SPOJ

问题描述:

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, ... , rp-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.

Input:5 2347 2340 0

题目题意:判断一个数是否为给定素数的原根。

题目分析:我们把p-1的因子全部打出来，然后去根据原根的性质判断一下，如果在p-1之前出现了a^x %p=1（x为p-1的因子）那么就不是

代码如下:

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define ll long longusing namespace std;const int maxn=1e3+100;int factor[maxn],fcnt;void get_factor(int n){    fcnt=0;    for (int i=1;i<=(int)sqrt(n);i++) {        if (n%i==0) {            if (i*i==n) {                factor[fcnt++]=i;            }            else {                factor[fcnt++]=i;                factor[fcnt++]=n/i;            }        }    }    sort (factor,factor+fcnt);}ll fast_pow(ll base,ll k,ll mod){    ll ans=1;    while (k) {        if (k&1)            ans=ans*base%mod;        base=base*base%mod;        k>>=1;    }    return ans%mod;}int main(){    int p,n;    while (scanf("%d%d",&p,&n)!=EOF) {        if (p==0&&n==0) break;        get_factor(p-1);        for (int i=1;i<=n;i++) {            int a;            scanf("%d",&a);            bool flag=true;            for (int i=0;i<fcnt-1;i++) {                if (fast_pow(a,factor[i],p)==1) {                    flag=false;                    break;                }            }            if (flag) puts("YES");            else puts("NO");        }    }    return 0;}