Prime Test

Time Limit: 6000MS Memory Limit: 65536K

Total Submissions: 29193 Accepted: 7392

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 < 2^54).

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

Source

POJ Monthly

题目大意：T组数据，对于输入的N，若N为素数，输出"Prime"，否则输出N的最小素因子

思路：由于N的规模为2^54所以普通的素性推断果断过不了。

要用Miller Rabin素数測试来做。

而若N不为素数，则须要对N进行素因子分解。由于N为大数，考虑用Pollar Rho整数分解来做。

#include
     
      #include
      
       #include
       
        #include
        
         #define MAX_VAL (pow(2.0,60))//miller_rabbin素性測试//__int64 mod_mul(__int64 x,__int64 y,__int64 mo)//{//    __int64 t;//    x %= mo;//    for(t = 0; y; x = (x<<1)%mo,y>>=1)//        if(y & 1)//            t = (t+x) %mo;////    return t;//}__int64 mod_mul(__int64 x,__int64 y,__int64 mo){    __int64 t,T,a,b,c,d,e,f,g,h,v,ans;    T = (__int64)(sqrt(double(mo)+0.5));    t = T*T - mo;    a = x / T;    b = x % T;    c = y / T;    d = y % T;    e = a*c / T;    f = a*c % T;    v = ((a*d+b*c)%mo + e*t) % mo;    g = v / T;    h = v % T;    ans = (((f+g)*t%mo + b*d)% mo + h*T)%mo;    while(ans < 0)        ans += mo;    return ans;}__int64 mod_exp(__int64 num,__int64 t,__int64 mo){    __int64 ret = 1, temp = num % mo;    for(; t; t >>=1,temp=mod_mul(temp,temp,mo))        if(t & 1)            ret = mod_mul(ret,temp,mo);    return ret;}bool miller_rabbin(__int64 n){    if(n == 2)        return true;    if(n < 2 || !(n&1))        return false;    int t = 0;    __int64 a,x,y,u = n-1;    while((u & 1) == 0)    {        t++;        u >>= 1;    }    for(int i = 0; i < 50; i++)    {        a = rand() % (n-1)+1;        x = mod_exp(a,u,n);        for(int j = 0; j < t; j++)        {            y = mod_mul(x,x,n);            if(y == 1 && x != 1 && x != n-1)                return false;            x = y;        }        if(x != 1)            return false;    }    return true;}//PollarRho大整数因子分解__int64 minFactor;__int64 gcd(__int64 a,__int64 b){    if(b == 0)        return a;    return gcd(b, a % b);}__int64 PollarRho(__int64 n, int c){    int i = 1;    srand(time(NULL));    __int64 x = rand() % n;    __int64 y = x;    int k = 2;    while(true)    {        i++;        x = (mod_exp(x,2,n) + c) % n;        __int64 d = gcd(y-x,n);        if(1 < d && d < n)            return d;        if(y == x)            return n;        if(i == k)        {            y = x;            k *= 2;        }    }}void getSmallest(__int64 n, int c){    if(n == 1)        return;    if(miller_rabbin(n))    {        if(n < minFactor)            minFactor = n;        return;    }    __int64 val = n;    while(val == n)        val = PollarRho(n,c--);    getSmallest(val,c);    getSmallest(n/val,c);}int main(){    int T;    __int64 n;    scanf("%d",&T);    while(T--)    {        scanf("%I64d",&n);        minFactor = MAX_VAL;        if(miller_rabbin(n))            printf("Prime\n");        else        {            getSmallest(n,200);            printf("%I64d\n",minFactor);        }    }    return 0;}