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：不是素数的数p，且a^p对p取模等于a，输出yes，其他的输出no。
直接暴力，不要打表，打表数组开不到1e9，暴力不会超时
#include<stdio.h>
#include<math.h>
#define ll long long
ll Pow(ll a,ll b)
{
ll res=1;
ll c=b;
while(b>0)
{
if(b&1) res=res*a%c;
a=a*a%c;
b>>=1;
}
return res;
}
ll su(ll n)
{
for(ll i=2;i*i<=n;i++)
{
if(n%i==0)
return 0;
}
return 1;
}
int main()
{
ll p,a,flag=0;
while(~scanf("%lld%lld",&p,&a))
{
if(a==0&&p==0) break;
if(su(p)==0) flag+=1;
if(flag!=0&&Pow(a,p)==a) printf("yes
");
else printf("no
");
flag=0;
}
return 0;
}