362 - 18,000 Seconds Remaining

[Mushfiqur Rahman]

I could'nt find out the the logical or other types of error. I checked it for many input. For those it's 100% ok! But still it's getting WA..!
But Why?

If anybody can give the answer i will be very grateful to him.

Code: Select all

Removed after got AC

[Kallol]

well, I think, the problem is in the type casting.
u have casted "float" in one place in ur code...try with making it double.
and instead of dividing by 5 , divide by 5.0.
hope this will help u
u may visit http://online-judge.uva.es/board/viewtopic.php?t=5222
and i thing there are more threads for this problem here . U may post in those thread ,too.

[Mushfiqur Rahman]

Dear Kallol,
Thanks to you for your great help. Now I got AC. When I changed my all variable into long int instead of doubel and then casting in double then it got AC. I sow your post after got AC. Then I checked my code with your tips but it got WA. So one should follow this suggestion that try to use all variable in long type and then cast it into double.

I am grateful to you.

[N@$!M]

Here is My Code For Which I Have got P.E.

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

int main(void)
{  
    double tb,takenTotal,takenPrev5,take;
    long t,dataset = 1;
    bool tag = false;

    scanf("%lf",&tb);
    while(tb!=0)
    {
        if(tag == false) tag = true;
        else printf("\n");
        ...
        ...
        ...
        ...
        ...      
                dataset++;
    }
    return 0;
}

.........CODE REMOVED

[Jan]

There should be a blank line after every case. But your code doesn't print a blank line after the last case.

Hope it helps.

[alamgir kabir]

I have got it accepted but very much surprised!!!!!!!!!
Ii's a very much technical and fraction error.
so everyone be careful to handle fraction errors.
I type cast the double into long
Thanks all

[ligregni]

Hi, I'm desesperate, I need help!,

I have revewed all the output sentences, I tried all the cases I've found and they are OK!,

I also tried printing after each case (including last one) "\n\n", or printing after each case (including last) "\n", but printing after each case (except the last) "\n" (all of this is printing two \n after all cases exept the last, where I only print one) or printing "\n\n" after all cases except last one (who is "empty" after him),

I tried printing all of this combinations after scanning the # of bytes or before that (I need your help giving me the answer of the \n issue, it will help a lot),

I am almost sure the Math issue is correct (for avoid ceil(), I made the same calculus whith a double and an integer, if they are not the same, I print the integer (long actually) plus 1),

I remember you I've tried with all cases I've found and some extrem I thinked (for example, 10000 bytes, and in the first 5 seconds just one byte has been transferred, and after that one stalled, and then all the bytes less one tranfers, and calculus the time remaining),

I also checked all the ':' and that stuff,

I'm getting crazy with this problem,

I know it's OK, but UVa doesn't think the same, and I'm pretty sure it is by any little thing, but I don't know, what, so, here is my code:

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/* 18,000 Seconds Remaining */

#include <stdio.h>

int main ()
{
/* FILE *in = freopen ("c.in", "r", stdin); /**/

	long b=0, B=0, c=0;

	scanf("%i", &b);

	while (b!=0)
	{
		c++;

/*		if (c!=1)
			printf("\n");

*/
		long sec=0, time[5];

		printf("Output for data set %d, %d bytes:\n", c, b);

		B=0;

		while (B<b)
		{
			scanf("%i", &time[sec%5]);

			B += time[sec%5];

			sec++;

			if (sec%5==0)
			{
				long suma=0, x=0;

				double t=0.0;

				while(x++<5)
					suma += time[x-1];

				t = suma/5.0;



				double d=0;

				if (t!=0)
				{
					x = (b-B)/t;

					d = (b-B)/t;

					if (x!=d)
						x++;
				}

				printf("   Time remaining: ");

				if (t==0)
					printf("stalled\n");
				else
					printf("%i seconds\n", x);
			}
		}

		printf("Total time: %i seconds\n", sec);

		scanf("%i", &b);

		if (b!=0)
			printf("\n");



	}

	return 0;
}

This version of my code is the one who prints \n after all cases and another \n after all cases except last one.

