[UVA][Math] 10642 - Can You Solve It?
Can
You Solve It?
Input: standard input
Output: standard output
Time
Limit: 1 second
First take a look at the following picture. In this picture, each circle has a coordinate according to Cartesian Coordinate System. You can move from one circle to another following the path denoted by forward arrow symbols. To go from a source circle to a destination circle, total number of step(s) needed = ( number of intermediate circles you pass + 1 ). For example, to go from ( 0 , 3 ) to ( 3 , 0 ) you have to pass two intermediate circles ( 1 , 2 ) and ( 2 , 1 ). So, in this case, total number of steps needed is 2 + 1 = 3. In this problem, you are to calculate number of step(s) needed for a given source circle and a destination circle. You can assume that, it is not possible to go back using the reverse direction of the arrows.
Input
The first line in the input is the number of test cases n
( 0 < n <= 500 ) to handle. Following there are n lines
each containing four integers ( 0 <= each integer <=100000 ) the
first pair of which represents the coordinates of the source circle and the other
represents that of the destination circle. The coordinates are listed in a form
( x , y ).
Output
For each pair of integers your program should output the case number first and
then the number of step(s) to reach the destination from the source.
You may assume that it is always possible to reach the destination circle from the
source circle.
Sample
Input
3
0 0 0 1
0 0 1 0
0 0 0 2
Sample Output
Case 1: 1
Case 2: 2
Case 3: 3
#include <stdlib.h>
long long L(long long x, long long y) {
int sum = 0, n;
sum = y*(y+1)/2;
n = x;
sum += (2*(2+y)+n-1)*n/2;
return sum;
}
int main() {
int n, C = 0;
scanf("%d", &n);
while(n--) {
long long x1, x2, y1, y2;
scanf("%lld %lld %lld %lld", &x1, &y1, &x2, &y2);
long long t1, t2, Ans;
t1 = L(x1, y1);
t2 = L(x2, y2);
Ans = t1-t2;
if(Ans < 0) Ans *= -1;
printf("Case %d: %lld\n", ++C, Ans);
}
return 0;
}