BestCoder#16 A-Revenge of Segment Tree

Revenge of Segment Tree

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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

[align=left]Output[/align]
For each test case, output the answer mod 1 000 000 007.

[align=left]Sample Input[/align]

2
1
2
3
1 2 3

[align=left]Sample Output[/align]

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.

这道题是学长提供的思路，不过当时做的时候没注意的超Int 看来大家都喜欢这样出题 卡人。
当时还用java写，各种超时。。。

核心就是每个数字在子串出现的次数遵循 i * (n-i+1)(这里的i是从 1 开始的)

x*i%mod*(n-i+1)%mod;
x 是 每个数字， mod = 1000000007;
一开始没取模，就是这个式子超int