Pseudoprime numbers（快速幂模+素性测试）

Pseudoprime numbers

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, 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 2
10 3
341 2
341 3
1105 2
1105 3
0 0

Sample Output

no
no
yes
no
yes
yes

题意：

给定p和a，判断p是不是以a为基础的伪素数

即不是素数，但满足以a为底数时的费马小定理公式 ap≡a (mod p) a p ≡ a   ( m o d   p )

分析：

直接快速幂取模判断是否满足上面式子，且不是素数则是伪素数

如果知道费马小定理的人都知道费马小定理的公式其实是写成

ap−1≡1 (mod p) a p − 1 ≡ 1   ( m o d   p )

但是不能用这个式子判断是否满足，wa了

满足这个式子需要一个条件那就是p不能整除a

但是题目告诉了a < p所以我也不知道为什么了。。。希望有大佬解释一下

两面同乘a是为了得到一个对所有整数a都成立的陈述，这种形式使用起来更加方便

题目也给的第一个式子那就用第一个式子判断吧。。

code：

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int S = 8;
ll mul_mod(ll a,ll b,ll mod){
    a %= mod;
    b %= mod;
    ll ans = 0;
    while(b){
        if(b & 1){
            ans += a;
            if(ans >= mod) ans -= mod;
        }
        a <<= 1;
        if(a >= mod) a -= mod;
        b >>= 1;
    }
    return ans;
}

ll q_pow(ll a,ll b,ll mod){
    ll ans = 1;
    a %= mod;
    while(b){
        if(b & 1){
            ans = mul_mod(ans,a,mod);
        }
        a = mul_mod(a,a,mod);
        b >>= 1;
    }
    return ans;
}

bool check(ll a,ll n,ll x,ll t){
    ll ret = q_pow(a,x,n);
    ll last = ret;
    for(int i = 1; i <= t; i++){
        ret = mul_mod(ret,ret,n);
        if(ret == 1 && last != 1 && last != n-1) return true;
        last = ret;
    }
    if(ret != 1) return true;
    return false;
}

bool Miller_Rabin(ll n){
    if(n < 2) return false;
    if(n == 2) return true;
    if((n & 1) == 0) return false;
    ll x = n - 1;
    ll t = 0;
    while((x & 1) == 0){
        x >>= 1;
        t++;
    }
    for(int i = 0; i < S; i++){
        ll a = rand() % (n-1) + 1;
        if(check(a,n,x,t))
            return false;
    }
    return true;
}
int main(){
    ll p,a;
    while(scanf("%lld%lld",&p,&a) != EOF){
        if(p == 0 && a == 0) break;
        if(q_pow(a,p,p) == a && !Miller_Rabin(p)){
            printf("yes\n");
        }
        else printf("no\n");
    }
    return 0;
}