SPOJ PROOT Primitive Root （数学找规律）

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 2347 2340 0Output:YESNOYESNO

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.

题解：

给一个n，m,m组数据，每组数据的k，表示一个数字，表示以k为底，下标从1到n-1，数值mod n如果不重复，就输出yes，否则no

思路：

打表出所有n-1的因子，假设当前因子之一有个x，看k的x次方mod n是否为1，为1就是no，遍历完还没找到1就是yes

代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#include<stdio.h>using namespace std;#define ll long longll quick(ll x,ll y,ll m){    int ans=1;    while(y!=0)    {        if(y%2==1)        {            ans=(ans*x)%m;        }        x=(x*x)%m;        y/=2;    }    return ans;}int ans;int p[1005];int main(){    ll i,j,n,m,d;    while(scanf("%lld%lld",&n,&m)!=EOF)    {        if(n==0&&m==0)            break;        ans=0;        ll t=n-1;        ll temp=sqrt(t);        for(i=2;i<=temp;i++)        {            if(t%i==0)            {                ll s=t/i;                if(s==i)                {                    p[ans]=i;                    ans++;                }                else                {                    p[ans]=i;                    p[ans+1]=s;                    ans+=2;                }            }        }        while(m--)        {            int tag=1;            scanf("%lld",&d);            for(i=0;i<ans;i++)            {                if(t%p[i]==0)                {                    if(quick(d,p[i],n)==1)                    {                        tag=0;                        goto loop;                    }                }            }            loop:;            if(tag)                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}