POJ1811-Prime Test-素数测试+Pollard rho因数分解

题意：

给你一个数n(n <= 2^54)，判断n是不是素数，如果是输出Prime，否则输出n最小的素因子

参考：先用miller_rubin素数测试判断是否为素数，是则用Pollard rho因数分解得到最小素因子

以下代码是网上找的，前面的稳定在1S+，后面的1/3概率TLE,但是run time 稳定在500S

稳定1S+代码：

#include <cstdio>#include <cstdlib>#include <ctime>typedef __int64 LL;int T;LL n, s;LL factor[110000];LL mods(LL x, LL y, LL n){x %= n;y %= n;LL tmp = 0;while(y){if(y & 1)tmp = (tmp + x) % n;x = (x << 1) % n;y >>= 1;}return tmp;}LL pow(LL x, LL y, LL n){x %= n;LL tmp = 1;while(y){if(y&1) tmp = mods(tmp, x, n);x = mods(x, x, n);y >>= 1;}return tmp;}int judge(LL tmp, LL n, LL m, LL t){//n - 1 == m * 2^ tLL tp = m;LL v = pow(tmp , m, n);LL last = v;for(LL i = 1; i <= t; i++){v = mods(v, v, n);if(v == 1){if(last != 1 && last != n-1)return 0;}last = v;tp <<= 1;}if(v == 1)return 1;return 0;}int miller_rubin(LL x, int k){LL cnt = 0;LL m = x - 1;while(!(m & 1)) cnt++, m >>= 1;while(k--){LL tmp = rand()%(x-1) + 1;if(!judge(tmp, x, m, cnt)){return 0;}}return 1;}LL gcd(LL a, LL b){return b == 0?a: gcd(b, a%b);}LL f(LL x, LL n, LL c){return (mods(x, x, n) + c) % n;}LL poll(LL n, LL c){if(!(n & 1))return 2;LL x = rand() % n;LL y = x;LL i = 1;LL k = 2;while(1){i++;x = f(x, n, c);LL d = gcd(y - x + n, n);if(d != 1 && d != n)return d;if(y == x)return n;if(i == k){y = x;k += k;}}}void find(LL n){if(miller_rubin(n,5)){factor[s++] = n;return;}LL p = n;while(p >= n) p = poll(p, (LL)(rand()%(n-1)+1));find(p);find(n/p);}int main(){srand(time(NULL));scanf("%d",&T);while(T--){scanf("%I64d",&n);if(n == 1)break;s = 0;find(n);if(s == 1){printf("Prime\n");continue;}LL tp = 10000000000LL;for(int i = 0; i < s; i++){if(tp > factor[i])tp = factor[i];}printf("%I64d\n",tp);}return 0;}

大几率TLE,但是ac time 500ms+

//此代码提交有1/3几率TLE，但是AC时间稳定在500S-#include<iostream>#include<ctime>#include<cstdlib>#include<cmath>#include<algorithm>#define MAX (pow(2.0, 60))                               //标记最大值#define C 240#define TIME 12                                                 //Miller测试次数using namespace std;__int64 MIN;__int64 gcd(__int64 a, __int64 b) //计算a和b的最大公约数 {    if (b == 0)        return a;    return gcd(b, a % b);}__int64 mod_mult(__int64 a, __int64 b, __int64 n) //计算(a*b) mod n{    __int64 s = 0;    a = a % n;    while (b) {        if (b & 1) {            s += a;            if (s >= n)                s -= n;        }        a = a << 1;        if (a >= n)            a -= n;        b = b >> 1;    }    return s;}__int64 mod_exp(__int64 a, __int64 b, __int64 n) //计算(a^b) mod n{    __int64 d = 1;    a = a % n;    while (b >= 1) {        if (b & 1)            d = mod_mult(d, a, n);        a = mod_mult(a, a, n);        b = b >> 1;    }    return d;}bool Wintess(__int64 a, __int64 n) //以a为基对n进行Miller测试并实现二次探测{    __int64 m, x, y;    int i, j = 0;    m = n - 1;    while (m % 2 == 0) //计算(n-1)=m*(2^j)中的j和m,j=0时m=n-1,不断的除以2直至n为奇数    {        m = m >> 1;        j++;    }    x = mod_exp(a, m, n);    for (i = 1; i <= j; i++) {        y = mod_exp(x, 2, n);        if ((y == 1) && (x != 1) && (x != n - 1)) //二次探测            return true; //返回true时,n是合数        x = y;    }    if (y != 1)        return true;    return false;}bool miller_rabin(int times, __int64 n) //对n进行s次的Miller测试{    __int64 a;    int i;    if (n == 1)        return false;    if (n == 2)        return true;    if (n % 2 == 0)        return false;    srand(time(NULL));    for (i = 1; i <= times; i++) {        a = rand() % (n - 1) + 1;        if (Wintess(a, n))            return false;    }    return true;}__int64 Pollard(__int64 n, int c) //对n进行因字分解,找出n的一个因子,注意该因子不一定是最小的{    __int64 i, k, x, y, d;    srand(time(NULL));    i = 1;    k = 2;    x = rand() % n;    y = x;    while (true) {        i++;        x = (mod_mult(x, x, n) + c) % n;        d = gcd(y - x, n);        if (d > 1 && d < n)            return d;        if (y == x) //该数已经出现过,直接返回即可            return n;        if (i == k) {            y = x;            k = k << 1;        }    }}void get_small(__int64 n, int c) //找出最小的素数因子{    __int64 m;    if (n == 1)        return;    if (miller_rabin(TIME, n)) //判断是否为素数    {        if (n < MIN)            MIN = n;        return;    }    m = n;    while (m == n) //找出n的一个因子        m = Pollard(n, c--);    get_small(m, c); //二分查找    get_small(n / m, c);}int main() {    int total;    __int64 n;    scanf("%d", &total);    while (total--) {        scanf("%I64d", &n);        MIN = MAX;        if (miller_rabin(TIME, n))            printf("Prime\n");        else {            get_small(n, C);            printf("%I64d\n", MIN);        }    }    return 0;}