一.说明 主要是对字符串的字段进行hash取模 二.修改配置文件config-sharding.yaml,并重启服务 # # Licensed to the Apache Software Foundation (ASF) under one or more # contributor license agreements. See the NOTICE file distributed with # this work for additional information regarding
一.简要说明 以下配置实现了: 1.分库分表 2.每一个分库的读写分离 3.读库负载均衡算法 4.雪花算法,生成唯一id 5.字段取模 二.配置项 # # Licensed to the Apache Software Foundation (ASF) under one or more # contributor license agreements. See the NOTICE file distributed with # this work for additional informa
(转自:http://www.jb51.net/article/54947.htm) 本文实例汇总了C语言实现的快速幂取模算法,是比较常见的算法.分享给大家供大家参考之用.具体如下: 首先,所谓的快速幂,实际上是快速幂取模的缩写,简单的说,就是快速的求一个幂式的模(余).在程序设计过程中,经常要去求一些大数对于某个数的余数,为了得到更快.计算范围更大的算法,产生了快速幂取模算法.我们先从简单的例子入手:求abmodc 算法1.直接设计这个算法: ; ;i<=b;i++) { ans = ans
Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that
题意:给出一大数K(4 <= K <= 10^100)与一整数L(2 <= L <= 106),K为两个素数的乘积(The cryptographic keys are created from the product of two primes) 问构成K的最小素数是否绝对小于L,若是,则输出BAD p,p为最小素数,否则输出GOOD; 分析:从小到大枚举1~10^6内的素数p,while(p<L)时,判断K是否能被p整除,若能则证明构成K的最小素数绝对小于L,反之则大于L
小明系列故事——师兄帮帮忙 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 3502 Accepted Submission(s): 894 Problem Description 小明自从告别了ACM/ICPC之后,就开始潜心研究数学问题了,一则可以为接下来的考研做准备,再者可以借此机会帮助一些同学,尤其是漂亮的师妹.这不,班里
二分求幂 int getMi(int a,int b) { ; ) { //当二进制位k位为1时,需要累乘a的2^k次方,然后用ans保存 == ) { ans *= a; } a *= a; b /= ; } return ans; } 快速幂取模运算 公式: 最终版算法: int PowerMod(int a, int b, int c) { ; a = a % c; ) { = = )ans = (ans * a) % c; b = b/; a = (a * a) % c; } retur