11481 - Arrange the Numbers

[Ron]

need help...!!!
i am stuck since contest time, but i have no ability to solve it myself.
please, give me some hints.

[Robert Gerbicz]

Use inclusion-exclusion principle.

[andmej]

Is this DP?

[Naani]

Yep, I solved it during the contest using a recurrence relation.

Hint : Think of how to get a recurrence relation to find the number of derangements of n numbers.

[srrajesh]

Use inclusion-exclusion principle.

Can you explain a little more?
Inclusion and exclusion on what?

[Monsoon]

on which elements won't move:
first we choose which k elements won't move at first m positions (m choose k)
then we want to count how many there are permutations of n' = n - k elements that all of m' = m - k first elements will move
we use in/out
+ all permutations
- all permutations that have at least two elements that won't move
+ all permutations thath have at least three elements that won't move

and so on

[mmonish]

I tried to solve this prob but getting WA.Anyone please check the following I/O..
Input:

Code: Select all

15
1000 500 200
15 5 3
117 111 6
958 54 54
666 654 55
545 21 12
878 500 465
900 250 150
88 54 12
20 10 8
18 18 18
12 5 4
545 500 1
42 32 12
6 5 4

My Output:

Code: Select all

corected after AC...
Case 1: 762643459
Case 2: 27967972
Case 3: 998177293
Case 4: 531446229
Case 5: 980126982
Case 6: 648615184
Case 7: 329986694
Case 8: 9598617
Case 9: 920587914
Case 10: 125855874
Case 11: 1
Case 12: 176400
Case 13: 399786634
Case 14: 387022056
Case 15: 5

[Ron]

Input :

Code: Select all

2
1000 500 200
15 5 3

My Ac's output :

Code: Select all

Case 1: 762643459
Case 2: 27967972

[DP]

@ srrajesh,

Use inclusion-exclusion principle.

Can you explain a little more?
Inclusion and exclusion on what?

See the following links:
http://answers.yahoo.com/question/index ... 200AAXSD4I
http://mathworld.wolfram.com/Derangement.html ("Nicholas Bernoulli also solved the problem using the inclusion-exclusion principle")

[mmonish]

thx Ron..
silly mistake in my coding.now i get AC..
i solved it with dp.

[newkid]

Whats wrong with my thinking -

From first m we need to choose k, thats mCk
for remaining m - k positions we can't use the first remaining m - k numbers..
So we fill those positions with from end n - m number.
The way we can do that is (n-m)P(m-k)
Now we have remaining n-m positions and n-m numbers
We can fill that with (n-m)!

So in my thinking the solution to (n,m,k) is simply mCk * (n-m)P(m-k) * (n-m)!

I know I am missing something. Can anyone verify
Thanks