A. Candy Bags
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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
2
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 }

334A Candy Bags的更多相关文章

  1. codeforces 334A - Candy Bags

    忘了是偶数了,在纸上画奇数画了半天... #include<cstdio> #include<cstring> #include<cstdlib> #include ...

  2. Candy Bags

    读懂了题就会发现这是个超级大水题 Description Gerald has n younger brothers and their number happens to be even. One ...

  3. A. Candy Bags

    A. Candy Bags http://codeforces.com/problemset/problem/334/A   time limit per test 1 second memory l ...

  4. F - Candy Bags

    A. Candy Bags time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  5. codeforces A. Candy Bags 解题报告

    题目链接:http://codeforces.com/contest/334/problem/A 题意:有n个人,将1-n袋(第 i  袋共有 i  颗糖果,1<= i  <=n)所有的糖 ...

  6. Candy Distribution

    Kids like candies, so much that they start beating each other if the candies are not fairly distribu ...

  7. CodeForces Round 194 Div2

    A. Candy Bagstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputs ...

  8. Codeforces Round #194 (Div.1 + Div. 2)

    A. Candy Bags 总糖果数\(\frac{n^2(n^2+1)}{2}\),所以每人的数量为\(\frac{n}{2}(n^2+1)\) \(n\)是偶数. B. Eight Point S ...

  9. [LeetCode] Candy 分糖果问题

    There are N children standing in a line. Each child is assigned a rating value. You are giving candi ...

随机推荐

  1. 关于阿里云服务器安装了Apache开放80端口访问不了网页

    先用netstat -tlunp查看80端口是否打开,再关闭服务器的防火墙,可以用 systemctl status firewalld 查看防火墙状态  systemctl stop firewal ...

  2. 天天用SpringBoot居然还不知道它的自动装配的原理?

    引言 最近有个读者在面试,面试中被问到了这样一个问题"看你项目中用到了springboot,你说下springboot的自动配置是怎么实现的?"这应该是一个springboot里面 ...

  3. click的简单使用

    click的简单使用 先通过一个简单的例子来认知一下click把 import click @click.command() @click.option('-p', '--port', default ...

  4. 不错的网站压力测试工具webbench

    webbench最多可以模拟3万个并发连接去测试网站的负载能力,个人感觉要比Apache自带的ab压力测试工具好,安装使用也特别方便. 1.适用系统:Linux 2.前期准备:yum install ...

  5. jackson学习之四:WRAP_ROOT_VALUE(root对象)

    欢迎访问我的GitHub https://github.com/zq2599/blog_demos 内容:所有原创文章分类汇总及配套源码,涉及Java.Docker.Kubernetes.DevOPS ...

  6. cookie加密 当浏览器全面禁用三方 Cookie

    cookie加密    cookie  localstorage    区别 https://mp.weixin.qq.com/s/vHeRStcCUarwqsY7Y1rpGg 当浏览器全面禁用三方 ...

  7. Beating JSON performance with Protobuf https://auth0.com/blog/beating-json-performance-with-protobuf/

    Beating JSON performance with Protobuf https://auth0.com/blog/beating-json-performance-with-protobuf ...

  8. A1Z26 Cipher - Letter Number A=1 B=2 C=3 - Online Decoder, Translator https://www.dcode.fr/letter-number-cipher

    A1Z26 Cipher - Letter Number A=1 B=2 C=3 - Online Decoder, Translator https://www.dcode.fr/letter-nu ...

  9. 服务降级 托底预案 Nginx中使用Lua脚本检测CPU使用率,当达到阀值时开启限流,让用户排队

    https://mp.weixin.qq.com/s/FZAcQQAKomGEe95kln1HCQ 在京东我们是如何做服务降级的 https://mp.weixin.qq.com/s/FZAcQQAK ...

  10. 扒一扒ELF文件

    ELF文件(Executable Linkable Format)是一种文件存储格式.Linux下的目标文件和可执行文件都按照该格式进行存储,有必要做个总结. 目录 1. 链接举例 2. ELF文件类 ...