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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
#719529#513. Pizza Cuttingkevinyang#WA 0ms4276kbC++203.3kb2024-11-07 03:01:122024-11-07 03:01:12

Judging History

This is the latest submission verdict.

  • [2024-11-07 03:01:12]
  • Judged
  • Verdict: WA
  • Time: 0ms
  • Memory: 4276kb
  • [2024-11-07 03:01:12]
  • Submitted

answer

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

#define rep(i,a,b) for (int i=a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;


template <class T> int sgn(T x) { return (x > 0 )- (x < 0) ; }
template <class T>
struct Point {
    typedef Point P;
    T x, y;
    explicit Point(T x = 0, T y = 0) : x(x), y(y) {}
    bool operator<(P p) const {return tie(x,y) < tie(p.x,p.y); }
    bool operator== (P p) const {return tie(x,y)== tie(p.x,p.y); }
    P operator+(P p) const { return P(x+p.x,y+p.y); }
    P operator-(P p) const { return P(x-p.x,y-p.y); }
    P operator*(T d) const { return P(x*d,y*d); }
    P operator/(T d) const { return P(x/d,y/d); }
    T dot(P p) const { return x*p.x + y*p.y; }
    T cross(P p) const { return x*p.y - y*p.x; }
    T cross(P a, P b) const { return (a-*this).cross(b-*this); }
    T dist2() const { return x*x+y*y;}
    double dist() const { return sqrt((double)dist2()); }
    double angle() const { return atan2(y,x);}
    P unit() const { return *this/dist(); }
    P perp() const { return P(-y,x); }
    P normal() const { return perp().unit(); }
    P rotate(double a) const {
        return P(x*cos(a)-y*sin(a), x*sin(a)+y*cos(a));
    }
    friend ostream& operator<<(ostream& os, P p) {
        return os << "(" << p.x << "," << p.y << ")";
    }
};

typedef Point<double> P;
#define arg(p, q) atan2(p.cross(q), p.dot(q))
double circlePoly(P c, double r, vector<P> ps) {
    auto tri = [&](P p, P q) {
        auto r2 = r*r/2;
        P d = q-p;
        auto a  = d.dot(p)/d.dist2(), b = (p.dist2()-r*r)/d.dist2();
        auto det = a*a-b;
        if (det <= 0) return arg(p,q)*r2;
        auto s = max(0., -a-sqrt(det)), t = min(1., -a+sqrt(det));
        if (t < 0 || 1 <= s) return arg(p, q) * r2;
        P u = p+d*s, v=p+d*t;
        return arg(p,u) * r2 + u.cross(v)/2 + arg(v,q)*r2;
    };
    auto sum = 0.0;
    rep(i,0,sz(ps))
        sum += tri(ps[i] - c, ps[(i+1)%sz(ps)] - c);
    return sum;
}

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

    int DX, DY, X, Y;
    double R, PR;
    cin >>R >> DX >> DY >> X >> Y >> PR;

    while (X < R) X += DX;
    while (X >= R) X -= DX;
    while (Y < R) Y += DY;
    while (Y >= R) Y -= DY;

    double dx = DX;
    double dy = DY;
    P orig{0,0};
    auto eval_square = [&](double cx, double cy) -> double {
        vector<P> ps;
        ps.push_back(P{cx,cy});
        ps.push_back(P{cx+dx,cy});
        ps.push_back(P{cx+dx,cy+dy});
        ps.push_back(P{cx,cy+dy});
        double a = circlePoly(orig, R, ps);
        return a;
    };

    vector<double> areas;

    int cx = X, cy = Y;
    while (true) {
        while (true) {
            // eval square
            double  a = eval_square(cx, cy);
            if (a > 1e-9) areas.push_back(a);
            if (cx < -R) break;
            cx -= DX;
        }
        if (cy < -R) break;
        cy -= R;
        cx = X;
    }
    sort(areas.begin(), areas.end());

    int ans = 0;
    int N = areas.size()-1;
    const double EPS = 1e-6;
    double yy = areas.back();
    for (int i = 0; i < N; ++i) {
        double xx = areas[i];
        if (xx/yy <= PR + EPS) ++ans;
    }
    cout << ans << '\n';

}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 4252kb

input:

100 45 90 0 -20 0.1

output:

4

result:

ok single line: '4'

Test #2:

score: 0
Accepted
time: 0ms
memory: 4276kb

input:

100 45 90 0 -20 .999

output:

14

result:

ok single line: '14'

Test #3:

score: -100
Wrong Answer
time: 0ms
memory: 4176kb

input:

3 1 1 2 -2 .9

output:

4

result:

wrong answer 1st lines differ - expected: '12', found: '4'