[ACM] HDU 5086 Revenge of Segment Tree（全部连续区间的和）

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.

 

Source

BestCoder Round #16

解题思路：

给定n个数的数列，求全部连续区间的和。。最简单的一道题，智商捉急啊，没想到。。

。

。

对于当前第i个数（i>=1），我们仅仅要知道有多少个区间包含a[i]就能够了，答案是 i*(n-i+1)， i代表第i个数前面有多少个数，包含它自己。(n-i+1)代表第i个数后面有多少个数，包含它自己，然后相乘，代表前面的标号和后边的标号两两配对。

如图：

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上图数列取得不恰当，标号和数正好相等。比方数列1，4，2。4。5。图也是和上图一样的。

关键的是标号，而不是详细的数。

代码：

#include <iostream>
#include <stdio.h>
using namespace std;
#define ll long long
const ll mod=1000000007;
ll n;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&n);
        ll x;
        ll ans=0;
        for(ll i=1;i<=n;i++)
        {
            scanf("%I64d",&x);
            ans=(ans+i*(n-i+1)%mod*x%mod)%mod;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

注意输出 I64