Codeforces Round #245 (Div. 2) A - Points and Segments (easy)
水到家了
#include <iostream>
#include <vector>
#include <algorithm> using namespace std; struct Point{
int index, pos;
Point(int index_ = , int pos_ = ){
index = index_;
pos = pos_;
} bool operator < (const Point& a) const{
return pos < a.pos;
}
}; int main(){
int n,m, l,r;
cin >> n >> m;
vector<Point> points(n);
for(int i = ; i < n ; ++ i){
cin >> points[i].pos;
points[i].index = i;
}
sort(points.begin(),points.end());
for(int i = ; i < m; ++ i) cin >> l >> r;
vector<int> res(n,);
for(int i = ; i < n ; ++ i ){
if(i% == ) res[points[i].index] =;
}
cout<<res[];
for(int i = ; i <n ; ++ i ) cout<<" "<<res[i];
cout<<endl; }
Codeforces Round #245 (Div. 2) A - Points and Segments (easy)的更多相关文章
- Codeforces Round #245 (Div. 2) A. Points and Segments (easy) 贪心
A. Points and Segments (easy) Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/con ...
- Codeforces Round #501 (Div. 3) 1015A Points in Segments (前缀和)
A. Points in Segments time limit per test 1 second memory limit per test 256 megabytes input standar ...
- Codeforces Round #486 (Div. 3) D. Points and Powers of Two
Codeforces Round #486 (Div. 3) D. Points and Powers of Two 题目连接: http://codeforces.com/group/T0ITBvo ...
- Codeforces Round #245 (Div. 2)
版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/yew1eb/article/details/25609981 A Points and Segmen ...
- Codeforces Round #466 (Div. 2) -A. Points on the line
2018-02-25 http://codeforces.com/contest/940/problem/A A. Points on the line time limit per test 1 s ...
- Codeforces Round #319 (Div. 1) C. Points on Plane 分块
C. Points on Plane Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/576/pro ...
- Codeforces Round #466 (Div. 2) A. Points on the line[数轴上有n个点,问最少去掉多少个点才能使剩下的点的最大距离为不超过k。]
A. Points on the line time limit per test 1 second memory limit per test 256 megabytes input standar ...
- Codeforces Round #245 (Div. 1) 429D - Tricky Function 最近点对
D. Tricky Function Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 codeforces.com/problemset/problem/42 ...
- Codeforces Round #245 (Div. 1) B. Working out (简单DP)
题目链接:http://codeforces.com/problemset/problem/429/B 给你一个矩阵,一个人从(1, 1) ->(n, m),只能向下或者向右: 一个人从(n, ...
随机推荐
- SQL Server占用内存的认识
SQL Server占用的内存主要由三部分组成:数据缓存(Data Buffer).执行缓存(Procedure Cache).以及SQL Server引擎程序.SQL Server引擎程序所占用缓存 ...
- ***PHP中error_reporting()用法详解(含codeigniter框架中屏蔽错误提示的解决方案)
php中我们对错误的处理会常用到error_reporting函数了,大家可以看到最多的是error_reporting(E_ALL ^ E_NOTICE)了,这个到底什么意思呢,下面我来来看看. e ...
- 重温WCF之WCF传输安全(十三)(1)前期准备之证书制作(转)
转载地址:http://www.cnblogs.com/lxblog/archive/2012/09/12/2682372.html 一.WCF中的安全方式 说到安全就会涉及到认证,消息一致性和机密性 ...
- x86架构的android手机兼容性问题
x86架构的android手机兼容性问题 http://www.cnblogs.com/guoxiaoqian/p/3984934.html 自从CES2012上Intel发布了针对移动市场的Medf ...
- 谈谈Delphi中的类和对象3---抽象类和它的实例
四.抽象类和它的实例 Delphi中有一个类称为是抽象类,你不能天真的直接为它创建一个实例,如 var StrLst: TString; begin StrLst:= TString.Create; ...
- nodejs2
jade@1.11.0 严格注意缩进 extends layout block content h1= title p Welcome to #{title} - var a='abc'; p his ...
- C字符数组赋值(转)
举例如下: char a[10];1.定义的时候直接用字符串赋值char a[10]="hello";注意:不能先定义再给它赋值,如 char a[10]; a[10]=" ...
- HDU 5900 QSC and Master 区间DP
QSC and Master Problem Description Every school has some legends, Northeastern University is the s ...
- Comet:基于 HTTP 长连接的“服务器推”技术解析
原文链接:http://www.cnblogs.com/deepleo/p/Comet.html 一.背景介绍 传统web请求,是显式的向服务器发送http Request,拿到Response后显示 ...
- sqoop中,如果数据中本身有换行符,会导致数据错位
sqoop中,如果数据中本身有换行符,会导致数据错位: 解决办法: 在sqoop import时修改配置文件 sudo -u hive sqoop import --connect jdbc:mysq ...