hdu 5086 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.
 
题意：
计算一个整数序列所有子序列的和。
 
/*
	题目叫做【线段树的复仇】，真是看呆我，其实是道找规律的题目
	只要计算A[i]的出现次数，直接乘，然后相加就行。
	
	至于出现次数，通过1-6个数的序列的各数字出现次数，每一斜行都是呈倍数递增的，第一斜行是1 2 3 4 5 6，第二斜行是2 4 6 8 10...
	所以易得第n行的第i个数的出现次数为i*(n-i+1)。
*/
 
代码：
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define mod 1000000007
using namespace std;
__int64 d[447100];
int main()
{
	//freopen("C:\\test.txt","r",stdin);
	//freopen("C:\\res.txt","w",stdout);
	int t;
	__int64 i,n;
	__int64 sum;
	scanf("%d",&t);
	while(t--)
	{
		sum=0;
		scanf("%I64d",&n);
		for(i=0;i<n;i++)
			scanf("%I64d",&d[i]);
		for(i=0;i<n;i++)
		{
			sum=sum+((((i+1)*(n-i))%mod)*d[i])%mod;
			sum%=mod;
		}
		printf("%I64d\n",sum);
	}
	return 0;
}