运用费马小定理和二次探测定理进行素数测试
#include<iostream> #include<ctime> #include<cstdlib> #include<cmath> #include<algorithm> using namespace std; __int64 mod_exp(__int64 a, __int64 b, __int64 n) //计算(a^b) mod n { __int64 d = 1; a = a % n; while (b) { if (b & 1) d = a*d%n; a = a*a%n; b = b >> 1; } return d; } bool Wintess(__int64 a, __int64 n) //以a为基对n进行Miller测试并实现二次探测 { __int64 m, x, y; int i, j = 0; m = n - 1; while (m % 2 == 0) //计算(n-1)=m*(2^j)中的j和m,j=0时m=n-1,不断的除以2直至n为奇数 { m = m >> 1; j++; } x = mod_exp(a, m, n); for (i = 1; i <= j; i++) { y = mod_exp(x, 2, n); if ((y == 1) && (x != 1) && (x != n - 1)) //二次探测 return true; //返回true时,n是合数 x = y; } if (y != 1) return true; return false; } bool miller_rabin(int times, __int64 n) //对n进行times次的Miller测试 { __int64 a; int i; if (n == 1) return false; if (n == 2) return true; if (n % 2 == 0) return false; srand(time(NULL)); for (i = 1; i <= times; i++) { a = rand() % (n - 2) + 2; if (Wintess(a, n)) return false; }return true; } int main() { int T; cin>>T; while(T--) { int n; cin>>n; if(miller_rabin(5,n)) cout<<n<<" is a prim"<<endl; else cout<<n<<" is not a prim"<<endl; }return 0; }
方法二:
#include<iostream> #include<ctime> #include<cstdlib> #include<cmath> #include<algorithm> using namespace std; typedef __int64 L; void power(L a,L p,L n,L& result,bool& composite) { L x; if(p==0) { result=1; return; } power(a,p/2,n,x,composite); result=(x*x)%n; if(result==1&&(x!=1)&&(x!=n-1)) composite=true; if(p&1) result=(result*a)%n; } bool Miller_Rabin(L n,L times) { L a,result; bool composite=false; srand(time(NULL)); for(L i=1;i<=times;++i) { a=rand()%(n-3)+2; power(a,n-1,n,result,composite); if(composite||result!=1) return false; } return true; } int main() { int T; cin>>T; while(T--) { int n; cin>>n; if(Miller_Rabin(n,5)) cout<<n<<" is a prim"<<endl; else cout<<n<<" is not a prim"<<endl; }return 0; }
#include <iostream> #include <stdlib.h> using namespace std; typedef long long L; void power(L a,L p,L n,L& result ,bool& flag) { L x; if(p==0) { result=1; return; } power(a,p/2,n,x,flag); result=(x*x)%n;//二次探测 if((result==1)&&(x!=1)&&(x!=n-1)) flag=true; if(p&1) result=(result*a)%n; } int main() { int T; cin>>T; while(T--){ int times; L n; cin>>n>>times;//n是待测的数,times是测试的次数 if(n<=1||(n%2==0)) { cout<<n<<" is not a prime "<<endl; continue; } if(n==2||n==3) { cout<<n<<" is a prime "<<endl; continue; } bool flag=false; L result; for(int i=0;i!=times;++i) { L a=rand()%(n-3)+2;//产生2——n-2之间的随机数 power(a,n-1,n,result ,flag); } if(flag||result!=1) cout<<n<<" is not a prime"<<endl; else cout<<n<<" is a prime"<<endl; }return 0; }