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NOVICE43 - Problem 3 |
When I first learned backtracking I made a program to find all the permutations of the English alphabets in lexicographically increasing. Filled with elation I showed the program to Rohil. Rohil being someone who likes to do stuff off the league was not impressed and gave me the following variation of the problem help me to solve the problem:
You have to find the number of permutations of length N (1<=N<=11) such that at whenever an alphabet (say 'c' ) appears in the permutation all the alphabets smaller than 'c' should have appeared before it at least once. An alphabet is smaller than another if it appears before the other in the English alphabet. ‘a’ being the smallest and ‘z’ being the largest. For example when N=2 then aa, ab are the only valid permutations and ba, bb is invalid since in ba all the alphabets smaller than b have not appeared at least once before it. See example for further clarification.
Input
Line 1: T (number of test cases)
Line 2: n1
Line 3: n2
…
…
Output
Line 1: number of such permutations of length n1
…
...
Example
Input: 2 2 3 Output: 2 5
Explanation for N=3, the possible permutations are: abc, aba, abb, aab, aaa
Added by: | Mahesh Chandra Sharma |
Date: | 2011-03-01 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | NSIT Noivce contest #4 |
hide comments
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2013-06-22 03:25:59 technophyle
The answer for N=11 is 6785** |
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2013-05-22 05:23:50 Rishikesh Jha
why is this problem in classical ? Shift to tutorial. |
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2012-12-15 11:43:54 Shubham Somani
If u cant do the math, then do the recursion :D |
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2012-07-09 14:14:47 Romal Thoppilan
Then whats the answer for n=11 , even i,m getting 182905 ... ? |
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2012-01-10 11:25:20 saket diwakar
@Aseem no |
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2011-06-09 19:21:06 Aseem Kumar
isn't the answer for N=11 1829** ? |
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2011-04-23 19:18:02 :D
yes |
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2011-04-23 16:08:35 Ikhaduri
will the answer fit into a 64 bit integer? Last edit: 2011-11-21 14:40:40 |
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2011-03-04 11:51:12 Mahesh Chandra Sharma
NOVICE43 specify that it is the 3rd problem of 4th Novice contest in which problems are always named by their index. And it doesn't look like a creativity contest though :-) |
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2011-03-02 17:31:03 hendrik
There is already a Problem 3 (NOVICE23). Cant you be a little bit more creative? |