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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
#826698#9869. Horizon ScanningTravelerWA 1ms4192kbC++172.2kb2024-12-22 15:28:592024-12-22 15:28:59

Judging History

你现在查看的是最新测评结果

  • [2024-12-22 15:28:59]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:4192kb
  • [2024-12-22 15:28:59]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define int long long
using i64 = long long;
struct Point {
    i64 x;
    i64 y;
    Point(i64 x = 0, i64 y = 0) : x(x), y(y) {}
};

bool operator==(const Point &a, const Point &b) {
    return a.x == b.x && a.y == b.y;
}

Point operator+(const Point &a, const Point &b) {
    return Point(a.x + b.x, a.y + b.y);
}

Point operator-(const Point &a, const Point &b) {
    return Point(a.x - b.x, a.y - b.y);
}
ld len(const Point &a) {
    return sqrtl(a.x * a.x + a.y * a.y);
}
//cos
i64 dot(const Point &a, const Point &b) {
    return a.x * b.x + a.y * b.y;
}
//sin
i64 cross(const Point &a, const Point &b) {
    return a.x * b.y - a.y * b.x;
}

inline bool cmp(Point A, Point B) { return (A.x == B.x) ? (A.y < B.y) : (A.x < B.x); }

bool cmp1(Point a,Point b)
{
    if(atan2(a.y,a.x)!=atan2(b.y,b.x))
        return atan2(a.y,a.x)<atan2(b.y,b.x);
    else return a.x<b.x;
}
// void g(ld &a) {
//     a = max(a, ld(-0.999999999999999));
//     a = min(a, ld(0.999999999999999));
// }
Point O = {1, 0}, P = {-1, 0};
ld PI = fabs(atan2(abs(cross(O, P)), dot(O, P)));
ld get(Point a, Point b) {
    ld ans = fabs(atan2(abs(cross(a, b)), dot(a, b)));
    if(cross(a, b) >= 0) {
        return ans;
    } else {
        return 2.0 * PI - ans;
    }
}
int tot = 0;
void solve() {
    tot++;
    int n, k;
    cin >> n >> k;
    vector<Point> a(n);
    for(int i = 0; i < n; i++) {
        cin >> a[i].x >> a[i].y;
    }
    sort(a.begin(), a.end(), cmp1);
    vector<Point> b(2 * n);
    for(int i = 0; i < n; i++) {
        b[i] = b[i + n] = a[i];
    }
    ld ans = 0;
    for(int i = 0; i < n; i++) {
        int j = i + k;
        if(b[i].x * b[j].y == b[j].x * b[i].y) ans = fmax(ans, 2.0 * PI);
        ans = fmax(ans, get(b[i], b[j]));
    }
    cout << ans << "\n";
}           
    
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed << setprecision(10);
    int T = 1;
    cin >> T;
    while(T--) solve();

    return 0;
}

詳細信息

Test #1:

score: 0
Wrong Answer
time: 1ms
memory: 4192kb

input:

5
1 1
0 1
8 2
1 0
1 1
0 1
-1 1
-1 0
-1 -1
0 -1
1 -1
4 2
-1 1
0 1
0 2
1 1
4 2
-1000000000 0
-998244353 1
998244353 1
1000000000 0
3 1
0 1
0 2
0 -1

output:

6.2831853072
1.5707963268
5.4977871438
3.1415926546
6.2831853072

result:

wrong answer 5th numbers differ - expected: '3.1415927', found: '6.2831853', error = '1.0000000'