A. Rook, Bishop and King----思维题

A. Rook, Bishop and King

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8 × 8 table.
A field is represented by a pair of integers (r, c) — the number of the row and the number of the column (in a classical
game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:

A rook moves any number of fields horizontally or vertically.

A bishop moves any number of fields diagonally.

A king moves one field in any direction — horizontally, vertically or diagonally.


The
pieces move like that

Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (r1, c1) to
field (r2, c2)?
At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem.

Input

The input contains four integers r1, c1, r2, c2 (1 ≤ r1, c1, r2, c2 ≤ 8)
— the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.

You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8.

Output

Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (r1, c1) to
field (r2, c2).
If a piece cannot make such a move, print a 0 instead of the corresponding number.

Examples

input

4 3 1 6

output

2 1 3

input

5 5 5 6

output

1 0 1

题目链接：http://codeforces.com/contest/370/problem/A

题目的意思是说给你起点和终点，让你求象，车，王各走多少步。

车的话同行同列只需一步，否则需要两步，王的步数也很简单，max(abs(x1-x2),abs(y1-y2))，象的比较麻烦，象任意对角线都能走，如果两个本身在同一对角线，就是一步，如果不在，就要判断下是否在同一色块，是就是2，否则，就走不到，输出0。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace std;
int main(){
int x1,x2,y1,y2;
while(scanf("%d%d%d%d",&x1,&y1,&x2,&y2)!=EOF){
int a=1,b=2,c=max(abs(x1-x2),abs(y1-y2));
if(x1!=x2&&y1!=y2)
a=2;
if(abs(x1-x2)==abs(y1-y2))
b=1;
else if((x2+y2+(y1-x1))%2!=0)
b=0;
printf("%d %d %d\n",a,b,c);
}
return 0;
}