C++(g++ 7.5.0) 解法, 执行用时: 83ms, 内存消耗: 496K, 提交时间: 2023-07-29 17:47:50

#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
const int N=300013;
typedef long long ll;
const ld pi=acosl(-1.0);

ld L,dx,dy;
ll k;
ld x,y;
ll f(ld x){
    return (ll)fabsl(floorl(x));
}
bool chk(ld t){
    ld tx=x+dx*t,ty=y+dy*t;
    return f(ty/sqrtl(3)/L+tx/L+0.5)+f(ty*2/L/sqrtl(3))+f(ty/sqrtl(3)/L-tx/L+0.5)>=k;
}

void solve(){
    cin>>L>>x>>y>>dx>>dy>>k;
    ld l=0,r=1e10,mid,ans=l;
    for(int i=1;i<=100;i++)mid=(l+r)/2,chk(mid)?r=ans=mid:l=mid;
    cout<<ans<<'\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout<<fixed<<setprecision(8);
    int T=1;
    cin>>T;
    while(T--)solve();
}

C++ 解法, 执行用时: 30ms, 内存消耗: 704K, 提交时间: 2022-01-15 01:54:34

#include<bits/stdc++.h>
using namespace std;
int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    cout << fixed << setprecision(20);
    auto f = [&](double x)->long long{return x >= 0 ? x : -x + 1;};
    int T, L, x, y, vx, vy, k;
    for (cin >> T; T; T -= 1) {
        cin >> L >> x >> y >> vx >> vy >> k;
        double l = 0, r = 1e10;
        while (r - l > max(1.0, l) * 1e-4){
            double m = (l + r) / 2, tx = (x + vx * m) / L, ty = (y + vy * m) / L / sqrt(3);
            (f(2 * ty) + f(-tx - ty + .5) + f(tx - ty + .5) >= k ? r : l) = m;
        }
        cout << (l + r) / 2 << "\n";
    }
}