10882 - Koerner's Pub

[sohel]

Since the range is so small (100), I thought of a bruteforce solution but that gives wrong answer.

My method:
given
A intersection B
B intersectin C
A intersectin C

find A intersectin B intersection C... right?

my code:

Code: Select all

  Removed after getting AC..
  thanks to mf.
}

Can someone give some input/output... or prove my method wrong.

[mf]

The lower bound may be 0, for example in "34 33 33".

[sohel]

Thanks..
.. got AC.

Sometimes it is hard to think of the obvious !!
Well, for me anyways.

[_.B._]

Greetings!
Any I/O for spare?
No idea why the WA

Thanks in advance!

[sohel]

How do you check whether the given data is faulty?

[_.B._]

Ouch!, forgot to take some comments away after correction
From

Code: Select all

low = obligatorios + (a[0] / 2); // + (a[0] % 2);

to

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low = obligatorios + (a[0] / 2) + (a[0] % 2);

Well, had other mistake.
By the way, I considered data faulty if:
- Higher value is > than 100.
- Lower value is < than the intersection of the other 2 values.

Keep posting!

[lovemagic]

i get WA.
can smbody plz tell me what's the output of this input?

71 60 60
72 60 60
70 60 60
72 60 6
10 10 10
34 33 33
33 35 35
0 0 1
1 1 1
34 34 34
100 100 80
80 80 100

[lovemagic]

at last got AC.....

i did a mistake & that was:

i overlooked there would be some friday when Ai != Ci != Mi.

[ahm.kam_92]

Hi everyone ...

It's my first time here.
i hope it's helpful

i had coded this problem and got WA ...
may anyone suggest any tricky input or check my code ?
Thanks in advance

that is my code:

Code: Select all

removed.

[helloneo]

Try this case..

Input

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12
71 60 60 
72 60 60 
70 60 60 
72 60 6 
10 10 10 
34 33 33 
33 35 35 
0 0 1 
1 1 1 
34 34 34 
100 100 80 
80 80 100

My output is ..

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Case #1: Between 46 and 60 times.
Case #2: Between 46 and 60 times.
Case #3: Between 45 and 60 times.
Case #4: The records are faulty.
Case #5: Between 0 and 10 times.
Case #6: Between 0 and 33 times.
Case #7: Between 2 and 33 times.
Case #8: Between 0 and 0 times.
Case #9: Between 0 and 1 times.
Case #10: Between 1 and 34 times.
Case #11: The records are faulty.
Case #12: Between 80 and 80 times.

[ahm.kam_92]

Oops!

seems that i am stupid .. i have done this conditions to be faulty and that is wrong:

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if(min(a,b)>c){printf("Case #%d: The records are faulty.\n",i+1);continue;}
if(min(a,c)>b){printf("Case #%d: The records are faulty.\n",i+1);continue;}
if(min(b,c)>a){printf("Case #%d: The records are faulty.\n",i+1);continue;}

i hope that is the only bug .. i will try it now.
Thanks very much.

[ahm.kam_92]

still WA

i dunno what is still wrong it is passing all the cases u have suggested.

[omar_elruby]

check case 7 again, your program gives a wrong output, it's just a small mistake ...

[ahm.kam_92]

it is AC thanks