hdu5496 Beauty of Sequence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 383    Accepted Submission(s): 167

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

这题看了好长时间题解，终于理解了。因为子序列太多，所以可以考虑每一个元素贡献的价值，对于每个数, 我们只算它出现在连续相同元素的第一个时的贡献, 这样会使计算简便很多. 假设当前的数是a[i], 那么i后面的数可以随便选有2^(n-i)种. 考虑a[i]前面的数, 要么一个不选, 要么选择的最后一个数和a[i]不同, 那么我们只要把前面出现过的i的位置记录下来，分别为b,c,d,..那么总的个数为2^(i-1)-2^(b-1)-2^(c-1)-...这样就可以算出来了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
#define maxn 100070
#define MOD 1000000007
map<int,int>mp;
map<int,int>::iterator it;
ll a[maxn],cishu[maxn];
ll kuaisumi(ll a,ll b,int c)
{
  ll ans = 1;
  a=a%c;
  while(b>0)
  {
      if(b%2==1)
      ans=(ans*a)%c;
      b=b/2;
      a=(a*a)%c;
  }
  return ans;
}
ll mul(ll x){
    return kuaisumi(2,x,MOD);
}

int main()
{
    ll m,i,j,T;
    ll n;
    scanf("%lld",&T);
    while(T--)
    {
        scanf("%lld",&n);
        for(i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        memset(cishu,0,sizeof(cishu));
        mp.clear();
        ll ans=0;
        for(i=1;i<=n;i++){
            if(!mp.count(a[i])){
                ans=(ans+a[i]*mul(n-1))%MOD;
            }
            else{
                ans=(ans+ a[i]*( mul(i-1)-mp[a[i] ])%MOD*mul(n-i)%MOD  )%MOD;
            }
            mp[a[i] ]=(mp[a[i] ]+mul(i-1) )%MOD;

        }
        printf("%lld
",(ans+MOD)%MOD);

    }
    return 0;
}