杭电 2852 树状数组+二分

题目：

KiKi's K-Number

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1314    Accepted Submission(s): 565

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 <e <100000), means press element e into Container.

If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container  

If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.

 

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!
   
   
   
   

ac代码：

//树状数组的应用。题目里给了3中操作，插入，删除和查询
//插入和删除可以归结为一类操作，用树状数组插入时，每个
//元素的值增加1即可，删除时，每个元素的值减1即可。查询
//操作时，用到了树状数组的查询和二分的方法。设比a大的第
//k大的元素，则设total=num[a+1]+...+num[M],若total比k大，
//则折半，时间复杂度为log(n);
#include <iostream>
#include <cstdio>
using namespace std;
const int M=100002;
int num[M];
int lowbit(int x){
  return x&(-x);
}//lowbit
void add(int pos,int value){
	while(pos<M){
	  num[pos]+=value;
	 // printf("num[%d]=%d\n",pos,num[pos]);
	  pos+=lowbit(pos);
	}
}//add
int sum(int x){
  int total=0;
  while(x>0){
    total+=num[x];
	x-=lowbit(x);
  }
 // printf("total=%d\n",total);
  return total;
}//sum
int find(int x,int y){
  int newsum=sum(x);
  int leftside=x+1;//最左为x+1
  int rightside=M-1;//最右为M-1
  int ans=M;
  int total=0;
  while(leftside<=rightside){
	int pos=(leftside+rightside)>>1;
    total=sum(pos)-newsum;
	if(total>=y){
	  rightside=pos-1;
	  if(pos<ans)
		  ans=pos;
	}//if
	else
		leftside=pos+1;
  }//while
  return ans;
}//find
int main(){
	freopen("1.txt","r",stdin);
	int n;
	while(~scanf("%d",&n)){
		int type;
		int x,y;
		for(int i=0;i<M;++i)
		   num[i]=0;
		while(n--){
		  scanf("%d",&type);
		  if(type==0)
		  {
		    scanf("%d",&x);
			add(x,1);
		  }//if
		  else if(type==1){
		    scanf("%d",&x);
			if(sum(x)-sum(x-1)==0)
				printf("No Elment!\n");
			else
				add(x,-1);
		  }//else if
		  else{
		    scanf("%d%d",&x,&y);
			int count=find(x,y);
			if(count==M)
				printf("Not Find!\n");
			else
				printf("%d\n",count);
		  }//else
		}//while
	}//while
  return 0;
}//main