hdu 2852(二分列挙+ツリー配列)

KiKi's K-Number
Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2254    Accepted Submission(s): 1024
Problem Description
For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations.
Push: Push a given element e to container
Pop: Pop element of a given e from container
Query: Given two elements a and k, query the kth larger number which greater than a in container;
Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?
 
Input
Input some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values:
If p is 0, then there will be an integer e (0 If p is 1, then there will be an integer e (0 If p is 2, then there will be two integers a and k (0   Output For each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".   Sample Input 5 0 5 1 2 0 6 2 3 2 2 8 1 7 0 2 0 2 0 4 2 1 1 2 1 2 2 1 3 2 1 4   Sample Output No Elment! 6 Not Find! 2 2 4 Not Find!   Source 2009 Multi-University Training Contest 4 - Host by HDU   Recommend gaojie         在这个问题上,指定了数个单曲,指定了几种操作(3种).                      0.挿入数eを表す。                      1.削除数eを表示する。                      2.提问比e很大,表k号的小数。         本来应该先读所有的操作,打算处理離散化.数据0#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int MAXN=100000+100; int C[MAXN]; int Lowbit[MAXN]; //C[i]=a[i-lowbit(i)+1]+…+a[i],下表1より //Lowbit[i]=i&(i^(i-1));またはLowbit[i]=i&(-i); //1.クエリー int QuerySum(int p) //元配列の下付き1-pの要素の和を問い合わせる { int nSum=0; while(p>0) { nSum+=C[p]; p-=Lowbit[p]; } return nSum; } //2.修正+初期化 void Modify(int p,int val) //元の配列の下の表はpの元素+valで、C[]配列の中の部分の元素の値の変化を招きます { while(p<=MAXN-10) { C[p]+=val; p+=Lowbit[p]; } } //************************************************************ int Binary_search(int l,int r,int k) { int low,high,mid,tmp; low=l;high=r;tmp=QuerySum(l); while(low<=high) { mid=(low+high)>>1; if(QuerySum(mid)-tmp<k) low=mid+1; else high=mid-1; } return low; } int main() { int n,op,a,b,i; for(i=1;i<MAXN;i++) Lowbit[i]=i&(-i); while(~scanf("%d",&n)) { memset(C,0,sizeof(C)); while(n--) { scanf("%d",&op); if(0==op) { scanf("%d",&a); a++; Modify(a,1); } else if(1==op) { scanf("%d",&a); a++; if(QuerySum(a)-QuerySum(a-1)) { //printf("********succeed"); Modify(a,-1); } else printf("No Elment!"); } else { scanf("%d%d",&a,&b); a++; if(QuerySum(MAXN-10)-QuerySum(a)<b) printf("Not Find!"); else printf("%d",Binary_search(a,MAXN-10,b)-1); } } } return 0; }