hdu5496

Beauty of Sequence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 668    Accepted Submission(s): 297

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

 

参考链接：

官方题解

http://blog.sina.com.cn/s/blog_61533c9b0100fa7w.html

http://blog.csdn.net/one_piece_hmh/article/details/48897927

我猜到了开头，却没有猜到结尾。想到了可能是求每个数的贡献，但是却没有推出求每个数贡献的简便方法。还是做题少，不思考。

遇到很到的数列求和，或是什么的，有可能是通过求每个数的贡献来求解最后答案。

（首尔府年糕紫菜包饭的菜米饭好吃又便宜）

已ac的代码：

#include<stdio.h>
#include<string.h>
#include<map>
#include<iostream>
using namespace std;
#define N 100010
#define MOD 1000000007

int num[N];
map<int,int> mapex;//傻了吧，其实一维的就够了，把前面的以相同的数为结尾的都加起来，省时间又省空间。

long long int km(long long int x,long long int y){
    long long int ans=1;

    while(y){
        if(y%2){
            ans=ans*x%MOD;
        }
        y=y/2;
        x=x*x%MOD;
    }

    return ans;
}

int main(){
    int t;
    int n;
    long long int ans;

    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);

        for(int i=0;i<n;i++){
            scanf("%d",&num[i]);
        }

        mapex.clear();//记住，是clear，不是erase
        ans=0;
        for(int i=0;i<n;i++){
            //printf("%d %lld %lld\n",mapex[num[i]],km(2,n-i-1),km(2,i));
            long long int tempfro=((km(2,i)-mapex[num[i]])%MOD+MOD)%MOD;
            ans=(ans+tempfro*km(2,n-1-i)%MOD*num[i]%MOD)%MOD;
            mapex[num[i]]=(mapex[num[i]]+km(2,i))%MOD;
        }

        printf("%lld\n",ans);
    }

    return 0;
}