【POJ1811】Prime Test
【题目大意】
若n是素数，输出“Prime”，否则输出n的最小素因子，（n<=2^54）
【题解】
和bzoj3667差不多，知识这道题没那么坑。
直接上Pollord_Rho和Rabin_Miller就行了。
1 /************* 2 POJ 1811 3 by chty 4 2016.11.7 5 *************/ 6 #include<iostream> 7 #include<cstdio> 8 #include<cstdlib> 9 #include<cstring> 10 #include<ctime> 11 #include<cmath> 12 #include<algorithm> 13 using namespace std; 14 typedef long long ll; 15 const ll INF=1000000000000000000LL; 16 const ll prime[10]={2,3,5,7,11,13,17,19,23}; 17 ll T,minn; 18 inline ll read() 19 { 20 ll x=0,f=1; char ch=getchar(); 21 while(!isdigit(ch)) {if(ch=='-') f=-1; ch=getchar();} 22 while(isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();} 23 return x*f; 24 } 25 ll gcd(ll a,ll b) {return !b?a:gcd(b,a%b);} 26 ll mul(ll x,ll y,ll mod) {return ((x*y-(ll)(((long double)x*y+0.5)/mod)*mod)%mod+mod)%mod;} 27 ll fast(ll a,ll b,ll mod) {ll sum=1;for(;b;b>>=1,a=mul(a,a,mod))if(b&1)sum=mul(sum,a,mod);return sum;} 28 bool Rabin_Miller(ll p,ll a) 29 { 30 if(p==2) return 1; 31 if(!(p&1)||p==1) return 0; 32 ll d=p-1; 33 while(!(d&1)) d>>=1; 34 ll m=fast(a,d,p); 35 if(m==1) return 1; 36 for(;d<p;d<<=1,m=mul(m,m,p)) if(m==p-1) return 1; 37 return 0; 38 } 39 bool isprime(ll x) 40 { 41 for(ll i=0;i<9;i++) 42 { 43 if(prime[i]==x) return 1; 44 if(!Rabin_Miller(x,prime[i])) return 0; 45 } 46 return 1; 47 } 48 void Pollord_Rho(ll x) 49 { 50 if(isprime(x)) {minn=min(minn,x); return;} 51 ll c=3; 52 while(1) 53 { 54 ll x1(1),x2(1),i(1),k(2); 55 while(1) 56 { 57 x1=(mul(x1,x1,x)+c)%x; 58 ll d=gcd(x1>x2?x1-x2:x2-x1,x); 59 if(d>1&&d<x) 60 { 61 Pollord_Rho(d); 62 Pollord_Rho(x/d); 63 return; 64 } 65 if(x1==x2) break; 66 if(++i==k) x2=x1,k<<=1; 67 } 68 c++; 69 } 70 } 71 void solve(ll x) 72 { 73 if(isprime(x)) {puts("Prime"); return;} 74 minn=INF; 75 Pollord_Rho(x); 76 printf("%lld\n",minn); 77 } 78 int main() 79 { 80 freopen("cin.in","r",stdin); 81 freopen("cout.out","w",stdout); 82 T=read(); 83 while(T--){ll x=read();solve(x);} 84 return 0; 85 }
