题目链接：

Beauty of Sequence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 813    Accepted Submission(s): 379

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：

一个数列的美丽值为这个数列合并相邻的且相同的数后这个序列的和,现在给一个序列,求所有子序列的美丽值得和;

思路：

套路题,dp[i]表示以第i个为结尾的所有的序列的美丽值得和,那么就可以递推了

if a[j]==a[i],dp[i]+=dp[j];

else dp[i]+=dp[j]+num[j]*a[i];其中j属于[1,i-1],num[j]表示以j结尾的序列的个数;

这样会超时啊,用前缀和优化,然后开个map记录每个值结尾的序列的个数;

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
const LL mod=1e9+7;
LL dp[maxn],num,sum[maxn];
map<int,LL>mp;
int n,a[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        mp.clear();
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        sum[1]=a[1];num=1;mp[a[1]]=1;
        for(int i=2;i<=n;i++)
        {
            dp[i]=(a[i]+sum[i-1])%mod;
            dp[i]=(dp[i]+(num-mp[a[i]]+mod)%mod*a[i])%mod;
            mp[a[i]]=(mp[a[i]]+num+1)%mod;
            num=(num*2+1)%mod;
            sum[i]=(sum[i-1]+dp[i])%mod;
        }
        printf("%lld\n",sum[n]);
    }
    return 0;
}