Sequence II HDU - 5919（主席树）

Mr. Frog has an integer sequence of length n, which can be denoted as a1,a2,⋯,ana1,a2,⋯,an There are m queries.

In the i-th query, you are given two integers lili and riri. Consider the subsequence ali,ali+1,ali+2,⋯,ariali,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 as p(i)1,p(i)2,⋯,p(i)kip1(i),p2(i),⋯,pki(i) (in ascending order, i.e.,p(i)1<p(i)2<⋯<p(i)kip1(i)<p2(i)<⋯<pki(i)).

Note that kiki is the number of different integers in this subsequence. You should output p(i)⌈ki2⌉p⌈ki2⌉(i)for the i-th query.
Input
In the first line of input, there is an integer T (T≤2T≤2) denoting the number of test cases.

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

There are two integers lili and riri 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 to l‘i,r‘i(1≤l‘i≤n,1≤r‘i≤n)li‘,ri‘(1≤li‘≤n,1≤ri‘≤n). As a result, the problem became more exciting.

We can denote the answers as ans1,ans2,⋯,ansmans1,ans2,⋯,ansm. Note that for each test case ans0=0ans0=0.

You can get the correct input li,rili,ri from what you read (we denote them as l‘i,r‘ili‘,ri‘)by the following formula:
li=min{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}
li=min{(li‘+ansi−1) mod n+1,(ri‘+ansi−1) mod n+1}

ri=max{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}
ri=max{(li‘+ansi−1) mod n+1,(ri‘+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,⋯,pmp1,p2,⋯,pm”, where x is the case number (starting from 1) and p1,p2,⋯,pmp1,p2,⋯,pm is the answer.
Sample Input
2
5 2
3 3 1 5 4
2 2
4 4
5 2
2 5 2 1 2
2 3
2 4
Sample Output
Case #1: 3 3
Case #2: 3 1

题意：长度为n的序列和m次询问，每次询问是l到r之间按每个数第一次出现的位置排序，问中位数的位置是什么。
思路：因为我们要记录每个数第一次出现的位置。之前bzoj有一个类似的题，bzoj题目的网址
但是这个题需要记录第一次出现的位置，所以我们要从后往前遍历数组，如果之前出现过，就将之前的位置清空，在当前位置记录下来。这个操作在主席树里操作。对于l~r，我们可以算出来这段区间里有多少个数，然后就在转化成了求区间l ~r内第k个数的位置。
代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;const int maxx=2e5+100;
struct node{int l;int r;int sum;
}p[maxx*40];
int a[maxx];
int n,m,tot,root[maxx];inline int build(int l,int r)
{int cur=++tot;p[cur].sum=0;if(l==r) return cur;int mid=l+r>>1;p[cur].l=build(l,mid);p[cur].r=build(mid+1,r);return cur;
}
inline int update(int rot,int l,int r,int pos,int v)
{int cur=++tot;p[cur]=p[rot];p[cur].sum+=v;if(l==r) return cur;int mid=l+r>>1;if(pos<=mid) p[cur].l=update(p[rot].l,l,mid,pos,v);else p[cur].r=update(p[rot].r,mid+1,r,pos,v);return cur;
}
inline int queryI(int rot,int l,int r,int L,int R)//求区间l~r的数字种类数
{if(l<=L&&R<=r) return p[rot].sum;int mid=L+R>>1;int sum=0;if(l<=mid) sum+=queryI(p[rot].l,l,r,L,mid);if(mid<r) sum+=queryI(p[rot].r,l,r,mid+1,R);return sum;
}
inline int queryII(int rot,int L,int R,int num)//求第num个数
{if(L==R) return L;int ret=p[p[rot].l].sum;int mid=L+R>>1;if(num<=ret) return queryII(p[rot].l,L,mid,num);else return queryII(p[rot].r,mid+1,R,num-ret);
}
int main()
{int t,k=0,l,r;scanf("%d",&t);while(t--){tot=0;scanf("%d%d",&n,&m);map<int,int> mp;for(int i=1;i<=n;i++) scanf("%d",&a[i]);root[n+1]=build(1,n);for(int i=n;i>=1;i--){if(mp.find(a[i])==mp.end()) root[i]=update(root[i+1],1,n,i,1);else{int x=update(root[i+1],1,n,mp[a[i]],-1);//x是将之前标记消了的版本，第i个版本在x的基础上建立起来root[i]=update(x,1,n,i,1);}mp[a[i]]=i;}printf("Case #%d:",++k);int ans=0;while(m--){scanf("%d%d",&l,&r);l=(l+ans)%n+1;r=(r+ans)%n+1;if(l>r) swap(l,r);int kk=(queryI(root[l],l,r,1,n)+1)>>1;ans=queryII(root[l],1,n,kk);printf(" %d",ans);}printf("\n");}return 0;
}

努力加油a啊，(^o)/~

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