I get WA at 0.190 secs aprox, so I know it could be almost at the end.

aprox 5 of my WA was because I forgot to comment the FILE *in line, but I have trained myself to avoid that.

THANKS

Sergio Ligregni

[kbr_iut]

AC,,,code deleted..
Its too pathetic that,,,when I used ceil(double vale) i got WA
when i used ceil(double value+1e-7)i got WA
when I used (int)(double value+.9999999)i got AC....but why?
so think everyone how much one should b careful calculating floating point numbers!

[mukit]

To Morning :
change the line :

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aver /= 5;

into

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aver /= 5.0;

And you should have print the output lines into proper format(spaces).
Like :

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cout << " Time remaining: " << ceiling((size - sum) / aver) << " seconds" << endl; 

would be

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cout << "   Time remaining: " << ceiling((size - sum) / aver) << " seconds" << endl;

Hope this will work.

Mukit,
CSE,JU

[Obaida]

To "Morning" you should use "return 0;" in "int main" function.
Else you can use void main.

[fR0D]

can somebody please point out why am i getting WA?

Code: Select all

Got AC.
Changed the line
printf("   Time remaining: %.0f seconds\n",ceil(remainingBytes/((float)transferedBytes/5.0)));
to
printf("   Time remaining: %.0f seconds\n",ceil(remainingBytes/((double)transferedBytes*0.2)));
and GOT AC.

i tried all test cases on forum and got expected answer still WA.

[wam3483]

OMFG ok, I've been working at this for 4 hours. I've joined this forum just because it seems like people round here know thier stuff (for the most part :].) I'm a java coder. I feel I'm pretty knowledgable, but Problem 362: 18,000 seconds left has just got me stumped.

Here's my code:

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import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int index = 0;
		int totalBytes = scan.nextInt();
		while (totalBytes > 0) {
			System.out.println("Output for data set " + (++index) + ", "
					+ totalBytes + " bytes:");

			int bytesRead = 0, seconds = 0, count = 0;
			double transWindow = 0;
			//while more bytes to read
			while (totalBytes > bytesRead) {
				++seconds;
				++count;
				int temp = scan.nextInt();
				transWindow += temp;
				bytesRead += temp;
				if (count == 5) {
					if (transWindow > 0) {
						System.out.println("   Time remaining: "
								+ (int) Math.ceil((totalBytes - bytesRead)
										/ (transWindow / 5.0)) + " seconds");
					} else {
						System.out.println("   Time remaining: stalled");
					}
					count =0;
					transWindow=0;
				}
			}
			System.out.println("Total time: " + seconds + " seconds\n");
			totalBytes = scan.nextInt();
		}
	}
}

Like I said, I get Time Limit Exceeded inside the 2nd "while" loop. I don't know why. I know it isn't infinte... I've tested a huge number of test cases against this code and I honestly can't find a thing wrong with it.

Please help me...

[Jan]

Use existing threads.

[kbr_iut]

u can use that technic posted immediate before u.
suppose
double d contains ur result. u need to round (to ceil)
just use this
long=tem=(int)(d+.9999999);
print tem;

hope it will help.

[abid_iut]

why WA it is giving every output right
here is the code:

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

int main()
{
	long size,time,count,sum,n=1,byte,line=0;
	double aver,remainsecond;
	while(scanf("%ld",&size)==1&& size){
			if(line)printf("\n");
			line=1;
	printf("Output for data set %ld, %ld bytes:\n",n++,size);
	sum=0;time=0;count=0;aver=0;
	while(sum!=size){
		scanf("%ld",&byte);
		sum+=byte;
		aver+=byte;
		count++;
		time++;
		if(count==5){
			count=0;
			if(aver==0)printf("   Time remaining: stalled\n");
			else {
				remainsecond=(size-sum)/(aver/5.0);
				printf("   Time remaining: %.0lf seconds\n",ceil(remainsecond));
			}
			aver=0;
		}
	}
	printf("Total time: %ld seconds\n",time);
	}
	return 0;
}