PROBLEM LINKS
[Practice][1] [Contest][2]DIFFICULTY
EASYMEDIUMPREREQUISITES
sorting, factorials, inverse modulo a prime, basic combinatoricsPROBLEM
The problem is, given count[1..n], the number of chocolates to be included from each village, we need to pack the chocolates into minimum number of boxes of a given size S. No two chocolates from different villages should be places in same box. Given that different villages get different colored boxes, find the total number of distinct gift packs possible. Two gift packs are considered same if and only if they have the same order of colored boxes, even in reverse. "ABC" is same as "ABC" and "CBA".EXPLANATION
First lets see how we can solve the problem of uniquely counting distinct gift packs by taking care of reverse order too. Suppose there are B1 boxes of type 1, B2 boxes of type 2 , … Bk boxes of type K, then the total number of gift packs possible can be found using the following multinomial.
SUM = B1 + B2 + … + Bk
Total possible ways, T = ( SUM! / ( B1! B2! B3! … Bk! ) )
But T may include duplicate items, where two items are same if reversed. Consider “AABBC”. T counts both “AABBC” and “CBBAA” differently and we need to take care of this.
Let P = total number of palindromic strings, where a string is palindrome if its exactly same as its reversed string.
Let U = total number of nonpalindromic strings
Our desired answer = ( U + P ) , but the value of T = ( 2 * U + P ), as it counts each nonpalindromic
string twice. Finding the number of nonpalindromic strings directly may not be easy, so instead we will
find the value of P, the number of palindromic strings. A palindromic string is uniquely defined by its first half, so we will half the counts of each character and find the number of possible halves. Note that P = 0, if the number of characters having odd count > 1 , as only one character can take the middle position in case of odd count.
SUM2 = B1/2 + B2/2 + … + Bk/2
Total number of palindromes, P = ( SUM2! / ( (B1/2)! (B2/2)! … (Bk/2)! ) )
If we have the values of T and P, we can find the desired answer = U + P = ( T + P ) / 2
So the problem is reduced to finding the values of T and P. Till this part, its a common approach. For further main part of the solution, refer any of setter’s or tester’s approach below.
SETTER'S SOLUTION
Can be found [here][3]APPROACH
Consider count* = 20 and find the number of boxes needed, if size S varies from 1 to 20.
Size of each box S  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20 
Number of boxes Bi  20  10  7  5  4  4  3  3  3  2  2  2  2  2  2  2  2  2  2  1 
The main idea is, instead of finding the value Bi for each size of the box S, we only visit the indices where there is a change in the number of boxes Bi required. Once we have this, we can scan the array linearly and propagate the values among the indices in between. The good part of it is, we mark the changes for each count* in the same array and propagate the values in O(n) time, once for all. So, given count* , here is how we can visit only the changes.
Each size in the interval [lo…hi] requires x boxes
Initially, lo = count*, hi = count*, x = 1;
while(lo>1)
{
hi = lo  1;
x = (n+hi1)/hi;
lo = (n+x1)/x;
}
We maintain the SUM array and DEN array, for finding the numerator and denominator respectively, as shown in the equation for T above. Initially, for all indices i, SUM* = 0 and DEN* = 1. These arrays store the change in the values from previous index.
In the table above, B[2] = 10 and B[3] = 7
We store the change as SUM[3] += 3 and DEN[3] *= 10!/7!
After noting all these changes for all the K types of boxes, a linear prefix sum is run on the array SUM, so that SUM* now has the desired numerator as shown in equation for T above. Similarly, we run a linear prefix product on the array DEN, so that DEN* has the desired denominator as shown in equation for T above. A similar approach is used to find the arrays SUM2[] , DEN2[] and ODDCOUNT[] , to find the value of P. Once we have these arrays, each query can be answered in constant time.
TESTER'S SOLUTION
Can be found [here][4]APPROACH
(needs further improvement)Here we describe the method for counting palindromic strings P and finding the value of T is similar.
Sort the queries at first, such that S[1] < S[2] < ... < S[Q].
Suppose, at first S = 1, and it is increasing.
Let
SUM = sum of count*/2 (integer division)
ODD = the number of odd elements of count
FACT = SUM! / (count\[1]/2)! / (count\[2]/2)! / ... / (count[N]/2)! be precalculated.
Then, the answer for S = 1 is FACT if ODD ≤ 1, 0 if ODD ≥ 2.
Let k = ceil(count* / j), then when S goes from k1 to k, the number of boxes corresponding to count* will go from j+1 to j.
Precalculate all time of decreasing the number of box, and sort this array.
After that, all we just need to simulate as follows.
When count* decreases,
if(count* is even)
{
FACT will be multiplied inverse_of(SUM) * count*
SUM will be SUM1
ODD will be ODD+1
count* will be count*1
}
else
{
ODD will be ODD1
count* will be count*1
}

We will try to explain the tester’s approach with more clarity soon. Feel free to add your own approach till then