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ID	题目	提交者	结果	用时	内存	语言	文件大小	提交时间	测评时间
#733705	#9576. Ordainer of Inexorable Judgment	lqh2024	Compile Error	/	/	C++20	2.4kb	2024-11-10 20:34:23	2024-11-10 20:34:25

Judging History

你现在查看的是最新测评结果

[2024-12-23 14:23:26]
hack成功，自动添加数据
(/hack/1303)

[2024-12-06 11:32:56]
hack成功，自动添加数据
(/hack/1271)

[2024-11-14 21:58:28]
hack成功，自动添加数据
(/hack/1181)

[2024-11-10 20:34:25]
评测

测评结果：Compile Error
用时：0ms
内存：0kb

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[2024-11-10 20:34:23]
提交

answer

#include <bits/stdc++.h>
#include <quadmath.h>
using namespace std;
#define int long long
#define double long double
struct point {
    double x, y;
};

signed main() {
    cin.tie(0) -> sync_with_stdio(0);

    int n;
    double x0, y0, d, t;
    cin >> n >> x0 >> y0 >> d >> t;

    vector<point> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i].x >> a[i].y;
    }

    double mx = 0, mi = 1e9;
    double PI = acos((double)(-1));
    auto get = [&](double k, double fen, point a) -> double {
        double theta = atan2(k, fen);
        // cout << "atan " << theta << "\n";
        // if (theta < 0) theta = PI - theta;
        // cout << "atan " << theta << "\n";
        // if (theta * 2 < PI && a.x < 0) {
        //     theta += PI;
        // } 
        // if (theta * 2 > PI && a.x > 0) {
        //     theta += PI;
        // }
        if (theta * 2 < PI && a.x < 0) {
            theta += PI;
        }  
        if (theta * 2 > -PI && a.x < 0) {
            theta += PI;
        }
        // cout << "atan " << theta << "\n";
        return theta;
    };

    cerr << (double)atan2q(1, 1);

    for (int i = 1; i <= n; i++) {
        auto [x, y] = a[i];
        if (abs(abs(x) - d) < 1e-9) {
            mx = max(mx, max(PI / 2 + (a[i].y < 0 ? PI : 0), get((y * y - d * d), (2 * x * y), a[i])));
            mi = min(mi, min(PI / 2 + (a[i].y < 0 ? PI : 0), get((y * y - d * d), (2 * x * y), a[i])));
        } else {
            double delta = 4 * x * x * y * y - 4 * (d * d - x * x) * (d * d - y * y);
            double k1 = (-2 * x * y * (d * d - x * x) * 2 + sqrt(delta));
            double k2 = (-2 * x * y * (d * d - x * x) * 2 - sqrt(delta));
            mx = max({mx, get(k1, (d * d - x * x) * 2, a[i]), get(k2, (d * d - x * x) * 2, a[i])});
            mi = min({mi, get(k1, (d * d - x * x) * 2, a[i]), get(k2, (d * d - x * x) * 2, a[i])});
        }
    }

    auto get_ans = [&](double t) -> double {
        double cnt = floor(t / (2 * PI)), res = 0;
        t -= cnt * 2 * PI;
        if (mx - mi > PI) {
            res += cnt * (2 * PI - mx + mi);
            res += min(mi, t);
        } else {
            res += cnt * (mx - mi);
            res += max<double>(0, min(mx, t) - mi);
        }

        return res;
    };
    
    cout << fixed << setprecision(12) << get_ans(atan2(y0, x0) + t) - get_ans(atan2(y0, x0)) << "\n";
}

源文件, 语言: C++20

詳細信息

/usr/bin/ld: /tmp/cc51xsNP.o: in function `main':
answer.code:(.text.startup+0x187): undefined reference to `atan2q'
collect2: error: ld returned 1 exit status