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;
}

1. Use __int64
2. Here is some fragment of my code(and the most significant):
   for(x=0;x<r;x++)
     k+=ceil(sqrt((double)(r-x)*(r+x))); <----- be careful!!