题目

洛谷

做法

\(i≤x≤j,a[i]<\frac{a[x]}{a[j]}\)

考虑\(a[x]\)的贡献，单调栈预处理\(L,R\)能作为最大值的区间

枚举一端点，仅需另一端点满足条件即可，启发式枚举端点

另一端点丢到树状数组里随便乱搞，由于是一个区间的要差分一下

My complete code

#include<bits/stdc++.h>
#include<vector>
using namespace std;
typedef long long LL;
const LL maxn=1e5+9;
struct node{
    LL val,x;
}que[maxn]; vector<LL> Q[maxn];
LL n,cnt;
LL L[maxn],R[maxn],a[maxn],b[maxn];
inline void Fir_LR(){
    LL tail(0);
    for(LL i=1;i<=n;++i){
        while(tail && que[tail].val<a[i]) --tail;
        L[i]=(tail?que[tail].x+1:1);
        que[++tail]=(node){a[i],i};
    }
    tail=0;
    for(LL i=n;i>=1;--i){
        while(tail && que[tail].val<=a[i]) --tail;
        R[i]=(tail?que[tail].x-1:n);
        que[++tail]=(node){a[i],i};
    }
}
inline void Fir_Q(){
    for(LL i=1;i<=n;++i)
        if(i-L[i]<=R[i]-i){
            for(LL j=L[i];j<=i;++j)
                Q[R[i]].push_back(a[i]/a[j]),
                Q[i-1].push_back(-a[i]/a[j]);
        }else{
            for(LL j=i;j<=R[i];++j)
                Q[i].push_back(a[i]/a[j]),
                Q[L[i]-1].push_back(-a[i]/a[j]);
        }
}
struct Tree_k{
    LL tree[maxn];
    inline LL Lowbit(LL x){ return x&-x; }
    inline void Add(LL x,LL val){
        for(;x<=cnt;x+=Lowbit(x)) tree[x]+=val;
    }
    inline LL Query(LL x){
        LL ret(0);
        for(;x;x-=Lowbit(x)) ret+=tree[x]; return ret;
    }
}Tk;
inline void Fir_S(){
    LL ret(0);
    for(LL i=1;i<=n;++i){
        Tk.Add(a[i],1);
        for(LL j=0;j<Q[i].size();++j){
            LL val(Q[i][j]);
            LL x(abs(val)>=b[cnt]?cnt:upper_bound(b+1,b+1+cnt,abs(val))-b-1);
            ret+=(val>0?1:-1)*Tk.Query(x);
        }
    }
    cout<<ret;
}
int main(){
    cin>>n;
    for(LL i=1;i<=n;++i) cin>>a[i],b[i]=a[i];
    Fir_LR();
    Fir_Q();
    sort(b+1,b+1+n), cnt=unique(b+1,b+1+n)-b-1;
    for(LL i=1;i<=n;++i) a[i]=lower_bound(b+1,b+1+cnt,a[i])-b;//,cout<<a[i]<<' ';cout<<endl;
    Fir_S();
}