Revenge of Segment Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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.

Source

BestCoder Round #16

简单题！！

刚开始看的时候没想法，就不想做了，后来学长发来代码也没看，不过这回排名88，288，488有机械键盘，然后我猥琐的把学长的代码提交了O(∩_∩)O哈哈~

后来自己又看了看，还挺简单的嘛！唉:-(

题意：求一个序列的所有连续子序列的序列和的和。

思路：考虑每个数出现在多少个子序列之中，假设第i个数为Ai，区间为[ L , R ] （1 <=  L <= i ,  i <= R  <= n  ）。那么包含Ai的区间满足i*（n-i+1）。累加i*（n-i+1）?A[i]就可以了。

注意了，这个  (i*(n-i+1))%maxn  公式里要全部定义为long  long，  不然会WA，所以以后有long long就最好全部定义为long  long

AC代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#define LL long long
using namespace std;

const LL maxn = 1000000007;

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int n;
		LL a, ans=0;
		scanf("%d", &n);
		for(LL i=1; i<=n; i++)
		{
			scanf("%I64d", &a);
			LL tmp = ((i*(n-i+1))%maxn);
			ans += ((a*tmp) % maxn);
			ans %= maxn;
		}
		printf("%I64d\n", ans);
	}
	return 0;
}