[数论]POJ 3641/HOJ 2700 Pseudoprime numbers 快速幂

传送门：Pseudoprime numbers

Pseudoprime numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6014 Accepted: 2404

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

3 210 3341 2341 31105 21105 30 0

Sample Output

nonoyesnoyesyes

Source

Waterloo Local Contest, 2007.9.23

解题报告：

此题题意是，如果p是素数输出no，如果p不是素数，判断a^p%p==a是否成立，如果成立输出yes,否则输出no。

解法：快速幂+素数判定

代码如下：

#include<iostream>#include<cstdio>using namespace std;long long pow_mod(long long a,long long b,long long n){    long long res=1;    while(b){        if(b&1) res=res*a%n;        a=a*a%n;        b>>=1;    }    return res;}long long isprime(long long n){    if(n==2)        return 1;    if(n<=1||n%2==0)        return 0;    long long j=3;    while(j*j<=n){        if(n%j==0)            return 0;        j+=2;    }    return 1;}int main(){    long long p,a;    while(scanf("%lld%lld",&p,&a)==2&&(p||a)){        if(isprime(p))            printf("no\n");        else{            if(pow_mod(a,p,p)==a)                printf("yes\n");            else                printf("no\n");        }    }    return 0;}