HDU5086——Revenge of Segment Tree(BestCoder Round #16)

Revenge of Segment Tree

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.

题目大意：

　　　　给定一个序列，求所有子序列的和，包括其本身。

解题思路：

　　　　求每个数出现的次数，cI即可。sum=c1*a1+c2*a2+...+cn*an

　　　　易推出公式：ci=i*(N-i+1)

　　　　故可以在Ｏ(n)时间内解决问题。

　　　　ps:需要MOD,为保证不超longlong，最好每一步都进行MOD。

Code：

/*************************************************************************
    > File Name: BestCode#16_1001.cpp
    > Author: Enumz
    > Mail: 369372123@qq.com
    > Created Time: 2014年11月01日 星期六 17时43分53秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<list>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<bitset>
#include<climits>
#define MAXN 447010
#define MOD 1000000007
using namespace std;
long long t;
int main()
{
    int T;
    cin>>T;
    while (T--)
    {
        int N;
        long long sum=0;
        scanf("%d",&N);
        for (int i=1;i<=N;i++)
        {
            scanf("%I64d",&t);
            sum+=t%MOD*i%MOD*(N-i+1)%MOD;
            sum=sum%MOD;
        }
        cout<<sum%MOD<<endl;
    }
    return 0;
}