[快速幂] POJ-3641 Pseudoprime numbers

Pseudoprime numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 10782 Accepted: 4660

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

题意:给出两个数a，p.若p是素数输出no,否则判断a的p次方对p取模之是不是等于a

//已AC代码#include<iostream>#include<cstdio>#include<cctype>#include<cstring>#include<string.h>#include<math.h>typedef long long int ll;using namespace std;ll power(ll a, ll b, ll mod){    ll res=1;    while(b>0)    {        if(b&1)            res=res*a%mod;        b=b>>1;        a=a*a%mod;    }    return res;}bool isPrime( int num ){    if(num ==2|| num==3 )        return 1 ;    if(num %6!= 1&&num %6!= 5)        return 0 ;    int tmp =sqrt( num);    for(int i= 5;i <=tmp; i+=6 )    if(num %i== 0||num %(i+ 2)==0 )        return 0 ;    return 1 ;    }int main(){    int p,a;    while(~scanf("%d%d",&p,&a)&&(a+p))    {    /*cout<<"a="<<a<<endl;    cout<<"p="<<p<<endl;        int y1=power(a,p,p);        int y2=a%p;        cout<<"mod1:"<<y1<<endl;        cout<<"mod2:"<<y2<<endl;        cout<<"a是素数？"<<isPrime(a)<<endl;*/        if((a==(power(a,p,p)))&&!isPrime(p))        {            puts("yes");        }        else            puts("no");    }}