### An Experiment by Penny

Below given c++ code is for CRAN01 spoj
or An experiment by penny.

Here to find
the logic you can test it for some small test by pen & paper. For myself I have
checked for approx. 20 test cases then I got the logic .

Logic is
maximum corner distance from given point is the minimum time to fill the entire
box.

For example I am taking a grid of 7 x 7 and starting position as (4,5).

Here in the
above table there are four corners coloured as yellow green black & blue
and the starting position is red coloured box. Now the corners coloured with yellow
and black are at maximum distances = 7 So here answer will be 7 .I get this
logic from observation so if you want to clear your logic please use pen &
paper and draw some small test cases or use brute-force.

#include <bits/stdc++.h> using namespace std; int main() { int t; scanf("%d",&t); while(t--) { int n,m,x,y; scanf("%d%d%d%d",&n,&m,&x,&y); int r1,r2,r3,r4,result; r1 = abs(x-1) + abs(y-1); r2 = abs(x-1) + abs(y-m); r3 = abs(x-n) + abs(y-1); r4 = abs(x-n) + abs(y-m); result = max(r1,max(r2,max(r3,r4))); printf("%d\n",result); } return 0; }

here also python 2.7 code is for CRAN01 spoj
or An experiment by penny.

import sys,math t=int(sys.stdin.readline()) while t: n,m = map(int,sys.stdin.readline().split()) x,y = map(int,sys.stdin.readline().split()) r1 = abs(x-1) + abs(y-1) r2 = abs(x-1) + abs(y-m) r3 = abs(x-n) + abs(y-1) r4 = abs(x-n) + abs(y-m) result = max(r1,max(r2,max(r3,r4))) sys.stdout.write(str(result)) sys.stdout.write("\n") t=t-1