105]={0,0},num=0,map[1005][1005]={0,0};
void Wasd(int a,int b,int c)
{
    int x,y,i,j;
    x=a+c-1;
    y=b+c-1;
    for(i=a;i<=x;i++)
        for(j=b;j<=y;j++)
            map[i][j]=0;
}
void Basd(int a,int b,int c)
{
    int x,y,i,j;
    x=a+c-1;
    y=b+c-1;
    for(i=a;i<=x;i++)
        for(j=b;j<=y;j++)
            map[i][j]=1;
}
void Tasd(int a,int b,int c)
{
    int x,y,i,j;
    x=a+c-1;
    y=b+c-1;
    for(i=a;i<=x;i++)
        for(j=b;j<=y;j++)
            if(map[i][j]==1)
            {
                ans[num]++;
            }
    num++;
}
int main()
{
    int n,k,l,i,j,size;
    scanf("%d",&n);
    scanf("%c",&m);
    for(i=0;i<n;i++)
    {
        gets(arr[i]);
        brr[i][0]=strlen(arr[i]);
    }
    for(i=0;i<n;i++)
    {
        k=1;
        for(j=4;j<brr[i][0];j++)
        {
            if(arr[i][j]>='0'&&arr[i][j]<='9')
            {
                if(arr[i][j+1]>='0'&&arr[i][j+1]<='9')
                {
                    brr[i][k++]=(arr[i][j]-'0')*10+(arr[i][j+1]-'0');
                    if(arr[i][j+2]>='0'&&arr[i][j+2]<='9')
                    {
                        brr[i][k-1]=100;
                        j++;
                    }
                    j++;
                }
                else
                {
                    brr[i][k++]=arr[i][j]-'0';
                }
            }
        }
    }
    for(l=0;l<n;l++)
    {
        if(arr[l][0]=='W')
            Wasd(brr[l][1],brr[l][2],brr[l][3]);
        if(arr[l][0]=='B')
            Basd(brr[l][1],brr[l][2],brr[l][3]);
        if(arr[l][0]=='T')
            Tasd(brr[l][1],brr[l][2],brr[l][3]);
    }
    for(i=0;i<num;i++)
        printf("%d\n",ans[i]);
    return 0;
}