MST1 - Minimum Step To One
Problem statement is very easy. On a positive integer, you can perform any one of the following 3 steps.
- Subtract 1 from it. (n = n - 1)
- If it is divisible by 2, divide by 2. (if n % 2 == 0, then n = n / 2)
- If it is divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3)
Given a positive integer n and you task is find the minimum number of steps that takes n to one.
Input
The input contains an integer T (1 ≤ T ≤ 100) number of test cases. Second line input is N (0 < N ≤ 2*107) that indicates the positive number.
Output
For each case, print the case number and the minimum number of steps.
Example
Input: 3 1 4 7 Output: Case 1: 0 Case 2: 2 Case 3: 3
Explanation
- For N = 1, output: 0
- For N = 4, output: 2 (4 /2 = 2 /2 = 1)
- For N = 7, output: 3 ( 7 -1 = 6 /3 = 2 /2 = 1)
hide comments
cbdavide:
2017-07-28 23:16:31
First DP solution, I feel really, really happy :)
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ayush_1996:
2017-07-15 09:49:52
My First DP :D
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Omar:
2017-06-14 07:02:14
I solved it recursive + memo (top-down) in 0.48 sec. , and this is my code <snip> Last edit: 2023-03-16 12:13:20 |
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ekjot_kaur281:
2017-06-12 20:22:49
Got 3 WA because i didnt give space after colon |
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Nafis Islam:
2017-05-17 16:21:06
Top down dp works :D |
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cyanide_mind:
2017-04-08 20:03:21
My First DP :) |
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samarjeet27:
2016-04-15 12:04:16
Why does Global DP Array gives AC while Local Array gives WA ? |
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darkhire21:
2016-04-10 08:22:57
TOP down TLE ...!! |
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gohanssj9:
2016-02-24 15:33:12
first dp :D |
Added by: | Shipu Ahamed |
Date: | 2013-05-11 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |