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poj1811 Prime Test，随机素数测试

2024-07-31 12:00:07   222人阅读

Prime Test
Time Limit: 6000MS Memory Limit: 65536K
Total Submissions: 24514 Accepted: 5730
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


Miller_Rabin 算法


#include <cstdio>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

// rand(void)返回一个[0,RAND_MAX]之间的随机整数。
//random(n) 返回一个[0,n]之间的随机整数
ll random(ll n) {
    return (ll)((double)rand()/RAND_MAX*n + 0.5);
}

inline ll mul_mod(ll a, ll b, ll c) {
    ll res = 0;
    a %= c;
    b %= c;
    for(; b; b>>= 1, a=(a<<1)%c) {
        if(b&1) res = (res+a)%c;
    }
    return res;
}

ll pow_mod(ll a, ll b, ll c) {
    ll res = 1;
    for(; b; b>>=1, a=mul_mod(a,a,c)) {
        if(b&1) res = mul_mod(res, a, c);
    }
    return res;
}

bool check(ll a, ll n, ll x, ll t) {
    ll ret = pow_mod(a, x, n);
    ll last = ret;
    for(int i=1; i<=t; ++i) {
        ret = mul_mod(ret, ret, n);
        if(ret==1 && last!=1 && last!=n-1) return true;
        last = ret;
    }
    if(ret!=1) return true;
    else return false;
}

const int N = 8;
bool miller_rabin(ll n) {
    if(n<2) return false;
    if(n==2) return true;
    if( (n&1)==0) return false;
    ll x = n-1;
    ll t = 0;
    while( (x&1)==0 ) {
        x >>= 1;
        t++;
    }

    for(int i=0; i< N; ++i) {
        ll a = random(x-2) + 1;
        if( check(a,n,x,t) )
            return false;
    }
    return true;
}

ll factor[100];
int tol;

ll gcd(ll a, ll b) {
        return b?gcd(b,a%b): a>=0?a:-a;
}

ll pollard_rho(ll x, ll c) {
    ll i=1, k=2;
    ll x0 = random(x-2) + 1;
    ll y = x0;
    while(1) {
        i++;
        x0 = (mul_mod(x0,x0,x) + c) % x;
        ll d = gcd(y-x0, x);
        if(d != 1 && d != x) return d;
        if(y == x0) return x;
        if(i == k) {
            y = x0;
            k += k;
        }
    }
}

void findfac(ll n, int k) {
    if(n==1) return ;
    if(miller_rabin(n)) {
        factor[tol++] = n;
        return ;
    }
    ll p = n;
    int c = k;
    while(p >= n) p = pollard_rho(p, c--);
    findfac(p, k);
    findfac(n/p, k);
}

int main() {
    int T;
    ll n;
   // srand(time(NULL));
   //POJ上G++要去掉这句话,一个忧伤的故事。。
    scanf("%d", &T);
    while(T--) {
        scanf("%lld", &n);
        if(miller_rabin(n)) {
            printf("Prime\n");
        } else {
            tol = 0;
            findfac(n, 107);
            ll ans = factor[0];
            for(int i=1; i<tol; ++i)
                if(ans > factor[i])
                        ans = factor[i];
            printf("%lld\n", ans);
        }
    }
    return 0;
}


poj1811 Prime Test，随机素数测试


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