MemSQL Start[c]UP 2.0 - Round 1（无聊练手B题）

浏览数：29 / 时间：2015年06月12日

You are given a rectangular grid of lattice points from (0, 0) to (n, m) inclusive. You have to choose exactly 4 different points to build a polyline possibly with self-intersections and self-touching. This polyline should be as long as possible.

A polyline defined by points p1, p2, p3, p4 consists of the line segments p1 p2, p2 p3, p3 p4, and its length is the sum of the lengths of the individual line segments.

Input

The only line of the input contains two integers n and m (0 ≤ n, m ≤ 1000). It is guaranteed that grid contains at least 4 different points.

Output

Print 4 lines with two integers per line separated by space — coordinates of points p1, p2, p3, p4 in order which represent the longest possible polyline.

Judge program compares your answer and jury‘s answer with 10 - 6 precision.

Sample test(s)

input

1 1

output

input

0 10

output

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说的是给你一个矩形区域，让你画三条线，怎样画使得线最长

根据n和m的大小，有四种画法，一是X型，一是斜对角的Z型

#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    if(n==0)
    {
        printf("0 1\n0 %d\n0 0\n0 %d\n",m,m-1);
    }
    else if(m==0)
    {
        printf("1 0\n%d 0\n0 0\n%d 0\n",n,n-1);
    }
    else if(m>=n)
    {
        if(2*sqrt(m*m+n*n)+m<sqrt(m*m+n*n)+2*sqrt(m*m+(n-1)*(n-1))) printf("1 0\n%d %d\n0 0\n%d %d\n",n,m,n-1,m);
        else printf("0 0\n%d %d\n%d 0\n0 %d\n",n,m,n,m);
    }
    else
    {
        if(2*sqrt(m*m+n*n)+n<sqrt(m*m+n*n)+2*sqrt((m-1)*(m-1)+n*n)) printf("0 1\n%d %d\n0 0\n%d %d\n",n,m,n,m-1);
        else printf("0 0\n%d %d\n0 %d\n%d 0\n",n,m,m,n);
    }
    return 0;
}