*HDU2852 树状数组(求第K小的数)

KiKi's K-Number

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3864    Accepted Submission(s): 1715

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!

 

Source

2009 Multi-University Training Contest 4 - Host by HDU

 

题意：

有三种操作：0 a，将a加入容器，1 a 将a从容器中删去，2 a b 求大于a的第b个数。

代码：

//求大于a的第k个数就是求第sum(a)+k小的数。二分求第k小数。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int A[100005];
int m;
int lowbit(int x)
{
    return x&(-x);
}
void add(int id,int val)
{
    while(id<=100000)
    {
        A[id]+=val;
        id+=lowbit(id);
    }
}
int sum(int id)
{
    int s=0;
    while(id>0)
    {
        s+=A[id];
        id-=lowbit(id);
    }
    return s;
}
int main()
{
    int a,b,p;
    while(scanf("%d",&m)!=EOF)
    {
        memset(A,0,sizeof(A));
        for(int i=0;i<m;i++)
        {
            scanf("%d",&p);
            if(p==0)
            {
                scanf("%d",&a);
                add(a,1);
            }
            else if(p==1)
            {
                scanf("%d",&a);
                if(sum(a)-sum(a-1)==0)
                printf("No Elment!
");
                else add(a,-1);
            }
            else if(p==2)
            {
                scanf("%d%d",&a,&b);
                int tem=sum(a);
                if(sum(100000)-tem<b)
                {
                    printf("Not Find!
");
                    continue;
                }
                tem+=b;
                int lef=1,rig=100000,mid;
                while(lef+1<rig)        //二分结束条件！！！！！！！
                {
                    mid=(lef+rig)>>1;
                    int num=sum(mid);
                    if(num>=tem) rig=mid;
                    else lef=mid;
                }
                printf("%d
",rig);
            }
        }
    }
    return 0;
}