CodeForces 803A Maximal Binary Matrix-【思维+构造】

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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 putk 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：给一个n*n的矩阵，在矩阵中加入k个1，使矩阵为对称矩阵，并且从上到下从左到右都是最大的

解题：直接构造矩阵  但要注意第二个for要从 i 开始

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){int n,k;int a[110][110];while(scanf("%d%d",&n,&k)!=EOF){memset(a,0,sizeof(a));if(n*n < k){printf("-1\n");continue;}for(int i = 1; i <= n; i++){for(int j = i; j <= n; j++)//j从i 开始  {if(k == 0)break;if(i == j){a[i][j]=1;//对角线 从上往下 k--;}else{if(k == 1)//当 k=1 时 在对角线两边不能形成对称结构 continue;//不是break；当为1时 可以排在对角线上 a[i][j]=a[j][i]=1;//对角线两边；k-=2; }if(k == 0)break;}if(k == 0)break;}for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++)printf("%d%c",a[i][j],j == n ?'\n':' '); }}return 0;}