poj 1811

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题目描述：

大质数(特别大)的判定质数和分解质数.

题解：

Miller_Rabin 算法进行素数测试 和 pollard_rho 算法进行质因素分解

重点：

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <ctype.h>#include <limits.h>#include <cstdlib>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#include <set>#include <bitset>#include <time.h>#define CLR(a) memset(a, 0, sizeof(a))#define REP(i, a, b) for(ll i = a;i < b;i++)#define REP_D(i, a, b) for(ll i = a;i <= b;i++)typedef long long ll;using namespace std;ll n;const ll S = 8;ll mul_mod(ll a, ll b, ll c){    a %= c;    b %= c;    ll res = 0;    ll tmp = a;    while(b)    {        if(b&1)        {            res = res + tmp;            if(res >= c)                res -= c;        }        tmp <<= 1;        if(tmp >= c)            tmp -= c;        b >>= 1;    }    return res;}ll pow_mod(ll x, ll n, ll c){    ll res = 1;    ll tmp = x%c;    while(n)    {        if(n&1)        {            res = mul_mod(res, tmp, c);        }        n >>= 1;        tmp = mul_mod(tmp, tmp, c);    }    return res;}ll check(ll a, ll n, ll x, ll t){    ll res = pow_mod(a, x, n);    ll last = res;    for(ll i=1; i<= t; i++)    {        res = mul_mod(res, res, n);        if(res==1&&(last != 1 && last != n-1))            return 1;        last = res;    }    if(res!=1)        return 1;    return 0;}ll miller_rabin(ll n){    if(n < 2)        return 0;    if(n==2)        return 1;    if((n&1)==0)        return 0;    ll x=n-1;    ll t = 0;    while((x&1)==0)    {        t++;        x>>=1;    }    srand(time(NULL));    for(ll i=1; i<=S; i++)    {        ll a = rand()%(n-1)+1;        if(check(a, n, x, t))        {            return 0;        }    }    return 1;}ll gcd(ll a, ll b){    if(a < 0)        a= -a;    if(b < 0)        b = -b;    //a = abs(a);    //b = abs(b);    if(b==0)        return a;    return gcd(b, a%b);}long long pollard_rho(long long x,long long c){    long long i = 1, k = 2;    srand(time(NULL));    long long x0 = rand()%(x-1) + 1;    long long y = x0;    while(1)    {        i ++;        x0 = (mul_mod(x0,x0,x) + c)%x;        long long d = gcd(y - x0,x);        if( d != 1 && d != x)return d;        if(y == x0)return x;        if(i == k)        {            y = x0;            k += k;        }    }}ll factor[110], tot;void findfac(ll n, ll k){    if(n==1)        return;    if(miller_rabin(n))    {        factor[tot++]= n;        return;    }    ll p = n;    ll c= k;    while(p >= n)    {        p = pollard_rho(p, c);        c--;    }    findfac(p, k);    findfac(n/p, k);}void solve(){    if(miller_rabin(n))    {        printf("Prime\n");        return;    }    else    {        tot= 0;        findfac(n, 107);    }    sort(factor, factor + tot);    printf("%I64d\n", factor[0]);}int main(){  //  freopen("2Bin.txt", "r", stdin);    //freopen("3Bout.txt", "w", stdout);    ll t;    scanf("%I64d", &t);    while(t--)    {        scanf("%I64d", &n);        solve();    }    return 0;}