想着费马定理和二次探测定理就能随手推了。
做一次是log2n的。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long int ll; 4 ll T,n; 5 ll qpow(ll x,ll y,ll mod) 6 { 7 ll ans=1,base=x; 8 while(y) 9 { 10 if(y&1)ans=ans*base%mod; 11 base=base*base%mod; 12 y>>=1; 13 } 14 return ans; 15 } 16 bool check(ll p,ll a,ll m,ll q) 17 { 18 if(a==p)return 1; 19 ll base=qpow(a,m,p); 20 ll last=base; 21 for(int i=1;i<=q;++i) 22 { 23 base=base*base%p; 24 if(base==1&&last!=1&&last!=p-1)return 0; 25 last=base; 26 } 27 if(base!=1)return 0; 28 return 1; 29 } 30 bool miller(ll n) 31 { 32 if(n==1)return 0; 33 if(n==2)return 1; 34 if(n%2==0)return 0; 35 ll x=n-1,sum=0; 36 while(x%2==0){++sum;x/=2;} 37 if(!check(n,2,x,sum))return 0; 38 if(!check(n,3,x,sum))return 0; 39 if(!check(n,5,x,sum))return 0; 40 if(!check(n,7,x,sum))return 0; 41 if(!check(n,11,x,sum))return 0; 42 if(!check(n,13,x,sum))return 0; 43 if(!check(n,17,x,sum))return 0; 44 return 1; 45 } 46 int main() 47 { 48 ios::sync_with_stdio(false); 49 cin>>T; 50 while(T--) 51 { 52 cin>>n; 53 cout<<n<<" "<<(miller(n)==1?"Yes":"No")<<endl; 54 } 55 return 0; 56 }
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