CodeChef DISTNUM2 Easy Queries 节点数组线段树

Description

You are given an array A consisting of N positive integers. You have to answer Q queries on it of following type:

• l r k : Let S denote the sorted (in increasing order) set of elements of array A with its indices between l and r. Note that set Scontains distinct elements (i.e. no duplicates).
  You need to find k^th number in it. If such a number does not exist, i.e. the S has less than k elements, output -1.

All the indices in the queries are 1-based.

Input

The first line of input contains two space separated integers N and Q denoting the number of elements in A, and the number of queries, respectively.

The second line of input contains N space separated integers denoting the array A.

Each of the next Q lines contains five integers a_i, b_i, c_i, d_i, k_i.
We will generate l_i, r_i indices for this query as follows:

Let answer for i - 1th query equal ansi - 1. 
For 0th query ans0 = 0. 
Define li = (ai x max(ansi - 1, 0) + bi) mod N + 1, 
ri = (ci x max(ansi-1, 0) + di) mod N + 1. 
If li > ri, then swap li and ri.
	

Output

For each query, output the answer to the query in a single line. If such a number doesn't exist, output -1.

Constraints

• 1 ≤ N, Q ≤ 10⁵
• 1 ≤ A_i ≤ 10⁹
• 0 ≤ a_i, b_i, c_i, d_i ≤ N
• 1 ≤ l_i ≤ r_i ≤ N
• 1 ≤ k_i ≤ N

Example

Input:
4 4
3 2 1 2
0 1 0 3 2
2 0 0 3 4
1 2 1 3 2
2 0 0 3 3

Output:
2
-1
2
3

Input:
10 10
9 10 6 3 8 4 9 6 4 10
0 2 0 9 3
1 9 1 3 3
1 8 1 0 3
1 2 1 7 2
1 6 1 2 3
1 4 1 3 1
1 6 1 6 1
1 4 1 8 1
1 9 1 3 3
1 9 1 2 1

Output:
6
9
10
4
6
3
10
4
6
4

Subtasks

• Subtask #1 (10 points) : Q x N ≤ 10⁷
• Subtask #2 (20 points) : k_i = 1
• Subtask #3 (30 points) : a_i = 0, c_i = 0
• Subtask #4 (40 points) : Original constraints

Explanation

Example #1:

Query 1. Sorted set of elements : {1, 2}. Second number in this is 2.

Query 2. Sorted set of elements : {1, 2, 3}. Fourth number doesn't exist, hence answer is -1.

Query 3. Sorted set of elements : {1, 2}. Second number in this set is 2.

Query 4. Sorted set of elements : {1, 2, 3}. Third number in this set is 3.

 

题意：

　　给定长度为N的序列A，其中每个元素都有正整数。

　　你需要回答Q个询问：

　　　　l,r,k:记s为序列 A下标在l到r之间的元素按照升序排列得到的序列(重复元素只留一个)。

　　　　你需要求出其第k个元素的值，如果包含小于k个元素，则输出-1.

　　　　下标从1开始编号

题解：

　　线段树，每个节点保存不含重复元素的动态数组

　　查询的时候二分就OK 复杂度O( q*logn*logn)

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 1e5+10, M = 2e2+11, inf = 2e9, mod = 1e9+7;
typedef long long ll;
int n, q;
ll ar[N],num[N];
vector< ll > da[4 * N];
void merges(vector<ll> &a, vector<ll> &b, vector<ll> &c)
{
    int lenb = 0 , lenc = 0;
    while(lenb < b.size() && lenc < c.size()) {
        if(b[lenb] == c[lenc]) {
            a.push_back(b[lenb]);
            lenb++, lenc++;
        }else {
        if(b[lenb] < c[lenc]) {
            a.push_back(b[lenb++]);
        } else a.push_back(c[lenc++]);

        }
    }
    while(lenb < b.size()) {
        a.push_back(b[lenb++]);
    }
    while(lenc < c.size()) {
        a.push_back(c[lenc++]);
    }
}

void build(int k,int l,int r) {
    if(r == l) {
        da[k].push_back(ar[l]);
        return ;
    }
    build(k<<1,l,(l+r)/2);build(k<<1|1,(r+l)/2+1,r);
    merges(da[k],da[k<<1],da[k<<1|1]);
}
ll query(int i,int j,ll x,int k,int l,int r) {
    if(i==l&&j==r) return upper_bound(da[k].begin(),da[k].end(),x) - da[k].begin();
    else {
        int mid  = (l+r)>>1;
        if(j<=mid) return query(i,j,x,k<<1,l,mid);
        else if(i>mid) return query(i,j,x,k<<1|1,mid+1,r);
        else return query(i,mid,x,k<<1,l,mid)+query(mid+1,j,x,k<<1|1,mid+1,r);
    }
}

ll solve(int l,int r,int k) {
    int lb = 1, rb = n, ans = 1;
    while(lb<=rb) {
        int mid = (lb+rb)>>1;
        if(query(l,r,num[mid],1,1,n)>=k) rb = mid-1, ans = mid;
        else lb = mid + 1;
      //  cout<<1<<endl;
    }
    if(query(l,r,num[ans],1,1,n)<k) {
        return -1;
    }
    else return num[ans];
}
int main()
{
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++) scanf("%lld",&ar[i]), num[i] = ar[i];
    sort(num+1,num+n+1);
    build(1,1,n);
    ll pre = 0;
    for(int i=1;i<=q;i++) {
        ll a,b,c,d,k;
        scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&k);
        int l = (a*max(pre,0ll)+b) % n + 1;
        int r = (c*max(pre,0ll)+d) % n + 1;
        printf("%d
",pre = solve(l,r,k));
    }
}