codeforce1029B B. Creating the Contest(简单dp,简单版单调栈)
1 second
256 megabytes
standard input
standard output
You are given a problemset consisting of nn problems. The difficulty of the ii-th problem is aiai. It is guaranteed that all difficulties are distinct and are given in the increasing order.
You have to assemble the contest which consists of some problems of the given problemset. In other words, the contest you have to assemble should be a subset of problems (not necessary consecutive) of the given problemset. There is only one condition that should be satisfied: for each problem but the hardest one (the problem with the maximum difficulty) there should be a problem with the difficulty greater than the difficulty of this problem but not greater than twice the difficulty of this problem. In other words, let ai1,ai2,…,aipai1,ai2,…,aip be the difficulties of the selected problems in increasing order. Then for each jj from 11 to p−1p−1 aij+1≤aij⋅2aij+1≤aij⋅2 should hold. It means that the contest consisting of only one problem is always valid.
Among all contests satisfying the condition above you have to assemble one with the maximum number of problems. Your task is to find this number of problems.
The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of problems in the problemset.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — difficulties of the problems. It is guaranteed that difficulties of the problems are distinct and are given in the increasing order.
Print a single integer — maximum number of problems in the contest satisfying the condition in the problem statement.
10
1 2 5 6 7 10 21 23 24 49
4
5
2 10 50 110 250
1
6
4 7 12 100 150 199
3
Description of the first example: there are 1010 valid contests consisting of 11 problem, 1010 valid contests consisting of 22 problems ([1,2],[5,6],[5,7],[5,10],[6,7],[6,10],[7,10],[21,23],[21,24],[23,24][1,2],[5,6],[5,7],[5,10],[6,7],[6,10],[7,10],[21,23],[21,24],[23,24]), 55 valid contests consisting of 33 problems ([5,6,7],[5,6,10],[5,7,10],[6,7,10],[21,23,24][5,6,7],[5,6,10],[5,7,10],[6,7,10],[21,23,24]) and a single valid contest consisting of 44 problems ([5,6,7,10][5,6,7,10]).
In the second example all the valid contests consist of 11 problem.
In the third example are two contests consisting of 33 problems: [4,7,12][4,7,12] and [100,150,199][100,150,199].
已知序列是单调递增的,找最长序列满足 前一项*2>=后一项,用单调栈(单调定义为:前一项*2>=后一项)实现即可。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define maxn 110
#define maxm 10010
#define inf 0x3f3f3f
using namespace std;
stack<int>s;
int main()
{
while(!s.empty())
s.pop();
int n;
scanf("%d",&n);
int mm=;
int sum=;
for(int i=;i<=n;i++)
{
int x;
scanf("%d",&x);
if(s.empty())
{
sum=;
s.push(x);
}
else
{
while(!s.empty ())
{
int y=s.top();
if(y*>=x)//符合条件就放进去
{
s.push(x);
sum++;
break;
}
else//否则把前一项踢了
{
s.pop();
sum--;
}
}
if(s.empty ())
{
sum=;
s.push(x);
}
}
if(sum>mm)
mm=sum;
}
printf("%d\n",mm);
return ;
}
codeforce1029B B. Creating the Contest(简单dp,简单版单调栈)的更多相关文章
- 【dp 状态压缩 单调栈】bzoj3591: 最长上升子序列
奇妙的单调栈状压dp Description 给出1~n的一个排列的一个最长上升子序列,求原排列可能的种类数. Input 第一行一个整数n. 第二行一个整数k,表示最长上升子序列的长度. 第三行k个 ...
- 「10.11」chess(DP,组合数学)·array(单调栈)·ants(莫队,并茶几)
菜鸡wwb因为想不出口胡题所以来写题解了 A. chess 昨天晚上考试,有点困 开考先花五分钟扫了一边题,好开始肝$T1$ 看了一眼$m$的范围很大,第一反应矩阵快速幂?? $n$很小,那么可以打$ ...
