KiKi's K-Number

KiKi's K-Number

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 211 Accepted Submission(s): 88

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

 

Recommend

gaojie

/*
树状数组基本应用，查询的时候用二分，二分少了一个等号wa了一下午真心心累
*/
#include<bits/stdc++.h>
#define lowbit(x) x&(-x)
#define N 100005
using namespace std;
int c[N];
int build(int x,int val)//包括存元素，删元素
{
    while(x<N)
    {
        c[x]+=val;
        x+=lowbit(x);
    }
}
int findx(int x)
{
    int cur=;
    while(x>)
    {
        cur+=c[x];
        x-=lowbit(x);
    }
    return cur;
}
int queuy(int p,int k)
{
    int l=p+,r=N-,mid,m;
    int now=findx(p);
    int flag=-;
    while(l<=r)
    {
        //cout<<"l="<<l<<" r="<<r<<endl;
        m=(l+r)/;
        mid=findx(m)-now;
        if(mid>=k&&findx(m-)-now<k)//敲好找到元素个数是k的
        {
            flag=m;
            break;
        }
        else if(mid<k)
            l=m+;
        else
            r=m-;
    }
    return flag;
}
int n,op,a,b;
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF)
    {
        memset(c,,sizeof c);
        for(int ca=;ca<n;ca++)
        {
            scanf("%d",&op);
            if(op==)
            {
                scanf("%d",&a);
                build(a,);
            }
            else if(op==)
            {
                scanf("%d",&a);
                int s=findx(a)-findx(a-);//判断这个位置是不是有数
                if(s==)
                    puts("No Elment!");
                else
                    build(a,-);
            }
            else if(op==)
            {
                scanf("%d%d",&a,&b);
                int s=queuy(a,b);
                if(s==-)
                    puts("Not Find!");
                else
                    printf("%d\n",s);
            }
        }
    }
    return ;
}