Divisible Group Sums Given a list of N numbers you will be allowed to choose any M of them. So you can choose in NCM ways. You will have to determine how many of these chosen groups have a sum, which is divisible by D. Input Input starts with an inte…

题目问从N个数中取出M个数,有多少种取法使它们的和能被D整除. dp[i][j][k]表示,前i个数取出j个数模D的余数为k的方案数 我用“我为人人”的方式来转移,就从i到i+1转移,对于第i+1个数有取和不取两种选择,然后确定j和k这两个维度的情况. 另外题目说数字是32位有符号整数,所以是会出现负数的...模D之后加D再模D就行了. #include<cstdio> #include<cstring> using namespace std; ][][]; int main()…

题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1125 题意: 给你n个数,q次询问,每次询问问你取其中m个数是d的整数倍的方案数. 题意: dp[i][j][k] 表示前i个数, %d=j, 取了k个的方案数. ID SUBMISSION TIME PROBLEM SOURCE CPU MEMORY VERDICT 839200 2016-10-15 14:59:00 1125 - Divisible Group Sums…

Divisible Group Sums Given a list of N numbers you will be allowed to choose any M of them. So you can choose in NCM ways. You will have to determine how many of these chosen groups have a sum, which is divisible by D. Input Input starts with an inte…

Given a list of N numbers you will be allowed to choose any M of them. So you can choose in NCM ways. You will have to determine how many of these chosen groups have a sum, which is divisible by D. Input Input starts with an integer T (≤ 20), denotin…

称号:给你n数字.免去m一个,这使得他们可分割d.问:有多少种借贷. 分析:dp,D01背包. 背包整数分区. 首先.整点d.则全部数字均在整数区间[0,d)上: 然后,确定背包容量,最大为20*10 = 200,计算二维01背包: 最后,求出全部能整除d的整数取法的和就可以. 说明:注意使用long long防止溢出:注意数据中的负数. #include <algorithm> #include <iostream> #include <cstdlib> #inclu…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

发信人: fennec (fennec), 信区: Algorithm 标 题: acm 总结 by fennec 发信站: 吉林大学牡丹园站 (Wed Dec 8 16:27:55 2004) ACM总结(fennec) 其实在北京比赛完的时候,我就想写了,不过还是早了点,直到上海比赛结束,大家的心中都不是太好受.郭老师有句话:你们这样做也是对的,不成功就成仁.让我的心也能安慰了不少. 我是从大一下学期开始接触ACM的,那时候我们学校才刚刚起步,siyee,wjiang师兄可以说是我的领路人了…

提高自己的实力, 也为了证明, 开始板刷lightoj,每天题量>=1: 题目的类型会在这边说明,具体见分页博客: SUM=54; 1000 Greetings from LightOJ [简单A+B] 1001 Opposite Task  [简单题] 1002 Country Roads[搜索题] 1003 Drunk[判环] 1004 Monkey Banana Problem [基础DP] 1006 Hex-a-bonacci[记忆化搜索] 1008 Fibsieve`s Fantabu…

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2…

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 动态规划 前缀和求余 日期 题目地址:https://leetcode.com/problems/subarray-sums-divisible-by-k/ 题目描述 Given an array A of integers, return the number of (contiguous, non-empty) subarrays that ha…

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2…

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2…

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2…

题目如下: Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5,…

分析 这题场上前缀和都想出来了,然后就没有然后了...哭惹.jpg 前缀和相减能够得到任意一段连续区间的和,然后他们取余\(K\)看余数是否为0就能得到.这是朴素的遍历算法.那么反过来说,如果两个前缀和都对\(K\)余\(p\),那么他们相减一定能够被K整除. 我们就这么统计就可以了,然后一个简单的求和即可.想到不难,难的是想到. 代码 class Solution { public int subarraysDivByK(int[] A, int K) { int[] prefix = new…

前缀和(prefix sum/cumulative sum)的应用. 还用了一个知识点: a≡b(mod d) 则 a-b被d整除. 即:a与b对d同余,则a-b被d整除. class Solution(object): def subarraysDivByK(self, A, K): P = [0] for x in A: P.append((P[-1] + x) % K) count = collections.Counter(P) return sum(v*(v-1)/2 for v in…

