Cards Sorting
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
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-th from top card in the deck.
Output
Print the total number of times Vasily takes the top card from the deck.
Examples
input
46 3 1 2
output
7
input
11000
output
1
input
73 3 3 3 3 3 3
output
7
Note
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】从上往下遍历所有的卡片。如果当前卡片上的数字跟当前堆中的最小值相等，则将该卡片扔点，否则将该卡片放到最低端，问一共遍历多少次。
【分析】模拟一遍，从最小的开始找，二分出可以衔接上一次位置的当前起始位置，遍历到此轮最后一个地方，再遍历前面没有遍历的地方，用树状数组来统计。
#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;
}