B - B

B - B

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Submit

Status
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

---

1. 题意：给定数a,p使a^p%p=a，其中p不是素数。此题介绍的就是关于伪素数的知识，题中的公式即费马小定理。
2. 思路：用快速幂求余数，用素数判定法判断是否为素数。
3. 失误：英文题，理解有点模糊，刚开始没注意p不能是素数。
4. 代码如下：

#include<cstdio>
#include<cmath>

__int64 quickpow(__int64 base,__int64 m,__int64 mod)//快速幂取模 
{
    __int64 ans=1;

    while(m>0)
    {
        if(m%2==1)
        {
            ans=(ans*base)%mod;
            --m;
        }

        base=(base*base)%mod;
        m/=2;
    }

    return ans;
}


__int64 prime(__int64 n)//是否是素数 
{
    __int64 i;

    if(n%2==0)//一步就可以筛除一半的数，用数论的知识优化代码，提高速度
    return 0;

    for(i=2;i<= sqrt(n);++i)
    {
        if(n%i==0)
        return 0;
    }

    return 1;
}

int main()
{
    __int64 a,p,c;
    while(scanf("%I64d %I64d",&p,&a)!=EOF&&(a||p))
    {
        c=quickpow(a,p,p);

        if(c==a&& (prime(p)==0))
        printf("yes\n");
        else
        printf("no\n");

    }
    return 0;
}