CF 689D

689D - Friends and Subsequences

题意:

• 大致跟之前题目一样,用ST表维护a[]区间max，b[]区间min,找出多少对(l,r)使得maxa(l,r) == minb(l,r)
• 切题的感觉很爽唉
• 同样而二分查找,找最小和最大下标满足条件
• cf中%I64d, 一般是%lld

代码:

#include<bits/stdc++.h>
#define ll long long
const int  maxn=200010;
int sta[maxn][18];
int stb[maxn][18];
int a[maxn];
int b[maxn];
ll res;
int n;
void build(){
    int maxl=floor(log2(n));
    for(int i=1;i<=n;i++){
        sta[i][0]=a[i];
        stb[i][0]=b[i];
    }
    int mul=1;
    for(int j=1;j<=maxl;j++){
        for(int i=1;i<=n&&(i+mul)<=n;i++){
            sta[i][j]=std::max(sta[i][j-1],sta[i+mul][j-1]);
            stb[i][j]=std::min(stb[i][j-1],stb[i+mul][j-1]);
        }
        mul=mul*2;
    }
}
int check(int x,int y){
    int len=floor(log2(y-x+1));
    int maxa=std::max(sta[x][len],sta[y-(1<<len)+1][len]);
    int minb=std::min(stb[x][len],stb[y-(1<<len)+1][len]);
    //printf("db x:%d y:%d maxa: %d minb %d
",x,y,maxa,minb);
    return maxa-minb;
}
// find min() key=0
//find max() key=0
int b1(int begin,int end){
    int l=begin,r=end,m;
    while(l<r){
        m=l+((r-l)>>1);
        if(check(begin,m)<0) l=m+1;
        else r=m;
    }
    if(check(begin,l)==0) return l;
    return -1;
}
int b2(int begin,int end){
    int l=begin,r=end,m;
    while(l<r){
        m=l+((r-l+1)>>1);
        if(check(begin,m)<=0) l=m;
        else r=m-1;
    }
    if(check(begin,l)==0) return l;
    return -1;
}
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]);
    }
    build();

    res=0;
    for(int i=1;i<=n;i++){
        if(b1(i,n)==-1) continue;
        int r1=b1(i,n);
        int r2=b2(i,n);
        res+=(r2-r1+1);
    }
    printf("%I64d
",res);
}