Codeforces B. Fox and Minimal path

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本文介绍了一种算法，用于构造一个简单无向图，该图包含指定数量的从顶点1到顶点2的最短路径。通过将输入值k转换为二进制并构建分层图，确保了输出图中精确的最短路径数量。

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将k换成2进制，构造分层的图

Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."

Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?

Input

The first line contains a single integer k (1 ≤ k ≤ 109).

Output

You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.

The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 ton.

The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.

Sample test(s)

input
2

output
4
NNYY
NNYY
YYNN
YYNN

input
9

output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN

input
1

output
2
NY
YN

Note

In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.

In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.

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

using namespace std;

int K,Bit[33],bi;

void getbit(int x)
{
	bi=0;
	while(x)
	{
		if(x&1) Bit[bi]++;
		x>>=1;bi++;
	}
}

int G[1200][1200],st=41;

void Add_line(int a,int b)
{
	G[a][b]=G[b][a]=true;
}

void doit(int& a1,int& b1,int& a2,int& b2,int& st,int n)
{
	for(int i=0;i<n;i++)
	{
		if(i==0) a1=st+1,b1=st;
		if(i==n-1) a2=st+1,b2=st;
	
		if(i+1<n)
		{
			Add_line(st,st+3);
			Add_line(st,st+2);
			Add_line(st+1,st+2);
			Add_line(st+1,st+3);
		}
		st+=2;
	}
}

int main()
{
	scanf("%d",&K);
	K--;
	getbit(K);
	//主线 1--3--4---...---40---2
	for(int i=4;i<=40;i++) Add_line(i,i-1);
	Add_line(1,3);Add_line(40,2);

	//支线
	for(int i=bi-1;i>=0;i--)
	{
		if(Bit[i])
		{
			if(i)
			{
				int a1,b1,a2,b2;
				doit(a1,b1,a2,b2,st,i);
				Add_line(1,a1);Add_line(1,b1);
				Add_line(3+i,a2);Add_line(3+i,b2);
			}
			else
			{
				Add_line(3,st);Add_line(5,st);
			}
		}
	}

	printf("1000\n");
	for(int i=1;i<=1000;i++)
	{
		for(int j=1;j<=1000;j++)
		{
			if(G[i][j]) putchar('Y');
			else putchar('N');
		}
		putchar(10);
	}

	return 0;
}