【构造】Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals) B. High Load
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- No 'Access-Control-Allow-Origin' Ajax跨域访问解决方案
No 'Access-Control-Allow-Origin' header is present on the requested resource. 当使用ajax访问远程服务器时,请求失败,浏 ...
- C# IEqualityComparer 使用方法 Linq Distinct使用方法
创建 IEqualityComparer的接口类必须实现Equals和GetHashCode方法 public class TipComparer : IEqualityComparer<Tip ...
- HTML JS文字闪烁实现(项目top.htm分析)
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <!-- saved from ur ...
- Linux 入门记录:十三、Linux 扩展权限
一.默认权限 每一个终端都有一个 umask 属性,是用来确定新建文件或目录的默认权限的“掩码”(mask 有“掩码”的含义,至于 u,后面说). Linux 中一般有默认的权限掩码,使用命令 uma ...
- shell中的IFS和$*变量
本文转载自http://blog.chinaunix.net/uid-22566367-id-381955.html 自我记录内容.在工程中遇到了相关内容的shell脚本.在此处记录 STRING1= ...
- 【hihocoder】sam1-基本概念
这题有毒…… 原本只是想复习下sam,于是写…… 后来发现自己傻了不知道怎么维护endpos…… 一气之下直接kmp拉倒,mdzz UPD:现在我好像会维护endpos了…… #include< ...
- CMDB (后台管理) CURD 插件
查 a. 基本实现 <!DOCTYPE html> <html lang="en"> <head> <meta charset=" ...
- python_day6学习笔记
一.Logger模块 logging.basicConfig函数 可通过具体参数来更改logging模块默认行为,可用参数有 filename: 用指定的文件名创建FiledHandler(后边会具体 ...
- 【hdoj_2079】选课时间(母函数)
题目:http://acm.hdu.edu.cn/showproblem.php?pid=2079 此题采用母函数的知识求解,套用母函数模板即可: http://blog.csdn.net/ten_s ...
- MySQL 的数据存储引擎
MySQL的存储引擎 InnoDB: MySQL5.5之后的默认存储引擎. 采用MVCC来支持高并发,并且实现了四个标准的隔离级别(默认可重复读). 支持事务,支持外键.支持行锁.非锁定读(默认读取操 ...