Some Chinese Remainder Theorem code

CRT.cpp

#include <iostream>
#include <string>
#include <vector>

int gcd(int u, int v) {
	int t;
	while (v) {
		t = u;
		u = v;
		v = t % v;
	}
	return u < 0 ? -u : u; /* abs(u) */
}

bool check(int *c, int *values, int length)
{
	for (int i = 0; i < length; i++)
		if (values[i] <= c[i])
			return false;

	for (int i = 0; i < length - 1; i++)
		for (int j = i + 1; j < length; j++)
		{
			auto g = gcd(values[i], values[j]);
			if (c[i] % g != c[j] % g)
				return false;
			//if (gcd(values[i], values[j]) != 1)
			//	return false;
		}
	return true;
}

unsigned long long chirem(int *c, int *m, int n) {
	int i, j;
	unsigned long long x(0);
	std::vector<unsigned long long> M(n), b(n);
	for (i = 0; i < n; ++i)
		if (c[i] < m[i])
			c[i] %= m[i];

	for (i = 0; i < n; ++i) {	// Gets M's
		if (i == 0) {
			M[0] = m[1];
			//M.push_back(m[1]);
			j = 1;
		}
		else {
			M[i] = m[0];
			//M.push_back(m[0]);
			j = 0;
		} 
		while (++j < n)
			if (i != j)
				M[i] *= m[j];
	}
	for (i = 0; i < n; ++i) {	// Gets b's
		j = 0;
		while ((M[i] * ++j) % m[i] != 1 && j > 0) {}
		if (j < 0)
			return -1;
		else
			b[i] = j;
	}
	for (i = 0; i < n; ++i)
		x += c[i] * b[i] * M[i];	// Finds a number that works
	return x % (M[0] * m[0]);	// Finds a number less than PI m.
}

unsigned long long get_score(unsigned long long z1, unsigned long long z2, int i, int j, int k)
{
	auto s = std::to_string(z1).size() + std::to_string(z2).size();
	if (i != 0)
		s += std::to_string(i).size() + 1;
	if (j != 1)
		s += std::to_string(j).size() + 3;
	if (k != 0)
		s += std::to_string(k).size() + 2;
	return s;
}

void foo(int c1[], int c2[])
{
	int v[] = { 'M', 'D', 'C', 'L', 'X', 'V', 'I' };
	int n = _countof(v);

	unsigned long long best_score = ~0LL;
	int best_i, best_j, best_k;

	for (int i = -1000; i < 1000; i++)
		for (int j = 1; j < 2; j++)
			for (int k = -1000; k < 1000; k++)
			{
				int s[_countof(v)];

				for (int t = 0; t < n; t++)
				{
					char ch = v[t];
					ch += i;
					ch *= j;
					ch += k;
					s[t] = ch;
				}
					//s[t] = (v[t] + i) * j + k;

				if (check(c1, s, n)/* && check(c2, s, n)*/)
				{
					auto z1 = chirem(c1, s, n);
					auto z2 = chirem(c2, s, n);
					auto score = get_score(z1, z2, i, j, k);
					if (score < best_score)
					{
						best_score = score;
						best_i = i;
						best_j = j;
						best_k = k;

						std::cout << "yay " << i << ' ' << j << ' ' << k << ' ' << ' ' << z1 << ' ' << z2 << ' ' << score << std::endl;
					}
				}

			}

}

void bar(int c1[])
{
	int v[] = { 'M', 'D', 'C', 'L', 'X', 'V', 'I' };
	int n = _countof(v);
	
	for (auto k = 1 ; k < 1000000000; k++)
	{
		auto ok = true;
		for (int i = 0; i < n; i++)
			if (k % v[i]%7 != c1[i])
			{
				ok = false;
				break;
			}
			if (ok)
				std::cout << k << std::endl;
	}
}

int main()
{
	for (auto c = "MDCLXVI"; *c; c++)
	{
		auto v = (*c&15) * 2 +5;
		auto e2 = 833109565 % v;
		auto d = 19147339 % v;
		auto k = 1; for (; d--; k*=5);
		std::cout << *c << ' ' << (1 << e2) *k << std::endl;
	}

	int c1[] = { 3, 6, 2, 5, 1, 4, 0 };
	int c2[] = { 0, 0, 0, 0, 0, 0, 0 };
	bar(c1);

	//for (auto c = "MDCLXVI"; *c; c++)
	//{
	//	auto v = *c * 2 - 105;
	//	auto e2 = 257708304605 % v;
	//	auto d = 401329224880 % v;
	//	std::cout << *c << ' ' << (1 << e2) - d << std::endl;
	//}

	//int c1[] = { 24, 12, 28, 14, 6, 3, 0 };
	//int c2[] = { 10, 9, 7, 6, 4, 3, 0 };
	//foo(c1, c2);
}