10947 - Bear with me, again..

[Tamagodzi]

Thnx mjf ...

0.150 seconds after that without using optimized io

Official solution uses a non-logical algorithm, that says that you can't go from one island to another if they overlap. We're going to change the input, so that no 2 islands intersect.

And a negative island number is logical? ...

0.123 after switching to scanf/printf from cin cout ^^

[little joey]

Adrian will probably say something like: SCH3155 D0NN3RW3773R v#RD4MM73 SCHW31N3R31!

[Adrian Kuegel]

Adrian will probably say something like: SCH3155 D0NN3RW3773R v#RD4MM73 SCHW31N3R31!

Yes, something like that
Well, I had two "real" WA on that problem anyway, since first I had a bug in my formula.

[DongJoo Kim]

Somthing funny about this problem is that I actually got ACC before rejudge although I didn't submit the code with wrong formula.

In my code, the distance formula was wrong

it was...

dist = (x1-y1)*(x1-y1) + (x2-y2)*(x2-y2) ;



I have no idea Y I got AC... at that time.

It was very amazing

[Per]

This problem definitly needs re-rejudging (and I want my perfect score 6 AC for 6 submissions for the contest back...).

It seems the problem is rejudged now, there are a lot more people with 6 solved, you being one of them. Still lots of TLEs though, so I guess they still have a lot more I/O than during the actual contest. Oh well.

For some reason they seem to have rejudged F as well, my solution where I accidentally typed "int isprime[10000010];" rather than "int isprime[1000010];" and (rightfully) got MLE has been rejudged to AC. Strange.

[sclo]

I found that after the rejudge, my O(n^3) solution is TLE, but O(n^2) is AC.

[arjun2304]

Code: Select all

#include<iostream>
#include<cstdio>
#include<cmath>
#include<climits>

using namespace std;

int xc[110],yc[110],r[110];
double a[110][110];

double min(double x,double y)
{
       if(x>y)return y;
       return x;
}
double max(double x,double y)
{
       if(x>y)return x;
       return y;
}
double distance(int x1,int y1,int x2,int y2)
{
       return sqrt(((x1-x2)*(x1-x2))+((y1-y2)*(y1-y2)));
}
int main()
{
    int i,j,n,sx,sy,sr,dx,dy,dr;
    double k,m;
    while(scanf("%lf%lf",&k,&m)==2)
    {
    cin>>xc[0]>>yc[0]>>r[0]>>dx>>dy>>dr;
    cin>>n;
    for(i=1;i<=n;i++)
    scanf("%d%d%d",&xc[i],&yc[i],&r[i]);
    xc[n+1]=dx;yc[n+1]=dy;r[n+1]=dr;
    for(i=0;i<=n+1;i++)
    {
    for(j=0;j<=n+1;j++)
    {
    a[i][j]=INT_MAX;
}
a[i][i]=0;
}
for(i=0;i<=n+1;i++)
{
for(j=i+1;j<=n+1;j++)
{
a[i][j]=a[j][i]=distance(xc[i],yc[i],xc[j],yc[j])-r[i]-r[j];
}
}
for(k=0;k<=n+1;k++)
{
for(i=0;i<=n+1;i++)
{
for(j=0;j<=n+1;j++)
{
a[i][j]=min(a[i][j],max(a[i][k],a[k][j]));
}
}
}
if(a[0][n+1]>(k*m)||a[0][n+1]==INT_MAX)
printf("Larry and Ryan will be eaten to death.\n");
else
printf("Larry and Ryan will escape!\n");
}
return 0;
}

    

[brianfry713]

Compile Error:

Code: Select all

code.cpp: In function 'int main()':
code.cpp:57:31: error: invalid types 'double [110][double]' for array subscript
code.cpp:57:36: error: invalid types 'double [110][110][double]' for array subscript

[DD]

This problem is easy enough so you should not get W.A.