Primitive Root 原根

最新推荐文章于 2024-02-08 22:43:13 发布

Big_Rui

最新推荐文章于 2024-02-08 22:43:13 发布

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该博客介绍了在密码学中，尤其是Diffie-Hellman密钥交换过程中，素数和原根的重要作用。原根是指对于给定素数p，其指数r生成的模p下的幂次序列是不同的。任务是编写程序判断给定的r是否为素数p的原根，以便于Cryptography Experts Group (CEG)建立相关的密钥交换系统。

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题意：给定模P，n个c,判断c是否是p的原根，

《数论概论》中“幂模p与原根”一章中有提到阶的概念：如果gcd(c,p)=1，则a模p的阶是指使得

a^d=1（mod p）的最小指数d(d>=1)；例如2、3、4、5、6模7的阶分别是3、6、3、6、2。

重要性质：一个数a模p的阶e总能整除p-1。所以可以枚举p-1的所有因子factor

（不包括p-1），如果存

在小于p-1的因子满足a^factor=1(mod p) 则说明a模p的阶不是p-1；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
int a[10000];
int mod,cnt;

void pr(int p)
{
     a[0]=1;cnt=0;
     for(int i=2;i<=(int)sqrt(p);i++)
     {
           if(p%i==0)
           {
               a[++cnt]=i;
               a[++cnt]=p/i;
           }
     }
}

ll powmod(ll x,ll n)
{
    if(n==0) return 1%mod;
    ll temp=powmod(x,n>>1);
    temp=temp*temp%mod;
    if(n&1) temp=temp*x%mod;
    return temp;
}

int main()
{
    int n;
    while(scanf("%d%d",&mod,&n)&&mod||n)
    {
        int c;
        for(int i=1;i<=n;i++)
        {
            bool tag=false;
            memset(a,0,sizeof(a));
            scanf("%d",&c);
            pr(mod-1);
            for(int j=0;j<=cnt;j++)
            {
                if(powmod(c,a[j])==1)
                {
                    tag=true;
                    break;
                }
            }
            if(powmod(c,mod-1)!=1)  tag=false;
            if(tag) printf("NO\n");
            else printf("YES\n");
        }
    }
    return 0;
}

In the field of Cryptography, prime numbers play an important role. We are interested in a scheme called "Diffie-Hellman" key exchange which allows two communicating parties to exchange a secret key. This method requires a prime number p and r which is a primitive root of p to be publicly known. For a prime number p, r is a primitive root if and only if it's exponents r, r², r³, ... , r^p-1 are distinct (mod p).

Cryptography Experts Group (CEG) is trying to develop such a system. They want to have a list of prime numbers and their primitive roots. You are going to write a program to help them. Given a prime number p and another integer r < p , you need to tell whether r is a primitive root of p.

Input

There will be multiple test cases. Each test case starts with two integers p ( p < 2 ³¹) and n (1 ≤ n ≤ 100 ) separated by a space on a single line. p is the prime number we want to use and n is the number of candidates we need to check. Then n lines follow each containing a single integer to check. An empty line follows each test case and the end of test cases is indicated by p=0 and n=0 and it should not be processed. The number of test cases is atmost 60.

Output

For each test case print "YES" (quotes for clarity) if r is a primitive root of p and "NO" (again quotes for clarity) otherwise.

Example

Input:
5 2
3
4

7 2
3
4

0 0


Output:
YES
NO
YES
NO

Explanation

In the first test case 3¹, 3², 3³ and 3⁴ are respectively 3, 4, 2 and 1 (mod 5). So, 3 is a primitive root of 5.

4¹, 4² , 4³ and 4⁴ are respectively 4, 1, 4 and 1 respectively. So, 4 is not a primitive root of 5.