hdu 5086(递推)

Revenge of Segment Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1541    Accepted Submission(s): 552

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

 

题意:求一个序列所有的连续子序列之和。

题解:假设序列为 1 2 3

那么合法序列有:

1 第一项

1 2 第二项

2

1 2 3 第三项

2 3

3

dp[i]代表第i项 那么我们可以看出 dp[i] = dp[i-1]+i*a[i]

最终答案累加即可。

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long LL;
const int mod = 1000000007;
const int N = 447005;
int n;
LL a[N];
LL dp[N];
int main()
{

    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        dp[1]  = a[1];
        for(int i=2;i<=n;i++){
            dp[i] = (dp[i-1] + (i*a[i])%mod)%mod;
        }
        LL ans = 0;
        for(int i=1;i<=n;i++){
            ans = (ans+dp[i])%mod;
        }
        printf("%lld
",ans);
    }
    return 0;
}