Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1087
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1087
Queen
By Fábio Dias Moreira 
Brazil
Timelimit: 1
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 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.
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.
| Input Sample | Output Sample |
|
4 4 6 2 |
1 |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
int main ()
{
unsigned short x1, y1, x2, y2;
while (true)
{
scanf("%hd %hd %hd %hd", &x1, &y1, &x2, &y2);
if (x1 == 0 && y1 == 0 && x2 == 0 && y2 == 0)
break;
if (x1 == x2 && y1 == y2)
printf("0\n");
else if (abs(x2 - x1) == abs(y2 - y1))
printf("1\n");
else if (x1 == x2 || y1 == y2)
printf("1\n");
else
printf("2\n");
}
}
Input
3 5 3 5
5 5 4 3
0 0 0 0
Output
0
2
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <stdio.h>
int main(){
int x1,y1,x2,y2;
while(true){
scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
if(x1 == 0 && y1 == 0 && x2 == 0 && y2 == 0) break;
if(x1 == x2 && y1 == y2) printf("0\n");
else if((x2-x1) == -(y2-y1) || -(x2-x1) == -(y2-y1) || -(x2-x1) == (y2-y1) || (x2-x1) == (y2-y1)) printf("1\n");
else if(x1 == x2 || y1 == y2) printf("1\n");
else printf("2\n");
}
return 0;
}
Input
3 5 3 5
5 5 4 3
0 0 0 0
Output
0
2
#3 Code Example with Java Programming
Code -
Java Programming
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
Scanner entrada = new Scanner(System.in);
int x1 , y1 , x2 , y2 , qtdMov;
while(true) {
x1 = entrada.nextInt();
y1 = entrada.nextInt();
x2 = entrada.nextInt();
y2 = entrada.nextInt();
if(x1 == 0 && y1 == 0 && x2 == 0 && y2 == 0)
break;
if(x1 == x2 && y1 == y2)
System.out.printf("0\n");
else if ((x2-x1) == -(y2-y1) || -(x2-x1) == -(y2-y1) || -(x2-x1) == (y2-y1) || (x2-x1) == (y2-y1))
System.out.printf("1\n");
else if(x1 == x2 || y1 == y2)
System.out.printf("1\n");
else
System.out.printf("2\n");
}
}
}
Input
3 5 3 5
5 5 4 3
0 0 0 0
Output
0
2
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const { readFileSync } = require("fs")
const input = readFileSync("/dev/stdin", "utf8")
.split("\n")
.map((line) => line.split(" ").map((value) => Number.parseInt(value, 10)))
function main() {
const responses = []
for (const [X1, Y1, X2, Y2] of input) {
if ([X1, Y1, X2, Y2].every((crd) => crd == 0)) break
if (X1 == X2 && Y1 == Y2) responses.push("0")
else if ((X1 != X2 && Y1 == Y2) || (X1 == X2 && Y1 != Y2) || Math.abs(X1 - X2) == Math.abs(Y1 - Y2)) responses.push("1")
else responses.push("2")
}
console.log(responses.join("\n"))
}
main()
Input
3 5 3 5
5 5 4 3
0 0 0 0
Output
0
2
#5 Code Example with Python Programming
Code -
Python Programming
from math import fabs
aux=2
while True:
x1,y1,x2,y2=map(int,input().split())
if x1==y1==x2==y2==0:break
if x1==x2 and y1==y2:print(0)
elif((x2-x1) == -(y2-y1) or -(x2-x1) == -(y2-y1) or -(x2-x1) == (y2-y1) or (x2-x1) == (y2-y1)):print("1")
elif x1==x2 or y1==y2:print(1)
else:print(2)
Input
3 5 3 5
5 5 4 3
0 0 0 0
Output
0
2