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

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heyang | We have carefully selected several similar problems for you:

显然枚举全部区间是不可能的，我们得找找规律什么的，能够发现，设全部数的和是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)

再枚举几个就能够找到规律

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)的时间里就完毕了计算。注意用__int64以及取模

#include #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 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;}

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