제출 #531726

#제출 시각아이디문제언어결과실행 시간메모리
531726zaneyuIzbori (COCI22_izbori)C++14
110 / 110
90 ms18800 KiB
/*
5
2 2 1 2 3
*/ 
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
 
//#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<ll,ll>;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e17;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]+1<<((i!=sz(V)-1)?" ":"");
    return out;
}
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=2e5+5;
int arr[maxn],tmp[maxn];
pii a[maxn];
vector<pii> v[maxn];
int mp[2*maxn];
int main(){
	ios::sync_with_stdio(false),cin.tie(0);
    int n;
    cin>>n;
    REP(i,n) cin>>arr[i],a[i].f=arr[i],a[i].s=i;
    sort(a,a+n);
    int cnt=0;
    REP(i,n){
    	if(i and arr[a[i].s]!=arr[a[i-1].s]) ++cnt;
    	tmp[a[i].s]=cnt;
    }
    ++cnt;
    REP(i,n) arr[i]=tmp[i];
   	for(int i=0;i<n;){
   		int p=i;
   		while(i<n and arr[i]==arr[p]) ++i;
   		v[arr[p]].pb({p,i});
   	}
    ll ans=0;
    REP(i,cnt){
    	int pv=0;
    	int cur=n,mn=n,tot=0,up=n;
    	v[i].pb({n,n});
    	vector<int> vv;
    	mp[cur]++,vv.pb(cur);
	    for(auto x:v[i]){
	        int z=x.f-pv;
	        if(z){
	            while(cur-1>mn and z){
	                cur--;
	                tot-=mp[cur];
	                ans+=tot;
	                --z;
	            }
	            if(z){
	                tot=0,cur-=z,mn=cur;
	            }
	            mp[cur]++,vv.pb(cur);
	        }
	        for(int j=x.f;j<x.s;j++){
		        ++cur;
		        tot+=mp[cur-1];
		        ans+=tot;
		        if(j!=x.s-1) mp[cur]++,vv.pb(cur);
		        else if(cur>=up) mp[cur]++,vv.pb(cur);
		        if(cur<up) mp[cur]++,vv.pb(cur);
		        MXTO(up,cur);
	    	}
	    	pv=x.s;
	    }
	    for(int x:vv) mp[x]--;
	}
    cout<<ans;
}
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