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

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本文介绍了一种构造简单无向图的方法，确保从指定起点到终点的最短路径数量为预设值K，适用于编程竞赛题目设计。通过将K转换为二进制并逐一构建子图来实现目标路径数。

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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 G_ij 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: G_ii = 'N' and G_ij = G_ji 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

Output

4
NNYY
NNYY
YYNN
YYNN

Input

Output

8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN

Input

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）：

3、那么按照上述过程写出来代码即可、

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int map[1005][1005];
int a[50];
int main()
{
    int k;
    while(~scanf("%d",&k))
    {
        int cont=0;
        while(k)
        {
            a[cont++]=k%2;
            k/=2;
        }
        int now=3;
        map[1][3]=map[3][1]=1;
        for(int i=0;i<cont-1;i++)
        {
            map[now][now+1]=1;
            map[now+1][now]=1;
            map[now][now+2]=1;
            map[now+2][now]=1;
            map[now+1][now+3]=1;
            map[now+3][now+1]=1;
            map[now+2][now+3]=1;
            map[now+3][now+2]=1;
            now+=3;
        }
        int tmp=now+1;
        map[now][2]=1;
        map[2][now]=1;
        for(int i=0;i<(cont-2)*2;i++)
        {
            map[tmp][tmp+1]=1;
            map[tmp+1][tmp]=1;
            tmp++;
        }
        map[tmp][now]=1;
        map[now][tmp]=1;
        int pos=now+1;
        now=tmp;
        int contz=1;
        for(int i=0;i<cont;i++)
        {
            if(a[i]==1)
            {
                map[contz*3][pos]=1;
                map[pos][contz*3]=1;
            }
            contz++;
            pos+=2;
        }
        printf("%d\n",now);
        for(int i=1;i<=now;i++)
        {
            for(int j=1;j<=now;j++)
            {
                if(map[i][j]==1)
                {
                    printf("Y");
                }
                else printf("N");
            }
            printf("\n");
        }
    }
}