Codechef DISTNUM2 Easy Queries node Array segment tree

Description

You is given an array A consisting of N positive integers. 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 kth number in it. If such a number does not exist, i.e. the S have less than K elements, output-1.

All the indices in the queries is 1-based.

Input

The first line of input contains, 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 ai, bi, ci, di, ki.
We'll generate li, ri indices for this query as follows:

i-1th ansi-1 0th ANS0 li = (aimax (ansi-1, 0)biNri = (cimax (ansi-1, 0 )diNliriliri.

Output

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

Constraints

• 1 ≤ N, Q ≤
• 1 ≤ Ai ≤ 109
• 0 ≤ ai, bi, ci, di ≤ N
• 1 ≤ li ≤ ri ≤ N
• 1 ≤ ki ≤ N

Example

Input:4 2 1 1 0 3 0 0 3 2 1 3 0 0 3 3Output:2-123Input:10 109 10 6 3 8 4 9 6 4 2 0 9 to 9 1 3 to 8 1 0 to 2 1 7 6 1 2 4 1 3 6 1 6 each 4 1 8 9 1 3 9Output1 10464

Subtasks

• Subtask #1 (points): Q x N ≤107
• Subtask #2 (points): ki = 1
• Subtask #3 (points): ai = 0, ci = 0
• Subtask #4 (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.

Test instructions

A sequence of n for a given length, where each element has a positive integer.

You need to answer Q for a question:

L,r,k: A sequence in which the elements in the sequence a (s) are labeled in ascending order from L to R (only one of the repeating elements is left).

You need to ask for the value of its K-element, and output-1 if it contains less than k elements.

Subscript numbered starting from 1

Exercises

Segment tree, where each node holds a dynamic array with no repeating elements

Query when the two points OK Complexity O (Q*LOGN*LOGN)

#include <iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>#include<vector>using namespacestd;Const intN = 1e5+Ten, M = 2e2+ One, inf = 2e9, mod = 1e9+7; typedefLong Longll;intN, q;ll ar[n],num[n];vector< ll > da[4*N];voidMerges (vector<ll> &a, vector<ll> &b, Vector<ll> &c) {    intLenB =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++]); } ElseA.push_back (c[lenc++]); }    }     while(LenB <b.size ()) {A.push_back (B[lenb++]); }     while(Lenc <c.size ()) {A.push_back (C[lenc++]); }}voidBuildintKintLintr) {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 (intIintJ,ll x,intKintLintr) {if(I==L&AMP;&AMP;J==R)returnUpper_bound (Da[k].begin (), Da[k].end (), X)-Da[k].begin (); Else {        intMid = (l+r) >>1; if(J<=mid)returnQuery (i,j,x,k<<1, L,mid); Else if(I>mid)returnQuery (i,j,x,k<<1|1, mid+1, R); Else returnQuery (i,mid,x,k<<1, L,mid) +query (mid+1,j,x,k<<1|1, mid+1, R); }}ll Solve (intLintRintk) {intLB =1, RB = n, ans =1;  while(lb<=RB) {        intMid = (LB+RB) >>1; if(Query (L,r,num[mid),1,1, N) >=k) RB = mid-1, ans =mid; ElseLB = mid +1; //cout<<1<<endl;    }    if(Query (L,r,num[ans),1,1, N) <k) {return-1; }    Else returnNum[ans];}intMain () {scanf ("%d%d",&n,&q);  for(intI=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(intI=1; i<=q;i++) {ll a,b,c,d,k; scanf ("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&k); intL = (A*max (PRE,0LL) +b)% n +1; intR = (C*max (PRE,0LL) +d)% n +1; printf ("%d\n", pre =solve (l,r,k)); }}

Codechef DISTNUM2 Easy Queries node Array segment tree