Revenge of Segment Tree(思维)

Revenge of Segment Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1796    Accepted Submission(s): 666

 

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.

题意 : 求所有区间的和 .

题解: 对于当前第i 个数 ,我们只需要计算出一共在多少个子区间中出现a[i] , 再用    C(a[i]) 表示a[i] 在所有子区间的出现次数. 答案是C(a[i]) =   i*(n-i+1)， i代表第i个数前面有多少个数，包括它自己，(n-i+1)代表第i个数后面有多少个数，包括它自己，然后相乘，代表前面的标号和后边的标号两两配对 .

参考 : https://blog.csdn.net/sr_19930829/article/details/40707173

如图：

 

#include <iostream>
#include <cstdio>
using namespace std ;
typedef long long LL ;
const LL mod = 1000000007 ;
int main(){
	int T  ;
	cin >>T ;
	while(T--){
		LL n  ;
		LL x  ;
		LL ans = 0 ;
		scanf("%I64d",&n);
		for(int i = 1 ; i<=n ;i++ ){
			scanf("%I64d",&x);
			ans = (ans +i*(n-i+1)%mod*x%mod)%mod ;
		}
		printf("%I64d\n",ans);
		
	}
	
	return 0 ;
}