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：
1 /*************************************************************************
2 > File Name: BestCode#16_1001.cpp
3 > Author: Enumz
4 > Mail: [email protected]
5 > Created Time: 2014年11月01日 星期六 17时43分53秒
6 ************************************************************************/
7
8 #include<iostream>
9 #include<cstdio>
10 #include<cstdlib>
11 #include<string>
12 #include<cstring>
13 #include<list>
14 #include<queue>
15 #include<stack>
16 #include<map>
17 #include<set>
18 #include<algorithm>
19 #include<cmath>
20 #include<bitset>
21 #include<climits>
22 #define MAXN 447010
23 #define MOD 1000000007
24 using namespace std;
25 long long t;
26 int main()
27 {
28 int T;
29 cin>>T;
30 while (T--)
31 {
32 int N;
33 long long sum=0;
34 scanf("%d",&N);
35 for (int i=1;i<=N;i++)
36 {
37 scanf("%I64d",&t);
38 sum+=t%MOD*i%MOD*(N-i+1)%MOD;
39 sum=sum%MOD;
40 }
41 cout<<sum%MOD<<endl;
42 }
43 return 0;
44 }