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 ≤ 1,000,000,000 and 1 < a < p, determine whether or not
p is a base-a pseudoprime.
Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing
p and a. For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".
Input
Output
Sample Input
3 2 10 3 341 2 341 3 1105 2 1105 3 0 0
Sample Output
no no yes no yes yes
HINT
题意：
给定p，a两个数字2<=p<=1000000000,1<a<p;若p为素数，则输出no，否则（若a^p%p=a 输出yes，否则输出no）
代码如下：
#include <iostream>
#include <cmath>
using namespace std;
bool isPrime(long long n)
{
long long i;
for (i=2;i<=sqrt(n);i++)
{
if (n%i==0)
return false;
}
return true;
}
long long mod(long long a,long long n,long long m)
{
long long d=1,t=a;
while (n>0) //通过n来求a^p%p;
{
if (n%2==1) //当n为奇数时先乘一个a并%m;
d=(d*t)%m;
n/=2;
t=(t*t)%m; //缩短计算过程
}
return d;
}
int main()
{
long long a,p;
while (cin>>p>>a)
{
if (a==0&&p==0)
break;
if (isPrime(p))
cout<<"no"<<endl;
else
{
if (a==mod(a,p,p))
cout<<"yes"<<endl;
else
cout<<"no"<<endl;
}
}
return 0;
}
运行结果：
这道题是去年新秀赛的一道英语题，本来昨天晚上就应该完成的，但昨天晚上下雨，为了从机房尽早赶回宿舍测试没通过就没再去考虑它了。今天念了它一天。和同学一交流才发现，英语果然还是硬伤，用上了百度翻译结果还是弄错了题意，p是素数的话应该是输出no，我却认为是输出yes。。。心好累。。。