Pseudoprime numbers---费马小定理

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p = a (mod p). That is, if we raise a to the p^th 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

题解：

/*通过 p的值判断，若p为素数，就断定不是伪素数，若p不是素数，则判断式子(a^n)%p==a;若相等，
则输出yes,否则输出no;由于数据较大，故容易超时，费马小定理；*/

难受的是sqrt函数返回的是double型，在类型转换的时候poj一直提示compile error,浪费了我好多时间

#include<iostream>
#include<string.h>
#include<math.h>
#define max 0x3f3f3f3f
#define ll long long
#define mod 1000000007
using namespace std;
ll vis[100005];
ll cnt = 0;
int isprime(ll p)
{
    int x=(double)sqrt(p)+0.5;
    for(ll i=2;i<=x;i++)
        if(p%i==0)
            return 0;
        return 1;
}

ll f(ll a, ll b, ll n)  //定义函数，求a的b次方对n取模
{
    ll t, y;
    t = 1;
    y = a;
    while (b != 0)
    {
        if ((b & 1) == 1)
            t = t * y%n;
        y = y * y%n;
        b = b >> 1;
    }
    return t;
}
int main()
{
    ll a, p;
    while (cin >> p >> a)
    {
        if (a == 0 && p == 0)
            break;
        else
        {
            if (isprime(p))
                cout << "no" << endl;
            else if (a == f(a, p, p))
                cout << "yes" << endl;
            else
                cout << "no" << endl;
        }
    }
    return 0;
}

标签：Pseudoprime,费马,no,ll,base,numbers,Input,yes,define
来源： https://www.cnblogs.com/-citywall123/p/10686651.html