codeforces 232A Cycles (构建图，贪心+模拟）

                                                  Cycles

Description

描述

John Doe started thinking about graphs. After some thought he decided that he wants to paint an undirected graph, containing exactly kcycles of length 3.

A cycle of length 3 is an unordered group of three distinct graph vertices a, b and c, such that each pair of them is connected by a graph edge.

John has been painting for long, but he has not been a success. Help him find such graph. Note that the number of vertices there shouldn't exceed 100, or else John will have problems painting it.

Input

输入

A single line contains an integer k (1 ≤ k ≤ 105) — the number of cycles of length 3 in the required graph.

Output

输出

In the first line print integer n (3 ≤ n ≤ 100) — the number of vertices in the found graph. In each of next n lines print n characters "0" and "1": the i-th character of the j-th line should equal "0", if vertices i and j do not have an edge between them, otherwise it should equal "1". Note that as the required graph is undirected, the i-th character of the j-th line must equal the j-th character of the i-th line. The graph shouldn't contain self-loops, so the i-th character of the i-th line must equal "0" for all i.

Sample Input

Input

1

Output

3011101110

Input

10

Output

50111110111110111110111110

贪心手法。点最多一百个，那么，要让每个点的作用都发挥出来。所有，可以模拟每次加入点所带来的影响，来确定，两个点是否要连接。

#include <iostream>#include<cstdio>using namespace std;int ans,k,ss;int gra[103][103];int main(){    int n,i,j,k;    scanf("%d",&n);    gra[1][2]=gra[2][1]=1;//最少存在一条边    for(i=3;i<=100;i++)//依次增加顶点    {        for(j=1;j<i;j++)        {            ss=0;//三元环个数            for(k=1;k<j;k++)             if(gra[k][j]&&gra[k][i]) ss++;            if(n>=ss)            {                n-=ss;                gra[i][j]=gra[j][i]=1;            }            if(n==0)break;        }        if(n==0)break;    }       n=i;       printf("%d\n",n);       for(i=1;i<=n;i++)       {           for(j=1;j<=n;j++)           printf("%d",gra[i][j]);           printf("\n");       }       return 0;}