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

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 withn 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 tok?

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 exactlyk shortest paths between vertex 1 and vertex 2 of the graph.

The first line must contain an integer n. Then adjacency matrixG with n rows andn columns must follow. Each element of the matrix must be 'N' or 'Y'. IfG_ij is 'Y', then graphG 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: G_ii = 'N' andG_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

2

Output

4NNYYNNYYYYNNYYNN

Input

9

Output

8NNYYYNNNNNNNNYYYYNNNNYYYYNNNNYYYYNNNNYYYNYYYYNNNNYYYYNNNNYYYYNNN

Input

1

Output

2NYYN

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");        }    }}