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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.用樹狀數組的方法，因爲只有N次操作，但是不知道操作數範圍，所以要先離散化一下，那麼就要先記錄；
2.有兩種操作：push pop 在pop的時候怎樣找到位置在(n+1)/2的數呢。。。肯定不能排序 位置在那個地方的數的特點是前面存在一定數量比他小的，因爲是實時的操作，那麼用樹狀數組記錄某個
後記：我覺得這道題很強大。
首先剛開始看的時候我沒有想到，後來聽某牛說用樹狀數組可以解，就試着想了下