hdu1905 Pseudoprime numbers （快速幂+素数筛法）

Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2529    Accepted Submission(s): 1055

Problem 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 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 ≤ 1,000,000,000 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

 

Author

Gordon V. Cormack

 

Source

2008-1杭电公开赛(非原创)

 

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解析：注意这不是完全版的费马小定理，所以不能用a^(p-1)%p==1 来判断。

代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long LL;
const int maxn=1e5;
bool f[maxn+10];
int p[maxn+10];

LL pow_mod(LL a,LL b)
{
LL ans=1,x=a,c=b;
while(b)
{
if(b&1)ans=(ans*a)%c;
a=(a*a)%c,b>>=1;
}
return ans==x;
}

bool is_prime(LL x)
{
LL i,k=(LL)sqrt(x*1.0);
for(i=1;i<=p[0] && p[i]<=k;i++)
if(x%p[i]==0)return 0;
return 1;
}

int main()
{
int i,j;LL a,b;
for(i=2;i<=maxn;i++)
{
if(!f[i])p[++p[0]]=i;
for(j=1;j<=p[0] && i*p[j]<=maxn;j++)
{
f[i*p[j]]=1;
if(i%p[j]==0)break;
}
}
while(scanf("%I64d%I64d",&b,&a),b)
if(pow_mod(a,b) && !is_prime(b))
printf("yes\n");
else
printf("no\n");
return 0;
}