Discussion @ 1303. Minimal Coverage @ Timus Online Judge

mastersobg WA 4 [8] // Problem 1303. Minimal Coverage 15 Apr 2008 14:24

Could anybody help me? I have WA at 4th test!

Vladimir Plyashkun [USU] Re: WA 4 [2] // Problem 1303. Minimal Coverage 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

Smilodon_am [Obninsk INPE] Re: WA 4 [1] // Problem 1303. Minimal Coverage 18 Sep 2012 21:54

This test helped me:
4
0 4
-5 0
3 4
-4 4
0 0
Answer is:
1
-4 4

yyh0001 Re: WA 4 // Problem 1303. Minimal Coverage 20 Sep 2012 16:32

Is the answer :
1
0 4
right?
i have wa 4 too,and i am confused about the output principles.

Marin Shalamanov Re: WA 4 [4] // Problem 1303. Minimal Coverage 30 Jan 2013 21:37

This test helped me:
10
-5 1
1 14
-5 2
2 3
3 19
0 0

The answer is:
2
-5 1
1 14

Now, let's fight with WA5 :D

Clockware Re: WA 4 [2] // Problem 1303. Minimal Coverage 24 Feb 2013 20:13

why not this:

2
-5 2
1 14

?

Actually, what if there are several possibilites to cover?

Erik Kvam Re: WA 4 // Problem 1303. Minimal Coverage 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

alp Re: WA 4 // Problem 1303. Minimal Coverage 7 Apr 2014 21:17

It is right answer.

tepamid Re: WA 4 // Problem 1303. Minimal Coverage 5 Sep 2019 14:45

This test helped me too.
At last got AC. My solution is similar to task 1987.

Edited by author 05.09.2019 16:47

Discussion of Problem 1303. Minimal Coverage