B. Fox and Minimal path
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

整数N拆写成二进制。

如 15 = 2^3 + 2^2 + 2^1 +2^0 ;

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <vector>
#include <set>
#include <math.h>
#include <map>
using namespace std;
typedef long long LL ; int pow2[] ;
vector<int> ans ;
vector<int> ::iterator it ;
map<int ,int>my_hash ;
int shortest ;
int id ;
const int S = ;
const int T = ;
bool grid[][] ; void make_line(int dot , int type){
vector < pair<int ,int> > me ;
vector < pair<int ,int> > ::iterator it ;
me.clear() ;
for(int i = ; i <= dot ; i++){
me.push_back(make_pair(id,id+)) ;
id += ;
}
if(type == && dot != shortest){
my_hash.clear() ;
for(int i = dot + ; i <= shortest ; i++){
me.push_back(make_pair(id,id)) ;
my_hash[i] = id ;
id += ;
}
}
it = me.begin() ;
grid[S][it->first] = grid[S][it->second] = ;
grid[it->first][S] = grid[it->second][S] = ;
if(type == && dot != shortest){
it = me.end() - ;
grid[it->first][T] = grid[it->second][T] = ;
grid[T][it->first] = grid[T][it->second] = ;
}
else {
it = me.end() - ;
int v = (dot == shortest ? T : my_hash[dot+]) ;
grid[it->first][v] = grid[it->second][v] = ;
grid[v][it->first] = grid[v][it->second] = ;
}
for(int i = ; i < me.size() ; i++){
int u1 = me[i-].first ;
int u2 = me[i-].second ;
int v1 = me[i].first ;
int v2 = me[i].second ;
grid[u1][v1] = grid[u1][v2] = ;
grid[u2][v1] = grid[u2][v2] = ; grid[v1][u1] = grid[v1][u2] = ;
grid[v2][u1] = grid[v2][u2] = ;
}
} void output(){
int n = id - , i , j ;
cout<<n<<endl ;
for(i = ; i <= n ; i++){
for(j = ; j <= n ; j++)
printf("%s",grid[i][j]?"Y":"N") ;
puts("") ;
}
} int main(){
int i , k ,sum = ;
pow2[] = ;
for(i = ; i <= ; i++)
pow2[i] = pow2[i-] * ;
cin>>k ;
if(k == ){
puts("") ;
puts("NY") ;
puts("YN") ;
return ;
}
ans.clear() ;
for(i = ; i >= ; i--){
if(k >= pow2[i]){
ans.push_back(i) ;
k -= pow2[i] ;
}
}
memset(grid,,sizeof(grid)) ;
shortest = *ans.begin() ;
id = ;
make_line(ans[ans.size()-] ,) ;
for(i = ans.size() - ; i >= ; i--)
make_line(ans[i] , ) ;
output() ;
return ;
}

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