hdu5086——Revenge of Segment Tree

Revenge of Segment Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 383    Accepted Submission(s): 163

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

 

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显然枚举所有区间是不可能的，我们得找找规律什么的，可以发现，设所有数的和是sum， S1(区间长度为1）的是sum，S2 = 2 * sum - (a1 + an)
S3 = 3 * sum - (2 * a1 + a2 + 2 *an + a1)
再枚举几个就可以找到规律

所以，总的和里，从左往右看 a1出现了(n-1)*n/2次，a2是(n - 2)*(n - 1)/2次........................
从右往左看，an出现了(n-1)*n/2次，an-1是(n - 2)*(n - 1)/2次........................

所以在O(n)的时间里就完成了计算，注意用__int64以及取模

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

__int64 a[447100];
__int64 b[447100];
const __int64 mod = 1000000007;

int main()
{
	int t, n;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		__int64 ans = 0, x;
		__int64 sum = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%I64d", &x);
			b[i] = x;
			a[i] = (__int64)(n - i) * (1 + n - i) / 2 % mod;
			sum += x;
			sum %= mod;
		}
		for (int i = 1; i <= n; i++)
		{
			a[i] = (__int64)a[i] * b[i] % mod;
		}
		for (int i = n; i >= 1; i--)
		{
			a[i] += (__int64)(i - 1) * i  / 2 % mod * b[i] % mod;
		}
		ans = (__int64) n * (n + 1) / 2 % mod * sum % mod;
		for (int i = 1; i <= n; i++)
		{
			ans -= a[i];
			ans %= mod;
			if (ans < 0)
			{
				ans += mod;
			}
			ans %= mod;
		}
		printf("%I64d\n", ans);
	}
	return 0;
}