Codeforces B. Fox and Minimal path-布布扣-bubuko.com

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; }