HDU1905 Pseudoprime numbers

最新推荐文章于 2018-12-12 13:41:41 发布

最新推荐文章于 2018-12-12 13:41:41 发布 · 82 阅读

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本文介绍了一种基于费马小定理的伪素数判断算法，并通过C++实现了一个简单的程序来验证特定条件下整数是否为基apseudoprime伪素数。该算法适用于2到1,000,000,000范围内的整数。

Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 633Accepted Submission(s): 258

Problem 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 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 ≤ 1,000,000,000 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

【解题思路】这是一道不错的数论题目，可以学到很多东西，比如说将指数很大的数进

行二进制处理，还有就是判断素数和伪素数。最有用的知识点是：

1、如果p是奇数，则有(a.^p)mod(m)==((a%m)*(a.^(p-1)%m)%m成立；

2、如果p是偶数，则有a.^p==a.^(p/2)*a.^(p/2)。

也许可以说这两个结论人人都知道，但用起来就不是那么一回事了。

#include<iostream> using namespace std; int plist[100001],pcount=0; int prime(int n){ int i; if ((n!=2&&!(n%2))||(n!=3&&!(n%3))||(n!=5&&!(n%5))||(n!=7&&!(n%7))) return 0; for (i=0;plist[i]*plist[i]<=n;i++) if (!(n%plist[i])) return 0; return n>1; } void initprime(){ int i; for (plist[pcount++]=2,i=3;i<100000;i++) if (prime(i)) plist[pcount++]=i; } __int64 _mode(__int64 a,__int64 m) { __int64 ans=1,exp=m; while(exp) { if(exp%2==1) ans=(ans*a)%m; a=(a*a)%m; exp=exp/2; } return ans; } int main() { __int64 pp,a; int i,j; initprime(); scanf("%I64d %I64d",&pp,&a); while(1) { if(pp==0 &&a==0) break; if(prime(pp)==1) { cout<<"no"<<endl; } else { if(_mode(a,pp)==a) cout<<"yes"<<endl; else cout<<"no"<<endl; } scanf("%I64d %I64d",&pp,&a); } return 0; }