让你构造一棵树(给定了总结点数和总的叶子数),使得直径最小。

就先弄个菊花图(周围一圈叶子,中间一个点),然后平均地往那一圈放其他的点即可。

#include<cstdio>
using namespace std;
int n,K,Ks[200010],x[200010],y[200010],ans,e;
int main(){
scanf("%d%d",&n,&K);
for(int i=1;i<=K;++i){
Ks[i]=1;
}
int now=2;
for(;;){
for(int i=1;i<=K;++i,++now){
if(now>n){
goto OUT;
}
x[++e]=Ks[i];
y[e]=now;
Ks[i]=now;
if(i==1){
++ans;
}
else if(i==2){
++ans;
}
}
}
OUT:
printf("%d\n",ans);
for(int i=1;i<=e;++i){
printf("%d %d\n",x[i],y[i]);
}
return 0;
}

【构造】Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals) B. High Load的更多相关文章

  1. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) D. High Load 构造

    D. High Load 题目连接: http://codeforces.com/contest/828/problem/D Description Arkady needs your help ag ...

  2. Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals)

    Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals) A.String Reconstruction B. High Load C ...

  3. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) E. DNA Evolution 树状数组

    E. DNA Evolution 题目连接: http://codeforces.com/contest/828/problem/E Description Everyone knows that D ...

  4. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 828E) - 分块

    Everyone knows that DNA strands consist of nucleotides. There are four types of nucleotides: "A ...

  5. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) D 构造烟花

    D. High Load time limit per test 2 seconds memory limit per test 512 megabytes input standard input ...

  6. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals)

    题目链接:http://codeforces.com/contest/828 A. Restaurant Tables time limit per test 1 second memory limi ...

  7. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) C. String Reconstruction 并查集

    C. String Reconstruction 题目连接: http://codeforces.com/contest/828/problem/C Description Ivan had stri ...

  8. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) A,B,C

    A.题目链接:http://codeforces.com/contest/828/problem/A 解题思路: 直接暴力模拟 #include<bits/stdc++.h> using ...

  9. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem D (Codeforces 828D) - 贪心

    Arkady needs your help again! This time he decided to build his own high-speed Internet exchange poi ...

随机推荐

  1. 【ALB学习笔记】基于多线程方式的串行通信接口数据接收案例

    基于多线程方式的串行通信接口数据接收案例 广东职业技术技术学院  欧浩源 1.案例背景 在本博客的<[CC2530入门教程-06]CC2530的ADC工作原理与应用>中实现了电压数据采集的 ...

  2. fileinput 小计(显示历史上传图片)

    今天又需要,要求在选中某条记录后显示历史上传图片 上传控件是fileinput.js 想法:界面有上传图片的控件,重新加载控件,并加入历史上传图片地址 实现代码: var filepathArray ...

  3. spring-retry 重试机制

    业务场景 应用中需要实现一个功能: 需要将数据上传到远程存储服务,同时在返回处理成功情况下做其他操作.这个功能不复杂,分为两个步骤:第一步调用远程的Rest服务逻辑包装给处理方法返回处理结果:第二步拿 ...

  4. mysql in/no in/like

    % 任意字符 _ 任意一个字符 in (value,......) 在这里 not in (value,......) 不在这里 mysql> select 'a' not in (1,2,3, ...

  5. Mojo_1_第一个简单例子

    use Mojolicious::Lite; #根目录,Get方法打开 #正接显示文本text get '/' => sub{ my $service = shift; $service-> ...

  6. CTF线下赛AWD模式下的生存技巧

    作者:Veneno@Nu1L 稿费:200RMB 投稿方式:发送邮件至linwei#360.cn,或登陆网页版在线投稿 原文:https://www.anquanke.com/post/id/8467 ...

  7. python基础===单元测试unittest

    ''' 编写一个名为Employee 的类,其方法__init__()接受名.姓和年薪,并 将它们都存储在属性中.编写一个名为give_raise()的方法,它默认将年薪增加5000 美元,但也能够接 ...

  8. 可以高度定制的代理服务器anyproxy

    简介 anyproxy是一款可以高度定制的代理服务器,基于nodejs. 特征 支持https明文代理 支持低网速模拟 支持二次开发,可以用javascript控制代理的全部流程,搭建前端个性化调试环 ...

  9. [ python ] 类的组合

    首先,使用面向对象是一个人狗大战的实例: class Person: def __init__(self, name, hp, aggr, sex): self.name = name self.hp ...

  10. [PAT] 1143 Lowest Common Ancestor(30 分)1145 Hashing - Average Search Time(25 分)

    1145 Hashing - Average Search Time(25 分)The task of this problem is simple: insert a sequence of dis ...