题目链接:http://poj.org/problem?id=3641
题目描述:
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
题目大意:
输入两个整数p和a,如果p是素数,输出no,否则判断(a^p)%p是否与a相等,相等就输出yes,否则no
AC代码:
/*
快速幂取模
by_Superxd
*/
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#define MOD 1000000000+7
#define LL long long
using namespace std;
LL m,n;
LL su(LL x)
{
if(x==2)
return 1;
for(LL i=2;i<=sqrt(x);i++)
{
if(x%i==0)
return 1;
}
return 0;
}
LL quckmod(LL x,LL y,LL z)//快速幂取模x^z
{
LL cnt=1;
while(y)
{
if(y%2==1)
{
cnt*=x;
cnt%=z;
}
x*=x;
x%=z;
y/=2;
}
return cnt;
}
int main()
{
while(1)
{
cin>>m>>n;
if(n==0&&m==0)
break;
if(!su(m))
cout<<"no"<<endl;
else
{
LL k=n;
LL p=quckmod(n,m,m);
if(p==k)
cout<<"yes"<<endl;
else
cout<<"no"<<endl;
}
}
return 0;
}