POJ3641(快速幂）判断a^p=a(modp)是否成立

作者：shangce | 来源：互联网 | 2023-10-13 11:10

DescriptionFermatstheoremstatesthatforanyprimenumberpandforanyintegera

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p = a (mod p). That is, if we raise a to the p^th 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

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
 2 #include
 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\n");
34         else
35         {
36             if(f(a,p) == a) printf("yes\n");
37             else 
38             printf("no\n");
39         }
40         
41     }
42 }