求原根

  1 /*
  2   题意：求一个数x的原根
  3   题解：x=p1^e1 * p2^e2 * p3^e3 * pm^em
  4         依次判断 a^((x-1)/pi) == 1 mod x 是否等于1,如果都不等于1那么a是原根。
  5   时间：2018.07.26
  6 */
  7 
  8 #include <bits/stdc++.h>
  9 using namespace std;
 10 
 11 typedef long long LL;
 12 const int MAXN = 100005;
 13 const LL MOD7 = 1e9+7;
 14 
 15 LL pow(LL a,LL b, LL P)
 16 {
 17     // printf("a=%lld b=%lld P=%lld\n",a,b,P);
 18     LL res=1LL;
 19     LL ans=a%P;
 20     while (b)
 21     {
 22         if (b&1) res=res*ans%P;
 23         ans=ans*ans%P;
 24         b>>=1;
 25     }
 26     return res;
 27 }
 28 
 29 
 30 int prime[MAXN];
 31 int cnt;
 32 
 33 void initPrime()
 34 {
 35     cnt=0;
 36     for (int i=2;i<MAXN;++i)
 37     {
 38         if (!prime[i]) prime[cnt++]=i;
 39         for (int j=0;j<cnt&&i*prime[j]<MAXN;++j)
 40         {
 41             prime[i*prime[j]]=1;
 42             if (i%prime[j]==0) break;
 43         }
 44     }
 45 }
 46 
 47 int fac[200];
 48 int num[200];
 49 int fnum;
 50 
 51 void Fac(int x)
 52 {
 53     fnum=0;
 54     for (int i=0;i<cnt&&prime[i]<=x;++i)
 55     {
 56         int tmp=0;
 57         while (x%prime[i]==0)
 58         {
 59             ++tmp;
 60             x/=prime[i];
 61         }
 62         if (tmp)
 63         {
 64             fac[fnum] = prime[i];
 65             num[fnum++] = tmp;
 66         }
 67     }
 68     if (x!=1)
 69     {
 70         fac[fnum]=x;
 71         num[fnum++]=1;
 72     }
 73 }
 74 
 75 bool check(int a,int x)
 76 {
 77     for (int i=0;i<fnum;++i)
 78     {
 79         if (pow((LL)a,(LL)(x-1)/fac[i],(LL)x)==1LL) return false;
 80     }
 81     return true;
 82 }
 83 
 84 int x;
 85 
 86 
 87 int main()
 88 {
 89 #ifndef ONLINE_JUDGE
 90    //  freopen("test.txt","r",stdin);
 91 #endif // ONLINE_JUDGE
 92     initPrime();
 93     while (scanf("%d",&x)!=-1)
 94     {
 95         Fac(x-1);
 96         for (int i=2;i<=x-1;++i)
 97         {
 98             if (check(i, x))
 99             {
100                 printf("%d\n",i);
101                 break;
102             }
103         }
104         // printf("%d\n",x);
105     }
106     return 0;
107 }