Problem 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 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 ≤ 1,000,000,000 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
Author
Gordon V. Cormack
Source
思路
就是判断\(a^p\%p==a\)，计算\(a^p\)可以用快速幂的方法，快速幂本质也是二分不断加速
代码
#include<bits/stdc++.h>
using namespace std;
typedef __int64 ll;
bool isprime(ll x)
{
for(int i=2;i<sqrt(x);i++)
if(x%i==0)
return false;
return true;
}//判断是否为质数
ll quickpower(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;
}//返回a^b%c的结果
int main()
{
int a,p;
while(cin>>p>>a)
{
if(p==0 && a==0) break;
if(isprime(p))
cout << "no" << endl;
else
{
int ans_power = quickpower(a,p,p);
if(ans_power==a)
cout << "yes" << endl;
else
cout << "no" << endl;
}
}
return 0;
}