10060 - A Hole to Catch a Man

[10153EN]

I want to ask if there's mistakes in the sample input? It states that the line of input is in the form of T X0 Y0 X1 Y2 ... X0 Y0

However, for the sample input:

2
2 0 0 0 10 5 15 12 10 10 0 0 0
5 0 0 5 100 100 0 0 0
5 3
1
2 0 0 10 0 10 10 0 10
5 2
0

The second last line isn't of this format. Do I understand wrongly? If not, can anyone give the actual sample input data for this case? Thx.

[10153EN]

Problem solved. Just to add the first point at the end of the line~

Thx for reading~

[ljb]

the problem say:
Ti X0 Y0 X1 Y1 X2 Y2 … … Xn Yn X0 Y0

Where Ti is the thickness of the sheet, and Xi Yi are the coordinates of corner points. The line ends with co-ordinate of the first point.

but the second input does not end with the x0 y0
1
2 0 0 10 0 10 10 0 10 <-should be 0?
5 2

i got wrong answer ,55555555555

[cytse]

I think that line should be
2 0 0 10 0 10 10 0 10 0 0

[jpfarias]

Man, I thought I've got this problem, but it appears that something is wrong...

My approach to solve this problem was summing up all the volumes of the pieces and then dividing by the volume of the man hole. What's wrong?

Thanks, JP!

[Whinii F.]

It's likely you rounded off your result, but you must floor() it to get AC.

[CodeMaker]

Hi.......really this one is tough any help plz? i m getting WRONG ANSWER. A lot actually

Code: Select all

#include<cstdio>
#include<cmath>
struct point
{
   double x,y;
};

point p[10];
int n;

double area(point a,point b,point c)
{
   double area;

   area=a.x*(b.y-c.y)+b.x*(c.y-a.y)+c.x*(a.y-b.y);
   return 0.5*fabs(area);
}

int main()
{
   double sum,pt,r,t,total,pi=2*acos(0.0);
   int k;
//   freopen("in.in","r",stdin);
   while(scanf("%d",&n)==1 && n)
   {
      total=0;
      for(k=0;k<n;k++)
      {
         scanf("%lf%lf%lf%lf%lf",&pt,&p[0].x,&p[0].y,&p[1].x,&p[1].y);

         sum=0;
         while(scanf("%lf%lf",&p[2].x,&p[2].y) && (p[2].x!=p[0].x || p[2].y!=p[0].y))
         {
            sum+=area(p[0],p[1],p[2]);
            p[1]=p[2];
         }

         total+=pt*sum;
      }
      scanf("%lf%lf",&r,&t);
      printf("%.0lf\n",floor(total/(pi*r*r*t)));
   }
   return 0;
}






[shahriar_manzoor]

The formula u r using is only acceptable for convex polygon. On the other hand the input polygons may be concave. For concave polygons some or the triangle areas must be negative if you want to get the true result.

[CodeMaker]

Thank you, Now I have solved this problem and also thanks to my friend Dip who helped me to find the precision error.

[asif_rahman0]

Hello I'm facing problem to understand the problem 10060.Can anybody help me?

[daveon]

Hi,

The problem is basically this:
Given a volume L1,L2,...,LN and a volume of a man hole MH,
how many times can MH go into L1+L2+...+LN or
answer = floor( (L1+L2+...+LN) / MH );

[mohiul alam prince]

Hi

I am getting wa in this problem but don't know why. I have trying to solve
this problem by this equation

Code: Select all

 
(x1y2 + x2y3 + x3y4 + ............. + xny1) - (y1x2 + y2x3 + y3x4 +.......... ynx1)


can any body help me.

Here is my code

Code: Select all

#include <stdio.h>
#include <string.h>
#include <math.h>

#define pi acos(-1)

char Table[100024];
double x[20005];
double y[20005];

int main() {

   int i, j;
   int N;
   
   //freopen("D:\\in.txt", "r", stdin);
   while (scanf("%d", &N) == 1) {
      if (N == 0) break;

      gets(Table);
      double ans1 = 0, ans2 = 0, ans = 0;
      for (j = 1; j <= N; j++) {
         gets(Table);
         
         char *p;
         p = strtok(Table, " ");
         int n = 1;
         int m = 1;
         double t;
         sscanf(p, "%lf", &t);
         while (1) {
            p = strtok(NULL, " ");
            if (p == NULL) break;
            sscanf(p, "%lf", &x[n++]);
            p = strtok(NULL, " ");
            sscanf(p, "%lf", &y[m++]);
         }
         ans1 = ans2 = 0.0;         
         for (i = 1; i < n - 1; i++) {
            ans1 += ((x[i] * y[i + 1]) - (y[i] * x[i + 1]));
         }
         ans1 += ((x[n] * y[1]) - (y[n] * x[1]));
         
         ans += fabs(0.5 * (ans1) * t);
      }
      double area;
      double r, s;
      scanf("%lf %lf", &r, &s);
      area = 1.0 * s * pi * r * r;
      printf("%.0lf\n", floor(ans / area + .5));
   }
   return 0;
}



Thanks
MAP

[sohel]

Why are you using gets to take input..

each polygon may be given in more than one line !!

[mohiul alam prince]

Sohel
Thanks for your reply.

My solution has other problems i have fixed this problem and got
AC.

MAP.

[chops]

please give me some I/O.
Thanks in advance.