| 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, 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
Source
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 ;
}