USTH-第三次测试-1004-Revenge of Segment Tree

Revenge of Segment Tree
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 23 Accepted Submission(s) : 7
[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.[/hint]

[align=left]Source[/align]
BestCoder Round #16

根据出现次数，一句话分布规律：
ans = (ans + (i * (N - i + 1) % mod) * (x % mod)) % mod;
就是这么简单的次数规律，跟线段数没有半毛钱关系；

S：

#include <stdio.h>
#include <iostream>

using namespace std;

long long N;
const long long mod = 1000000007;

int main()
{
int t;

scanf("%d", &t);

while(t--)
{
long long x;
long long ans = 0;

scanf("%I64d", &N);
for(long long i = 1; i <= N; i++)
{
scanf("%I64d", &x);
ans = (ans + (i * (N - i + 1) % mod) * (x % mod)) % mod;
}
printf("%I64d\n", ans);
}

return 0;
}