D. Tablecity

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/575/problem/D

Description

There was a big bank robbery in Tablecity. In order to catch the thief, the President called none other than Albert – Tablecity’s Chief of Police. Albert does not know where the thief is located, but he does know how he moves.

Tablecity can be represented as 1000 × 2 grid, where every cell represents one district. Each district has its own unique name “(X, Y)”, where X and Y are the coordinates of the district in the grid. The thief’s movement is as

Every hour the thief will leave the district (X, Y) he is currently hiding in, and move to one of the districts: (X - 1, Y), (X + 1, Y),(X - 1, Y - 1), (X - 1, Y + 1), (X + 1, Y - 1), (X + 1, Y + 1) as long as it exists in Tablecity.

Below is an example of thief’s possible movements if he is located in district (7,1):

Albert has enough people so that every hour he can pick any two districts in Tablecity and fully investigate them, making sure that if the thief is located in one of them, he will get caught. Albert promised the President that the thief will be caught in no more than 2015 hours and needs your help in order to achieve that.

Input

There is no input for this problem.

Output

The first line of output contains integer N – duration of police search in hours. Each of the following N lines contains exactly 4 integersXi1, Yi1, Xi2, Yi2 separated by spaces, that represent 2 districts (Xi1, Yi1), (Xi2, Yi2) which got investigated during i-th hour. Output is given in chronological order (i-th line contains districts investigated during i-th hour) and should guarantee that the thief is caught in no more than 2015 hours, regardless of thief’s initial position and movement.

  • N ≤ 2015
  • 1 ≤ X ≤ 1000
  • 1 ≤ Y ≤ 2

Sample Input


Sample Output


HINT

 

题意

有一个1000*2的空间,里面有一个小偷,每次小偷可以往四周走一格,你并不知道小偷在哪儿

现在要求让你安排一种巡查方案,每次巡查两个房间,要求不管小偷采取什么方案,你都能抓住他

题解:

往左扫一遍,然后再往右扫一遍

很经典的一个面试题,大概就是奇偶性的关系吧,保证自己和老鼠会是同奇偶

代码:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
printf("%d\n",);
for(int i=;i<=;i++)
printf("%d %d %d %d\n",i,,i,);
for(int i=;i>=;i--)
printf("%d %d %d %d\n",i,,i,);
return ;
}

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