Codeforces 389D Fox and Minimal path【构造+二进制思维】好题！

转自：http://blog.csdn.net/mengxiang000000/article/details/53072307

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 to n.

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.

Examples

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.

题目大意：

让你构造出来一个图，里边包含N个点，并且保证从1到2的最短路的条数为K个，求一个可行解。

N必须小于等于1000。

思路：

1、一开始的思路是将给出的数K因子分解，但是想到如果K是一个很大的素数，那么结果是不可行的，因为题干要求点的数量小于等于1000.

2、然后考虑，无论一个数多大，都能用二进制数来表示，那么我们考虑将输入进来的K先转成二进制数。

比如21:10101

那么我们考虑将其弄成2^0+2^2+2^4即可，对应我们很容易搞出来2^4的图：



那么那么我们2^0，直接从3连出来一条分路，只要保证从1-3-分路-2的长度和从1到2的最短路长度相同即可：



那么2^2同理，我们从9分离出来一条分路即可（因为从1到9上边的路有4种走法，那么再从9连入20，那么就多出来4种走法）：

!

#include <cstdio>
#include <cstring>
using namespace std;
int f[1005][1005];
int main()
{
int k;
scanf("%d",&k);
int w[39],cnt=0;
while(k)
{
w[cnt++]=k%2;
k/=2;
}
f[1][3]=f[3][1]=1;
int now=3;
for(int i=1; i<cnt; i++)
{
f[now][now+1]=f[now+1][now]=1;
f[now][now+2]=f[now+2][now]=1;
f[now+1][now+3]=f[now+3][now+1]=1;
f[now+2][now+3]=f[now+3][now+2]=1;
now+=3;
}
int tmp=now+1;
f[now][2]=f[2][now]=1;
for(int i=0; i<(cnt-2)*2; i++)
{
f[tmp][tmp+1]=f[tmp+1][tmp]=1;
tmp++;
}
f[tmp][now]=f[now][tmp]=1;
int pos=now+1;
now=tmp;
int t=1;
for(int i=0; i<cnt; i++)
{
if(w[i]==1)
{
f[t*3][pos]=f[pos][t*3]=1;
}
t++;
pos+=2;
}
printf("%d\n",now);
for(int i=1; i<=now; i++)
{
for(int j=1; j<=now; j++)
if(f[i][j])
printf("Y");
else printf("N");
puts("");
}
return 0;
}