CodeForces 38B Chess

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本文探讨了一个有趣的国际象棋问题：在标准的8×8棋盘上，放置一个战车和一个骑士，确保两者互不吃掉的情况下，计算可以额外放置多少个骑士而不违反国际象棋规则。通过编程求解这一问题，并给出具体实现。

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Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.

Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.

Input

The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.

Output

Print a single number which is the required number of ways.

Sample test(s)

Input

a1
b2

Output

Input

a8
d4

Output

38

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int tp[10][10];
int dir[8][2]={2,1,2,-1,-2,1,-2,-1,1,2,1,-2,-1,2,-1,-2};

int check(int x,int y){
    if(x>=1&&x<=8&&y>=1&&y<=8){
        return 1;
    }
    return 0;
}

int main(){
    char a[4],b[4];
    int x1,y1,x2,y2,sum;
    while(~scanf("%s %s",a,b)){
        memset(tp,0,sizeof(0));
        sum = 0;
        x1=a[0]-'a'+1;
        y1=a[1]-'0';
        x2=b[0]-'a'+1;
        y2=b[1]-'0';
        tp[x1][y1]=tp[x2][y2]=1;
        for(int k=0;k<8;k++){   //马能吃的位置
            int x0=x2+dir[k][0];
            int y0=y2+dir[k][1];
            if(check(x0,y0)){
                tp[x0][y0]=2;
            }
        }
        for(int k=0;k<8;k++){   //放马能吃到车的位置
            int x0=x1+dir[k][0];
            int y0=y1+dir[k][1];
            if(check(x0,y0)){
                tp[x0][y0] = 2;
            }
        }
        for(int i=1;i<=8;i++){  //车能吃的位置
            tp[x1][i]=3;
            tp[i][y1]=3;
        }
        for(int i=1;i<=8;i++){
            for(int j=1;j<=8;j++){
                if(!tp[i][j]){
                    sum++;
                }
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}