alpercakan's blog

Floating Precision Problem
By alpercakan, history, 9 years ago, In English

For the problem UVa-579 I get AC if I use ((h*60)*0.5)+(m*0.5)-(m*6) for calculation, but if I use (h*30+m/2)-(m*6) I get WA. I know that it's something about precision problem (like comparison), but answer is printed with 3 digits after decimal point; so I didn't think this kind of change in calculation would be a problem. I don't know what's the best way for this kind of calculations. How should I calculate?

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9 years ago, # |
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Do you read h and m as integers or as floats?

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9 years ago, # |
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In the second formula m/2 would evaluate to an integer if m is of integer type. To avoid that either change the type of m or use m/2.0 which will show that you require floating type result :)

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    9 years ago, # ^ |
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    Sorry, I typed that wrong. Actually I use 2.0 but I get WA

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      9 years ago, # ^ |
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      Can you post the actual program. That might help making it clear what the issue is.

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        9 years ago, # ^ |
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        #include <cstdio>
#include <vector>

using namespace std;

int main()
{
	int h, m;
	vector<double> r;
	while (scanf(" %d:%d", &h, &m) == 2)
	{
		if (h == 0 && m == 0) break;
		
		double d1 = 30*h + m/2.0, d2 = (6 * m);

		d1 -= d2;

		if (d1 < 0) d1 *= -1;
		if (d1 > 180) d1 = 360 - d1;
                ///if (d1 == 180) d1 = 0;

		r.push_back(d1);
	}

	for (int i = 0; i < r.size(); ++i)
		printf("%.3f\n", r[i]);

	return 0;
}
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          9 years ago, # ^ |
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          Can you also post the code that got AC?

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            9 years ago, # ^ |
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            double diff = ((h*60)*0.5)+(m*0.5)-(m*6);

if (diff < 0) diff *= -1;
if (diff > 180) diff = 360 - diff ;
///if (diff == 180) diff = 0;

r.push_back(diff );

            Rest is the same.

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              9 years ago, # ^ |
              ← Rev. 2   Vote: I like it 0 Vote: I do not like it

              You must have had a mistake somewhere else. I get AC for both versions of your code.

              Also, there is no reason for the code to be imprecise -- each value is either an integer or an integer_and_a_half, and those can be represented exactly in a double. There are no precision errors anywhere.

              My best guess at the moment is that Enchom was right and that the version that got WA actually used integer division by accident.

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          9 years ago, # ^ |
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          can u try this one (h*30+(m*0.5))-(m*6) , I wonder

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