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…
-->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…
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…
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…
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--; }…
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…
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…
链接:传送门 题意:题目给出费马小定理:Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). 我们知道Miller-Rabin素数测试的算法原理就是基于费马小定理的,因为我们在测试底数的时候只是随机一些 a ,所以可能有的合数就脸一白通过了测试,于是就产生了伪素数这一概念,现在给你一对 p and a,判断 p 是否是以 a 为基的伪素数 思路:对于素数来说是不…
Problem Description Xinlv wrote some sequences on the paper a long time ago, they might be arithmetic or geometric sequences. The numbers are not very clear now, and only the first three numbers of each sequence are recognizable. Xinlv wants to know…
题意: 思路: 对于每个幂次方,将幂指数的二进制形式表示,从右到左移位,每次底数自乘,循环内每步取模. #include <cstdio> typedef long long LL; LL Ksm(LL a, LL b, LL p) { LL ans = 1; while(b) { if(b & 1) { ans = (ans * a) % p; } a = (a * a) % p; b >>= 1; } return ans; } int main() { LL p, a…
#include <cstdio> typedef long long ll; int quick_pow(ll a,ll b,ll mod){ ll ans=; ))ans=(ans*a)%mod; return ans; } int main(){ int z,m,h,a,b,ans; for(scanf("%d",&z);z--;){ scanf(; while(h--)scanf("%d%d",&a,&b),ans=(an…
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…
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 Descriptions 费马定理指出,对于任意的素数 p 和任意的整数 a > 1,满足 ap = a (mod p) .也就是说,a的 p 次幂除以 p 的余数等于 a .p 的某些 (但不是很多) 非素数的值,被称之为以 a 为底的伪素数,对于某个 a 具有该特性.并且,某些 Carmichael 数,对于全部的 a 来说,是以 a为底的伪素数. 给定 2 < p ≤ 1000000000 且 1 < a < p ,判断 p 是否为以 …
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…
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…
A sequence of numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4550 Accepted Submission(s): 1444 Problem Description Xinlv wrote some sequences on the paper a long time ago, they might…
Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5934 Accepted: 3461 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
