Pseudoprime numbers
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8682 | Accepted: 3645 |
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
题目分析：主要在于素数的判断和快速幂算法
1 #include <iostream>
2 using namespace std;
3
4 int prime(long long a)
5 {
6 int i;
7 if(a == 2)
8 return 1;
9 for(i = 2; i*i<=a; i++)
10 if(a%i == 0)
11 return 0;
12 return 1;
13 }
14
15 long long mod(long long a,long long b,long long c)
16 {
17 long long ans = 1;
18 a=a%c;
19 while(b>0)
20 {
21 if(b%2==1) ans=(ans*a)%c;
22 b=b/2;
23 a=(a*a)%c;
24 }
25 return ans;
26 }
27 int main()
28 {
29 long long a,p;
30
31 while(cin >> p >> a)
32 {
33 if(p==0 && a==0) break;
34 long long ans;
35 if(prime(p))
36 cout << "no" << endl;
37 else
38 {
39 ans = mod(a,p,p);
40 if(ans == a)
41 cout << "yes" << endl;
42 else
43 cout << "no" << endl;
44 }
45 }
46
47 return 0;
48 }