prime tan challenge

prime_tan.cpp

#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>

const int MAXT = 1e7;

using namespace std;

typedef long long ll;

double pi = acos(-1.0);
double tann(double x)
{
	return tan(x);
}

void solve()
{
	vector<pair<double,int>> tab;
	for (int i = 0; i < MAXT; ++i)
	{
		tab.push_back(pair<double, int>(fmod((double)i,pi), i));
	}
	sort(tab.begin(), tab.end());
	for (double i = 0; i < 9.007e8; ++i)
	{
		double ii = i*MAXT;
		double r = pi-fmod(ii + pi/2, pi);
		if (r < 0)
			r += pi;
		vector<pair<double,int>>::iterator fi = lower_bound(tab.begin(), tab.end(), pair<double, int>(r, 0));
		int f = 0;
		if (fi != tab.end())
			f = fi - tab.begin();
		int j = f;
		int cnt = 0;
		while (tann(ii+tab[j].second) >= 0)
		{
			j = (j + 1) % MAXT;
			++cnt;
		}
		while (tann(ii+tab[j].second) < 0)
		{
			j = (j + MAXT - 1) % MAXT;
			++cnt;
		}
		//printf("(%lf %lf)\n", r, tab[j].first);
		for(;;)
		{
			++cnt;
			pair<double, int> &v = tab[j];
			double t = tann(ii + v.second);
			if (ii+v.second-1e-7 < t)
			{
				printf("%d %lf %d %lf %lf\n", j, v.first, v.second, t, ii+v.second);
				fflush(stdout);
			}
			if (t < ii)
				break;
			j = (j + MAXT - 1) % MAXT;
		}
		//if (cnt > 20)
		//	printf("===%lf %d\n", ii, cnt);
	}
}

int main()
{
	solve();
	return 0;
}