Miller-Rabin素数检测算法 其基于以下两个定理. Fermat小定理 若n是素数,则∀a(a̸≡0(modn))\forall a(a \not\equiv 0 \pmod{n})∀a(a̸​≡0(modn)),有an−1≡1(modn)a^{n-1} \equiv 1 \pmod{n}an−1≡1(modn). 二次探测定理 若n是素数,则x2≡1(modn)x^2 \equiv 1 \pmod{n}x2≡1(modn)只有平凡根x=±1x=\pm1x=±1,即x=1,x=n−1x=…

一些前置知识可以看一下我的联赛前数学知识 如何判断一个数是否为质数 方法一:试除法 扫描\(2\sim \sqrt{n}\)之间的所有整数,依次检查它们能否整除\(n\),若都不能整除,则\(n\)是质数,否则\(n\)是合数. 代码 bool is_prime(int n){ if(n<2) return 0; int m=sqrt(n); for(int i=2;i<=m;i++){ if(n%i==0) return 0; } return 1; } 方法二.线性筛 用 \(O(n)\)…

#include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<cstdlib> #include<ctime> #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #…

Prime Test Time Limit: 6000MS Memory Limit: 65536K Total Submissions: 29193 Accepted: 7392 Case Time Limit: 4000MS Description Given a big integer number, you are required to find out whether it's a prime number. Input The first line contains the num…

GCD & LCM Inverse Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9756Accepted: 1819 Description Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a and b.…

Prime Test Time Limit: 6000MS Memory Limit: 65536K Total Submissions: 29193 Accepted: 7392 Case Time Limit: 4000MS Description Given a big integer number, you are required to find out whether it's a prime number. Input The first line contains the num…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6664    Accepted Submission(s): 3997 Problem Description Eddy's interest is very extensive, recently he is interested in prime…

遇到了一个题: Description: Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states: Every even integer greater than 2 can be expressed as the sum of two primes. The actual verification…

关于素数的基本介绍请参考百度百科here和维基百科here的介绍 首先介绍几条关于素数的基本定理: 定理1:如果n不是素数,则n至少有一个( 1, sqrt(n) ]范围内的的因子 定理2:如果n不是素数,则n至少有一个(1, sqrt(n) ]范围内的素数因子 定理3:定义f(n)为不大于n的素数的个数,则 f(n) 近似等于 n/ln(n) (ln为自然对数) ,具体请参考here 求不超过n的素数                         本文地址 算法1:埃拉托斯特尼筛法,该算法的…

判断正整数p是否是素数 方法一 朴素的判定   …