HDU 5919 （还算基础的主席树）



                                   Sequence II

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1045    Accepted Submission(s): 251

Problem Description

Mr. Frog has an integer sequence of length n, which can be denoted asa1,a2,⋯,an There are m queries.

In the i-th query, you are given two integers li and ri. Consider the subsequence ali,ali+1,ali+2,⋯,ari.

We can denote the positions(the positions according to the original sequence) where an integer appears first in this subsequence asp(i)1,p(i)2,⋯,p(i)ki (in ascending order, i.e.,p(i)1<p(i)2<⋯<p(i)ki).

Note that ki is the number of different integers in this subsequence. You should output p(i)⌈ki2⌉for the i-th query.

 

Input

In the first line of input, there is an integer T (T≤2) denoting the number of test cases.

Each test case starts with two integers n (n≤2×105) and m (m≤2×105). There are n integers in the next line, which indicate the integers in the sequence(i.e.,a1,a2,⋯,an,0≤ai≤2×105).

There are two integers li and ri in the following m lines.

However, Mr. Frog thought that this problem was too young too simple so he became angry. He modified each query tol‘i,r‘i(1≤l‘i≤n,1≤r‘i≤n). As a result, the problem became more exciting.

We can denote the answers as ans1,ans2,⋯,ansm. Note that for each test case ans0=0.

You can get the correct input li,ri from what you read (we denote them as l‘i,r‘i)by the following formula:

li=min{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}

ri=max{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}

 

Output

You should output one single line for each test case.

For each test case, output one line “Case #x: p1,p2,⋯,pm”, where x is the case number (starting from 1) and p1,p2,⋯,pm is the answer.

 

Sample Input

25 23 3 1 5 42 24 45 22 5 2 1 22 32 4

 

Sample Output

Case #1: 3 3Case #2: 3 1
Hint
 

题意：

求给定一个区间,在对前边一个答案进行运算后得到一个新的区间,然后问这个区间里面有k个不同的数,把它们第一次出现的位置从小到大排序,问第k/2个位置是什么;

题解：

好像是强制在线，一看要解决的问题，求区间不同数的个数和区间第K大，主席树没跑，因为他要的是第一次出现的位置，我们从后向前插入。

代码：

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <map>using namespace std;const int MAX=2e5+500;int a[MAX],root[MAX];map<int,int >mp;struct{    int l,r,sum;} T[MAX*40];int cnt=0;void update(int l,int r,int &x,int y,int pos,int v){    T[++cnt]=T[y];    x=cnt;    T[cnt].sum+=v;    if(l==r)        return ;    int mid=(r+l)>>1;    if(pos<=mid)    {        update(l,mid,T[cnt].l,T[y].l,pos,v);    }    else        update(mid+1,r,T[cnt].r,T[y].r,pos,v);}int query(int rt,int l,int r,int L,int R){    if(L<=l&&R>=r) return T[rt].sum;    int mid=(l+r)>>1;    int ans=0;    if(L<=mid)        ans+=query(T[rt].l,l,mid,L,R);    if(R>mid)        ans+=query(T[rt].r,mid+1,r,L,R);    return ans;}int he(int rt,int l,int r,int k){    if(l==r) return l;    int m=(l+r)>>1;    int tmp=T[T[rt].l].sum;    if(k<=tmp)        return he(T[rt].l,l,m,k);    else        return he(T[rt].r,m+1,r,k-tmp);}void init(){    mp.clear();    cnt=0;//忘记初始化cnt，bug了很长时间。    memset(root,0,sizeof(root));}int main(){    int n,t,q;    int cas=1;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&q);        for(int i=1; i<=n; i++) scanf("%d",&a[i]);        init();        T[n+1].l=T[n+1].r=T[n+1].sum=0;        for(int i=n; i>=1; i--)        {            if(mp[a[i]]==0)                update(1,n,root[i],root[i+1],i,1);            else            {                int tmp;                update(1,n,tmp,root[i+1],mp[a[i]],-1);                update(1,n,root[i],tmp,i,1);            }            mp[a[i]]=i;        }        printf("Case #%d:",cas++);        int ans=0;        while(q--)        {            int x,y,l,r;            scanf("%d%d",&x,&y);            x=(x+ans)%n+1;            y=(y+ans)%n+1;            l=min(x,y);            r=max(x,y);            int k=(query(root[l],1,n,l,r)+1)>>1;            // cout<<k<<endl;            ans=he(root[l],1,n,k);            printf(" %d",ans);        }        printf("\n");    }    return 0;}