POJ 1811 Prime Test
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 32534 | Accepted: 8557 | |
| Case Time Limit: 4000MS | ||
Description
Given a big integer number, you are required to find out whether it's a prime number.
Input
The first line contains the number of test cases T (1 <= T <= 20 ), then the following T lines each contains an integer number N (2 <= N < 254).
Output
For each test case, if N is a prime number, output a line containing the word "Prime", otherwise, output a line containing the smallest prime factor of N.
Sample Input
2 5 10
Sample Output
Prime 2
1 /*遇上一个题目不同，这个题目是输出最小的质因子*/ 2 #include<iostream> 3 using namespace std; 4 #include<cstdio> 5 #define S 10 6 #include<cstdlib> 7 #include<ctime> 8 #define ll long long 9 ll cas, maxz; 10 ll read() 11 { 12 ll ans=0;char c; 13 c=getchar(); 14 while(c<'0'||c>'9') c=getchar(); 15 while(c>='0'&&c<='9') 16 { 17 ans=ans*10+c-'0'; 18 c=getchar(); 19 } 20 return ans; 21 } 22 ll quick_mul_mod(ll a,ll b,ll c)//a*b%c 23 { 24 ll ret=0; 25 a%=c;b%=c; 26 while(b) 27 { 28 if(b&1) 29 { 30 ret+=a; 31 ret%=c; 32 b--; 33 } 34 a<<=1; 35 a%=c; 36 b>>=1; 37 } 38 return ret; 39 } 40 ll gcd(ll a,ll b) 41 { 42 if(a==0) return 1; 43 if(a<0) return gcd(-a,b); 44 if(b==0) 45 return a; 46 return gcd(b,a%b); 47 } 48 ll Pollard_rho(ll x,ll c) 49 { 50 ll x1=rand()%(x-1)+1; 51 ll x2=x1; 52 int i=1,k=2; 53 while(1) 54 { 55 i++; 56 x1=(quick_mul_mod(x1,x1,x)+c)%x; 57 ll d=gcd(x2-x1,x); 58 if(d!=1&&d!=x) return d; 59 if(x2==x1) return x; 60 if(i==k) 61 { 62 x2=x1; 63 k+=k; 64 } 65 } 66 67 } 68 ll quick_mod(ll a,ll b,ll c)//ji suan a^b%c 69 { 70 ll ans=1; 71 a%=c; 72 while(b) 73 { 74 if(b&1) 75 { 76 b--; 77 ans=quick_mul_mod(ans,a,c); 78 } 79 b>>=1; 80 a=quick_mul_mod(a,a,c); 81 } 82 return ans; 83 } 84 bool Miller_rabin(ll n) 85 { 86 if(n==2) return true; 87 if(n<=1||!(n&1)) return false; 88 ll u=n-1,t=0; 89 while(!(u&1)) 90 { 91 u>>=1; 92 t++; 93 } 94 for(int i=0;i<S;++i) 95 { 96 ll x=rand()%(n-1)+1; 97 x=quick_mod(x,u,n); 98 for(int i=1;i<=t;++i) 99 { 100 ll y=quick_mul_mod(x,x,n); 101 if(y==1&&x!=1&&x!=n-1) 102 return false; 103 x=y; 104 } 105 if(x!=1) return false; 106 } 107 return true; 108 } 109 void findpri(ll n) 110 { 111 if(n<=1) return; 112 if(Miller_rabin(n)) 113 { 114 if(n!=0) 115 maxz=min(maxz,n); 116 return; 117 } 118 ll p=n; 119 while(p==n) 120 p=Pollard_rho(p,rand()%(n-1)+1); 121 findpri(p); 122 findpri(n/p); 123 } 124 int main() 125 { 126 srand(time(0)); 127 cas=read(); 128 while(cas--) 129 { 130 maxz=(1<<31)-1;/*不知道为什么这个赋初值的最大值，只有赋值为小于等于(1<<31)-1才对，我的maxz明明是long long型的啊*/ 131 ll n=read(); 132 findpri(n); 133 if(Miller_rabin(n)) 134 printf("Prime "); 135 else printf("%lld ",maxz); 136 } 137 return 0; 138 }