Codeforces 417 C
| Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
One day, at the "Russian Code Cup" event it was decided to play football as an out of competition event. All participants was divided into nteams and played several matches, two teams could not play against each other more than once.
The appointed Judge was the most experienced member — Pavel. But since he was the wisest of all, he soon got bored of the game and fell asleep. Waking up, he discovered that the tournament is over and the teams want to know the results of all the matches.
Pavel didn't want anyone to discover about him sleeping and not keeping an eye on the results, so he decided to recover the results of all games. To do this, he asked all the teams and learned that the real winner was friendship, that is, each team beat the other teams exactly ktimes. Help Pavel come up with chronology of the tournir that meets all the conditions, or otherwise report that there is no such table.
Input
The first line contains two integers — n and k (1 ≤ n, k ≤ 1000).
Output
In the first line print an integer m — number of the played games. The following m lines should contain the information about all the matches, one match per line. The i-th line should contain two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi). The numbers ai and bi mean, that in the i-th match the team with number ai won against the team with number bi. You can assume, that the teams are numbered from 1 ton.
If a tournir that meets the conditions of the problem does not exist, then print -1.
Sample Input
3 1
3
1 2
2 3
3 1
Source
#include <iostream>
#include <stdio.h>
typedef long long LL;
using namespace std;
int main()
{
int i,n,m,k,j;
cin>>n>>k;
if (n-<*k)
cout<<-<<endl;
else
{
cout<<n*k<<endl;
for (i=;i<=n;i++)
{
for (j=;j<k;j++)
printf("%d %d\n",i,(i+j)%n+);
}
}
return ;
}
Codeforces 417 C的更多相关文章
- Codeforces 417 D. Cunning Gena
按monitor排序,然后状压DP... . D. Cunning Gena time limit per test 1 second memory limit per test 256 megaby ...
- Codeforces Round #417 (Div. 2) D. Sagheer and Kindergarten(树中判祖先)
http://codeforces.com/contest/812/problem/D 题意: 现在有n个孩子,m个玩具,每次输入x y,表示x孩子想要y玩具,如果y玩具没人玩,那么x就可以去玩,如果 ...
- Codeforces Round #417 (Div. 2) B. Sagheer, the Hausmeister
http://codeforces.com/contest/812/problem/B 题意: 有n层楼,每层楼有m个房间,1表示灯开着,0表示灯关了.最两侧的是楼梯. 现在每从一个房间移动到另一个房 ...
- Codeforces Round #417 (Div. 2) B. Sagheer, the Hausmeister —— DP
题目链接:http://codeforces.com/problemset/problem/812/B B. Sagheer, the Hausmeister time limit per test ...
- [Codeforces Round#417 Div.2]
来自FallDream的博客,未经允许,请勿转载,谢谢. 有毒的一场div2 找了个1300的小号,结果B题题目看错没交 D题题目剧毒 E题差了10秒钟没交上去. 233 ------- A.Sag ...
- Codeforces Round #417 (Div. 2)-A. Sagheer and Crossroad
[题意概述] 在一个十字路口 ,给定红绿灯的情况, 按逆时针方向一次给出各个路口的左转,直行,右转,以及行人车道,判断汽车是否有可能撞到行人 [题目分析] 需要在逻辑上清晰,只需要把所有情况列出来即可 ...
- Codeforces Round #417 (Div. 2) C. Sagheer and Nubian Market
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...
- Codeforces Round #417 C. Sagheer and Nubian Market
C. Sagheer and Nubian Market time limit per test 2 seconds memory limit per test 256 megabytes O ...
- Codeforces Round #417 B. Sagheer, the Hausmeister
B. Sagheer, the Hausmeister time limit per test 1 second memory limit per test 256 megabytes Som ...
随机推荐
- 高斯混合聚类及EM实现
一.引言 我们谈到了用 k-means 进行聚类的方法,这次我们来说一下另一个很流行的算法:Gaussian Mixture Model (GMM).事实上,GMM 和 k-means 很像,不过 G ...
- Erlang第二课 ---- bit串
Erlang是被设计来用在电信设备中的,这意味着需要处理大量的二进制数据.也正因为如此,Erlang把binary和binary string提升到了一个相当高的位置,提供了极为丰富的操作机制.当然, ...
- 网页设计师常用的PHOTOSHOP插件
Photoshop是网页设计师常用的一个非常重要而强大的工具,可以让网页设计师的工作高效便捷的进行,也为设计师们的天马行空提供了实际技术实现.一般我们的网页设计师设计完成后,需要将其转换制作成网页形式 ...
- 中国天气网-天气预报接口api
中国天气网地址:http://www.weather.com.cn 请求服务 : 查询实时天气信息 http://www.weather.com.cn/data/sk/101110101.html 在 ...
- linux学习之用户管理
用户管理是在root用户下进行相关操作的 1.配置文件路径: 保存用户信息的文件:/etc/passwd 保存密码的文件:/etc/shadow 保存用 ...
- mysql 为字段增加主键
alter table 表名 add primary key(`字段`)
- 编译过程中,termcap.h 文件找不到路径 licli.a终于生成
编译过程中,termcap.h 文件找不到路径 查看是linux 源码下找不到termcap.h文件 安装了所有关于*cap*的源码包也不起作用 今天终于解决了这个问题,搜 ...
- shell 删除文件下的* (copy).jpg备份文件
shell编程中, 在for, while循环中为什么不用(), {} 不是没有; 而是因为(), {}做了其他用途: (): 执行命令组, 注意这个命令组是新开一个子shell中执行, 因此,括号 ...
- linux命令别名的使用
语 法:alias[别名]=[指令名称] 1,查看该用户下的别名: alias 2,有的系统里没有ll这个命令,原因就是没有定义ll='ls -l --color=tty'这个别名 如果想永久生效,就 ...
- Hadoop之 hdfs 系统
一.NameNode维护着2张表: 1.文件系统的目录结构,以及元数据信息 2.文件与数据块列表的对应关系 存放在fsimage中,在运行的时候加载到内存中的. 操作日志写到edits中 二.Da ...