11494 - Queen

The Problem

The game of Chess has several pieces with curious movements. One of them is the Queen, which can move any number of squares in any direction: in the same line, in the same column or in any of the diagonals, as illustrated by the figure below (black dots represent positions the queen may reach in one move):

The great Chess Master Kary Gasparov invented a new type of chess problem: given the position of a queen in an empty standard chess board (that is, an 8 x 8 board) how many moves are needed so that she reaches another given square in the board?

Kary found the solution for some of those problems, but is having a difficult time to solve some others, and therefore he has asked that you write a program to solve this type of problem.

The Input

The input contains several test cases. The only line of each test case contains four integers X₁, Y₁, X₂ and Y₂ (1 ≤ X₁, Y₁, X₂, Y₂ ≤ 8). The queen starts in the square with coordinates (X₁, Y₁), and must finish at the square with coordinates (X₂, Y₂). In the chessboard, columns are numbered from 1 to 8, from left ro right; lines are also numbered from 1 to 8, from top to bottom. The coordinates of a square in line X and column Y are (X, Y).

The end of input is indicated by a line containing four zeros, separated by spaces.

The Output

For each test case in the input your program must print a single line, containing an integer, indicating the smallest number of moves needed for the queen to reach the new position.

西洋棋中有几个子的走法满特别的，其中一个就是皇后。她可以循垂直、水平、或对角线的方向随她走几格，如下图(黑点表示皇后可以一步走到的格子)：


我们的问题是：在标准的西洋棋空棋盘(8 x 8 棋盘) 上摆一个皇后，它要走几步才能走到某个特定的格子？

Input

输入档包含了好几笔测试资料。每笔测试资料只有一行，其中含有整数X1, Y1, X2 及Y2 (1 ≤ X1, Y1, X2, Y2 ≤ 8)。皇后从座标(X1, Y1) 的格子开始，必须在座标(X2, Y2) 的格子结束。在棋盘中的行由左至右编号为1 到8，列则由上至下编号为1 到8。位于第X 列第Y 行的格子其座标为(X, Y)。

输入的结尾以四个由空白隔开的0 来表示。

请参考Sample Input。

Output

对于每笔测试你的程式要印出一行，该行含有一个整数，表示皇后要走到新的位置至少需要几步。

Sample Input

4 4 6 2
3 5 3 5
5 5 4 3
0 0 0 0

Sample Output

1
0
2
解题思路：国际象棋中女王的下法，类似于数学坐标问题，要注意只走1步的条件，横竖直走没有问题，问题是斜着走的条件“(x1-x2)%(y1-y2)==0”

#include<stdio.h>
int main()
{int x1,x2,y1,y2;
while(scanf("%d%d%d%d",&x1,&y1,&x2,&y2)!=EOF){
if((x1==0&&x2==0)&&(y1==0&&y2==0))
break;
if(x1==x2&&y1==y2)
printf("0\n");
else if((y1==y2)||(x1==x2))
printf("1\n");
else if(((x1-x2)/(y1-y2)==-1||(x1-x2)/(y1-y2)==1)&&(x1-x2)%(y1-y2)==0)
printf("1\n");
else printf("2\n");
}
return 0;
}