Codeforces Round #424 (Div. 2) E. Cards Sorting(树状数组)

E. Cards Sorting
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
4
6 3 1 2
output
7
input
1
1000
output
1
input
7
3 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.
题目；n张牌叠成一摞。初始最上面是1，每张牌有权值，每次对于最上面的牌来说，如果它是现在权值最小的，丢出去，否则将此牌放到最下面。问多少次操作后n张牌都没有了。
题解：我们先预处理出不同权值牌的所在位置，用树状数组记录每个位置是否有牌，用last标记上一次结束操作的位置。具体操作看代码。
代码：

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<vector>#include<queue>#include<algorithm>#include<math.h>#include<map>using namespace std;typedef long long int ll;typedef pair<int,int>pa;const int N=1e5+10;const int MOD=1e9+7;const ll INF=1e18;int read(){    int x=0;    char ch = getchar();    while('0'>ch||ch>'9')ch=getchar();    while('0'<=ch&&ch<='9')    {        x=(x<<3)+(x<<1)+ch-'0';        ch=getchar();    }    return x;}/************************************************************/struct node{    int x,id;    bool  operator <(const node &t)const    {        if(x!=t.x) return x<t.x;        return id<t.id;    }}a[N];int n;vector<int>p[N];int sum[N];int lowbit(int x){    return x&(-x);}int add(int x,int val){    while(x<=N)    {        sum[x]+=val;        x+=lowbit(x);    }}ll query(int x){    ll res=0;    while(x)    {        res+=sum[x];        x-=lowbit(x);    }    return res;}int main(){    cin>>n;    memset(sum,0,sizeof(sum));    for(int i=1;i<=n;i++)    {        cin>>a[i].x;        a[i].id=i;    }    sort(a+1,a+1+n);    int sz=0;    p[++sz].push_back(a[1].id);    for(int i=2;i<=n;i++)    {        if(a[i].x!=a[i-1].x) sz++;        p[sz].push_back(a[i].id);    }    for(int i=1;i<=n;i++) add(i,1);    ll ans=0;    int last=0;    for(int i=1;i<=sz;i++)    {        int pos=upper_bound(p[i].begin(),p[i].end(),last)-p[i].begin();        int cnt=p[i].size();        int en=p[i][cnt-1];        if(pos>0) ans+=query(n)-query(last);        else            ans+=query(en)-query(last);        for(int j=cnt-1;j>=pos;j--)        {            add(p[i][j],-1);            p[i].pop_back();        }        if(pos>0) i--,last=0;        else  last=en==n?0:en;    }    cout<<ans<<endl;}