HDU 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 ${a_1,a_2,...,a_n}$. 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 $10^9+7$.

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 $n$ $(1 \le n \le 10^5)$, indicating the size of the sequence. The following line contains $n$ integers $a_1,a_2,...,a_n$, denoting the sequence $(1 \le a_i \le 10^9)$.

The sum of values $n$ for all the test cases does not exceed $2000000$.

Output

For each test case, print the answer modulo $10^9+7$

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

找到递推方法然后就好了。。。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn = 300005;
const LL base = 1e9 + 7;
int T, n, m, a[maxn], b[maxn], c[maxn];
LL now, bef, f[maxn], sum[maxn], tot;

int main()
{
  scanf("%d", &T);
  while (T--)
  {
    scanf("%d", &n);    tot = 0;
    for (int i = 1; i <= n; i++)
    {
      scanf("%d", &a[i]);
      b[i] = a[i];
      sum[i] = f[i] = 0;
    }
    sort(b + 1, b + n + 1);
    m = unique(b + 1, b + n + 1) - b;
    now = 1;
    for (int i = 1; i <= n; i++) c[i] = lower_bound(b + 1, b + m, a[i]) - b;
    for (int i = 1; i <= n; i++)
    {
      bef = (now - f[c[i]] + base) % base*a[i] % base;
      (f[c[i]] += now) %= base;
      (sum[c[i]] += tot + bef) %= base;

      ((tot <<= 1) += bef) %= base;
      (now <<= 1) %= base;
    }
    printf("%I64d\n", tot);
  }
  return 0;
}