ZIO 2018 answers and cutoff discussion

general
,
21

Hey Guys, good exam.
Here are my answers

  1. 74, 67, 65
  2. 3, 2, 4
  3. 19, 16, 85
  4. 49, 261, (wrong answer for sure :P)
22

Can someone confirm the answers?

23

my answers are…

  1. 74, 74, 65

  2. 2, 2, 3

  3. cant solve it

  4. 61, 321, 2311

24

I think I’ve done well myself. Was very worried in the morning. My answers:

  1. 74
  2. 67
  3. 65
  4. 3
  5. 2
  6. 2
  7. 20
  8. 16
  9. Forgot
  10. 49
  11. 261
  12. 1651
25

Does anybody have 2 as their answer in 4th and 6th? ( I got 2 but now I don’t remember how. I think I made a mistake or something)

1 Like
26

Hey my answers were as follows

1) 74 2) 67 3) 65 4) 3 5) 2 6) 4 7) 28 8) 9 9) 51 10) 49 11) 201 12) 957

I am getting 45. My first two questions and the first part of the last question is correct. I am in class 10th can someone please tell me the expected cutoff.

27
  1. 74, forgot, 65
  2. 3, 2, 4
  3. 20, forgot, forgot
  4. 49, 261, forgot
28

@saioneer3683 damn , we have the same answers for the last problem, let’s hope ours is right, can make all the difference!

1 Like
29

My answers
74
67
65
3
3
4
20
15
94
36
120
360

How did you solve 7,8,9 and 10,11,12
For 7,8,9 I used brute force
For 10,11,12 I used permutations and combinations

30

My Answers (To the best of my Recollection):

  1. 74, 67, 65
  2. 3, 2, 4
  3. 20, 16, 96
  4. 49, 261, 1531

The Questions were easy. Guess The cutoff should be 55/50

Logic for the Questions were:

  1. Group together values at indices with same remainder, Pick max value of each of K groups, Add.
  2. Greedily fill each position, values closest to 0 first (break into chunks of 10) and fill rest from end.
  3. Dynamic Program. Solve for last k elements, for any update add for all Good sub-strings ans of next.
  4. Write n as sum of no.s < 3. For each, count all permutations and divide by factorial of number of equal-sized groups.
31

@siddharth2000 I think its the same algo used in dividing a string into least no of palindromes I set the rightmost 1 position to value of 1 and then kept moving leftward and assume there is a partition there.Then if th selected position is 1 then the number of ways from this position is 1+(number of ways from next position) , if the selected positions value is 0 then find out the smallest possible continuous good string from this position and set its value to 1+(value of position just after the position of the smallest continuous good string). Answer is the sum of these numbers.I can’t say that it is completely correct as I figured this out only after 173 mins :stuck_out_tongue:

32

@animesh1309 how many cases did you make in last problem (1st subtask)?

33

@devildudeja I thought of solving it recursively as substrings would repeat themselves as I kept partitioning from right to left but still decided to solve it by brute force

34

@devildudeja I thought of solving it recursively as substrings would repeat themselves as I kept partitioning from right to left but still decided to solve it by brute force

35

can anyone knows the questions on Q.2 … Plzz give.

36

For Q.)6 I got answer as 2…What about you guys???

37

Does anyone remember the questions, including the values?

38

my answers were-
1.74
2.67
3.65
4.3
5.2
6.4
7,8,9.dont know
10.23
11.63
12.179
what do u thik will be cut off for 10 std?

39

How is Q.) 6 answer coming as 6?

40

when will the result come?