POJ 3641 Pseudoprime numbers

Pseudoprime numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8550 Accepted: 3590

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

英语非常难读（好多从句OTZ）……读懂题就没问题了！

题意：费马定理：任意素数p和任意整数a(a>1)有如下性质：a的p次方除以p的余数等于a。

有一些同样具有这个性质的非素数p被称为基数a的伪素数。（所有符合条件的a都被称为卡迈克尔数）

现在给定两个数p和a，问p是否是a的伪素数。是的话输出“yes”，反之输出“no”

题解：先用快速幂判定任意的p是否符合费马定理，然后再判断p是不是素数~

大概后台给的数据比较少，我直接暴力判定素数也没TLE……最好还是打个表吧=。=（感觉我这个代码很容易被hack啊）

#include<stdio.h>#include<string.h>#include<math.h>__int64 qp(__int64 a,__int64 p){__int64 ans=1;__int64 mod=p;while(p){if(p&1)ans=ans*a%mod;a=a*a%mod;p>>=1;}return ans;}int main(){__int64 p,a,i;while(scanf("%I64d%I64d",&p,&a),a||p){int flag=0;__int64 temp=qp(a,p);if(temp==a){for(i=2;i<sqrt(p);i++){if(p%i==0){flag=1;break;}}if(flag)printf("yes\n");elseprintf("no\n");}elseprintf("no\n");}return 0;}