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整天做一些模板题感觉药丸

设\(val_i\)表示第\(i\)个位置的值

看到区间最大值考虑最值分治。对于当前的区间\([l,r]\)，找到区间最大值\(mid\)，递归\([l,mid-1]\)和\([mid+1,r]\)，然后考虑pair\((i,j)(i \in [l,mid] , r \in [mid,r])\)的贡献。

扫\([l,mid]\)和\([mid,r]\)中较短的一段区间，那么对于扫到的一个位置\(i\)，它的贡献就是另一段区间中值小于等于\(\lfloor \frac{val_{mid}}{val_i} \rfloor\)的位置的数量，使用主席树维护即可。

#include<bits/stdc++.h>
//this code is written by Itst
using namespace std;int read(){int a = 0; char c = getchar();while(!isdigit(c)) c = getchar();while(isdigit(c)){a = a * 10 + c - 48;c = getchar();}return a;
}#define ll long long
#define lb lower_bound
#define ub upper_bound
const int _ = 1e5 + 7;
namespace segTree{int lch[_ * 20] , rch[_ * 20] , sum[_ * 20] , cntN;#define mid ((l + r) >> 1)int modify(int x , int l , int r , int tar){int t = ++cntN;sum[t] = sum[x] + 1; lch[t] = lch[x]; rch[t] = rch[x];if(l == r) return t;if(mid >= tar) lch[t] = modify(lch[t] , l , mid , tar);else rch[t] = modify(rch[t] , mid + 1 , r , tar);return t;}int query(int x , int l , int r , int R){if(r <= R) return sum[x];int sum = query(lch[x] , l , mid , R);if(mid < R) sum += query(rch[x] , mid + 1 , r , R);return sum;}
}
int rt[_] , val[_] , lsh[_] , ST[21][_] , logg2[_] , N , cntL;int cmp(int a , int b){return val[a] > val[b] ? a : b;}void init(){logg2[0] = -1;for(int i = 1 ; i <= N ; ++i){ST[0][i] = i;logg2[i] = logg2[i >> 1] + 1;}for(int i = 1 ; 1 << i <= N ; ++i)for(int j = 1 ; j + (1 << i) - 1 <= N ; ++j)ST[i][j] = cmp(ST[i - 1][j] , ST[i - 1][j + (1 << (i - 1))]);
}int qST(int l , int r){int t = logg2[r - l + 1];return cmp(ST[t][l] , ST[t][r - (1 << t) + 1]);
}ll solve(int l , int r){if(l > r) return 0;if(l == r) return lsh[val[l]] == 1 ? 1 : 0;int Mid = qST(l , r) , w = lsh[val[Mid]];ll sum = solve(l , Mid - 1) + solve(Mid + 1 , r);int L = l , R = Mid , rgeL = Mid , rgeR = r;if(Mid - l > r - Mid){swap(rgeL , L); swap(rgeR , R);}while(L <= R){int num = ub(lsh + 1 , lsh + cntL , w / lsh[val[L]]) - lsh - 1;if(num)sum += segTree::query(rt[rgeR] , 1 , cntL , num) - segTree::query(rt[rgeL - 1] , 1 , cntL , num);++L;}return sum;
}int main(){
#ifndef ONLINE_JUDGEfreopen("in","r",stdin);//freopen("out","w",stdout);
#endifN = read();for(int i = 1 ; i <= N ; ++i)val[i] = lsh[i] = read();sort(lsh + 1 , lsh + N + 1);cntL = unique(lsh + 1 , lsh + N + 1) - lsh;for(int i = 1 ; i <= N ; ++i){val[i] = lb(lsh + 1 , lsh + cntL , val[i]) - lsh;rt[i] = segTree::modify(rt[i - 1] , 1 , cntL , val[i]);}init();cout << solve(1 , N);return 0;
}