poj3641 pseudoprime numbers – 快速幂运算模板

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

定义伪素数满足a^p = a (mod p),给出a和p的值，问a^p是否为伪素数。

题解

用快速幂运算计算出a^p%p的值之后和a比较即可。

代码

#define ll long long
using namespace std;
ll (ll a,ll b,ll c){
    ll ans=1;
    while(b){
        if(b&1) ans = (ans*a)%c;
        a = (a*a)%c;
        b>>=1;
    }
    return ans;
}
ll quick_mul(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1) ans = ans*a;
        a = a*a;
        b>>=1;
    }
    return ans;
}
bool judge_prime(ll n){
    for(int i=2;i*i<=n;++i){
        if(n%i==0)
            return false;
    }
    return true;
}
int main(){
    int p,a;
    while(~scanf("%d%d",&p,&a)){
        if(p==0 && a==0) break;
        ll temp = quick_mul(a,p);
        if(!judge_prime(p) && a==quick_mod(a,p,p)){
            printf("yesn");
        }
        else
            printf("non");
    }
}

poj3641 pseudoprime numbers – 快速幂运算模板

题目连接

Description

Input

Output

Sample Input

Sample Output

题意

题解

代码

近期文章

近期评论

标签

热门

文章归档

分类目录

功能