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
本题用到快速幂，素数判定、二者结合；
题意：输入两个数p,a.先判断p是否为素数，如果是，输出no。否则，再判断a的p次方取余p是否为a，是则yes，反之
则no。
#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;
typedef long long ll; int isprime(ll n) { if(n<=3) return n>1; int k; k=sqrt(n); if(n%6!= 1 && n%6!=5) return 0; for(int i=5;i<=k;i+=6) { if(n%i==0 || n%(i+2)==0) return 0; } return 1; } ll qpow(ll a, ll n,ll mod)//计算a^n % mod { ll re = 1; while(n) { if(n & 1)//判断n的最后一位是否为1 re = (re * a) % mod; n >>= 1;//舍去n的最后一位 a = (a * a) % mod;//将a平方 } return re; } int main() { ll p,a; while(cin>>p>>a&&a&&p) { if(isprime(p)) cout<<"no"<<endl; else { if(a==qpow(a,p,p)) cout<<"yes"<<endl; else cout<<"no"<<endl; } } return 0; }
typedef
typedef long long ll;
快速幂模板
ll qpow(ll a, ll n,ll mod)//计算a^n % mod
{
ll re = 1;
while(n) { if(n & 1)//判断n的最后一位是否为1 re = (re * a) % mod; n >>= 1;//舍去n的最后一位 a = (a * a) % mod;//将a平方 } return re;
}
质数判定模板
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int isprime(ll n)
{
if(n<=3) return n>1; int k; k=sqrt(n); if(n%6!= 1 && n%6!=5) return 0; for(int i=5;i<=k;i+=6) { if(n%i==0 || n%(i+2)==0) return 0; } return 1; }
注意输入用cin，用scanf会wa