Codeforces-C-Grid game(思维)

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本文介绍了一款基于4x4网格的游戏，玩家需按序列放置2x1或1x2的瓷砖，目标是在不交叉的情况下放置所有瓷砖并删除满行或满列。通过特定策略，文章提供了一个示例解决方案，展示了如何避免交叉并完成游戏。

You are given a 4x4 grid. You play a game — there is a sequence of tiles, each of them is either 2x1 or 1x2. Your task is to consequently place all tiles from the given sequence in the grid. When tile is placed, each cell which is located in fully occupied row or column is deleted (cells are deleted at the same time independently). You can place tile in the grid at any position, the only condition is that tiles (and tile parts) should not overlap. Your goal is to proceed all given figures and avoid crossing at any time.

Input

The only line contains a string ss consisting of zeroes and ones (1≤|s|≤10001≤|s|≤1000). Zero describes vertical tile, one describes horizontal tile.

Output

Output |s||s| lines — for each tile you should output two positive integers r,cr,c, not exceeding 44, representing numbers of smallest row and column intersecting with it.

If there exist multiple solutions, print any of them.

Example

input

Copy

output

Copy

1 1
1 2
1 4

Note

Following image illustrates the example after placing all three tiles:

Then the first row is deleted:

思路：我们可以建立两个数组，如果是0我们就把它放在第一列的两个位置，满了就消去，如果是1我们就放在最右边两列，满了就消去

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;
string str;
int main()
{
	int a[2][2]={1,1,3,1};
	int b[4][2]={1,3,2,3,3,3,4,3};
	cin>>str;
	int len=str.length();
	int s1=0;
	int s2=0;
	for(int t=0;t<len;t++)
	{
		if(str[t]=='0')
		{
			cout<<a[s1][0]<<" "<<a[s1][1]<<endl;
			s1++;
		}
	    if(str[t]=='1')
		{
			cout<<b[s2][0]<<" "<<b[s2][1]<<endl;
			s2++;
		}
		if(s1==2)
		{
			s1=0;
		}
		if(s2==4)
		{
			s2=0;
		}
	}
	
	
	return 0;
}