Revenge of Segment Tree

Problem Description
In computer science, a segment tree is a tree data structure for storing intervals, or segments. It allows querying which of the stored segments contain a given point. It is, in principle, a static structure; that is, its content cannot be modified once the structure is built. A similar data structure is the interval tree.
A segment tree for a set I of n intervals uses O(n log n) storage and can be built in O(n log n) time. Segment trees support searching for all the intervals that contain a query point in O(log n + k), k being the number of retrieved intervals or segments.
---Wikipedia
Today, Segment Tree takes revenge on you. As Segment Tree can answer the sum query of a interval sequence easily, your task is calculating the sum of the sum of all continuous sub-sequences of a given number sequence.
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence.
[Technical Specification]
1. 1 <= T <= 10
2. 1 <= N <= 447 000
3. 0 <= Ai <= 1 000 000 000
Output
For each test case, output the answer mod 1 000 000 007.
Sample Input
2 1 2 3 1 2 3
Sample Output
2 20
Hint
For the second test case, all continuous sub-sequences are [1], [2], [3], [1, 2], [2, 3] and [1, 2, 3]. So the sum of the sum of the sub-sequences is 1 + 2 + 3 + 3 + 5 + 6 = 20. Huge input, faster I/O method is recommended. And as N is rather big, too straightforward algorithm (for example, O(N^2)) will lead Time Limit Exceeded. And one more little helpful hint, be careful about the overflow of int.

题目大意：

　　　　给定一个序列，求所有子序列的和，包括其本身。

解题思路：

　　　　求每个数出现的次数，cI即可。sum=c1*a1+c2*a2+...+cn*an

　　　　易推出公式：ci=i*(N-i+1)

　　　　故可以在Ｏ(n)时间内解决问题。

　　　　ps:需要MOD,为保证不超longlong，最好每一步都进行MOD。

Code：

 /*************************************************************************

     > File Name: BestCode#16_1001.cpp

     > Author: Enumz

     > Mail: 369372123@qq.com

     > Created Time: 2014年11月01日 星期六 17时43分53秒

  ************************************************************************/

 #include<iostream>

 #include<cstdio>

 #include<cstdlib>

 #include<string>

 #include<cstring>

 #include<list>

 #include<queue>

 #include<stack>

 #include<map>

 #include<set>

 #include<algorithm>

 #include<cmath>

 #include<bitset>

 #include<climits>

 #define MAXN 447010

 #define MOD 1000000007

 using namespace std;

 long long t;

 int main()

 {

     int T;

     cin>>T;

     while (T--)

     {

         int N;

         long long sum=;

         scanf("%d",&N);

         for (int i=;i<=N;i++)

         {

             scanf("%I64d",&t);

             sum+=t%MOD*i%MOD*(N-i+)%MOD;

             sum=sum%MOD;

         }

         cout<<sum%MOD<<endl;

     }

     return ;

 }