CodeForces - 361A-Levko and Table （思维）

最新推荐文章于 2019-07-04 23:36:00 发布

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本文探讨了如何构造一个n×n的矩阵，使得每行和每列的元素之和等于给定值k。通过设置对角线元素为k，其余元素为0，成功生成了一个满足条件的美丽矩阵。

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Levko loves tables that consist of n rows and n columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals k.

Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them.

Input

The single line contains two integers, n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000).

Output

Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.

If there are multiple suitable tables, you are allowed to print any of them.

Examples

Input

2 4

Output

1 3
3 1

Input

4 7

Output

Note

In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.

In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements.

题解：这道题我最初的想法是用深搜来做，但是感觉放在第一题的位置肯定大材小用，然后不难发现只要输出任意符合要求的一组即可，只需要对角线是k即可

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;
int a[105][105];
int main() {
	int n,k;
	cin>>n>>k;
	for(int t=0; t<n; t++) {
		for(int j=0; j<n; j++) {
			if(t==j) {
				a[t][j]=k;
			}
		}
	}
	for(int t=0; t<n; t++) {
		for(int j=0; j<n; j++) {
			cout<<a[t][j]<<" ";
		}
		cout<<endl;

	}
	return 0;
}