codeforces 643 C - Count Triangles

题面

大意

就是找三角形的个数，三角形三边满足 A <= x <= B <= y <= C <= z <= D 数量级是5e5

思路

我的思路比较数学一点吧。在[A,B]遍历最小边x，然后在O(1)的时间内算出当前这个值对应的个数。对于x而言，只要有x + y > z这个不等式成立即可满足三角形成立，因为暗含(x <= y <= z)，考虑y两个极端的情况，y = B时，只要z满足 z <= x + B - 1即可，所以z的取值是[C, x + B - 1]，不过这里需要讨论:

1. x + B - 1 < c 此时取不到数
2. c <= x + B - 1 <= d 此时个数为[C, x + B - 1]
3. x + B - 1 > d 此时个数为[C, D]
   同理对于y = C可以这样讨论，这样讨论之后还是不能直接求解，还需要一个累加的过程，这里先埋个坑… 等之后有空补上QaQ

总体上时间复杂度O(n) 空间复杂度O(1)

AC代码

int main() {
	std::ios::sync_with_stdio(false);
	cin.tie(0);
#ifdef MOON97
	freopen("in.txt", "r", stdin);
#endif
	ll a,b,c,d;
	cin >>a>>b>>c>>d;
	ll res = 0;
	for (ll x = a; x <= b; x++) {
		ll a1 = 0;
		ll an = 0;
		int pos2 = x + c - 1;
		int pos1 = x + b - 1;
		if (d >= pos2) {
			if (c > pos1) {
				a1 = 1;
				an = pos2 - c + 1;
			}
			else {
				a1 = pos1 - c + 1;
				an = pos2 - c + 1;
			}
			res += (a1 + an) * (an - a1 + 1) / 2;
		}
		else if (d <= pos1){
			res += (d - c + 1) * (c - b + 1);
 		}
		else {
			if (c > pos1) {
				a1 = 1;
				an = pos2 - c + 1;
				res += (a1 + an) * (an - a1 + 1) / 2;
				an = pos2 - d;
				res -= (a1 + an) * (an - a1 + 1) / 2;
			}
			else {
				a1 = pos1 - c + 1;
				an = pos2 - c + 1;
				res += (a1 + an) * (an - a1 + 1) / 2;
				a1 = 1;
				an = pos2 - d;
				res -= (a1 + an) * (an - a1 + 1) / 2;
			}
		}
		//if (x == 1) cout << a1 << " " << an <<endl;
	}
	cout << res << endl;
	return 0;
}