I don't know what is meant with the wording "at least four digits after decimal point". Does it mean the first four digits have to be exact ? I got WA with the format "%.4f" (at leat in Python) and i got AC with a code using the "%.5f" format. But the later doesn't guarantee you get the first four digits correct. For instance if x=10000085 then y=sqrt(x)=3162.291099819876... and "%.5f" %y prints 3162.29110 giving only 3 right digits after decimal point. And the examples given in the sample don't help.

I've changed input reading method from "read line by line" to "read word by word", and got AC because of it. Which seems strange. Probably, test 9 is incorrect.

I suppose my error is either in the procedure of scanning/printing.
My program receives input and prints output using file pointers.

FILE *fi,*fo;

fi = stdin;
fo = stdout;

And the program reads/writes by fscanf/fprintf. If you have any knowledge concerning the input/output for this problem, please comment.

#include <iostream>


using namespace std;
int main()
{
   int a,b,i=0,x=0,y=0;
   cin>>a>>b;

       while(x+b<=a)
       {
           x=x+b;
           i++;
       }
       y=a-x;
       if(y>0)
       {
            i++;
       }
       while((y+b)<a)
       {
            y=y+b;
            i++;
       }
       i++;

   cout<<i;


return 0;

}

Solved numerically, but with complete botva. Is there some beautiful geometric solution?
If yes - any clues how to do it? Thanks.

#include<stdio.h>

int main(){
int m,n;
while(scanf("%d %d",&m,&n) == 2){
 if(m == 0 || n == 1){
  printf("yes\n");
 }else{
  printf("no\n");
 }
}
return 0;
}

почему возникает такая проблема? код вроде соответствует примеру, приведенному в руководстве. В Visual Studio 2010  все работает. что делать?

почему возникает такая проблема? код вроде соответствует примеру, приведенному в руководстве. В Visual Studio 2010  все работает. что делать?

Unable to clear test3.
please help

#include<stdio.h>
#include<math.h>
int main()
{
    float r;
        float sum=0.0;
    int i,n,nxt=0;
    scanf("%d %f",&n,&r);

        float dim[n][2];
        for(i=0; i<n;i++)
        {
            scanf("%f%f",&dim[i][0],&dim[i][1]);
        }

        for(i=0;i<n;i++)
        {
            if(i+1==n)
            nxt=n;
            else nxt=i+1;
            float y=pow((dim[i][0]-dim[nxt][0]),2);
            float z=pow((dim[i][1]-dim[nxt][1]),2);
            y=y+z;
sum+=sqrt(y);

        }
        sum+=2*3.14159*r;


    printf("%.02f\n",sum);
return 0;
}

First of all I suppose the graph is a tree (M=N-1 and all vertices are connected, see the text). Is it a tree? I choose one vertex (let it be 1) and consider it as root. I perform breadth first algorithm starting from the root and obtain vertices in increasing order of distances (in the same time I obtain the parent for each vertex). The result is the array C[1..N]. From N downto 1 I search for available C[i] and P[C[i]] (P[C[i]] is the parent for C[i]). I mark C[i] and P[C[i]] not to be used furthermore (unavailable). The edge [C[i],P[C[i]] is part of solution

and the final result is WA on test 7...
what's wrong ?

Edited by author 15.11.2005 20:32

I don't know why
Please help if you can



#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <malloc.h>


int main()
{
    char* Buffer =(char*) calloc(1, sizeof(*Buffer));
    int i=0,j, ch,k;
    double value =0.0f;
    int wasValueReadingStarted=0, NeedToBePrinted=1;

    ch = getchar();
    while (ch != EOF)
    {
        if(i)
        {
            Buffer = (char*)realloc(Buffer, (i+1)*sizeof(*Buffer));
        }
        Buffer[i]=ch;
        ch = getchar();
        i++;
    }

    j = i-1;

    while (j>=0)
    {
        ch = Buffer[j];
        if (ch ==' '  || ch == '\t' || ch == '\n')
        {
            k=0;
            if (wasValueReadingStarted && NeedToBePrinted)
            {
                printf("%.4f\n", sqrt(value));
                NeedToBePrinted=0;
                value = 0.0f;
            }
        }
        else
        {
            value += (ch-'0')*pow(10,k);
            wasValueReadingStarted = 1;
            NeedToBePrinted=1;
            k++;
        }
        j--;
    }
    if (wasValueReadingStarted && NeedToBePrinted)
                printf("%.4f\n", sqrt(value));
    return 0;
}

Edited by author 03.05.2013 17:36

main()
{
long n,sumperv=0,sumposled=0,perv,posled;
int flag=0;

scanf ("%ld",&n);
perv=(n+1)/1000;
posled=(n+1)%1000;
while (perv) {
      sumperv+=perv%10;
      perv/=10;
}
while (posled) {
      sumposled+=posled%10;
      posled/=10;
}
if (sumperv==sumposled)
   flag=1;

perv=(n-1)/1000;
posled=(n-1)%1000;
sumperv=0; sumposled=0;
while (perv) {
      sumperv+=perv%10;
      perv/=10;
}
while (posled) {
      sumposled+=posled%10;
      posled/=10;
}
if (sumperv==sumposled)
   flag=1;

if (flag==1)
   printf ("yes");
else
    printf ("no");



}

[code delete]

Edited by author 09.05.2013 15:31

Hi,

Does anyone have test 12 and it's output so that I can debug my program?

Edited by author 03.05.2013 12:20

Why TLE on 1 test?
It takes no time on example. Any piece of advice?

#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    vector<long long> n;
    long long a = 0;
    while( a != EOF)
    {
        scanf("%llu", &a);
        n.push_back(a);
    }
    for (int i = n.size() - 2; i >= 0; i--)
    {
        printf("%.4f\n", sqrt(long double(n.at(i))));
    }
    return 0;
}