Description
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 exactly n(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.
Sample Input
6 1 2 3 2 1 4 6 7 1 2 3 2
2
3 3 3 3 1 1 1
0
Hint
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.
如Hint所示,a为max[],b为min[],查找最大与最小区间的相同区间个数
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> using namespace std; int n,a[200005],b[200005]; long long ans; deque<int>x,y; int main() { scanf("%d",&n); for(int i=1; i<=n; i++) scanf("%d",&a[i]); for(int i=1; i<=n; i++) scanf("%d",&b[i]); for(int i=1,j=1; i<=n; i++) { while(!x.empty()&&a[x.back()]<=a[i]) x.pop_back(); while(!y.empty()&&b[y.back()]>=b[i]) y.pop_back(); x.push_back(i); y.push_back(i); while(j<=i&&a[x.front()]-b[y.front()]>0) { j++; while(!x.empty()&&x.front()<j) x.pop_front(); while(!y.empty()&&y.front()<j) y.pop_front(); } if(!x.empty()&&!y.empty()&&a[x.front()]==b[y.front()]) { ans+=min(x.front(),y.front())-j+1; } } printf("%lld",ans); }