大模拟真的好难呀

问题:读题不够细，因为一个\0A卡了一个半小时，能做出来已经很不错了！

Tip优化

1. 不用结构体，改用三维数组表示会方便很多
2. 字符串的处理不够熟悉，单纯为了模拟练习用的char*，改用C++的string会简单很多
3. …

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>

using namespace std;
const int N = 2000;
#define LL long long
int m, n ,p, q;

struct node{
    int R = 0;
    int G = 0;
    int B = 0;
}Node[N][N],Mean[N*N];

//像素点  R,G,B
//求矩阵范围内的像素均值  容斥原理
void getMean(int x1,int y1, int x2, int y2,int h)  // 左上角，右上角
{
    // printf("x1:%d,y1:%d,x2:%d,y2:%d\n",x1,y1,x2,y2);
    Mean[h].R = (Node[x2][y2].R - Node[x1-1][y2].R - Node[x2][y1-1].R + Node[x1-1][y1-1].R)/(p*q);
    Mean[h].G = (Node[x2][y2].G - Node[x1-1][y2].G - Node[x2][y1-1].G + Node[x1-1][y1-1].G)/(p*q);
    Mean[h].B= (Node[x2][y2].B - Node[x1-1][y2].B - Node[x2][y1-1].B + Node[x1-1][y1-1].B)/(p*q);

}

int HexToInt(char*a)
{
    int m = 0;
    int length = strlen(a);
    for(int i = 0; i < length; i++)
    {
        if(a[i]<='9' && a[i]>='0') m = m*16 + a[i]-'0';
        else if(a[i]<='z' && a[i]>='a') m = m*16 + a[i]-'a'+10;
        else if(a[i]<='Z' && a[i]>='A') m = m*16 + a[i]-'A'+10;

    }
    return m;
}


void charToRGB(node* nd, char* a)
{
    int length = strlen(a);
    char b[10];
    if(length == 1){
        for(int i =0; i<6;i++)
            b[i] = a[0];
    }
    else if(length == 3){
        for(int i = 0,m =0; i<6; i+=2)
        {
            b[i] = a[m];
            b[i+1] = a[m++];
        }
    }
    else memcpy(b,a,sizeof b);
    char r[3]={0};
    char g[3]={0};
    char b0[3]={0};
    strncpy(r,b,2);
    strncpy(g,b+2,2);
    strncpy(b0,b+4,2);
    nd->R = HexToInt(r);
    nd->G = HexToInt(g);
    nd->B = HexToInt(b0);
}


void RGBIntToStr(int a)
{
    if(a>=100)
    {
        int num1 = a/100;
        a = a%100;
        int num2 = a/10;
        int num3 = a%10;
        printf("\\x3%d\\x3%d\\x3%d",num1,num2,num3);
    }else if(a>=10)
    {
        int num1 = a/10;
        int num2 = a%10;
        printf("\\x3%d\\x3%d",num1,num2);
    }else{
        printf("\\x3%d",a);
    }
}


int printResult(int h,bool end,bool first)
{
    if(Mean[h].R == Mean[h-1].R && Mean[h].G == Mean[h-1].G && Mean[h].B == Mean[h-1].B && !first)
    {
        if(!end) printf("\\x20");
        return 0;
    }
    else if(Mean[h].R == 0 && Mean[h].G == 0 && Mean[h].B == 0)
    {
        if(first){
            printf("\\x20");
            return 0;

        }
        printf("\\x1B\\x5B\\x30\\x6D");
        if(!end) printf("\\x20");
        return 0;
    }
    printf("\\x1B\\x5B\\x34\\x38\\x3B\\x32\\x3B");
    RGBIntToStr(Mean[h].R);
    printf("\\x3B");
    RGBIntToStr(Mean[h].G);
    printf("\\x3B");
    RGBIntToStr(Mean[h].B);
    printf("\\x6D\\x20");
}


int main()
{
    cin>>m>>n>>p>>q;
    getchar();
/**
 * 
 *   m(y)------->
 * n(x)
 * |
 * |
 * |
 * V
 **/
    for(int i = 1; i <= n; i ++)
        for( int j = 1; j <= m; j ++)
        {
            char a[8];
            scanf("#%s\n",a);  // 处理字符到RGB
            charToRGB(&Node[i][j],a);
            Node[i][j].R = Node[i-1][j].R+Node[i][j-1].R-Node[i-1][j-1].R+Node[i][j].R;
            Node[i][j].G = Node[i-1][j].G+Node[i][j-1].G-Node[i-1][j-1].G+Node[i][j].G;
            Node[i][j].B = Node[i-1][j].B+Node[i][j-1].B-Node[i-1][j-1].B+Node[i][j].B;

        }




    // //处理输出
    int h=1;
    for(int i = 0; i < n/q ; i++)  //n/q 
    {
        bool first = true;
        for(int j = 0; j< m/p; j++) // m/p
        {
            int x1 = 1+q*i, y1 =1+j*p, x2 =x1+q-1 , y2 =y1+p-1;
            getMean(x1,y1,x2,y2,h);
            printResult(h,false,first);
            first = false;
            h++;
        }
        printResult(h,true,first);
        printf("\\x0A");

    }

    return 0;
}