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.
InputThe 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-th from top card in the deck.
OutputPrint the total number of times Vasily takes the top card from the deck.
Examples input Copy4output Copy
6 3 1 2
7input Copy
1output Copy
1000
1input Copy
7output Copy
3 3 3 3 3 3 3
7Note
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.
题意:给你n个数,每次取出当前第一个数,如果这个数是这些数中的最小值,则扔掉,否则放到最后
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int amn=1e5+5; 5 int a[amn]; 6 set<int> p[amn]; 7 set<int>::iterator k; 8 int main(){ 9 int n; 10 cin>>n; 11 for(int i=1;i<=n;i++){ 12 cin>>a[i]; 13 p[a[i]].insert(i); 14 } 15 sort(a+1,a+1+n); 16 ll tot=n,mid=0,ans=n; 17 for(int i=1;i<=n;i++){ 18 k=p[a[i]].lower_bound(mid); 19 if(k==p[a[i]].end()){ 20 ans+=tot; 21 mid=0; 22 k=p[a[i]].lower_bound(mid); 23 } 24 mid=*k; 25 p[a[i]].erase(k); 26 tot--; 27 } 28 printf("%lld\n",ans); 29 }