POJ 3641 Pseudoprime numbers 测试费马小定理伪素数

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点击打开链接 Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6018   Accepted: 2407 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 费马小定理:若p是素数且a是正整数,那么a^p=a(mod p) 若a是正整数,p是合数且满足a^p=a(mod p),那么称p为以a为基的伪素数。 给你p和a,让你判断p是不是伪素数。 //384K 16MS #include<stdio.h> #include<math.h> bool isprime(long long x)//判断x是不是素数,如果是素数,肯定不是伪素数 { if(x==1||x==2)return true; long long tmp=sqrt(x); for(long long i=2;i<=tmp;i++) if(x%i==0)return false; return true; } long long quick_mod(long long a,long long b,long long m)//快速幂求a^b%m { long long ans=1; while(b) { if(b&1){ans=(ans*a)%m;b--;} b/=2; a=a*a%m; } return ans; } int main() { long long p,a; while(scanf("%lld%lld",&p,&a),p|a) { if(isprime(p)){printf("no\n");continue;} long long mol=quick_mod(a,p,p); if(mol==a%p)printf("yes\n"); else printf("no\n"); } return 0; }

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