一.框架集成cordova 将cordova集成到现有框架中 一般cordova工程是通过CMD命令来创建一个工程并添加Android.ios等平台,这样的创建方式可以完整的下载开发过程中所需要的的插件.也是最方便和快捷一种方式.因此我们需要用这种方式将我们现有的框架放入到已建好的cordova工程中. 1. 创建我们需要的cordova工程(以手上某项目为例) CMD 命令: $ cordova create ZWYPhone com.centit.zwyphone ZWYPhone 后面三项…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28550   Accepted: 11828   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

E - Find The Multiple Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42035   Accepted: 17654   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

Problem   Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more…

Find The Multiple Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21851 Accepted: 8984 Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only t…

题意: 给定一个n,让求一个M,它是n个倍数并且在k进制之下 M的不同的数字最少. 思路: 这里用到一个结论就是任意两个数可以组成任何数的倍数.知道这个之后就可以用搜索来做了.还有一个问题就是最多找n+1个数,因为由鸽巢原理,这n+1个数当中模上n一定有一个一同的.所以他们一减就是答案.如果找到直接是它的倍数的话,就直接返回. 搜索时保存的是它的余数,如果余数为0 的时候直接返回.还有就是在搜索中并不是直接找余数相同的两个数.而是找余数为0的.当暴力不同元素个数为2的时候,这时候已经算是找余数刚…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 22576   Accepted: 9291   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

题目: Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than…

转载自:優YoU  http://user.qzone.qq.com/289065406/blog/1303946967 以下内容属于以上这位dalao http://poj.org/problem?id=1426 题意 给出一个整数n,(1 <= n <= 200).求出任意一个它的倍数m,要求m必须只由十进制的'0'或'1'组成. 分析 首先暴力枚举肯定是不可能的 1000ms 想不超时都难,而且枚举还要解决大数问题.. 要不是人家把这题放到搜索,怎么也想不到用BFS... 解题方法: B…

#include<stdio.h> #include<stdlib.h> #include<string.h> bool found; void DFS(unsigned __int64 t ,int n,int k) { if(found) return ;//如果已经发现了答案就没搜的必要了 if(t%n==0) {//发现答案,输出,标记变量该true printf("%I64u\n",t); found=true; return ; } if…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18012   Accepted: 7297   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

//Memory Time //2236K 32MS #include<iostream> using namespace std; ]; //保存每次mod n的余数 //由于198的余数序列是最长的 //经过反复二分验证,436905是能存储198余数序列的最少空间 //但POJ肯定又越界测试了...524286是AC的最低下限,不然铁定RE int main(int i) { int n; while(cin>>n) { if(!n) break; mod[]=%n; //初…

题意: 给一个数n,让你找出一个只有1,0,组成的十进制数,要求是找到的数可以被n整除. 用DFS是搜索 当前位数字 (除最高位固定为1),因为每一位都只有0或1两种选择,换而言之是一个双入口BFS. 用DFS也可用queue代替BFS也可. #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #include<queue> #include<alg…

http://poj.org/problem?id=1426 测试了一番,从1-200的所有值都有long long下的解,所以可以直接用long long 存储 从1出发,每次向10*s和10*s+1转移,只存储余数即可, 对于余数i,肯定只有第一个余数为i的最有用,只记录这个值即可 #include <cstdio> #include <cstring> #include <queue> using namespace std; const int maxn=222…

最近在处理SharePoint Office365的相关开发的时候发现了这样一个奇怪的现象: 无法通过API更新Editor field,只要已更新就会throw Exception,由于是Office365的Exception,无法单单从Exception中分析出问题原因,而且奇怪的是新创建的List也存在同样的问题. 试了半天也没找到解决办法,单单看field的SchemaXml也没发现什么比较特殊的地方. 无奈之下save了一个list template,上传到本地的SharePoint…

A - 棋盘问题:在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别.要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C. 解题思路:DFS,在这里有两个搜索方向,同时对每个位置的描述由xy坐标完成,第一次我尝试使用pair+vector保存棋盘位置,用两个数组描述放过棋子的行和列但是由于清除标记没做好WA了.这里是因为DFS搜索中状态转移没确定好,导致清楚标记复杂而出错,改为逐行递归逐列遍历.在这里…

Lucene搜索的时候就要构造查询语句,本篇就介绍下各种Query.IndexSearcher是搜索主类,提供的常用查询接口有: TopDocs search(Query query, int n);//find the top n hits for query TopDocs search(Query query, Filter filter, int n);// find the top n hits for query, applying filter if no-null Query q…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 27433   Accepted: 11408   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

简单的小功能,但是用起来还是蛮爽的.分享出来让更多的人有更快的开发效率,开开心心快乐编程. 如果你还没有使用过composer,你可就out了,看我的教程分享,composer简直就是必备神奇有木有.都说到这个点上了,我们赶紧使用composer进行安装吧. 不急,先来看看效果图是啥样的,不然都没心情没欲望看下去. 啥玩意,不感兴趣?继续看嘛,看完再操作一边才能觉得好在哪里. 有木有感觉很帅气,当然啦,远远不止,还很上档次用起来效果也是杠杠的有木有. 好了好了,抓紧时间安装,不然聊起来真是没完没…

配置文件位于%ES_HOME%/config/elasticsearch.yml文件中,用Editplus打开它,你便可以进行配置.   所有的配置都可以使用环境变量,例如:node.rack: ${RACK_ENV_VAR}  表示环境变量中有一个RACK_ENV_VAR变量. 下面列举一下elasticsearch的可配置项: 1. 集群名称,默认为elasticsearch:cluster.name: elasticsearch 2. 节点名称,es启动时会自动创建节点名称,但你也可进行配…

Mine Number Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Every one once played the game called Mine Sweeping, here I change the rule. You are given an n*m map, every element is a '*' representing a mine, or a '.' representing any other…

Description Roger Wilco is in charge of the design of a low orbiting space station for the planet Mars. To simplify construction, the station is made up of a series of Airtight Cubical Modules (ACM's), which are connected together once in space. One…

Xenia and Weights time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play wi…

Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8435    Accepted Submission(s): 5248 Problem Description There is a rectangular room, covered with square tiles. Each tile is colore…

DZY Loves Chessboard time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output DZY loves chessboard, and he enjoys playing with it. He has a chessboard of n rows and m columns. Some cells of the ches…

Saving Princess claire_ Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2354    Accepted Submission(s): 843 Problem Description Princess claire_ was jailed in a maze by Grand Demon Monster(GDM)…

POJ 1426 Find The Multiple 题意:给定一个整数n,求n的一个倍数,要求这个倍数只含0和1 参考博客:点我 解法一:普通的BFS(用G++能过但C++会超时) 从小到大搜索直至找到满足条件的数,注意最高位一定为1 假设 n=6  k即为当前所求的目标数,不满足条件则进一步递推 (i 为层数(深度),在解法二的优化中体现,此时可以不管) 1%6=1 (k=1) i=1 { (1*10+0)%6=4 (k=10) i=2 { (10*10+0)%6=4 (k=100) i=4…

来源:http://blog.csdn.net/zx13525079024/article/details/25310781 solrconfig.xml配置文件主要定义了SOLR的一些处理规则,包括索引数据的存放位置,更新,删除,查询的一些规则配置. 可以在tomcat的安装路径下找到这个文件C:\Program Files\Apache Software Foundation\Tomcat 8.0\solr\collection1\conf 1.datadir节点 <dataDir>${s…

Red and Black Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6762 Accepted Submission(s): 4284 Problem Description There is a rectangular room, covered with square tiles. Each tile is colored eit…