CF#803 A. Maximal Binary Matrix（贪心）

最新推荐文章于 2021-07-25 15:34:51 发布

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本篇介绍如何在给定的n*n零矩阵中填充k个1，使之成为关于主对角线对称且字典序最大的矩阵。文章包含输入输出样例及解决思路。

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
2 1

output
1 0 
0 0 

input
3 2

output
1 0 0 
0 1 0 
0 0 0 

input
2 5

output

-1 





题意：给出一个 n * n 的0矩阵，添加 k 个 1使其变为对称矩阵




#include <bits/stdc++.h>
using namespace std;

const int N = 100 + 10;
int a[N][N];

int main()
{
    int n, k;
    while(scanf("%d%d", &n, &k) == 2)
    {
        memset(a, 0, sizeof(a));
        if(n * n < k)
        {
            printf("-1\n");
            continue ;
        }
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(!a[i][j])
                {
                    if(k && i == j) k--, a[i][j] = 1;
                    else if(k >= 2) k -= 2, a[i][j] = a[j][i] = 1;
                }
                printf("%d%c", a[i][j], j < n - 1 ? ' ' : '\n');
            }
        }
    }
}