单求原根个数phi[phi[n]].
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 using namespace std; 7 typedef long long ll; 8 const int maxn = 1e5; 9 bool flag[maxn]; //标记数组 10 ll phi[maxn]; //欧拉函数值 11 int prime[maxn]; //同时得到素数筛 12 int cnt = 0; 13 void Get_phi(int n) 14 { 15 cnt = 0; 16 memset(flag,true,sizeof(flag)); 17 phi[1] = 1; 18 for(int i=2;i<=n;i++) 19 { 20 if(flag[i]) //素数 21 { 22 prime[cnt++] = i; 23 phi[i] = i-1; //素数的欧拉函数值是i-1 24 } 25 for(int j=0;j<cnt;j++) 26 { 27 if(i*prime[j]>n) 28 { 29 break; 30 } 31 flag[i*prime[j]] = false;//素数的倍数不是素数 32 if(i%prime[j]==0) //i%mod prime = 0,那么phi(i*p) = p*phi(i) 33 { 34 phi[i*prime[j]] = prime[j]*phi[i]; 35 break; 36 } 37 else phi[i*prime[j]] = (prime[j]-1)*phi[i];//i mod prime != 0, 那么 phi(i * prime) == phi(i) * (prime-1) 38 } 39 } 40 } 41 int main() 42 { 43 ll n; 44 Get_phi(maxn-10); 45 while(cin>>n) 46 { 47 cout<<phi[phi[n]]<<endl; 48 } 49 return 0;
求具体原根，一步一步来。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 using namespace std; 7 typedef long long ll; 8 const int maxn = 1e5; 9 //欧拉函数 10 ll euler(ll n){ //返回euler(n) 11 ll res=n,a=n; 12 for(ll i=2;i*i<=a;i++){ 13 if(a%i==0){ 14 res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出 15 while(a%i==0) a/=i; 16 } 17 } 18 if(a>1) res=res/a*(a-1); 19 return res; 20 } 21 //线性素数筛 22 int prime[maxn] = {0},num_prime = 0; 23 int isNotPrime[maxn] = {1,1}; 24 void is_prime(int N) 25 { 26 for(int i=2;i<N;i++) 27 { 28 if(!isNotPrime[i]) 29 { 30 prime[num_prime++] = i; 31 } 32 for(int j=0;j<num_prime&&i*prime[j]<N;j++) 33 { 34 isNotPrime[i*prime[j]] = 1; 35 if(!(i%prime[j])) 36 { 37 break; 38 } 39 } 40 } 41 return; 42 } 43 //分解质因数 44 int curprime[maxn]; 45 int cnt = 0; 46 void dividep(int n) 47 { 48 int t = sqrt(1.0*n); 49 for(int i=0;prime[i]<=t;i++) 50 { 51 if(n%prime[i]==0) 52 { 53 curprime[cnt++] = prime[i]; 54 while(n%prime[i]==0) n /= prime[i]; 55 } 56 } 57 if(n>1) curprime[cnt++] = n; 58 } 59 int quick_pow(int a,int n,int mod) 60 { 61 int res = 1; 62 while(n) 63 { 64 if(n&1) res = res*a%mod; 65 n /= 2; 66 a = a*a%mod; 67 } 68 return res; 69 } 70 int main() 71 { 72 ll n; 73 is_prime(maxn-10); 74 /*for(int i=0;i<num_prime;i++) 75 { 76 printf("%d ",prime[i]); 77 if(num_prime%10==0) 78 { 79 printf(" "); 80 } 81 }*/ 82 while(cin>>n) 83 { 84 cnt = 0; 85 dividep(n-1); 86 int ans = 0; 87 for(int i=2;i<n;i++) 88 { 89 int flag = 0; 90 for(int j=0;j<cnt;j++) 91 { 92 int t = (n-1)/curprime[j]; 93 if(quick_pow(i,t,n)==1) 94 { 95 flag = 1; 96 break; 97 } 98 } 99 if(!flag) 100 { 101 ans++; 102 } 103 } 104 cout<<ans<<endl; 105 } 106 return 0; 107 }