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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 |
---|---|---|---|---|---|---|---|---|---|
#162963 | #7110. Kuririn MIRACLE | ucup-team1196 | TL | 118ms | 3660kb | C++23 | 3.2kb | 2023-09-03 18:17:12 | 2023-09-03 18:17:13 |
Judging History
answer
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*/
#include <bits/stdc++.h>
using namespace std;
constexpr long double pi = acos(-1.0l);
constexpr long double eps = 1e-9;
#define hps eps*0.5
long double a, b, v, r, d;
long double f(long double x) {
return (long double)2.0l * r / v / (sqrtl(3.0l + sinl(x) * sinl(x)) - sinl(x));
}
long double simpson(long double l,long double r){
return (r-l)*(f(l)+f(r)+4.0l*f((l+r)/2.0l))/6.0l;
}
long double jf(long double l,long double r,long double ans,long double step){
long double mid=(1.0l*l+r)/2.0l;
long double fl=simpson(l,mid),fr=simpson(mid,r);
if(abs(fl+fr-ans)<15.0l*eps&&step<0.0l) return ans;
else return jf(l,mid,fl,step-1.0l)+jf(mid,r,fr,step-1.0l);
}
void solve()
{
srand(998244353);
cin >> v >> r >> d;
// cout << v << ' ' << r << ' ' << d << '\n';
long double l = 1.0l*pi/2.0l, rr = pi;
auto check = [&](long double mid) {
long double t = jf((long double)0.0l, mid, simpson(0.0l, mid), 0.0l);
long double dis = 2.0l * r + v * t;
//cout<<mid<<'\n';
//cout << dis << '\n';
long double delta = d - dis;
long double shengyu = pi - mid;
//cout << delta << '\n';
long double dd = sqrt(delta * delta + 4 * r * r - 4 * r * delta * cos(shengyu));
long double s1 = 2 * r * sin(shengyu) / dd;
long double s2 = (sqrtl(3.0l + sinl(mid) * sinl(mid)) - sinl(mid)) * cos(mid) / -2 ;
//cout<<shengyu<<" "<<s2<<"\n";
if(s2 < s1) {
return false;
} return true;
};
for(int i = 1; i <= 100; i++) {
long double mid = (l + rr) / 2;
if(check(mid)) {
rr = mid;
} else {
l = mid;
}
}
long double t = jf((long double)0.0l, rr, simpson(0, rr), 0);
long double shengyu = pi - rr;
long double dis = 2.0l * r + v * t;
long double delta = d - dis;
long double py = 2 *r*sin(shengyu);
long double px = 2 *r*cos(shengyu);
long double dd = sqrt(delta * delta + 4 * r * r - 4 * r * delta * cos(shengyu));
// cout << t << '\n';
long double t2 = d / v,he = (d - 4 * r ) / v;
cout << fixed << setprecision(10) << '\n';
if(t+dd /2 / v > he)
cout << t2 << '\n';
else
cout << t + dd / 2 / v<< '\n';
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int t;
cin >> t;
while(t--) {
solve();
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 2ms
memory: 3648kb
input:
1 2.00 3 30.0
output:
8.3105799566
result:
ok found '8.3105800', expected '8.3105799', error '0.0000000'
Test #2:
score: 0
Accepted
time: 118ms
memory: 3660kb
input:
100 8.66 6.05 71.59 8.44 8.67 82.55 4.23 9.68 24.74 6.63 9.95 7.45 1.2 7.27 97.55 2.08 2.94 31.61 8.48 5.56 20.16 7.35 5.55 52.48 1.35 8.15 51.23 3.88 9.02 72.37 7.03 1.97 57.03 1.78 4.44 93.44 5.5 6.11 52.94 1.09 9.64 69.01 1.01 3.54 16.78 1.63 4.55 44.06 5.82 5.51 5.92 5.62 2.93 76.01 4.75 4.43 73...
output:
4.4578371330 5.4883608307 5.8486997636 1.1236802413 43.2664800858 8.3054901571 2.3773584906 4.0154860009 37.9481481481 18.6520618557 4.1551236497 27.1789630139 9.6254545455 63.3119266055 16.6138613861 15.0939945133 1.0171821306 6.9496235167 8.1504303414 8.8584474886 1.7697642164...
result:
ok 100 numbers
Test #3:
score: -100
Time Limit Exceeded
input:
1000 7.52 6.68 80.67 5.34 6.82 45.75 3.57 6.35 22.35 9.55 3.65 31.38 9.37 5.73 40.42 1.83 8.38 82.97 4.86 6.13 9.45 3.88 5.34 84.19 6.49 4.15 85.24 6.51 7.23 42.97 3.0 5.98 57.53 9.35 2.76 69.07 6.76 9.16 91.24 2.98 9.89 49.2 9.16 3.85 66.14 1.84 3.03 37.43 8.01 5.04 50.98 4.05 9.86 4.76 5.4 5.49 60...
output:
5.7708826606 8.5674157303 6.2605042017 3.2858638743 4.3137673426 25.1764842491 1.9444444444 11.4062996263 6.8071912335 6.6006144393 10.7277126092 3.8001978533 7.4844843240 16.5100671141 3.7756970302 10.9157110785 3.5175276608 1.1753086420 6.0952874502 14.4431818182 8.4168241292 ...