HDU 3874 Necklace （树状数组+离线操作）

Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define the beautiful value of some interval  x,yx,y as F(x,y). F(x,y) is calculated as the sum of the beautiful value from the xth ball to the yth ball and the same value is ONLY COUNTED ONCE. For example, if the necklace is 1 1 1 2 3 1, we have F(1,3)=1, F(2,4)=3, F(2,6)=6. 

Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals  L,RL,R (1<=L<=R<=N) and you must tell her F(L,R) of them.

InputThe first line is T(T<=10), representing the number of test cases. 
  For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query. OutputFor each query, output a line contains an integer number, representing the result of the query. Sample Input

2
6
1 2 3 4 3 5
3
1 2
3 5
2 6
6
1 1 1 2 3 5
3
1 1
2 4
3 5

Sample Output

3
7
14
1
3
6

题解：

一开始毫无头绪，百度后发现有种东西叫离线操作，感觉说明白就是在插入数据之前先按照特殊情况处理一下数据，这题就是这样，首先将得到的询问区间按照右端点从小到大排序，如果相同按左区间从小到大排序，这样就避免了询问后面段的区间的时候由于前面段的数据被处理了而出错，然后就按照排好序的询问进行处理，插入到哪处理到哪，整个有点动态同时进行的意思，遇到之前出现过的数就删掉然后重新插入该位置就好了

代码：

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