传送门:洛谷 - P4755
思路分析
枚举$i$作为区间最大值的贡献,只需要枚举区间的一边,然后查询另一边有几个数满足
以权值较大为根来建立笛卡尔树,可以很好的维护$i$作为区间最大值时的左端点和有端点,查询利用主席树完成
AC代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 1e5+10;
#define int long long
struct node{
int l,r,sum;
}tree[N*40];
int rt[N],cnt;
#define ls(x) tree[x].l
#define rs(x) tree[x].r
#define mid (l+r>>1)
void pp(int x){
tree[x].sum = tree[ls(x)].sum + tree[rs(x)].sum;
}
void ins(int l,int r,int &x,int pre,int val){
x=++cnt;
tree[x] = tree[pre];
if(l==r){
tree[x].sum++;
return ;
}
if(val<=mid) ins(l,mid,ls(x),ls(pre),val);
else ins(mid+1,r,rs(x),rs(pre),val);
pp(x);
}
int query(int l,int r,int L,int R,int ql,int qr){
if(ql<=l && r<=qr){
return tree[R].sum - tree[L].sum;
}
int ans = 0;
if(ql<=mid) ans += query(l,mid,ls(L),ls(R),ql,qr);
if(qr>mid) ans += query(mid+1,r,rs(L),rs(R),ql,qr);
return ans;
}
namespace ditree{
struct node{
int ls,rs,val;
}tree[N];
int st[N],top=0,root,ans;
void build(int *a,int n){
for(int i=1;i<=n;i++){
int j=0;
while(top && a[st[top]]<a[i]) j = st[top],top--;
if(!top) root = i;
else tree[st[top]].rs = i;
tree[i].ls = j;
tree[i].val = a[i];
st[++top] = i;
}
}
void dfs(int x,int l,int r){
if(x-l<r-x) for(int i=l;i<=x;i++) ans += query(1,1e9,rt[x-1],rt[r],1,tree[x].val/tree[i].val);
else for(int i=x;i<=r;i++) ans += query(1,1e9,rt[l-1],rt[x],1,tree[x].val/tree[i].val);
if(tree[x].ls) dfs(tree[x].ls,l,x-1);
if(tree[x].rs) dfs(tree[x].rs,x+1,r);
}
};
int a[N];
signed main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
int n;
read(n);
for(int i=1;i<=n;i++) {
read(a[i]);
ins(1,1e9,rt[i],rt[i-1],a[i]);
}
ditree::build(a,n);
ditree::dfs(ditree::root,1,n);
printf("%lld\n",ditree::ans);
return 0;
}