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

21231 2 3

 

Sample Output

220
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

 

Recommend

heyang   |   We have carefully selected several similar problems for you:  5089 5088 5085 5084 5082 

 

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;}