Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25225   Accepted: 13757 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…
Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 40339    Accepted Submission(s): 17813 Problem Description A ring is compose of n circles as shown in diagram. Put natural num…
4 Values whose Sum is 0 Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 17875 Accepted: 5255 Case Time Limit: 5000MS Description The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how man…
POJ2739 Sum of Consecutive Prime Numbers 题目大意:给出一个整数,如果有一段连续的素数之和等于该数,即满足要求,求出这种连续的素数的个数 水题:艾氏筛法打表+尺取法区间推进,0ms水过(注意循环的终止条件) #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include &l…
题目大意 给定N,要求你计算用连续的素数的和能够组成N的种数 题解 先筛选出素数,然后暴力判断即可... 代码: #include<iostream> #include<cstring> using namespace std; #define MAXN 10000 ],cnt; ]; void get_prime() { cnt=; memset(check,false,sizeof(check)); ;i<=MAXN;i++) { if(!check[i]) prime[…
Prime Path Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14091 Accepted: 7959 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit r…
Bound Found Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 4056   Accepted: 1249   Special Judge Description Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration…
参考:https://www.cnblogs.com/baozou/articles/4481191.html #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; bool prime[N]; void primemap() { memset(prime,true,sizeof(prime)); prime[]=prime[]=false; ;i<N;i++…
the algorithm of three version below is essentially the same, namely, Kadane's algorithm, which is of O(n) complexity. https://en.wikipedia.org/wiki/Maximum_subarray_problem the evolution of the implementations is to remove redundancy and do what is…
Self Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 22101   Accepted: 12429 Description In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to…
Specialized Four-Digit Numbers Time Limit: 2 Seconds      Memory Limit: 65536 KB Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadeci…
1007. Maximum Subsequence Sum (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj} where 1 <= i <= j <= K. The …
Sum It Up Problem Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four different sums that eq…
errors made, boundary conditions, <= vs < , decreasing vs increasing , ++, –, '0'/'1' vs 0/1 prototype of sum, return the starting position of c-style string containing the sum, just like sprintf return number of characters successfully read. p1=sum…
A. Case of the Zeros and Ones time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing…
Fibonacci Again Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 43539 Accepted Submission(s): 20797 Problem Description There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n…
Sum Problem Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 338086    Accepted Submission(s): 85117 Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this pro…
Matrix Time Limit: 2 Seconds      Memory Limit: 65536 KB Given an n*n matrix A, whose entries Ai,j are integer numbers ( 0 <= i < n, 0 <= j < n ). An operation SHIFT at row i ( 0 <= i < n ) will move the integers in the row one position…
FatMouse and Cheese Problem Description FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse…
  精选常用: 1.  ctrl+shift+r:打开资源 这可能是所有快捷键组合中最省时间的了.这组快捷键可以让你打开你的工作区中任何一个文件,而你只需要按下文件名或mask名中的前几个字母,比如applic*.xml.美中不足的是这组快捷键并非在所有视图下都能用. 2. ctrl+o:快速outline 如果想要查看当前类的方法或某个特定方法,但又不想把代码拉上拉下,也不想使用查找功能的话,就用ctrl+o吧.它可以列出当前类中的所有方法及属性,你只需输入你想要查询的方法名,点击enter就…
一.功能说明: 可以解决类似 http://****/news 情形,Url路径支持正则匹配. 二.操作步骤: 1.增加URL重写模块: using System; using System.IO; using System.Text.RegularExpressions; using System.Web; using System.Xml; /// <summary> /// URL重写Module /// </summary> public class UrlRewriteM…
一.功能说明: 可以解决类似 http://****/news 情形,Url路径支持正则匹配. 二.操作步骤: 1.增加URL重写模块: using System; using System.IO; using System.Text.RegularExpressions; using System.Web; using System.Xml; /// <summary> /// URL重写Module /// </summary> public class UrlRewriteM…
Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2406 Appoint description: Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "…
人见人爱A+B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 30385    Accepted Submission(s): 20250 Problem Description HDOJ上面已经有10来道A+B的题目了,相信这些题目曾经是大家的最爱,希望今天的这个A+B能给大家带来好运,也希望这个题目能唤起大家对ACM曾经的热爱.…
1 JDBC连接数据库6步 Load the JDBC Driver Establish the Database Connection Create a Statement Object Execute a Query Process the Results Close the Connection 2 事务的4大特性 答:原子性A,一致性C,隔离性I,永久性D 3.select count(*) from student 和select count(id) from student 之间的区…
界面显示: 加密: 解密: 代码实现: public string EncryptString(string str)         {             #region 加密程序             char[] Base64Code = new char[] { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v',…
在一个Activity用完之后应该将之finish掉,但是,之前在学校里自己摸索着开发时并没有太注意这个问题,因为activity无论是否finish掉对功能的影响貌似都不是那么明显(这是读书时候的观点),因此也就没有仔细考虑过activity合适finish这个问题. 最近在公司的项目里遇到这个问题,于是就总结了一下. 假设我们有三个activity,分别叫做A.B.C 第一种情况: 我们先从A跳转到B再跳转到C,每次跳转完成之后都不finish当前的activity,那么到达C之后,如果我们…
How to Type Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6211    Accepted Submission(s): 2804 Problem Description Pirates have finished developing the typing software. He called Cathy to tes…
目录: [C#小知识]C#中一些易混淆概念总结--------数据类型存储位置,方法调用,out和ref参数的使用 [C#小知识]C#中一些易混淆概念总结(二)--------构造函数,this关键字,部分类,枚举 [C#小知识]C#中一些易混淆概念总结(三)--------结构,GC回收,静态成员,静态类 [C#小知识]C#中一些易混淆概念总结(四)---------解析Console.WriteLine() ----------------------------------分割线-----…
Linux下有一个方便的c语言程序排版工具,只要选择恰当的参数,可以轻易地使自己的程序具有统一的风格. 当然首先要安装indent,执行命令:apt-get install indent indent参数众多,可以自行查看man文档,下面是我个人使用的命令: indent -i2 -bli 0 -sob -npsl  input.c -o output.c 其中: -i n:设置缩排的格数 -bli n:设置{ }缩排的格数 -sob:删除多余的空白行 -npsl:程序类型与程序名称放在同一行…