HDU - 2852 树状数组查询第k大的数

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!

树状数组可以用来记录比一个数小的数有几个，配合二分查找，就能在(logn)^2的时间复杂度下求得第k大的数

#include<cstdio>
#include<map>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
const int num=100005;
int tree[num];
int mark[num];
int lowbit(int a)
{
    return a&-a;
}
void change(int a,int val)
{
    while(a<num){
        tree[a]+=val;
        a+=lowbit(a);
    }
}
int ask(int a)
{
    int sum=0;
    while(a>0){
        sum+=tree[a];
        a-=lowbit(a);
    }
    return sum;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=-1){
        memset(tree,0,sizeof(tree));
        memset(mark,0,sizeof(mark));
        for(int i=0;i<n;i++){
            int ch;
            scanf("%d",&ch);
            if(ch==0){
                int a;
                scanf("%d",&a);
                change(a,1);
                mark[a]++;
            }
            else if(ch==1){
                int a;
                scanf("%d",&a);
                if(mark[a]==0){
                    printf("No Elment!\n");
                    continue;
                }
                change(a,-1);
                mark[a]--;
            }
            else if(ch==2){
                int a,b;
                scanf("%d %d",&a,&b);
                int exc=ask(a);
                if(ask(num-1)-exc<b){
                    printf("Not Find!\n");
                    continue;
                }
                int l=1,r=num-1;
                while(l<r){
                    int bet=(l+r)/2;
                    if(ask(bet)-exc<b)
                        l=bet+1;
                    else
                        r=bet;
                }
                printf("%d\n",r);
            }
        }
    }
    return 0;
}