- BZOJ.3238.[AHOI2013]差异(后缀自动机 树形DP/后缀数组 单调栈)
题目链接 \(Description\) \(Solution\) len(Ti)+len(Tj)可以直接算出来,每个小于n的长度会被计算n-1次. \[\sum_{i=1}^n\sum_{j=i+1 ...
- Codeforces - 1033C - Permutation Game - 简单dp - 简单数论
https://codeforces.com/problemset/problem/1033/C 一开始觉得自己的答案会TLE,但是吸取徐州赛区的经验去莽了一发. 其实因为下面这个公式是 $O(nlo ...
- hdu 2084 数塔(简单dp)
题目 简单dp //简单的dp #include<stdio.h> #include<string.h> #include<algorithm> using nam ...
- The Preliminary Contest for ICPC China Nanchang National Invitational I. Max answer (单调栈+线段树)
题目链接:https://nanti.jisuanke.com/t/38228 题目大意:一个区间的值等于该区间的和乘以区间的最小值.给出一个含有n个数的序列(序列的值有正有负),找到该序列的区间最大 ...
- E. The Contest ( 简单DP || 思维 + 贪心)
传送门 题意: 有 n 个数 (1 ~ n) 分给了三个人 a, b, c: 其中 a 有 k1 个, b 有 k2 个, c 有 k3 个. 现在问最少需要多少操作,使得 a 中所有数 是 1 ~ ...
- codeforces Gym 100500H A. Potion of Immortality 简单DP
Problem H. ICPC QuestTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/a ...
- Codeforces Round #302 (Div. 2) C. Writing Code 简单dp
C. Writing Code Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/544/prob ...
随机推荐
- HDU1510 White rectangles( 乱搞 O(n^3) )题解
思路: 友谊赛的时候一直想到了,但是没想出来怎么遍历才能找到所有矩阵,卡住了. 这里讲一下完整思路:我们用一个num[i][j]表示第i行第j列每一列连续的白色格子数量,然后我们定义一个MIN,并且每 ...
- 【bzoj5170】Fable(树状数组)
题目传送门:https://www.lydsy.com/JudgeOnline/problem.php?id=5170 我们会发现,经过一轮冒泡后,若a[i]的前面有比它大的数,就一定会有一个被丢到后 ...
- codeforces 55 div2 C.Title 模拟
C. Title time limit per test 2 seconds memory limit per test 256 megabytes input standard input outp ...
- Vi/Vim三种模式
命令模式,底线命令模式,输入模式:
- 用训练好的caffemodel来进行分类
caffe程序自带有一张小猫图片,存放路径为caffe根目录下的 examples/images/cat.jpg, 如果我们想用一个训练好的caffemodel来对这张图片进行分类,那该怎么办呢? 如 ...
- 【Jmeter】配置不同业务请求比例,应对综合场景压测
背景 在进行综合场景压测时,遇到了如何实现不同的请求所占比例不同的问题. 有人说将这些请求分别放到单独的线程组下,然后将线程组的线程数按照比例进行配置. 这种方法不是很好,因为服务器对不同的请求处理能 ...
- Java 集合-Collections工具类
2017-11-05 23:41:53 Collections类 Collections类:Collections类是针对集合进行操作的工具类,都是静态方法. 常用方法: public static ...
- 构建工具 Ant、Maven和Gradle
构建工具的作用 依赖管理 测试,打包,发布 主流的构建工具 Ant:提供编译,测试,打包 Maven:在Ant的基础上提供了依赖管理和发布的功能 Gradle:在Maven的基础上使用Groovy管理 ...
- Android中Tablayout设置下划线宽度 和 dp和px之间进行相互转换
开发中遇到了一个问题,Tablayout设置下换线长度,看了点资料,分享给大家. 效果图: 直接贴代码(要在tabLayout添加完所有的tab后调用) public vo ...
- localStorage(本地存储)使用总结
1.https://www.cnblogs.com/st-leslie/p/5617130.html (localStorage使用总结)