POJ 3641 Pseudoprime numbers

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Pseudoprime numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7396 Accepted: 3050

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


数论水题。如果p是素数,不是伪素数,否则,如果满足a的p次方取余p等于a,则p是伪素数,剩下的情况均不是。


#include <iostream>#include <cstdio>using namespace std;bool is_prime(long long int m){ for(long long int i=2;i*i<=m;i++) if(m%i==0) return false; return true;}long long int mod_pow(long long int x,long long int n,long long int mod){long long int res=1;while(n>0){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}int main(void){  long long int p,a; while(scanf("%lld%lld",&p,&a)!=EOF&&(p||a)) {    if(is_prime(p))    printf("no\n");    else{    if(a==mod_pow(a,p,p))    printf("yes\n");    else    printf("no\n");} }}

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