传送门:poj3641 Pseudoprime numbers
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, 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:对于非素数p,如果存在a的p次幂对p取模等于a,那么这个数就是伪素数。
判断伪素数的步骤:1.判断p是否是素数?
2.判断a的p次幂对p取模是否等于a?由于这里p非常大,所以需要用到快速幂。
#include<iostream>
#include<cmath>
using namespace std;
bool isPrime(int x){
if(x==2) return true;
for(int i=2;i<=sqrt(x);i++){
if(x%i==0) return false;
}
return true;
}
long long PowerMod(long long a,long long b,long long c){
long long ans=1;
a=a%c;
while(b>0){
if(b&1) ans=(ans*a)%c;
b>>=1;
a=(a*a)%c;
}
return ans;
}
int main(){
int p,a;
while(cin>>p>>a){
if(p==0&&a==0) break;
if(isPrime(p)) cout<<"no"<<endl;
else if(PowerMod(a,p,p)==a) cout<<"yes"<<endl;
else cout<<"no"<<endl;
}
return 0;
}