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.
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