【poj 3641】 Pseudoprime numbers 【Waterloo Local Contest, 2007.9.23】

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 p^th power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

Output

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

3 210 3341 2341 31105 21105 30 0

Sample Output

nonoyesnoyesyes

这道题是判断p是否为一个a-伪素数的题目，可以用Miller测试解决，下面是代码：

#include<stdio.h>#include<time.h>#include<stdlib.h>#include<iostream>#define ll long longusing namespace std;ll random(ll n){return ((ll)(rand())*rand()*rand())%n+1;}ll read(){ll s=0;char c=getchar();while(c<'0'||c>'9'){c=getchar();}while(c>='0'&&c<='9'){s*=10;s+=c-'0';c=getchar();}return s;}ll q(ll a,ll b,ll k){a%=k;ll r=a,s=1;while(b){if(b&1){s*=r;s%=k;}r*=r;r%=k;b>>=1;}return s;}bool p(ll a,ll n){ll m=n-1;int k=0;while(!(m&1)){k++;m>>=1;}ll x=q(a,m,n);if(x==1||x==n-1){return 0;}while(k--){x=(x*x)%n;if(x==n-1){return 0;}}return 1;}bool check(ll n){int i;if(n==1||(!(n&1))){return 0;}if(n==2){return 1;}for(i=0;i<10;i++){ll x=random(n-2)+1;if(p(x,n)){return 0;}}return 1;}int main(){srand((unsigned)(time(NULL)));ll p,a;p=read();a=read();while(p||a){if(check(p)){printf("no\n");}else{if(q(a,p,p)==a){printf("yes\n");}else{printf("no\n");}}p=read();a=read();}return 0;}

注：随机次数越多，错误几率越小。