HDU 5919 Sequence II 主席树

Sequence II

Problem Description

 

Mr. Frog has an integer sequence of length n, which can be denoted as a1,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 as p(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 to l‘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

 

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

Hint

 

题意：

　　给定一个序列nn，有mm次查询，每次查询一个区间[l,r][l,r]，求区间中每一种数在区间中第一次出现的位置的中位数，强制在线。

题解：

　　强制在线

　　利用主席树求区间不同数的个数

　　这里有个技巧

　　倒着插入主席树

　　在寻找位置的中位数上就可以一个log解决了

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const int N = 2e5 + , M = 1e6, mod = 1e9+, inf = 2e9;

int T,n,q,a[N],l[N*],r[N*],v[N*],sz,root[N],ans[N],last[N];

void update(int x, int &y, int ll, int rr, int k,int c) {
     y = ++sz;
     v[y] = v[x] + c;
     r[y] = r[x];
     l[y] = l[x];
     if(ll == rr) return ;
     int mid = (ll+rr)>>;
     if(k <= mid) update(l[x], l[y], ll, mid, k,c);
     else update(r[x], r[y], mid + , rr, k,c);
}
int query(int x,int ll,int rr,int s,int t) {
    if (s <= ll && rr <= t) return v[x];
    int mid = ll + rr >> , res = ;
    if (s <= mid) res += query(l[x], ll, mid, s, t);
    if (t > mid) res += query(r[x], mid + , rr, s, t);
    return res;
}

int finds(int x,int ll,int rr,int k) {
        if(ll == rr) return ll;
        int mid = ll + rr >> ;
        if(v[l[x]] >= k) return finds(l[x],ll,mid,k);
        else return finds(r[x],mid+,rr,k-v[l[x]]);
}

int main() {
        int cas = ;
        scanf("%d",&T);
        while(T--) {
            scanf("%d%d",&n,&q);
            sz = ;
            memset(last,,sizeof(last));
            memset(v,,sizeof(v));
            memset(l,,sizeof(l));
            memset(r,,sizeof(r));
            ans[] = ;
            root[n+] = ;
            for(int i = ; i <= n; ++i) scanf("%d",&a[i]);
            for(int i = n; i >= ; --i) {
                int x = root[i] = ;
                update(root[i+],x,,n,i,);
                if(last[a[i]]) {
                    update(x,root[i],,n,last[a[i]],-);
                } else root[i] = x;
                last[a[i]] = i;
            }
            int L,R;
            for(int i = ; i <= q; ++i) {
                scanf("%d%d",&L,&R);
                L = ((L + ans[i-])%n) + ;
                R = ((R + ans[i-])%n) + ;
                if(L > R) swap(L,R);
                int sum = query(root[L],,n,L,R)  +  >> ;
                ans[i] = finds(root[L],,n,sum);
            }
            printf("Case #%d: ",++cas);
            for(int i = ; i < q; ++i)  printf("%d ",ans[i]);
            cout<<ans[q]<<endl;
        }
        return ;
}