[URI Online Judge] – 1087 – Queen

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.

Input

The input contains several test cases. The only line of each test case contains four integers X1, Y1, X2 andY2 (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 columnY are (X, Y).

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

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.

Sample Input Sample Output
4 4 6 2
3 5 3 5
5 5 4 3
0 0 0 0
1
0
2

solution here:

#include <cstdio>
#include <cmath>
using namespace std;
int main(){
 int x1, y1, x2, y2, d_x, d_y;
 while(scanf("%d %d %d %d", &x1, &y1, &x2, &y2) && x1 != 0 && y1 != 0 && x2 != 0 && y2 != 0){
 if (x1 > x2) d_x = x1 - x2;
 else d_x = x2 - x1;
if (y1 > y2) d_y = y1 - y2;
 else d_y = y2 - y1;
if (x1 == x2 && y1 == y2)
 printf("0\n");
 else if (x1 == x2 || y1 == y2 || d_x == d_y)
 printf("1\n");
 else
 printf("2\n");
 }
 return 0;
}
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