标签：c++ c语言 水题 algorithm namespace
Revenge of Segment Tree
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 205 Accepted Submission(s): 83
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
简单题！！
刚开始看的时候没想法，就不想做了，后来学长发来代码也没看，不过这回排名88，288，488有机械键盘，然后我猥琐的把学长的代码提交了O(∩_∩)O哈哈~
后来自己又看了看，还挺简单的嘛！唉:-(
题意：求一个序列的所有连续子序列的序列和的和。
思路：考虑每个数出现在多少个子序列之中，假设第i个数为Ai，区间为[ L , R ] （1 <= L <= i , i <= R <= n ）。那么包含Ai的区间满足i*（n-i+1）。累加i*（n-i+1）?A[i]就可以了。
注意了，这个 (i*(n-i+1))%maxn 公式里要全部定义为long long， 不然会WA，所以以后有long long就最好全部定义为long long
AC代码：
#include <cstdio>
#include <iostream>
#include <cstring>
#define LL long long
using namespace std;
const LL maxn = 1000000007;
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int n;
LL a, ans=0;
scanf("%d", &n);
for(LL i=1; i<=n; i++)
{
scanf("%I64d", &a);
LL tmp = ((i*(n-i+1))%maxn);
ans += ((a*tmp) % maxn);
ans %= maxn;
}
printf("%I64d\n", ans);
}
return 0;
}
HDU-5086-Revenge of Segment Tree （BestCoder Round #16）
标签：c++ c语言 水题 algorithm namespace
原文地址：http://blog.csdn.net/u014355480/article/details/40686767
