标签：acm 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
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
解题思路：
给定n个数的数列，求所有连续区间的和。。最简单的一道题，智商捉急啊，没想到。。。。
对于当前第i个数（i>=1），我们只要知道有多少个区间包含a[i]就可以了，答案是 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。
如图：
上图数列取得不恰当，标号和数正好相等，比如数列1，4，2，4，5，图也是和上图一样的。关键的是标号，而不是具体的数。
代码：
#include <iostream>
#include <stdio.h>
using namespace std;
#define ll long long
const ll mod=1000000007;
ll n;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%I64d",&n);
ll x;
ll ans=0;
for(ll i=1;i<=n;i++)
{
scanf("%I64d",&x);
ans=(ans+i*(n-i+1)%mod*x%mod)%mod;
}
printf("%I64d\n",ans);
}
return 0;
}
注意输出 I64[ACM] HDU 5086 Revenge of Segment Tree（所有连续区间的和）
标签：acm bestcoder round#16
原文地址：http://blog.csdn.net/sr_19930829/article/details/40707173
