题解：Pseudoprime numbers（伪素数）-优快云博客

根据Fermat定理，对于任意素数p和整数a>1，有ap ≡ a (mod p)。伪素数是满足这一性质的非素数值。给定2<p≤10^9且1<a<p，判断p是否为基数为a的伪素数。

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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

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long quickpow(long long a,long long b)
{
 long long c=1,m=b;
 while(b)
 {
  if(b%2) c=(c*a)%m;
  a=(a*a)%m;
  b/=2;	
 }
 return c;	
}

bool sushu(long long num)
{
 for(int i=2;i*i<num;i++)
  if(num%i==0) 
   return false;
 return true;
}
int main()
{
	long long p,a;
    while(cin>>p>>a&&(p||a))
    {
     if(sushu(p))
     {
     	cout<<"no"<<endl;continue;
	 }
	 if(quickpow(a,p)==a) cout<<"yes"<<endl;
	 else cout<<"no"<<endl;
	}
    return 0;
}