334A Candy Bags

最新推荐文章于 2024-01-16 10:43:30 发布

转载最新推荐文章于 2024-01-16 10:43:30 发布 · 190 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zmin/p/6881971.html

本文解析了一个关于糖果均匀分配的问题，通过编程找出规律并给出解决方案。使用C语言实现了一个程序，该程序可以为每个兄弟分配数量相等的糖果，确保所有兄弟获得的糖果总数相同。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Gerald has n younger brothers and their number happens to be even. One day he bought n² candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer k from 1 to n² he has exactly one bag with k candies.

Help him give n bags of candies to each brother so that all brothers got the same number of candies.

Input

The single line contains a single integer n (n is even, 2 ≤ n ≤ 100) — the number of Gerald's brothers.

Output

Let's assume that Gerald indexes his brothers with numbers from 1 to n. You need to print n lines, on the i-th line print n integers — the numbers of candies in the bags for the i-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to n². You can print the numbers in the lines in any order.

It is guaranteed that the solution exists at the given limits.

Examples

Input

Output

1 4
2 3

Note

The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.

这题很惭愧，看了一个小时没有看出规律T T。

问了朋友，恍然大悟。

这种找规律的题我真的真的找不出来啊太笨了。

本题的规律就是：对角线上的和。

如下图：

n=3;

n=4;

图片来源：http://liyishuai.blog.ustc.edu.cn/candy-bags/

代码如下：

 1 #include<cstdio>
 2 int main()
 3 {
 4     int n,i=1;
 5     scanf("%d",&n);
 6     for(int j=1;j<=n;j++)
 7     {
 8         i=j;
 9         while(i<=n*n)
10         {
11             printf("%d ",i);
12             if(i%n==0)       //在边界上 那下一步就要拐到下一行第一个 
13             i+=1;
14             else
15             i+=n+1;            //不在边界就斜着走到下一行
16         }
17         printf("\n");
18     }
19     return 0;
20 }