Codeforces 803 A. Maximal Binary Matrix （模拟

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本文探讨了一个编程挑战，即如何在一个n×n的矩阵中放置k个1，使得矩阵关于主对角线对称，并且在字典序上尽可能大。文章提供了一种解决方案，并附带了AC代码。

A. Maximal Binary Matrix

Description

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.

Sample Input

2 1

3 2

Sample Output

1 0 
0 0

1 0 0 
0 1 0 
0 0 0

题意

给您一个矩阵和一个数字k问您是否能完全填充
填充规则按照字典序（从上到下）和中轴线对称的方法

题解：

一直没理解字典序是什么意思后来想想就是按照从上到下的原则填充啊
~~我果然是个脑残~~

AC代码

#include <bits/stdc++.h>
using namespace std;
#define N 1000
int vis[N][N];
int main()
{
    int n, k;
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&k);
    bool flag = true;
    if(k > n*n) flag = false;
    else {
        for(int i = 0;i < n; i++) {
            for(int j = 0;j < n; j++) {
                if(vis[i][j]) continue;
                else if(i==j && k!=0) {
                    vis[i][j] = 1; k--;
                }
                else if (k>=2){
                    vis[i][j] = vis[j][i] = 1;
                    k -= 2;
                }   
            }
        }
        if(k) flag = false;
    }
    if(flag) {
        for(int i = 0;i < n; i++) {
            for(int j = 0;j < n; j++) {
                if(j==0) printf("%d",vis[i][j]);
                else printf(" %d",vis[i][j]);
            }
            printf("\n");
        }
    }
    else puts("-1");
return 0;
}