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,…
ios_base::sync_with_stdio(); cin.tie(); ], nxt[MAXM << ], Head[MAXN], ed = ; inline void addedge(int u, int v) { to[++ed] = v; nxt[ed] = Head[u]; Head[u] = ed; } #include<iostream> #include<cstdio> #include<cstdlib> #include<cst…
/* 矩阵乘法+快速幂. 一开始迷之题意.. 这个gcd有个规律. a b b c=a*x+b(x为常数). 然后要使b+c最小的话. 那x就等于1咯. 那么问题转化为求 a b b a+b 就是斐波那契了.... */ #include<iostream> #include<cstdio> #define MAXN 3 #define LL long long #define mod 1000000007 using namespace std; LL n; LL a[MAXN]…
1202: GCD Time Limit: 1 Sec  Memory Limit: 1280 MBSubmit: 201  Solved: 31[Submit][Status][Web Board] Description   Input The first line is an positive integer  T . (1<=T<= 10^3) indicates the number of test cases. In the next T lines, there are thre…
1.gcd int gcd(int a,int b){ return b?gcd(b,a%b):a; } 2.扩展gcd )extend great common divisor ll exgcd(ll l,ll r,ll &x,ll &y) { if(r==0){x=1;y=0;return l;} else { ll d=exgcd(r,l%r,y,x); y-=l/r*x; return d; } } 3.求a关于m的乘法逆元 ll mod_inverse(ll a,ll m){ l…
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…
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…
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 …
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…
输入a和p.如果p不是素数,则若满足ap = a (mod p)输出yes,不满足或者p为素数输出no.最简单的快速幂,啥也不说了. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; typedef long long ll; ll p,a; int whether(int p) { ; ;i*i<=p;i++) ) {…