| 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。
1 #include <iostream> 2 using namespace std; 3 typedef long long ll; 4 bool is_prime(ll x) 5 { 6 int i; 7 if (x == 2) return 1; 8 if (x == 1) return 0; 9 for (i = 2; i*i < x; i++) 10 { 11 if (x %i == 0) 12 return 0; 13 } 14 return 1; 15 } 16 int main() 17 { 18 ll a, p; 19 while (cin >> p>>a)//p=n 20 { 21 if (a == 0 && p == 0) break; 22 if (is_prime(p)) 23 { 24 cout << "no" << endl; 25 continue; 26 } 27 ll ans = 1; 28 ll k = p; 29 ll x = a; 30 while (p > 0) 31 { 32 if (p & 1) ans = (ans * a)%k; 33 a = (a * a)%k; 34 p >>= 1; 35 36 } 37 if (ans%k == x) cout << "yes" << endl; 38 else cout << "no" << endl; 39 } 40 return 0; 41 }