CodeForces 38B Chess

B. Chess

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 froma 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

a1b2

Output

44

Input

a8d4

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;}