2017.8.21-----二分！！！在树状数组中的巧妙运用

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
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!

 

 

 

大致意思是就是输入1是添加，输入0是移出，输入2是查询，2之后的a，b代表查询大于a的第b个数。

然后删除或查询如果时空的输出样例的那堆英文。

这道题让我彻底的知道了sum的来龙去脉。尤其是在二分的那一部分，理解不了sum是难以理解的。

就是相减的差值个数肯定要大于等于k的~~~

注意细节！！！！！

 

 

其实我本人对二分那个地方还是有很多疑惑的。。。。

 

下面是一个博客里的代码，重点理解他的二分：：：

 

//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
#include
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
      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; //这是二进制去掉最后一位的写法，其实相当于除以2
    total=sum(pos)-newsum;  //这是计算所在位置的和（个数）减去x位置的和（个数）是否大于y
    if(total>=y){
      rightside=pos-1;
      if(pos
          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
           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)  //如果等于M则说明这个数不存在。。
                printf("Not Find!\n");
            else
                printf("%d\n",count);
          }//else
        }//while
    }//while
  return 0;
}//main