HDU 2852 KiKi's K-Number 树状数组

KiKi's K-Number

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1593    Accepted Submission(s): 677

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!

 

忘记了初始化找了好久的BUG。。。。

这道题的操作有三种，插入和删除并到一个函数里完成 ，查询的操作要用到二分的方法。

#include 
#include 
#include 
#include 
using namespace std;
const int M = 100005;
int num[M];
int lowbit( int x )
{
    return x&(-x);    
}

void add( int x , int val )
{
    while( x < M )
    {
           num[x] += val;
           x += lowbit(x);
    }
}



int sum( int x )
{
    int ret = 0;
    while( x > 0 )
    {
           ret += num[x];
           x -= lowbit(x);
    }
    return ret;
}

int find( int pos , int k )
{
    int neww = sum(pos);
    int l , r , m;
    l = pos + 1;
    r = M - 1;
    int ans = M;
    int total = 0;
    while( l <= r )
    {
           m = ( l+r ) / 2;
           total = sum(m) - neww;
           if( total >= k )
           {
               r = m - 1;
               if( m < ans )
                   ans = m;
           }
           else
               l = m + 1;
    }
    return ans;
}

int main( )
{
    int n;
    int c,x,y;
    while( scanf("%d",&n) != EOF )    
    {
           memset( num , 0 , sizeof(num) );
           while( n-- ){
                  scanf("%d",&c);
                  if( c == 0 ){
                      scanf("%d",&x);
                      add( x , 1 ); 
                  }
                  else if( c == 1 ){
                       scanf("%d",&x);
                       if( sum(x) - sum(x-1) == 0 )
                           printf("No Elment!\n");
                       else
                           add( x , -1 );
                  }
                  else if( c == 2 ){
                       scanf("%d%d",&x,&y);
                       int ret = find( x , y );
                       if( ret == M )
                           printf("Not Find!\n");
                       else
                           printf("%d\n",ret);
                  }
           }
               
    }
    return 0;
}