Educational Codeforces Round 20 a Maximal Binary Matrix

最新推荐文章于 2018-09-20 23:35:44 发布

原创最新推荐文章于 2018-09-20 23:35:44 发布 · 191 阅读

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本文解析了CodeForces竞赛中的一道A级题目：如何在一个n阶矩阵中填入k个1，使得矩阵对称且字典序最大。通过判断k与n²的关系并采用特定填充策略来解决该问题。

题目链接:http://codeforces.com/contest/803/problem/A

You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.

One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.

If there exists no such matrix then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10⁶).

Output

If the answer exists then output resulting matrix. Otherwise output -1.

Examples

Input

Copy

2 1

Output

Copy

1 0 
0 0

Input

Copy

3 2

Output

Copy

1 0 0 
0 1 0 
0 0 0

Input

Copy

2 5

Output

Copy

-1

题面大意:给你一个n阶矩阵，初始化为0，然后让你填入k个1.要求填充之后的矩阵对称，并且从上到下，从左到右的字典序是最大的

思路:

开始判断k与n^2的关系。大于n^2，输出-1.否则二重循环扫描，每到一个点，判断是否为对角线上的点，如果是，那么填充计数器cnt+1,不过不是cnt+2,同时判断此时cnt与k的大小，比k大。那就将下一行的对角线上的元素变为1.然后输出矩阵

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
int g[105][105];
int main()
{
    int n,k;
   cin>>n>>k;
   int cnt=1;
   if(k>n*n)
    cout<<-1<<endl;
    else
    {
       for(int i=0;i<n;i++)
       {
           for(int j=i;j<n;j++)
           {
               if(cnt>k)
               {
                   break;
               }
               else
               {
                   if(i==j)
                  {

                     g[i][j]=1;
                     cnt++;
                  }
                  else
                  {
                     // cout<<cnt<<endl;
                   if(cnt+1<=k)
                   {
                       g[i][j]=1;
                       g[j][i]=1;
                       cnt+=2;
                   }
                   else
                   {
                       g[i+1][i+1]=1;
                       cnt++;
                   }
                  }
               }

           }
       }
       for(int i=0;i<n;i++)
       {
           for(int j=0;j<n;j++)
           {
               cout<<g[i][j]<<' ';
           }
           cout<<endl;
       }
    }
    return 0;
}