Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100?000, inclusive. It is possible that some cards have the same integers on them.

Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn‘t know where this card (or these cards) is.

You are to determine the total number of times Vasily takes the top card from the deck.

The first line contains single integer n (1?≤?n?≤?100?000) — the number of cards in the deck.

The second line contains a sequence of n integers a1,?a2,?...,?an (1?≤?ai?≤?100?000), where ai is the number written on the i

Print the total number of times Vasily takes the top card from the deck.

4
6 3 1 2

7

1
1000

1

7
3 3 3 3 3 3 3

7

In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2,?6,?3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6,?3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards.

【題意】從上往下遍歷所有的卡片。如果當前卡片上的數字跟當前堆中的最小值相等，則將該卡片扔點，否則將該卡片放到最低端，問一共遍歷多少次。

【分析】模擬一遍，從最小的開始找，二分出可以銜接上一次位置的當前起始位置，遍歷到此輪最後一個地方，再遍歷前面沒有遍歷的地方，用樹狀數組來統計。

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int N = 1e5+50;
int n;
int sum[N];
vector<int>vec[N];
void upd(int x,int add){
   for(int i=x;i<=100000;i+=i&(-i)){
       sum[i]+=add;
   }
}
int qry(int x){
    int ret=0;
    for(int i=x;i>=1;i-=i&(-i)){
        ret+=sum[i];
    }
    return ret;
}
int dis(int l,int r){
    if(l==-1)return qry(r);
    else if(l<r)return qry(r)-qry(l);
    else return qry(n)-qry(l)+qry(r);
}
int main(){
    scanf("%d",&n);
    for(int i=1,x;i<=n;i++){
        scanf("%d",&x);
        upd(i,1);
        vec[x].pb(i);
    }
    LL woqunimalegebi = 0;
    int pre=-1;
    for(int i=1;i<=100000;i++){
        if(!vec[i].size())continue;
        int pos=upper_bound(vec[i].begin(),vec[i].end(),pre)-vec[i].begin();
        for(int j=max(0,pos);j<vec[i].size();j++){
            woqunimalegebi+=dis(pre,vec[i][j]);
            pre=vec[i][j];
            upd(pre,-1);
        }
        for(int j=0;j<pos;j++){
            woqunimalegebi+=dis(pre,vec[i][j]);
            pre=vec[i][j];
            upd(pre,-1);
        }
    }
    printf("%lld\n",woqunimalegebi);
    return 0;
}

Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Cards Sorting（樹狀數組）