Pseudoprime numbers
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6044 | Accepted: 2421 |
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
Waterloo Local Contest, 2007.9.23
快速幂
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 6 using namespace std; 7 8 typedef long long ll; 9 10 int p,a; 11 12 bool judge() { 13 for(int i = 2; i * i <= p; i++) { 14 if(p % i == 0) return false; 15 } 16 17 return true; 18 } 19 bool mod_pow(ll x,ll n) { 20 ll res = 1; 21 while(n > 0) { 22 if(n & 1) res = res * x % p; 23 x = x * x % p; 24 n >>= 1; 25 } 26 27 return res == a; 28 } 29 30 int main() { 31 //freopen("sw.in","r",stdin); 32 33 while(~scanf("%d%d",&p,&a) && p && a) { 34 if(!judge() && mod_pow(a,p)) printf("yes "); 35 else printf("no "); 36 37 } 38 39 return 0; 40 }