Uva 10006 Carmichael Numbers (快速幂)】的更多相关文章

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…
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…
POJ3641 Pseudoprime numbers p是Pseudoprime numbers的条件: p是合数,(p^a)%p=a;所以首先要进行素数判断,再快速幂. 此题是大白P122 Carmichael Number 的简化版 /* * Created: 2016年03月30日 22时32分15秒 星期三 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring&g…
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…
POJ1995 Raising Modulo Numbers 计算(A1B1+A2B2+ ... +AHBH)mod M. 快速幂,套模板 /* * Created: 2016年03月30日 23时01分45秒 星期三 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #…
Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5532   Accepted: 3210 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
In a galaxy far far away there is an ancient game played among the planets. The specialty of the gameis that there is no limitation on the number of players in each team, as long as there is a captain inthe team. (The game is totally strategic, so so…
ZOJ2150 快速幂,但是用递归式的好像会栈溢出. #include<cstdio> #include<cstdlib> #include<iostream> #include<cmath> using namespace std; long long M,i; #define LL long long int _work(LL a,LL n) { LL ans=1; while(n){ if(n&1){ ans=(ans*a)%M; n--; }…
题意:给定 d , n , m (1<=d<=15,1<=n<=2^31-1,1<=m<=46340).a1 , a2 ..... ad.f(1), f(2) ..... f(d),求 f(n) = a1*f(n-1) + a2*f(n-2) +....+ ad*f(n-d),计算f(n) % m. 析:很明显的矩阵快速幂,构造矩阵, ,然后后面的就很简单了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1…