Submission #20008449

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Source Code Expand

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

#ifdef LOCAL
#define DEBUG(...) debug(#__VA_ARGS__, __VA_ARGS__)
#else
#define DEBUG(...) 6
#endif

template<typename T, typename S> ostream& operator << (ostream &os, const pair<T, S> &p) {return os << "(" << p.first << ", " << p.second << ")";}
template<typename C, typename T = decay<decltype(*begin(declval<C>()))>, typename enable_if<!is_same<C, string>::value>::type* = nullptr>
ostream& operator << (ostream &os, const C &c) {bool f = true; os << "["; for (const auto &x : c) {if (!f) os << ", "; f = false; os << x;} return os << "]";}
template<typename T> void debug(string s, T x) {cerr << s << " = " << x << "\n";}
template <typename T, typename... Args> void debug(string s, T x, Args... args) {for (int i=0, b=0; i<(int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++;
else if (s[i] == ')' || s[i] == '}') b--; else if (s[i] == ',' && b == 0) {cerr << s.substr(0, i) << " = " << x << " | "; debug(s.substr(s.find_first_not_of(' ', i + 1)), args...); break;}}

long long sqrtt(long long x) {
    long long l = 0, r = 2e9;
    while (l < r) {
        long long m = (l + r) / 2;
        if (m * m > x)
            r = m;
        else
            l = m + 1;
    }
    return l - 1;
}

long long ceilt(long long x, long long y) {
    if (y < 0) {
        x = -x;
        y = -y;
    }
    if (x < 0)
        return x / y;
    return (x + y - 1) / y;
}

long long floort(long long x, long long y) {
    if (y < 0) {
        x = -x;
        y = -y;
    }
    if (x >= 0)
        return x / y;
    return (x - y + 1) / y;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    long double x, y, r;
    cin >> x >> y >> r;
    long long xx = roundl(10000 * x), yy = roundl(10000 * y), rr = roundl(10000 * r);

    long long a = ceill(x - r), b = floorl(x + r), c = ceill(y - r), d = floorl(y + r), ret = 0;
    for (long long i=a; i<=b; i++) {
        long long sq = sqrtt(rr * rr - (10000 * i - xx) * (10000 * i - xx));
        long long e = max(ceilt(yy - sq, 10000), c), f = min(floort(yy + sq, 10000), d);
        ret += max(f - e + 1, 0LL);
    }
    cout << ret << "\n";

    return 0;
}

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