1.当设置group=1时: conv = nn.Conv2d(in_channels=, out_channels=, kernel_size=, groups=) conv.weight.data.size() 返回: torch.Size([, , , ]) 另一个例子: conv = nn.Conv2d(in_channels=, out_channels=, kernel_size=, groups=) conv.weight.data.size() 返回: torch.Size([,…

    group语句可以把具有相同键值的数据聚合在一起,与SQL中的group操作有着本质的区别,在SQL中group by字句创建的组必须直接注入一个或多个聚合函数.在Pig Latin中group和聚合函数之间没有直接的关系.     group关键字正如它字面所表达的:将包含了特定的键所对应的值的所有记录封装到一个bag中,之后,用户可以将这个结果传递给一个聚合函数或者使用它做其他一些处理.       触发reduce阶段   数据文件内容如下: [hadoop@vm1 ~]$ cat…

https://stackoverflow.com/questions/2404565/sql-server-difference-between-partition-by-and-group-by They're used in different places. group by modifies the entire query, like: select customerId, count(*) as orderCount from Orders group by customerId…

在上篇文章 .NET应用程序与数据库交互的若干问题 这篇文章中,讨论了一个计算热门商圈的问题,现在在这里扩展一下,假设我们需要从两张表中统计出热门商圈,这两张表内容如下: 上表是所有政区,商圈中的餐饮个数,名为FoodDistrict 下表是所有政区,商圈中的SPA个数,名为SPADistrict 现在要把这两张表,根据政区和商圈合并,然后相加Counts,根据Counts的总大小排序,统计热门商圈和热门政区. 在这里仅讨论合并的问题,以演示在SQLServer和C#中LINQ的实现方法: 通常…

一直以来都想写一点关于kafka consumer的东西,特别是关于新版consumer的中文资料很少.最近Kafka社区邮件组已经在讨论是否应该正式使用新版本consumer替换老版本,笔者也觉得时机成熟了,于是写下这篇文章讨论并总结一下新版本consumer的些许设计理念,希望能把consumer这点事说清楚,从而对广大使用者有所帮助. 在开始之前,我想花一点时间先来明确一些概念和术语,这会极大地方便我们下面的讨论.另外请原谅这文章有点长,毕竟要讨论的东西很多,虽然已然删除了很多太过细节的东…

适用场景:分组数据,为我们查找数据缩小范围. 说明:分配并返回对传入参数进行分组操作后的可枚举对象.分组:延迟 1.简单形式: var q = from p in db.Products group p by p.CategoryID into g select g; 语句描述:使用Group By按CategoryID划分产品. 说明:from p in db.Products 表示从表中将产品对象取出来.group p by p.CategoryID into g表示对p按CategoryI…

首先看下常见的攻击载荷,如下: select count(*),(floor(rand(0)*2))x from table group by x; 然后对于攻击载荷进行解释, floor(rand(0)*2) 查询表中内容大于等于3条会报错.一部分原因,因为floor(rand(0)*2)是有规律和固定的. 如果不理解,可以用数据库做下实验.select floor(rand(0)*2) from table ,数据量最好10条左右.你会发现一串0110110011--有规律的数字. 再然后s…

原文地址:http://www.stum.de/2008/02/06/querying-the-person-or-group-field-using-spquery/ Querying the “Person or Group” field using SPQuery (Update) One of the weak points of Sharepoint 2007 is documentation. If you want to query a “Person or Group” fiel…

order by 排序查询.asc升序.desc降序 示例: select * from 学生表 order by 年龄 ---查询学生表信息.按年龄的升序(默认.可缺省.从低到高)排列显示 也可以多条件排序. 比如 order by 年龄,成绩 desc 按年龄升序排列后.再按成绩降序排列 group by 分组查询. having 只能用于group by子句.作用于组内,having条件子句可以直接跟函数表达式.使用group by 子句的查询语句需要使用聚合函数. 示例: select…

分组语句必须和聚合函数在一起使用, group by子句负责将数据分成逻辑组,聚合函数对每一组进行统计计算 group by 必须放到 select 语句后面,如果select语句中有where子句,则group by必须放到where子句后面 select FAge  from T_Employee where FSubCompany ='beijing' group by FAge 需要分组的所有列都必须位于group by子句中,也就是没有出现在group by 子句中的列(聚合函数除外)…

GROUP BY和HAVING子句 GROUP BY子句 用于将信息划分为更小的组每一组行返回针对该组的单个结果 --统计每个部门的人数: Select count(*) from emp group by deptno; --根据部门分组,并统计 Select deptno, count(*) form emp group by deptno; select deptno, avg(sal) from emp group by deptno; --每个部门的平均工资 HAVING子句 用于指定…

You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k. Define a pair (u,v) which consists of one element from the first array and one element from the second array. Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) wit…