http://acm.hit.edu.cn/hoj/problem/view?id=3152 Dice My Tags (Edit) Source : Time limit : sec Memory limit : M Submitted : , Accepted : You have a dice with M faces, each face contains a distinct number. Your task is to calculate the expected number o
给你两个数字p,a.如果p是素数,并且ap mod p = a,输出“yes”,否则输出“no”. 很简单的板子题.核心算法是幂取模(算法详见<算法竞赛入门经典>315页). 幂取模板子: int pow_mod(int a,int n,int m) { ) ; , m); long long ans = (long long)x * x % m; ) ans = ans * a % m; return (int)ans; } 题目代码也比较简单,有一个坑点是如果用筛素数打表,数组开不了这么大
题目链接:https://ac.nowcoder.com/acm/contest/903/B 题意: 给你 q,n,p,求 q1+q2+...+qn 的和 模 p. 思路:一开始不会做,后面查了下发现有个等比数列求和的快速幂公式,附上链接https://www.cnblogs.com/yuiffy/p/3809176.html AC代码: #include<cstdio> #include<iostream> using namespace std; long long mod;
(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
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
D. The Child and Sequence At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks. Fortunately, Picks remembers how t
Sumdiv Sumdiv Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 15364 Accepted: 3790 Description Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of