Codeforces Round #416 (Div. 2) B. Vladik and Complicated Book
B. Vladik and Complicated Book
2 seconds
256 megabytes
standard input
standard output
Vladik had started reading a complicated book about algorithms containing n pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation P = [p1, p2, ..., pn], where pi denotes the number of page that should be read i-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation P from position l to position r inclusive, because she loves the order. For every of such sorting Vladik knows number x — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has px changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.
First line contains two space-separated integers n, m (1 ≤ n, m ≤ 104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.
Second line contains n space-separated integers p1, p2, ..., pn (1 ≤ pi ≤ n) — permutation P. Note that elements in permutation are distinct.
Each of the next m lines contains three space-separated integers li, ri, xi (1 ≤ li ≤ xi ≤ ri ≤ n) — left and right borders of sorted subsegment in i-th sorting and position that is interesting to Vladik.
For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise.
5 5
5 4 3 2 1
1 5 3
1 3 1
2 4 3
4 4 4
2 5 3
Yes
No
Yes
Yes
No
6 5
1 4 3 2 5 6
2 4 3
1 6 2
4 5 4
1 3 3
2 6 3
Yes
No
Yes
No
Yes
Explanation of first test case:
- [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes".
- [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No".
- [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes".
- [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes".
- [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No".
解题思路:
这道题简单的一笔,就是当时打的时候算了下复杂度懒得跟他多比比,直接暴力过了,结果被人出了个极限数据卡了时间复杂度,mmp!
被hack了,下面直接放代码,没啥讲的。以后不能瞎暴力了,要不迟早要翻水水
实现代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + ;
int n, m, a[N]; int main() {
cin>>n>>m;
for (int i = ; i <= n; i++) cin>>a[i];
while (m--) {
int l, r, x; cin>>l>>r>>x;
int cnt = ;
for (int i = l; i <= r; i++) {
if (a[i] < a[x]) cnt++;
}
if (x == l + cnt)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
}
Codeforces Round #416 (Div. 2) B. Vladik and Complicated Book的更多相关文章
- Codeforces Round #416 (Div. 2) D. Vladik and Favorite Game
地址:http://codeforces.com/contest/811/problem/D 题目: D. Vladik and Favorite Game time limit per test 2 ...
- Codeforces Round #416 (Div. 2) C. Vladik and Memorable Trip
http://codeforces.com/contest/811/problem/C 题意: 给出一行序列,现在要选出一些区间来(不必全部选完),但是相同的数必须出现在同一个区间中,也就是说该数要么 ...
- Codeforces Round #416 (Div. 2) A. Vladik and Courtesy【思维/模拟】
A. Vladik and Courtesy time limit per test 2 seconds memory limit per test 256 megabytes input stand ...
- 【分类讨论】【spfa】【BFS】Codeforces Round #416 (Div. 2) D. Vladik and Favorite Game
那个人第一步肯定要么能向下走,要么能向右走.于是一定可以判断出上下是否对调,或者左右是否对调. 然后他往这个方向再走一走就能发现一定可以再往旁边走,此时就可以判断出另一个方向是否对调. 都判断出来以后 ...
- 【动态规划】 Codeforces Round #416 (Div. 2) C. Vladik and Memorable Trip
划分那个序列,没必要完全覆盖原序列.对于划分出来的每个序列,对于某个值v,要么全都在该序列,要么全都不在该序列. 一个序列的价值是所有不同的值的异或和.整个的价值是所有划分出来的序列的价值之和. ...
- Codeforces Round #416 (Div. 2)(A,思维题,暴力,B,思维题,暴力)
A. Vladik and Courtesy time limit per test:2 seconds memory limit per test:256 megabytes input:stand ...
- Codeforces Round#416 Div.2
A. Vladik and Courtesy 题面 At regular competition Vladik and Valera won a and b candies respectively. ...
- Codeforces Round #416 (Div. 2) A+B
A. Vladik and Courtesy 2 seconds 256 megabytes At regular competition Vladik and Valera won a and ...
- Codeforces Round #416 (Div. 2)A B C 水 暴力 dp
A. Vladik and Courtesy time limit per test 2 seconds memory limit per test 256 megabytes input stand ...
随机推荐
- Django 学习 (第五部)
表设计: from django.db import models # Create your models here. # class Foo(models.Model): # name = mod ...
- LOJ2687 BOI2013 Vim 线头DP
传送门 多图警告!!! 一种很新奇的\(DP\),全网似乎只有一两篇题解-- 首先,序列中的一段\(e\)等价于在跳的过程中这一段\(e\)之后的一个字符必须要经过,并且在最后的答案中加上$2 \ti ...
- NOIP常见模板集合
Preface 这篇博客记录的是我联赛前虽然只有两天了的打板子记录. 只求真的能给我起到些作用吧,一般按照难度排序. 而且从这篇博客开始我会用H3的标题代替H4 为了节约篇幅,以下的代码一般均以cla ...
- [Spark][Python]获得 key,value形式的 RDD
[Spark][Python]获得 key,value形式的 RDD [training@localhost ~]$ cat users.txtuser001 Fred Flintstoneuser0 ...
- SQL行转列汇总 (转)
PIVOT 用于将列值旋转为列名(即行转列),在 SQL Server 2000可以用聚合函数配合CASE语句实现 PIVOT 的一般语法是:PIVOT(聚合函数(列) FOR 列 in (…) )A ...
- KVM虚拟机管理——资源调整
1. 概述2. 计算资源调整2.1 调整处理器配置2.2 调整内存配置3. 存储资源调整3.1 根分区扩展3.2 添加磁盘4. 网络资源调整 1. 概述 KVM在使用过程中,会涉及到计算(CPU,内存 ...
- 《Linux内核设计与实现》课本第四章学习总结
进程调度 4.1 多任务 多任务操作系统就是能同时并发的交互执行多个进程的操作系统. 多任务系统分为两种: 抢占式多任务:Linux提供了抢占式的多任务模式,由调度程序来决定什么时候停止一个进程的运行 ...
- 基于SSH框架的考勤管理系统的设计与实现
基于SSH框架的考勤管理系统的设计与实现
- Python学习笔记 -- 第四章
高阶函数 变量可以指向函数 f=abs f(-10) 10 变量f指向abs函数,直接调用abs()函数和调用f()完全相同 传入参数 变量可以指向函数,函数的参数可以接收另一个函数的参数,这种函数成 ...
- 大三上学期安卓一边学一边开始做一个自己觉得可以的项目 广商小助手App 加油
这项目构思好多 一个人一步一步来 一边做一边为后面应用铺设 广商小助手APP 设计出的软件登录场景 实现(算是可以) 界面大体出来了 界面点击方面也做了很多特效 上图其实点击各颜色后会出现各种图和反应 ...