代码如下:

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#include <stdio.h>
#include <time.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
typedef unsigned long long LL;
#define MAX (5000000)
bool bPrime[MAX];
void InitPrime()
{
int nMax = sqrt((double)MAX) + 1;
bPrime[0] = bPrime[1] = true;
for (int i = 2; i <= nMax; ++i)
{
if (!bPrime[i])
{
for (int j = 2 * i; j < MAX; j += i)
{
bPrime[j] = true;
}
}
}
}
LL multAndMod(LL a, LL b, LL n)
{
LL tmp = 0;
while (b)
{
if(b & 1)
{
tmp = (tmp + a) % n;
}
a = (a << 1) % n;
b >>= 1;
}
return tmp;
}
//计算a^u%n
LL ModExp(LL a, LL u, LL n)
{
LL d = 1;
a %= n;
while (u)
{
if (u & 1)
{
d = multAndMod(d, a, n);
}
a = multAndMod(a, a, n);
u >>= 1;
}
return d % n;
}
//判断nN是不是合数
bool Witness(LL a, LL nN)
{
LL u = nN - 1, t = 0;//将nN-1表示为u*2^t
while (u % 2 == 0)
{
t++;
u >>= 1;
}
LL x0 = ModExp(a, u, nN);//x是a^u
LL x1;
for (int i = 1; i <= t; ++i)
{
x1 = multAndMod(x0, x0, nN);
if (x1 == 1 && x0 != nN - 1 && x0 != 1)
{
return true;
}
x0 = x1;
}
if (x1 != 1)
{
return true;
}
return false;
}
//素数测试
bool MillerRabin(LL nN)
{
//if (nN < MAX)return !bPrime[nN];
const int TIME = 10;
for (int i = 0; i < TIME; ++i)
{
LL a = rand() % (nN - 1) + 1;
if (Witness(a, nN))
{
return false;
}
}
return true;
}
LL gcd(LL a, LL b)
{
if (a < b)swap(a, b);
while (b)
{
LL t = a;
a = b;
b = t % b;
}
return a;
}
//启发式寻找nN的因子
LL PollardRho(LL n, LL c)
{
LL i = 1, t = 2;
LL x, y;
LL ans;
srand(time(NULL));
y = x = rand() % n;
while(1)
{
i++;
x = (multAndMod(x, x, n) + c) % n;
ans = gcd(y - x, n);
if(ans > 1 && ans < n)
return ans;
if(x == y)
return n;
if(t == i)
{
y = x;
t <<= 1;
}
}
}
LL FindMin(LL nN, LL c)
{
//printf("nN:%I64u\n", nN);
if (MillerRabin(nN) || nN <= 1)
{
return nN;
}
LL p = nN;
while (p >= nN) p = PollardRho(p, c--);
if (p > 1)
p = FindMin(p, c);//分解p的最小因子
if (p < nN)
{
LL q = nN / p;
q = FindMin(q, c);//找到q的最小因子
p = min(p, q);
}
return p;
}
int main()
{
int nTest;
srand(time(NULL));
//InitPrime();
scanf("%d", &nTest);
while (nTest--)
{
LL nN;
scanf("%I64u", &nN);
if (nN > 2 && nN % 2 == 0)
{
printf("2\n");
}
else if (nN == 2 || MillerRabin(nN))
{
printf("Prime\n");
}
else
{
printf("%I64u\n", FindMin(nN, 181));
}
}
return 0;
}


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本文链接:http://xiaopengcheng.top/2012/09/24/poj 1811 Prime Test 数论 素数测试/