HDU 5204 Rikka with sequence (Bestcoder Round #37)

Rikka with sequenceTime limit:2000/1000 MS (java/others) Memory Limit: 65536/65536 K (java/others)
Total submission (s): 378 Accepted Submission (s): 75


Problem Descriptionas We know, Rikka is poor at math. Yuta is worrying on this situation, so he gives Rikka some math tasks to practice. There is one of the them:


Yuta has a sequence. Because the sequence is very self-willed (renxing) in first the sequence is empty. And then Yuta does n operations on the sequence, each operation is either of these:


1.ADD a number w into each gap of the sequence. For example if w=3 and the sequence before is "2 4", it'll 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 are "3 2 3 4 2", the answer would be 3.


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


It is too difficult for Rikka. Can you help her?
Inputthe First line contains one number n(n≤100000) . Each of the following n Lines describes an operation:if it is "1 W" it'll be the first type. Otherwise if it is "2 L R K", it'll be the the second type. (1≤W≤ ten 9 ,L≤R≤ ten )
R won't is larger than the length of the sequence

Outputfor each query operation, output one number–the answer.
Sample Input

61 31 12 2 3 21 22 3 5 22 1 4 4

Sample Output

323

official:

The question looks like a very powerful one. Actually, it's a flood problem. For an inquiry, consider this inquiry before the first
      
          i   
  
 
      
 Modify operation, the first position in the sequence is
     
      
           2i? 1  
 

      
     。 And then, within the scope of the inquiry, only
      
          -   
  
 
      
 The number, the violent dafa is good, the complexity of time
      
          o(nlOgR) 
  
 
      
 ，
      
          R   
  
 
      
 is the maximum subscript for the inquiry.

PS: Self-simulation will find that the number I want to insert, the number of inserts in the entire interval is 2^ (i-1), and in [1,r] also meet, the number of the number of inserts after a number must be twice times the number of previous inserts, or twice times +1, know this after, [l,r]=[1,r]-[1 , L-1], so sum[i]+=v* (x+1)/2; in [1,r] v is 1, in [1,l-1] v is-1, the Order of I at this time is the opposite of the insertion order.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace    std;const int maxn=110000;int a[maxn],b[maxn];long long sum[maxn];int len=69;void digt (Long long x,int v) {int i=0;        while (x) {sum[i]+=v* (x+1)/2;        X=X/2;    i++; }}bool cmp (int x,int y) {return a[len-x]<a[len-y];}    int main () {int t,op;    scanf ("%d", &t);    Long Long L,r,k,ans;        while (t--) {scanf ("%d", &op);        if (op==1) {scanf ("%d", &a[++len]);            } else {scanf ("%i64d%i64d%i64d", &l,&r,&k);            for (int i=0;i<=100;i++) sum[i]=0;            Digt (r,1);            Digt (l-1,-1);            for (int i=0;i<=69;i++) b[i]=i;            Sort (b,b+70,cmp);            ans=0;                for (int i=0;i<70;i++) {//cout<<i<< "" <<sum[b[i]]<<endl; if (ans+sum[b[i]]> =k) {printf ("%d\n", A[len-b[i]]);                Break                } else {ans+=sum[b[i]]; }}}} return 0;}

HDU 5204 Rikka with sequence (Bestcoder Round #37)