米勒-拉宾测试不适用于252097800623

我正在尝试编写miller-rabin测试。我发现了一些代码，例如：

https://www.sanfoundry.com/cpp-program-implement-miller-rabin-primality-test/https://www.geeksforgeeks.org/primality-test-set-3-miller-rabin/

当然，所有这些代码都适用于252097800623(这是素数(，但这是因为他们正在将其解析为int。当我在这些代码中将所有int改为long-long时，他们现在返回NO。我还根据另一篇文章编写了自己的代码，当我用11、101、17甚至1000000007这样的小数字测试它时，它也起了作用，但对2520978006 23这样的大数字进行了测试。我想写一个适用于从1到10^18的所有整数的程序

编辑

这是修改后的代码表第一个链接：

/* 
* C++ Program to Implement Milong longer Rabin Primality Test
*/
#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;

/* 
* calculates (a * b) % c taking long longo account that a * b might overflow 
*/
long long mulmod(long long a, long long b, long long mod)
{
long long x = 0,y = a % mod;
while (b > 0)
{
if (b % 2 == 1)
{    
x = (x + y) % mod;
}
y = (y * 2) % mod;
b /= 2;
}
return x % mod;
}
/* 
* modular exponentiation
*/
long long modulo(long long base, long long exponent, long long mod)
{
long long x = 1;
long long y = base;
while (exponent > 0)
{
if (exponent % 2 == 1)
x = (x * y) % mod;
y = (y * y) % mod;
exponent = exponent / 2;
}
return x % mod;
}

/*
* Milong longer-Rabin primality test, iteration signifies the accuracy
*/
bool Miller(long long p,long long iteration)
{
if (p < 2)
{
return false;
}
if (p != 2 && p % 2==0)
{
return false;
}
long long s = p - 1;
while (s % 2 == 0)
{
s /= 2;
}
for (long long i = 0; i < iteration; i++)
{
long long a = rand() % (p - 1) + 1, temp = s;
long long mod = modulo(a, temp, p);
while (temp != p - 1 && mod != 1 && mod != p - 1)
{
mod = mulmod(mod, mod, p);
temp *= 2;
}
if (mod != p - 1 && temp % 2 == 0)
{
return false;
}
}
return true;
}
//Main
int main()
{
long long iteration = 5;
long long num;
cout<<"Enter long longeger to test primality: ";
cin>>num;
if (Miller(num, iteration))
cout<<num<<" is prime"<<endl;
else
cout<<num<<" is not prime"<<endl;
return 0;
}