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
- linux文件描述符open file descriptors与open files的区别
一个文件被打开,也可能没有文件描述符,比如current working diretories,memory mapped files and executable text files ;losf可 ...
- mysql 源码安装
yum install -y gcc gcc-c++ autoconf libjpeg libjpeg-devel perl perl-CPAN libpng libpng-devel freetyp ...
- hdu 5033 单调栈 ****
看出来是单调栈维护斜率,但是不会写,2333,原来是和询问放在一起的 #include <iostream> #include <cstdio> #include <cs ...
- hdu 4731 2013成都赛区网络赛 找规律
题意:找字串中最长回文串的最小值的串 m=2的时候暴力打表找规律,打表可以用二进制枚举
- WPF线程(Step1)——Dispatcher
使用WPF开发时经常会遇上自己建立的线程需要更新界面UI内容,从而导致的跨线程问题. 异常内容: 异常类型:System.InvalidOperationException 异常描述: "S ...
- 【rqnoj378】 约会计划
题目描述 cc是个超级帅哥,口才又好,rp极高(这句话似乎降rp),又非常的幽默,所以很多mm都跟他关系不错.然而,最关键的是,cc能够很好的调解各各妹妹间的关系.mm之间的关系及其复杂,cc必须严格 ...
- barabasilab-networkScience学习笔记6-evolving networks
第一次接触复杂性科学是在一本叫think complexity的书上,Allen博士很好的讲述了数据结构与复杂性科学,barabasi是一个知名的复杂性网络科学家,barabasilab则是他所主导的 ...
- loadrunner通过C语言实现自定义字符出现次数截取对应字符串
void lr_custom_string_delim_save(char inputStr[500], char* outputStr, char *delim, int occrNo, int s ...
- Linux使用du和df查看磁盘和文件夹占用空间
df df可以查看一级文件夹大小.使用比例.档案系统及其挂入点,但对文件却无能为力. df -lh 参数 -h 表示使用「Human-readable」输出,也就是使用 GB.MB 等易读的格式. $ ...
- JDK BIO编程
网络编程的基本模型是Client/Server模型,也就是两个进程之间进行相互通信,其中服务端提供位置信息(绑定的IP地址和监听端口),客户端通过连接操作向服务端监听的地址发起连接请求,通过三次握手建 ...