CodeForces 630E-A rectangle

小灰灰 2022-08-21 01:17 16阅读 0赞

E. A rectangle

time limit per test

0.5 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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 (*x*1, *y*1) and (*x*2, *y*2), where *x*1 ≤ *x*2 and *y*1 ≤ *y*2, then all cells having center coordinates (*x*, *y*) such that *x*1 ≤ *x* ≤ *x*2 and *y*1 ≤ *y* ≤ *y*2 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 *x*2 - *x*1 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.

![f68ddb6334d8dabbf923af5e8aec8f073ce05a37.png][]

Input

The only line of input contains four integers *x*1, *y*1, *x*2, *y*2 ( - 109 ≤ *x*1 ≤ *x*2 ≤ 109,  - 109 ≤ *y*1 ≤ *y*2 ≤ 109) — the coordinates of the centers of two cells.

Output

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

Examples

input

1 1 5 5

output

解题思路：

这个所要算的矩形中，分短的和长的我们发现短的一个边的个数是边长/2（长和宽是一样的），而长的是（边长+1）/2，所以利用长和宽就能求出。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define LL __int64
    using namespace std;
    int main()
    {
    	LL x1,y1,x2,y2;
    	while(scanf("%I64d%I64d%I64d%I64d",&x1,&y1,&x2,&y2)!=EOF)
    	{
    		LL o,p;
    		o=x2-x1+1;
    		p=y2-y1+1;
    		LL ans;
    		ans=((p+1)/2)*((o+1)/2)+(p/2)*(o/2);	
    		printf("%I64d\n",ans);
    	}
    	return 0;
    }

[f68ddb6334d8dabbf923af5e8aec8f073ce05a37.png]: /images/20220731/4aae95a48c614a808f9b5e43222381f5.png