QOJ.ac
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 |
---|---|---|---|---|---|---|---|---|---|
#752858 | #3566. Cycling | gyydp123_LIM | 0 | 799ms | 4004kb | C++20 | 2.1kb | 2024-11-16 10:10:31 | 2024-11-16 10:10:31 |
Judging History
answer
//Start: 2024-11-16 09:22:38
#include<bits/stdc++.h>
#define For(i,j,k) for(int i=(j);i<=(k);++i)
#define ForDown(i,j,k) for(int i=(j);i>=(k);--i)
#define Debug(fmt, args...) fprintf(stderr,"(func %s, line #%d): " fmt, __func__, __LINE__, ##args),fflush(stderr)
#define debug(fmt, args...) fprintf(stderr,fmt,##args),fflush(stderr)
#define within :
#define LJY main
using namespace std;
typedef long long ll;
const int N=605;
const double inf=1e18,A=0.5,eps=1e-5;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
inline int read(){
char ch=getchar();int x=0,f=1;
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')
x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
return x*f;
}
double X;int m;
struct Light{double x,r,g;}a[N];
double f[11][N];
double solve(double A,double B,double C){
return (-B+sqrt(B*B-4*A*C))/(2*A);}
void ljy(){
For(i,1,m) scanf("%lf%lf%lf",&a[i].x,&a[i].r,&a[i].g);
For(j,0,10) For(i,0,X/10) f[j][i]=-inf;f[0][0]=0;
sort(a+1,a+1+m,[](Light a,Light b){return a.x<b.x;});
For(i,1,m) For(j,1,X/10) For(x,0,i-1) For(y,0,X/10) if(f[x][y]>=-eps){
double x2=a[i].x,t2=j/2*(a[i].r+a[i].g)+(j&1)*a[i].r;
double x1=a[x].x,t1=y/2*(a[x].r+a[x].g)+(y&1)*a[i].r,lim=f[x][y];
double dt=t2-t1,dx=x2-x1;
if(dt<0) continue;
double v0=(dx-dt*dt/2*A)/dt;
if(v0>lim) continue;
if(v0<0) t1=t2-sqrt(dx*2/A),v0=0;
bool flg=1;
For(k,x+1,i-1){
double dx=a[k].x-a[x].x;
double nt=t1+solve(A/2,v0,-dx);
double rst=nt-(int)(nt/(a[k].g+a[k].r))*(a[k].g+a[k].r);
if(rst<a[k].r-eps){flg=0;break;}
}if(flg) f[i][j]=max(f[i][j],sqrt(2*A*dx+v0*v0));
}double ans=inf;
For(i,0,m) For(j,0,X/10) if(f[i][j]>=-eps){
double dx=a[i].x,v0=f[i][j],t1=j/2*(a[i].g+a[i].r)+(j&1)*a[i].r;bool flg=1;
For(k,i+1,m){
double dx=a[k].x-a[i].x;
double nt=t1+solve(A/2,v0,-dx);
double rst=nt-(int)(nt/(a[k].g+a[k].r))*(a[k].g+a[k].r);
if(rst<a[k].r-eps){flg=0;break;}
}if(flg) ans=min(ans,t1+solve(A/2,v0,-(X-a[i].x)));
}printf("%.3lf\n",ans);
}
signed LJY(){while(scanf("%lf%d",&X,&m)==2) ljy();}
详细
Pretests
Final Tests
Test #1:
score: 0
Wrong Answer
time: 799ms
memory: 4004kb
input:
410.0 2 200.0 15.0 15.0 225.0 31.0 10.0 410.0 2 200.0 15.0 15.0 225.0 35.1 15.0 410.0 2 200.0 15.0 15.0 225.0 45.0 10.0 123.4 0 1234.5 0 1001.0 1 100.0 19.652502 10.581152 1002.0 1 800.0 20.98416 10.923976 1003.0 1 100.0 25.0 15.0 1005.0 1 500.0 60.0 15.0 1000.0 1 500.0 22.0 22.0 1103.0 10 101.234 1...
output:
41.497 52.623 57.213 22.217 70.271 63.277 63.309 68.340 78.682 84.524 66.423 428.492 315.578 162.419 91.789 97.892 99.779 148.498 2.000 2000000000.000 1000000000000000000.000 2000000000.000 2000000000.000 2000000000.000 2000000000.000 2000000000.000 2000000000.000 2000000000.000 2000000000.000 20000...
result:
wrong answer 20th lines differ - expected: '1685.830', found: '2000000000.000'