HDU 3641 Pseudoprime numbers（快速幂）

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11336		Accepted: 4891

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 p^th 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

Source

Waterloo Local Contest, 2007.9.23

 

 

题意：p不是素数，且a^p对p取模等于a，输出yes，其他的输出no。

题解：判断p是否是素数那部分直接蛮力求就好。

 #include <iostream>
 using namespace std;
 typedef long long ll;
 bool is_prime(ll x)
 {
     int i;
     if (x == ) return ;
     if (x == ) return ;
     for (i = ; i*i < x; i++)
     {
         if (x %i == )
             return ;
     }
     return ;
 }
 int main()
 {
     ll a, p;
     while (cin >> p>>a)//p=n
     {
         if (a ==  && p == ) break;
         if (is_prime(p))
         {
             cout << "no" << endl;
             continue;
         }
         ll ans = ;
         ll k = p;
         ll x = a;
         while (p > )
         {
             if (p & ) ans = (ans * a)%k;
             a = (a * a)%k;
             p >>= ;

         }
         if (ans%k == x) cout << "yes" << endl;
         else cout << "no" << endl;
     }
     return ;
 }