欢迎访问我的新博客:http://www.milkcu.com/blog/ 原文地址:http://www.milkcu.com/blog/archives/uva10006.html 原创:Carmichael Numbers - PC110702 作者:MilkCu 题目描述  Carmichael Numbers  An important topic nowadays in computer science is cryptography. Some people even think…

UVa 10006 - Carmichael Numbers An important topic nowadays in computer science is cryptography. Some people even think that cryptography is the only important field in computer science, and that life would not matter at all without cryptography. Alvar…

  Carmichael Numbers  An important topic nowadays in computer science is cryptography. Some people even think that cryptography is the only important field in computer science, and that life would not matter at all without cryptography. Alvaro is one…

题目链接:UVA10006 本来想直接打素数表,然后根据素数表来判断,结果一直超时,后来把素数表去掉,再在for循环中加判断才勉强过了. Some numbers that are not prime still pass the Fermat test with every number smaller than themselves. These numbers are called Carmichael numbers. 只要按着这两个条件判断即可. 具体看代码: #include<ios…

UVA10006 - Carmichael Numbers(筛选构造素数表+高速幂) 题目链接 题目大意:假设有一个合数.然后它满足随意大于1小于n的整数a, 满足a^n%n = a;这种合数叫做Carmichael Numbers. 题目给你n.然你推断是不是Carmichael Numbers. 解题思路:首先用筛选法构造素数表.推断n是否是合数,然后在用高速幂求a^2-a^(n - 1)是否满足上述的式子.高速幂的时候最好用long long ,防止相乘溢出. 代码: #include <…

-->Carmichael Numbers  Descriptions: 题目很长,基本没用,大致题意如下 给定一个数n,n是合数且对于任意的1 < a < n都有a的n次方模n等于a,这个数就是Carmichael Number. 输出The number n is a Carmichael number. n是素数 输出 n is normal. Input 多组输入,第一行给一个n (2 < n < 65000) .n = 0 表示输入结束并不需要处理 Output 对…

题意:给你一个数,让你判断是否是非素数,同时a^n%n==a (其中 a 的范围为 2~n-1) 思路:先判断是不是非素数,然后利用快速幂对每个a进行判断 代码: #include <iostream> #include <cmath> #include <cstdio> #include <algorithm> #define ll long long using namespace std; bool isprime(ll num) { ) return…

#include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> using namespace std; typedef long long LL; + ; //int prime[maxn + 1]; //第i个素数,保存区间内素数 bool is_prime[maxn]; //is_prime[i]为true表示i是素…

   当今计算机科学的一个重要的领域就是密码学.有些人甚至认为密码学是计算机科学中唯一重要的领域,没有密码学生命都没有意义. 阿尔瓦罗就是这样的一个人,它正在设计一个为西班牙杂烩菜饭加密的步骤.他在加密算法中应用了一些非常大的素数.然而确认一个非常大的数是不是素数并不是那么简单.一个费时的方法是用比这个数的平方根小的所有素数去除它,对于大整数来说,这样一定会毁掉这个杂烩菜饭的. 然而,一些很有信心耗时少的随机测试存在,其中一个就是费马测试. 在2和n-1之间随机选取一个数(n是我们要测试的数).…

 强伪素数 题目大意:利用费马定理找出强伪素数(就是本身是合数,但是满足费马定理的那些Carmichael Numbers) 很简单的一题,连费马小定理都不用要,不过就是要用暴力判断素数的方法先确定是不是素数,然后还有一个很重要的问题,那就是a和p是不互质的,不要用a^(p-1)=1(mod p)这个判据,比如4^6=4(mod 6),但是4^5=4(mod 6) #include <iostream> #include <functional> #include <algo…

题目连接 http://poj.org/problem?id=3641 Pseudoprime numbers Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but…

题目链接 题意:题目定义了Carmichael Numbers 即 a^p % p = a.并且p不是素数.之后输入p,a问p是否为Carmichael Numbers? 坑点:先是各种RE,因为poj不能用srand()...之后各种WA..因为里面(a,p) ?= 1不一定互素,即这时Fermat定理的性质并不能直接用欧拉定理来判定..即 a^(p-1)%p = 1判断是错误的..作的 #include<iostream> #include<cstdio> #include&l…

Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a ps…

Problem Description Fermat's theorem states that for any prime number p and for any integer a > 1, a^p == a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as…

Problem Description Fermat's theorem states that for any prime number p and for any integer a > 1, a^p == a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as…

Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11336   Accepted: 4891 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power…

Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes,…

Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10903   Accepted: 4710 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power…

Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes,…

Pseudoprime numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7954 Accepted: 3305 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and…

Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a ps…

题目链接:POJ 3641 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, know…

本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! Description Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the disc…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6044   Accepted: 2421 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power…

http://www.cnblogs.com/sxiszero/p/3618737.html 下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成.打牢基础,厚积薄发. 一.UVaOJ http://uva.onlinejudge.org 西班牙Valladolid大学的程序在线评测系统,是历史最悠久.最著名的OJ. 二.<算法竞赛入门经典> 刘汝佳  (UVaOJ  351道题) 以下部分内容摘自:http://sdkd…

(Step1-500题)UVaOJ+算法竞赛入门经典+挑战编程+USACO 下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成.打牢基础,厚积薄发. 一.UVaOJ http://uva.onlinejudge.org 西班牙Valladolid大学的程序在线评测系统,是历史最悠久.最著名的OJ. 二.<算法竞赛入门经典> 刘汝佳  (UVaOJ  351道题) 以下部分内容摘自:http://sdkdacm.5d6d.…

POJ 1852 Ants POJ 2386 Lake Counting POJ 1979 Red and Black AOJ 0118 Property Distribution AOJ 0333 Ball POJ 3009 Curling 2.0 AOJ 0558 Cheese POJ 3669 Meteor Shower AOJ 0121 Seven Puzzle POJ 2718 Smallest Difference POJ 3187 Backward Digit Sums POJ 3…

新年新气象,也希望新年可以挣大钱.不管今年年底会不会跟去年一样,满怀抱负却又壮志未酬.(不过没事,我已为各位卜上一卦,卦象显示各位都能挣钱...).已经上班两天了,公司大部分人还在休假,而我早已上班,估计今年我就是加班狗的命.(不说了,要坚强...) 以上扯淡已毕,下面言归正传. 这次的.NET加密解析系列中,前面已经讲解了散列加密.对称加密.数字签名三种加密方式,在这篇博文种,将会主要讲解非对称加密的原理,以及非对称加密在.NET种的应用. 一.非对称加密概述: 前面讲解过对称加密,对称加密中…

Given a prime P, 2 <= P < 2 31, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that B L == N (mod P) Input Read several lines of input, each…