Friends and Subsequences

Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their own, but together (with you) — who knows?

Every one of them has an integer sequences a and b of length n. Being given a query of the form of pair of integers (l, r), Mike can instantly tell the value of  while !Mike can instantly tell the value of .

Now suppose a robot (you!) asks them all possible different queries of pairs of integers (l, r) (1 ≤ l ≤ r ≤ n) (so he will make exactlyn(n + 1) / 2 queries) and counts how many times their answers coincide, thus for how many pairs  is satisfied.

How many occasions will the robot count?

Input

The first line contains only integer n (1 ≤ n ≤ 200 000).

The second line contains n integer numbers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the sequence a.

The third line contains n integer numbers b1, b2, ..., bn ( - 109 ≤ bi ≤ 109) — the sequence b.

Output

Print the only integer number — the number of occasions the robot will count, thus for how many pairs  is satisfied.

Examples
input
6
1 2 3 2 1 4
6 7 1 2 3 2
output
2
input
3
3 3 3
1 1 1
output
0
Note

The occasions in the first sample case are:

1.l = 4,r = 4 since max{2} = min{2}.

2.l = 4,r = 5 since max{2, 1} = min{2, 3}.

There are no occasions in the second sample case since Mike will answer 3 to any query pair, but !Mike will always answer 1.

分析:RMQ+二分;

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=2e5+;
const int dis[][]={,,-,,,-,,};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,p[maxn],a[][maxn],b[][maxn];
ll ans;
void init()
{
for(int i=;i<n;i++)p[i]=+p[i/];
for(int i=;i<;i++)
for(int j=;j+(<<i)-<n;j++)
a[i][j]=max(a[i-][j],a[i-][j+(<<(i-))]),b[i][j]=min(b[i-][j],b[i-][j+(<<(i-))]);
return;
}
int getma(int l,int r)
{
int x=p[r-l+];
return max(a[x][l],a[x][r-(<<x)+]);
}
int getmi(int l,int r)
{
int x=p[r-l+];
return min(b[x][l],b[x][r-(<<x)+]);
}
int getl(int now)
{
int l=now-,r=n;
while(r-l>)
{
int mid=(l+r)>>;
if(getma(now,mid)<getmi(now,mid))l=mid;
else r=mid;
}
return r;
}
int getr(int now)
{
int l=now-,r=n;
while(r-l>)
{
int mid=(l+r)>>;
if(getma(now,mid)<=getmi(now,mid))l=mid;
else r=mid;
}
return r;
}
int main()
{
int i,j,k,t;
scanf("%d",&n);
rep(i,,n-)scanf("%d",&a[][i]);
rep(i,,n-)scanf("%d",&b[][i]);
init();
rep(i,,n-)ans+=getr(i)-getl(i);
printf("%lld\n",ans);
return ;
}

Friends and Subsequences的更多相关文章

  1. codeforces 597C C. Subsequences(dp+树状数组)

    题目链接: C. Subsequences time limit per test 1 second memory limit per test 256 megabytes input standar ...

  2. [LeetCode] Distinct Subsequences 不同的子序列

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  3. Distinct Subsequences

    https://leetcode.com/problems/distinct-subsequences/ Given a string S and a string T, count the numb ...

  4. HDU 2227 Find the nondecreasing subsequences (DP+树状数组+离散化)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2227 Find the nondecreasing subsequences             ...

  5. Leetcode Distinct Subsequences

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  6. LeetCode(115) Distinct Subsequences

    题目 Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequen ...

  7. [Leetcode][JAVA] Distinct Subsequences

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  8. Distinct Subsequences Leetcode

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  9. 【leetcode】Distinct Subsequences(hard)

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  10. Codeforces Testing Round #12 C. Subsequences 树状数组

    C. Subsequences     For the given sequence with n different elements find the number of increasing s ...

随机推荐

  1. android 属性动画

    一直再追郭霖的博客和imooc上的一些新的视频,最近有讲到属性动画. 以下内容为博客学习以及imooc上视频资料的学习笔记: 在3.0之前比较常见的动画为tween动画和frame动画: tween动 ...

  2. ios简单更改系统TabBar的高度

    - (void)viewDidLayoutSubviews { [super viewDidLayoutSubviews]; CGRect frame = self.tabBar.frame; fra ...

  3. php curl调用相关api

    一.基本步骤 1.本次模拟是php的相关post请求,可通过CURLOPT_CUSTOMREQUEST设定相关POST.GET.PUT.DELETE相关适应REST API 2.相关重要的是curl_ ...

  4. [转]Publishing and Running ASP.NET Core Applications with IIS

    本文转自:https://weblog.west-wind.com/posts/2016/Jun/06/Publishing-and-Running-ASPNET-Core-Applications- ...

  5. JTAG上有多个设备时,该如何接呢?

    首先要了解JTAG管脚相关定义,具有JTAG口的芯片都有如下JTAG引脚定义(是相对芯片): TCK——测试时钟输入: TDI——测试数据输入: TDO——测试数据输出: TMS——测试模式选择,TM ...

  6. myeclipse复制项目

    一.myEclipse 复制后修改名称,访问不到项目 这是因为,你只是改了项目的名称,而没有改 下面是解决方法: 方法 1.右击你的项目,选择"properties",在" ...

  7. cornerstone 怎么使用

    Cornerstone的逻辑很清晰,界面打开后,左边栏上下分开,上面是working copies的列表,下面是REPOSITORIES的列表.常见的功能基本上跟windows一样,在上下文中可以得到 ...

  8. 19个心得 明明白白说Linux下的负载均衡

    [51CTO.com独家特稿]前言:作为一名Linux/unix系统工程师,这几年一直在涉及到对外项目,经手过许多小中型网站的架构,F5.LVS及Nginx接触的都比较多,我想一种比较通俗易懂的语气跟 ...

  9. sql server字段是逗号分割的id,关联明细表查询

    有时候一张表的一个字段是以逗号分割的一个字符串,分割的数字是明细表的主键id. 关联明细表查询可以这样做: ) ) --这是把areanos字段赋值给@areanos变量 set @areanos=' ...

  10. 转 : 如何用sys as sysdba权限连接数据库进行EXP/IMP

    使用sys as sysdba权限进行EXP/IMP与其它用户稍有不同,详细内容如下(摘自metalink) Applies to: Oracle Server - Enterprise Editio ...