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.

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:问如何将二进制矩阵排成字典序最大(需要主对角线对称)
解法:当然是这种啦,如果发现还有剩余1就输出-1
1 1 1 1 1 1 1 1 1...
1 1
1 1
1 .
1 .
 #include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=;
int x[][];
int n;
int num;
int main()
{
cin>>n>>num;
for(int i=;num>&&i<=n;i++)
{
x[i][i]=;
num--;
for(int j=i+;num>&&j<=n;j++)
{
x[i][j]=x[j][i]=;
num-=;
}
}
if(num>)
{
cout<<"-1"<<endl;
return ;
}
for(int i=;i<=n;i++)
{
for(int j=;j<=n;j++)
{
cout<<x[i][j]<<" ";
}
cout<<endl;
}
return ;
}

Educational Codeforces Round 20 A的更多相关文章

  1. Educational Codeforces Round 20

    Educational Codeforces Round 20  A. Maximal Binary Matrix 直接从上到下从左到右填,注意只剩一个要填的位置的情况 view code //#pr ...

  2. Educational Codeforces Round 20 D. Magazine Ad

    The main city magazine offers its readers an opportunity to publish their ads. The format of the ad ...

  3. Educational Codeforces Round 20 C(math)

    題目鏈接: http://codeforces.com/problemset/problem/803/C 題意: 給出兩個數n, k, 將n拆分成k個數的和,要求這k個數是嚴格遞增的,並且這k個數的g ...

  4. Educational Codeforces Round 20.C

    C. Maximal GCD time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  5. Educational Codeforces Round 20 C 数学/贪心/构造

    C. Maximal GCD time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  6. Educational Codeforces Round 20 C. Maximal GCD

    C. Maximal GCD time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  7. Educational Codeforces Round 20 B. Distances to Zero

    B. Distances to Zero time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  8. Educational Codeforces Round 20 A. Maximal Binary Matrix

    A. Maximal Binary Matrix time limit per test 1 second memory limit per test 256 megabytes input stan ...

  9. Educational Codeforces Round 20 E - Roma and Poker(dp)

    传送门 题意 Roma在玩一个游戏,一共玩了n局,赢则bourle+1,输则bourle-1,Roma将会在以下情况中退出 1.他赢了k个bourle 2.他输了k个bourle 现在给出一个字符串 ...

  10. Educational Codeforces Round 20 B

    Description You are given the array of integer numbers a0, a1, ..., an - 1. For each element find th ...

随机推荐

  1. HttpClient 认证

    第四章 HTTP认证 HttpClient提供对由HTTP标准规范定义的认证模式的完全支持.HttpClient的认证框架可以扩展支持非标准的认证模式,比如NTLM和SPNEGO. 4.1 用户凭证 ...

  2. win7 64位安装vs2013 出现'System.AccessViolationException的错误

    用管理员身份运行CMD,输入netsh winsock reset并回车(注意,必须是已管理员身份运行,这个重置LSP连接)

  3. JAVA学习之 Model2中的Servlet与.NET一般处理程序傻傻分不清楚

    时隔多日,多日合适吗,应该是时隔多月.我又想起了一般处理程序.这都是由于近期在实现的DRP系统中经经常使用到jsp+servlet达到界面与逻辑的分离.servlet负责处理从jsp传回的信息:每当这 ...

  4. poj3349(hash or violence)

    Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 38600   Accep ...

  5. 【bzoj4554】[Tjoi2016&Heoi2016]游戏

    现在问题有硬石头和软石头的限制 所以要对地图进行预处理 分行做,把有#隔开的*(x)形成联通块的存储下来. 分列作,把有#隔开的*(x)形成联通块的存储下来. 求出所有的行联通个数和列联通个数 作为二 ...

  6. Oracle Exception

    Oracle存储过程的异常处理 1.为了提高存储过程的健壮性,避免运行错误,当建立存储过程时应包含异常处理部分.2.异常(EXCEPTION)是一种PL/SQL标识符,包括预定义异常.非预定义异常和自 ...

  7. bzoj3137: [Baltic2013]tracks

    炸一看好像很神仙的样子,其实就是个sb题 万年不见的1A 但是我们可以反过来想,先选一个起点到终点的联通块,然后这联通块后面相当于就能够走了,继续找联通块 然后就能发现直接相邻的脚步相同的边权为0,否 ...

  8. leelazero and google colab

    https://github.com/gcp/leela-zero/blob/master/COLAB.md 左侧菜单展开,可以查看细节

  9. I.MX6 android 4.2 源码下载

    /************************************************************************* * I.MX6 android 4.2 源码下载 ...

  10. CodeForces19D:Points(线段树+set(动态查找每个点右上方的点))

    Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coo ...