2013
11-11

# PERMS

Count the number of permutations that have a specific number of inversions.

Given a permutation a1, a2, a3,…, an of the n integers 1, 2, 3, …, n, an inversion is a pair (ai, aj) where i < j and ai > aj. The number of inversions in a permutation gives an indication on how “unsorted” a permutation is. If we wish to analyze the average running time of a sorting algorithm, it is often useful to know how many permutations of n objects will have a certain number of inversions.

In this problem you are asked to compute the number of permutations of n values that have exactly k inversions.

For example, if n = 3, there are 6 permutations with the indicated inversions as follows:

123			0 inversions

132			1 inversion (3 > 2)

213			1 inversion (2 > 1)

231			2 inversions (2 > 1, 3 > 1)

312			2 inversions (3 > 1, 3 > 2)

321			3 inversions (3 > 2, 3 > 1, 2 > 1)

Therefore, for the permutations of 3 things

• 1 of them has 0 inversions
• 2 of them have 1 inversion
• 2 of them have 2 inversions
• 1 of them has 3 inversions
• 0 of them have 4 inversions
• 0 of them have 5 inversions
• etc.

The input consists one or more problems. The input for each problem is specified on a single line, giving the integer n (1 <= n <= 18) and a non-negative integer k (0 <= k <= 200). The end of input is specified by a line with n = k = 0.

For each problem, output the number of permutations of {1, …, n}with exactly k inversions.

3 0
3 1
3 2
3 3
4 2
4 10
13 23
18 80
0 0


1
2
2
1
5
0
46936280
184348859235088


//* @author:alpc12
import java.io.FileNotFoundException;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

public static void main(String[] args) throws FileNotFoundException {
new Main().run();

}

BigInteger [][] dp;

private void run() throws FileNotFoundException {
Scanner cin = new Scanner(System.in);
dp = new BigInteger[19][201];
for(int i = 0; i < 19; ++i) {
Arrays.fill(dp[i], BigInteger.ZERO);
}
dp[1][0] = BigInteger.ONE;
for(int i = 2; i < 19; ++i) {
for(int j = 0; j <= (i-1)*i/2 && j < 201; ++j) {
for(int k = j; k >= j-i+1 && k >= 0; --k) {
}