CodeForces - 630E A rectangle (数学规律)

CodeForces
- 630E
A rectangle

Time Limit: 500MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Developing tools for creation of locations maps for turn-based fights in a new game, Petya faced the following problem.

A field map consists of hexagonal cells. Since locations sizes are going to be big, a game designer wants to have a tool for quick filling of a field part with identical enemy units. This action will look like following: a game designer will select a rectangular
area on the map, and each cell whose center belongs to the selected rectangle will be filled with the enemy unit.

More formally, if a game designer selected cells having coordinates (x1, y1) and (x2, y2),
where x1 ≤ x2 andy1 ≤ y2,
then all cells having center coordinates (x, y) such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2 will
be filled. Orthogonal coordinates system is set up so that one of cell sides is parallel to OX axis, all hexagon centers have integer coordinates and for each integer x there
are cells having center with such x coordinate and for each integer y there
are cells having center with such y coordinate. It is guaranteed that difference x2 - x1 is
divisible by 2.

Working on the problem Petya decided that before painting selected units he wants to output number of units that will be painted on the map.

Help him implement counting of these units before painting.



Input

The only line of input contains four integers x1, y1, x2, y2 ( - 109 ≤ x1 ≤ x2 ≤ 109,  - 109 ≤ y1 ≤ y2 ≤ 109)
— the coordinates of the centers of two cells.

Output

Output one integer — the number of cells to be filled.

Sample Input

Input

1 1 5 5

Output

13

Source
Experimental Educational Round: VolBIT Formulas Blitz
//题意：
给你两个点的坐标，（x1,y1）,(x2,y2)，分别表示左下端点和右上端点，问在此矩形中一共可以将几个六边形涂黑（如图所示）。找规律题，就不多说了。。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 10010
#define M 1000000007
using namespace std;
int main()
{
ll x1,x2,y1,y2;
while(scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2)!=EOF)
{
ll ny=(y2-y1+2)/2;
ll nx=(x2-x1+1)/2;
ll sum=ny*(x2-x1+1)-nx;
printf("%lld\n",sum);
}
return 0;
}