// One-star problem
// Problem 10642 Can You Solve It?
/*
This program is written by
Prof. Chua-Huang Huang
Department of Information Engineering and Computer Science
Feng Chia University
Taichung, Taiwan
Disclaimer:
The programming problem is downloaded from UVa Online Judge (https://uva.onlinejudge.org/).
The program solution is provided for helping students to prepare Collegiate Programming
Examination (CPE). The author does not guarantee the program is completely correct to pass
UVa Online Judge platform or CPE examination platform.
This program is not intended for a student to copy only. He/She should practice the
programming problem himself/herself. Only use the program solution as a reference.
The author is not responsible if the program causes any damage of your computer or
personal properties.
No commercial use of this program is allowed without the author's written permission.
*/
#include
#include
#include
/* This problem is similar to UVA264 Count on Cantor.
Observe that we count the position of the circles along the ant-diagonal. The coordinate
(x, y) is on the (x+y)-th anti-diagonal and at the y-th element from the lower-right corner
of this diagonal. The first anti-diagonal has one element, the second anti-diagonal has
two elements, the third anti-diagonal has three elements, etc. These anti-diagonals form
an arithmetic progression sequence, 1, 2, 3, 4, .... That is, the k-th anti-diagonal
has k elements. The total number of circles in the first n anti-diagonals is n(n+1)/2.
Hence, to find the position of point (x, y), let n=x+y, there are n(n+1)/2 points on
the first n anti-diagonals. Since the circles travel from the lower-left corner,
the position of (x, y) is at the y-th element starting from the lower-right
corner of the (n+1)-st anti-diagonal.
Note that the output solution given in the problem is the result for circles traveling
from the upper-left corner downward. In this case point (x, y) is at the x-th element
from the upper-right corner.
For two circles (x1, y1) and (x2, y2), we calculate their positions starting from the
origin (0, 0), say, p1 and p2. Then the step(s) from (x1, y1) to (x2, y2) is |p1-p2|.
*/
int main(void) {
int cases;
int x1, y1, x2, y2;
int p1, p2;
int i;
scanf("%d", &cases); // Input the number of cases.
for (i=0; i