Pseudoprime numbers
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7000 | Accepted: 2855 |
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
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<iostream> 5 using namespace std; 6 #define max 1000000000 7 typedef long long ll; 8 bool isprime(int n){//判断素数，该题max过大，不适合用打表法 9 if(n==0||n==1){ 10 return false; 11 } 12 if(n==2){ 13 return true; 14 } 15 if(n%2==0) 16 return false; 17 int i=3,half=sqrt(n*1.0); 18 for(;i<=half;i+=2){ 19 if(n%i==0){ 20 return false; 21 } 22 } 23 return true; 24 } 25 ll work(ll a,ll b,ll k){//a^b mod k 快速幂模板 26 if(a==0){ 27 return 0; 28 } 29 if(b==0){ 30 return 1; 31 } 32 ll x=a%k,ans=1; 33 for(;b>0;b/=2){ 34 if(b%2){ 35 ans=(ans*x)%k;//这歩一定是会最后到达 36 } 37 x=x*x%k; 38 } 39 return ans; 40 } 41 int main() 42 { 43 int p,a; 44 while(cin>>p>>a&&p&&a){ 45 //cout<<isprime(p)<<endl; 46 if(!isprime(p)&&(work(a,p,p)==a%p)){//注意素数判断 47 cout<<"yes"<<endl; 48 } 49 else{ 50 cout<<"no"<<endl; 51 } 52 } 53 return 0; 54 }