CodeForces - 361A-Levko and Table （思维）

最新推荐文章于 2025-08-14 22:07:01 发布

转载最新推荐文章于 2025-08-14 22:07:01 发布 · 152 阅读

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原文链接：http://www.cnblogs.com/Staceyacm/p/10781879.html

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#数据结构与算法

本文介绍了一种简单有效的算法，用于解决寻找特定条件下的美丽矩阵问题。美丽矩阵是指矩阵中每一行和每一列的元素之和等于给定值k的矩阵。文章提供了一个示例代码，通过将对角线元素设置为k，其余位置为0，来快速生成符合条件的美丽矩阵。

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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;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781879.html