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QOJ
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 |
---|---|---|---|---|---|---|---|---|---|
#733516 | #9576. Ordainer of Inexorable Judgment | lqh2024 | WA | 0ms | 3888kb | C++23 | 2.0kb | 2024-11-10 19:35:21 | 2024-11-10 19:35:25 |
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 19:35:21]
- 提交
answer
#include <bits/stdc++.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;
const double PI = acos((double)(-1));
auto get = [&](double k, point a) -> double {
double theta = atan(k);
if (theta < 0) theta = PI / 2 - theta;
if (theta < PI / 2 && a.x < 0) {
theta += PI;
}
if (theta > PI / 2 && a.x > 0) {
theta += PI;
}
return theta;
};
for (int i = 1; i <= n; i++) {
auto [x, y] = a[i];
if (x == d) {
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 + sqrt(delta)) / 2 / (d * d - x * x);
double k2 = (-2 * x * y - sqrt(delta)) / 2 / (d * d - x * x);
mx = max({mx, get(k1, a[i]), get(k2, a[i])});
mi = min({mi, get(k1, a[i]), get(k2, 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;
};
if (t ==10000) cout << get_ans(atan2(y0, x0) + t) << " " << mx << " " << mi << "\n";
cout << fixed << setprecision(20) << get_ans(atan2(y0, x0) + t) - get_ans(atan2(y0, x0)) << "\n";
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3888kb
input:
3 1 0 1 1 1 2 2 1 2 2
output:
1.00000000000000000000
result:
ok found '1.0000000', expected '1.0000000', error '0.0000000'
Test #2:
score: 0
Accepted
time: 0ms
memory: 3884kb
input:
3 1 0 1 2 1 2 2 1 2 2
output:
1.57079632679489661926
result:
ok found '1.5707963', expected '1.5707963', error '0.0000000'
Test #3:
score: -100
Wrong Answer
time: 0ms
memory: 3852kb
input:
3 1 0 1 10000 1 2 2 1 2 2
output:
2500.71 1.5708 -0 2500.70775225747541781196
result:
wrong output format Extra information in the output file