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## Discussion of Problem 1303. Minimal Coverage

WA 4
Posted by M@STeR.SoBG 15 Apr 2008 14:24
Could anybody help me? I have WA at 4th test!
Re: WA 4
Posted by Vladimir Plyashkun [CSU] 3 May 2012 00:53
u may try this test:
2
1 2
0 1
0 0

so output should be:
2
0 1
1 2

when WA is
2
1 2
0 1
Re: WA 4
Posted by Smilodon_am [Obninsk INPE] 18 Sep 2012 21:54
This test helped me:
4
0 4
-5 0
3 4
-4 4
0 0
1
-4 4
Re: WA 4
Posted by yyh0001 20 Sep 2012 16:32
1
0 4
right?
i have wa 4 too,and i am confused about the output principles.
Re: WA 4
Posted by Marin Shalamanov 30 Jan 2013 21:37
This test helped me:
10
-5 1
1 14
-5 2
2 3
3 19
0 0

2
-5 1
1 14

Now, let's fight with WA5 :D
Re: WA 4
Posted by Clockware 24 Feb 2013 20:13
why not this:

2
-5 2
1 14

?

Actually, what if there are several possibilites to cover?
Re: WA 4
Posted by Erik Kvam 25 Feb 2013 04:53
I think that it does not matter which one you choose, as long as it has a minimum number of intervals. I guess the judge just tests whether the intervals you give cover the large interval.

Best regards, Erik

Edited by author 25.02.2013 04:56
Re: WA 4
Posted by alp 7 Apr 2014 21:17