Mex
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1892 Accepted Submission(s): 625
Problem Description
Mex is a function on a set of integers, which is universally used for impartial game theorem. For a non-negative integer set S, mex(S) is defined as the least non-negative integer which is not appeared in S. Now our problem is about mex function on a sequence.
Consider a sequence of non-negative integers {ai}, we define mex(L,R) as the least non-negative integer which is not appeared in the continuous subsequence from aL to aR, inclusive. Now we want to calculate the sum of mex(L,R) for all 1 <= L <= R <= n.
Input
The input contains at most 20 test cases.
For each test case, the first line contains one integer n, denoting the length of sequence.
The next line contains n non-integers separated by space, denoting the sequence.
(1 <= n <= 200000, 0 <= ai <= 10^9)
The input ends with n = 0.
Output
For each test case, output one line containing a integer denoting the answer.
Sample Input
3 0 1 3 5 1 0 2 0 1 0
Sample Output
5 24
题意：求所有区间mex的总和，定义mex[i，j]为区间[i，j]没有出现过的最小的数
思路：比较好的线段树题
先预处理出mex[1，i] (1=<i<=n)
然后预处理出每个位置 i 的下一个位置 next[i] ，假设现在位置的数为x，那么规定它的下一个位置为它后面离它最近的x的位置
然后 i 从1-n开始搞，每次删除i位置的数，然后在[ i+1，next[i]-1 ]这些位置找到满足位置p，且p位置的mex[i，p]值>value[i]
那么后面到next[i]-1的所有区间的mex都是大于value[i]的，因为mex满足单调性
然后就将mex[ p，next[i]-1 ]区间的值修改为value[i]
最后区间[i+1,n]求和就是以i+1为左端点，i+2~n为右端点的所有区间的mex和
涉及到的操作，区间更新，区间求和，区间求最大