题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5496
Beauty of Sequence
Problem Description
Sequence is beautiful and the beauty of an integer sequence is defined as follows: removes all but the first element from every consecutive group of equivalent elements of the sequence (i.e. unique function in C++ STL) and the summation of rest integers is the beauty of the sequence.
Now you are given a sequence A of n integers {a1,a2,...,an}. You need find the summation of the beauty of all the sub-sequence of A. As the answer may be very large, print it modulo 109+7.
Note: In mathematics, a sub-sequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example {1,3,2} is a sub-sequence of {1,4,3,5,2,1}.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤105), indicating the size of the sequence. The following line contains n integers a1,a2,...,an, denoting the sequence(1≤ai≤109).
The sum of values n for all the test cases does not exceed 2000000.
Output
For each test case, print the answer modulo 109+7 in a single line.
Sample Input
3
5
1 2 3 4 5
4
1 2 1 3
5
3 3 2 1 2
Sample Output
240
54
144
题意分析：
题目 是让我们求所有子序和的总和，并且在一个子序列中相邻且相等的数不重复累加（相当于当成一个数）。
题解：
当你一个问题想不通的时候，可以换一个角度来思考。
一开始直接想统计结果，但是明显统计量是天文数字，于是觉得是不是有什么规律，也没想出来，于是就想从反面思考这个问题，既然不能直接求和，那么能不能转而去求每个点对最后的ans的贡献呢。小试了一下，发现可行。
另外一个问题，在一个子序列中相邻相同点只考虑一次。可以选择在这样的子序列中只计算第一个点的贡献值。也就是，考虑一个点的贡献值，只要考虑那些包含它且在它前面没有与它相等的点的所有子序列，我们可以换个角度去求这样的子序列个数，统计所有包含改节点的子序列，然后减去不符合条件的子序列（具体看代码注释）
ac代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<map>
4 using namespace std;
5 typedef long long LL;
6
7 const int maxn = 1e5 + 10;
8 const int mod = 1e9 + 7;
9
10 int a[maxn];
11 int n;
12
13 int bin[maxn];
14 map<int, int> mymap;
15
16 void table() {
17 bin[0] = 1;
18 for (int i = 1; i < maxn; i++) bin[i] = bin[i - 1] * 2 % mod;
19 }
20
21 void init() {
22 mymap.clear();
23 }
24
25 int main() {
26 int tc;
27 table();
28 scanf("%d", &tc);
29 while (tc--) {
30 init();
31 scanf("%d", &n);
32 LL ans = 0;
33 for (int i = 1; i <= n; i++) {
34 scanf("%d", a + i);
35 //mymap[a[i]]表示在i之前,所有以a[i]结尾的子序列的数目
36 //bin[n-1]表示所有包含i的子序列，而mymap[a[i]]*bin[n-i]代表的就是不符合条件的子序列了。
37 LL tmp = ((bin[n - 1] - (LL)bin[n - i] * mymap[a[i]])%mod +mod) % mod;
38 ans = (ans + a[i] * tmp) % mod;
39 mymap[a[i]] += bin[i - 1];
40 mymap[a[i]] %= mod;
41 }
42 printf("%lld\n", ans);
43 }
44 return 0;
45 }