boj 1002

먼저 두 점이 같을 때 따로 처리해주고 거리의 제곱으로 판별하면 부호 신경쓸 것 없이 간단하다. 

#include <bits/stdc++.h>
using namespace std;


int main(void) {
	int T;
	cin >> T;
	while (T--) {
		int x1, y1, r1, x2, y2, r2;
		cin >> x1 >> y1 >> r1 >> x2 >> y2 >> r2;
		int d = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);

		if (d == 0) {
			if (r1 == r2) cout << "-1" << "\n";
			else cout << "0" << "\n";

		}

		else if (d > (r1 + r2) * (r1 + r2)) {
			cout << "0" << "\n";

		}

		else if (d == (r1 + r2) * (r1 + r2) || d == (r1 - r2) * (r1 - r2)) {
			cout << "1" << "\n";

		}

		else if (d > (r1 - r2) * (r1 - r2) && d < (r1 + r2) * (r1 + r2)) {
			cout << "2" << "\n";

		}
		else cout << "0" << "\n";

	}
}