codeforces-232【A贪心、思维、图】

题目链接：​​点击打开链接​​

A. Cycles

time limit per test

memory limit per test

input

output

kcycles of length 3.

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.

100, or else John will have problems painting it.

Input

k (1 ≤ k ≤ 105) — the number of cycles of length 3

Output

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.

Examples

input

1

output

3 011 101 110

input

10

output

5 01111 10111 11011 11101 11110

大意：要构成一个存在 k 个三元环的图，需要多少个点，输出顶点数 n，输出图。

思路：把每个点加入图中，找出三元环的数量，满足条件跳出循环。刚开始以为是组合数啊，直接 C( n，3 ) 啊，过了样例，就交一发，果断jj，zz。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
bool map[101][101];
int main()
{
  while(~scanf("%d",&n))
  {
    memset(map,0,sizeof(map));
    map[1][2]=map[2][1]=1; // 要构成三元环至少要有一条边 
    int ans;
    for(int i=3;i<=100;i++) // 题中说点最多为 100个 
    {  // 每次加入一个点 i 
      ans=i;
      for(int j=1;j<i;j++)
      {
        int cnt=0;
        for(int k=1;k<j;k++)
        {
          if(map[k][j]&&map[k][i])
            cnt++;
        }
        if(n>=cnt)
        {
          n-=cnt;
          map[i][j]=map[j][i]=1;
        }
        if(n==0)  break; // 满足条件 
      }
      if(n==0)  break;
    }
    printf("%d\n",ans);
    for(int i=1;i<=ans;i++)
    {
      for(int j=1;j<=ans;j++)
        printf("%d",map[i][j]);
      puts("");
    }
  }
  return 0;
}