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hdu 5086

2024-08-08 14:25:32   219人阅读

Revenge of Segment Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1175    Accepted Submission(s): 403


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.
 
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<cstdlib>#include<algorithm>#include<queue>#include<set>#include<vector>using namespace std;const long long MOD=1e9+7;#define LL long longLL a[447005],ans;int t,n;int main(){      scanf("%d",&t);      while(t--)      {            ans=0;            scanf("%d",&n);            for(int i=1;i<=n;i++)                  scanf("%lld",&a[i]);            for(int i=1;i<=n;i++)            {                  ans=(ans+(a[i]*i)%MOD*(n-i+1))%MOD;            }            printf("%lld\n",ans);      }      return 0;}

　　

hdu 5086


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