HDU 5204 Rikka with sequence

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:


Yuta have a sequence. Because the sequence is very self-willed(RenXing), at first the sequence is empty. And then Yuta do n operations on this sequence, each operation is either of these two types:


1.Add a number w into each gap of the sequence. For example if w=3 and the sequence before is “2 4”, it will be changed to “3 2 3 4 3”.
**after the first operation of the first type, there is only one number in the sequence**


2.Query the kth small number in the subsequence [L,R]. For example if k=2, L=2, R=4 and the sequence is “3 2 3 4 2”, the answer will be 3.


Yuta wants Rikka to tell him the answer of each query.


It is too difficult for Rikka. Can you help her?

 

Input

The first line contains one number  n(n≤100000) . Each of the following  n  lines describes an operation: if it is “1 w” it will be the first type. Otherwise if it is “2 L R k”, it will be the second type.  (1≤w≤109,L≤R≤1018)
R will not be larger than the length of the sequence

 

Output

For each query operation, output one number – the answer.

 

Sample Input

   
   
   
   

    
    
    
    6
1 3
1 1
2 2 3 2
1 2
2 3 5 2
2 1 4 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
2
3
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    因为每次加进来的数一定占据了当前的奇数位置，所以暴力的统计区间范围内的数字个数即可。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#include <cstdio>
#include <algorithm>
const int maxn = 200009;
using namespace std;
int n, a[maxn], m, tot;
long long L, R, l, r, K;

struct node
{
    int w;
    long long cnt;
    node(int w = 0, long long cnt = 0) :w(w), cnt(cnt){}
    bool operator<(const node &rhs)const{ return w < rhs.w; }
}p[maxn];

inline void solve()
{
    scanf("%I64d%I64d%I64d", &L, &R, &K);
    int time = 0;
    for (int t = tot - 1; L <= R; L = (L + 1) >>1, R = R >> 1, t--)
    {
        l = L; if (!(l & 1))l++;
        r = R; if (!(r & 1))r--;
        if (l <= r) p[time++] = node(a[t], (r - l) / 2 + 1);            
    }
    sort(p, p + time);
    for (int i = 0; K&&i < tot; i++)
    {
        K -= min(K, p[i].cnt);            
        if (!K) printf("%d\n", p[i].w);
    } 
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        tot = 0;
        while (n--)
        {
            scanf("%d", &m);
            if (m == 1) scanf("%d", &a[tot++]);
            else solve();
        }
    }
    return 0;
}