cf689d ST表RMQ+二分

/**
二分法，枚举左端点，向右寻找第一个最大值不等于最小值的端点
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define maxn 200050
int a[maxn],b[maxn],mx[maxn][30],mi[maxn][30],n;
long long ans;
void ST(){
    for(int i=1;i<=n;i++) mx[i][0]=a[i],mi[i][0]=b[i];
    for(int j=1;(1<<j)<=n;j++) 
        for(int i=1;i+(1<<j)-1<=n;i++){
            mx[i][j]=max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]);
            mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
        }
}
int query1(int L,int R){
    int k=log2(R-L+1);
    return max(mx[L][k],mx[R-(1<<k)+1][k]);
}
int query2(int L,int R){
    int k=log2(R-L+1);
    return min(mi[L][k],mi[R-(1<<k)+1][k]);
}

int main(){
    cin >> n;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<=n;i++) scanf("%d",&b[i]);
    ans=0;ST();
    for(int i=1;i<=n;i++){
        int l=i,r=n,r1=-1,r2=-1,tmp1,tmp2;
        while(l<=r){//先找左端点
            int mid=l+r>>1;
            tmp1=query1(i,mid);
            tmp2=query2(i,mid);
            if(tmp1<tmp2) l=mid+1;//左端点不够右
            else if(tmp1>tmp2) r=mid-1;//左端点右过头了
            else r1=mid,r=mid-1;//再往左找 
        }
        l=i,r=n;
        while(l<=r){
            int mid=l+r>>1;
            tmp1=query1(i,mid);
            tmp2=query2(i,mid);
            if(tmp1<tmp2) l=mid+1;//右端点不够右
            else if(tmp1>tmp2) r=mid-1;//右端点右过头了
            else r2=mid,l=mid+1;//再往右找 
        }
        if(r1==-1 || r2==-1) continue;
        ans+=r2-r1+1;
    }
    printf("%lld\n",ans);
    return 0;
}