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

hdu 5086(递推)的更多相关文章

  1. HDOJ(HDU).2044-2049 递推专题

    HDOJ(HDU).2044-2049 递推专题 点我挑战题目 HDU.2044 题意分析 先考虑递推关系:从1到第n个格子的时候由多少种走法? 如图,当n为下方格子的时候,由于只能向右走,所以有2中 ...

  2. HDU 2842 (递推+矩阵快速幂)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2842 题目大意:棒子上套环.第i个环能拿下的条件是:第i-1个环在棒子上,前i-2个环不在棒子上.每个 ...

  3. "红色病毒"问题 HDU 2065 递推+找循环节

    题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=2065 递推类题目, 可以考虑用数学方法来做, 但是明显也可以有递推思维来理解. 递推的话基本就是状态 ...

  4. Children’s Queue HDU 1297 递推+大数

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1297 题目大意: 有n个同学, 站成一排, 要求 女生最少是两个站在一起, 问有多少种排列方式. 题 ...

  5. hdu 2044-2050 递推专题

    总结一下做递推题的经验,一般都开成long long (别看项数少,随便就超了) 一般从第 i 项开始推其与前面项的关系(动态规划也是这样),而不是从第i 项推其与后面的项的关系. hdu2044:h ...

  6. ZOJ 3182 HDU 2842递推

    ZOJ 3182 Nine Interlinks 题目大意:把一些带标号的环套到棍子上,标号为1的可以所以操作,标号i的根子在棍子上时,只有它标号比它小的换都不在棍子上,才能把标号为i+1的环,放在棍 ...

  7. hdu 2604 递推 矩阵快速幂

    HDU 2604 Queuing (递推+矩阵快速幂) 这位作者讲的不错,可以看看他的 #include <cstdio> #include <iostream> #inclu ...

  8. hdu 4055 递推

    转自:http://blog.csdn.net/shiqi_614/article/details/7983298 题意:由数字1到n组成的所有排列中,问满足题目所给的n-1个字符的排列有多少个,如果 ...

  9. HDU 3123-GCC(递推)

    GCC Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Subm ...

随机推荐

  1. [bzoj3371][poj2009][Usaco2004 Mar]Moo University - Emergency Pizza Order 定制比萨饼

    标题这么长的..真是让感觉人头大脚轻. 贴题面先. Description     Moo大学的餐厅必须为$C(1\leq C\leq 1000)$头入学的奶牛新生定制比萨饼.比萨饼可以在比萨小屋订做 ...

  2. Java语言基础---两变量间的交换

    使用中间变量交换两个变量的值 int a = 10 , b = 11 , m; m = a; a = b; b = m; 不使用中间变量交换两个变量的值 int a = 10; int b = 11; ...

  3. python lamba表达式

    lambda函数也叫匿名函数,即,函数没有具体的名称. g=lambda x:x**2 def f(x): return x**2 lambda语句中,冒号前是参数,可以有多个,用逗号隔开,冒号右边是 ...

  4. PJSIP-PJLIB(samples) (the usage of the pjlib lib) (eg:string/I/O)

    Here are some samples about  PJLIB! PJLIB is the basic lib of PJSIP, so we need master the lib first ...

  5. [转]如何像Python高手(Pythonista)一样编程

    本文转自:http://xianglong.me/article/how-to-code-like-a-pythonista-idiomatic-python 最近在网上看到一篇介绍Pythonic编 ...

  6. 抓取网站访问者的QQ号码

    开源,是一种精神.但不开源,并不是没有精神,而可能是代码写得惨不忍睹,我属于后者.(首先申明:对代码提出意见可接受,虚心接受,但不能人身攻击啊!)     最近闲的蛋疼,喜欢到处看看做得好的站点, 莫 ...

  7. 【Minimum Window】cpp

    题目: Given a string S and a string T, find the minimum window in S which will contain all the charact ...

  8. 【志银】#define lowbit(x) ((x)&(-x))原理详解

    分析下列语句 #define lowbit(x) ((x)&(-x)) 可写成下列形式: int Lowbit(x) { return x&(-x); } 例1:x = 1 十进制转二 ...

  9. 201621123033 《Java程序设计》第4周学习总结

    1. 本周学习总结 1.1 写出你认为本周学习中比较重要的知识点关键词 父类 子类 继承 覆盖 抽象 1.2 尝试使用思维导图将这些关键词组织起来.注:思维导图一般不需要出现过多的字. 1.3 可选: ...

  10. PAT 甲级 1011 World Cup Betting

    https://pintia.cn/problem-sets/994805342720868352/problems/994805504927186944 With the 2010 FIFA Wor ...