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
如果p是素数，输出no；如果p不是素数，判断a^p对p取余是否等于a。
1 #include<cstdio> 2 #include<math.h> 3 __int64 f(__int64 a,__int64 b) 4 { 5 __int64 c=b,t=1; 6 while(b) 7 { 8 if(b % 2 != 0) 9 { 10 t=t*a%c; 11 } 12 a=a*a%c; 13 b/=2; 14 } 15 return t%c; 16 } 17 __int64 f2(__int64 a) 18 { 19 __int64 i; 20 if(a <= 1 || a % 2 == 0) return 0; 21 for(i=3;i<=sqrt(a);i++) 22 { 23 if(a % i == 0) return 0; 24 } 25 return 1; 26 } 27 int main() 28 { 29 30 __int64 p,a; 31 while(scanf("%I64d %I64d",&p,&a) && p && a) 32 { 33 if(f2(p) == 1) printf("no "); 34 else 35 { 36 if(f(a,p) == a) printf("yes "); 37 else 38 printf("no "); 39 } 40 41 } 42 }