2009_10-5_Practice1 E

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Special Container

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Design a container, which has two kinds of operation, push and pop.

Push: You should push the given number into the container.

Pop: Please find the middle number of the container. If these is n numbers in container, it will be the (n+1)/2-th num when sort increased. Then pop the number.

Input

The input contains one or more data sets. At first line of each input data set is an integer N (1<= N <= 100000) indicate the number of operations.

Then N lines follows, each line contains a number (0 or 1). "0" means a push operation, it's followed by an integer E. "1" means a pop operation.

You may assume all the numbers in the input file will be in the range of 32-bit integer.

Output

For each pop operation, you should print the integer popped. Please print "No Element!", if there is no number to pop. Please print a blank line after each data set.

 

Sample Input

9

0 10

0 -10

0 5

1

1

0 2

1

1

1

3

0 2

0 1

1

 

Sample Output

5

-10

2

10

No Element!

 

1

 

 

1.用树状数组的方法，因为只有N次操作，但是不知道操作数范围，所以要先离散化一下，那么就要先记录；

2.有两种操作：push pop 在pop的时候怎样找到位置在(n+1)/2的数呢。。。肯定不能排序 位置在那个地方的数的特点是前面存在一定数量比他小的，因为是实时的操作，那么用树状数组记录某个

 

后记：我觉得这道题很强大。

首先刚开始看的时候我没有想到，后来听某牛说用树状数组可以解，就试着想了下