CF - 803A. Maximal Binary Matrix - 思维+贪心

题目描述： A. Maximal Binary Matrix time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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 题意概述：给你一个n*n的矩阵，让你往里面填1，使得这个矩阵对称，然后输出这个矩阵字典序最大的那个矩阵，如果无法完成就输出-1，矩阵字典序就是每行都进行字典序比较 解题思路：带点贪心的构造题，首先k>n*N肯定是输出-1，剩下的，由于是要输出字典序最大的，那么肯定是尽可能把每一行都填满，但是要对称着填，对角线上消耗一个1，其余地方消耗两个1AC代码： #include <bits/stdc++.h> #define INF 0x3f3f3f3f #define maxn 100100 #define lson root << 1 #define rson root << 1 | 1 #define lent (t[root].r - t[root].l + 1) #define lenl (t[lson].r - t[lson].l + 1) #define lenr (t[rson].r - t[rson].l + 1) #define N 101 #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) using namespace std; const int mod = 1e9 + 7; typedef long long ll; typedef unsigned long long ull; int ma[N][N]; int n, k; void chan(int temp, int num) { if (k - num == 2) { ma[temp][temp] = 1; ma[temp + 1][temp + 1] = 1; return; } for (int i = temp; i <= n; i++) { ma[temp][i] = 1; ma[i][temp] = 1; num += 2; if (i == temp)num--; if (num == k)return; if (k - num == 1) { ma[temp + 1][temp + 1] = 1; return; } } return chan(temp + 1, num); } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif scanf("%d%d", &n, &k); if (k>n*n) { printf("-1"); return 0; } if (k != 0)chan(1, 0); for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { printf("%d ", ma[i][j]); } printf("\n"); } #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms.", _end_time - _begin_time); #endif return 0; }