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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
#826688#9869. Horizon ScanningTravelerWA 90ms4224kbC++172.3kb2024-12-22 15:22:472024-12-22 15:22:48

Judging History

你现在查看的是最新测评结果

  • [2024-12-22 15:22:48]
  • 评测
  • 测评结果:WA
  • 用时:90ms
  • 内存:4224kb
  • [2024-12-22 15:22:47]
  • 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;
    }
    if(tot == 42) {
        cout << (int)n << (int)k;
        for(auto [x, y] : a) {
            cout << (int)x << (int)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] == b[j]) 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: 100
Accepted
time: 0ms
memory: 4204kb

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
3.1415926536

result:

ok 5 numbers

Test #2:

score: -100
Wrong Answer
time: 90ms
memory: 4224kb

input:

10000
16 1
-10 -6
-5 -6
-4 9
-2 5
-2 10
1 -7
1 -5
1 6
3 1
4 -9
6 -10
6 -3
6 1
8 -5
8 -4
9 -4
17 4
-9 2
-8 -4
-8 -3
-8 -1
-6 -2
-6 -1
-6 8
-5 -8
-5 10
-4 8
-2 -8
4 -9
4 0
5 -3
8 -5
9 -2
10 10
10 6
-7 2
-4 6
-2 -7
-2 -1
-1 7
1 -9
1 8
3 -4
7 -4
9 -2
14 3
-9 10
-8 -10
-8 -8
-6 -7
-6 -5
-1 -7
-1 -2
0 -1
...

output:

1.6929914975
2.5748634361
4.6527582673
2.7726331074
5.7427658069
4.8576989910
3.4198923126
2.8127999621
6.2831853072
6.2831853072
5.1172807667
6.1467827028
3.8420890235
2.3424967168
3.4633432080
6.2831853072
5.9614347528
3.3247034709
5.2627749281
5.6724593428
1.6738779353
1.1141908549
2.4087775518
6...

result:

wrong output format Expected double, but "1918-101-9-2-710-6-6-4-2-41-32...12-94-6435-76-975946.2599336844" found