Problem:
Score : 600 points

Problem Statement

Let us define the oddness of a permutation p = {p_1,\ p_2,\ ...,\ p_n} of {1,\ 2,\ ...,\ n} as \sum_{i = 1}^n |i - p_i|.

Find the number of permutations of {1,\ 2,\ ...,\ n} of oddness k, modulo 10^9+7.

Constraints

All values in input are integers.

1 \leq n \leq 50

0 \leq k \leq n^2

Input

Input is given from Standard Input in the following format:

n k

Output

Print the number of permutations of {1,\ 2,\ ...,\ n} of oddness k, modulo 10^9+7.

Sample Input 1

3 2

Sample Output 1

2

There are six permutations of {1,\ 2,\ 3}. Among them, two have oddness of 2: {2,\ 1,\ 3} and {1,\ 3,\ 2}.

Sample Input 2

39 14

Sample Output 2

74764168

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Globalization;
using System.IO;
using System.Linq;
using System.Text;

class Program
{
    private int N, K;
    private int[] A;

    static void Main(string[] args)
    {
        new Program().Solve();
    }

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        K = sc.NextInt();

        var dp = new ModInt[N + 1, N + 1, K + 1];

        dp[0, 0, 0] = 1;

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j <= i; j++)
            {
                for (int k = 0; k <= K; k++)
                {
                    // i個目まで見て　保留がj個　奇妙さがk
                    // i個目　確定
                    if (k + j * 2 <= K)
                        dp[i + 1, j, k + j * 2] += dp[i, j, k] * (1 + 2 * j);
                    if (j + 1 <= N && k + (j + 1) * 2 <= K)
                        dp[i + 1, j + 1, k + (j + 1) * 2] += dp[i, j, k];
                    if (j - 1 >= 0 && k + (j - 1) * 2 <= K)
                        dp[i + 1, j - 1, k + (j - 1) * 2] += dp[i, j, k] * (j) * (j);
                }
            }
        }
        Console.WriteLine(dp[N, 0, K]);
    }
}


#region ModInt
/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt
{
    /// <summary>
    /// 剰余を取る値．
    /// </summary>
    public const long Mod = (int)1e9 + 7;

    /// <summary>
    /// 実際の数値．
    /// </summary>
    public long num;
    /// <summary>
    /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
    /// </summary>
    /// <param name="n">インスタンスが持つ値</param>
    /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
    public ModInt(long n) { num = n; }
    /// <summary>
    /// このインスタンスの数値を文字列に変換します．
    /// </summary>
    /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
    public override string ToString() { return num.ToString(); }
    public static ModInt operator +(ModInt l, ModInt r) { l.num += r.num; if (l.num >= Mod) l.num -= Mod; return l; }
    public static ModInt operator -(ModInt l, ModInt r) { l.num -= r.num; if (l.num < 0) l.num += Mod; return l; }
    public static ModInt operator *(ModInt l, ModInt r) { return new ModInt(l.num * r.num % Mod); }
    public static implicit operator ModInt(long n) { n %= Mod; if (n < 0) n += Mod; return new ModInt(n); }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(ModInt v, long k) { return Pow(v.num, k); }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(long v, long k)
    {
        long ret = 1;
        for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
            if ((k & 1) == 1) ret = ret * v % Mod;
        return new ModInt(ret);
    }
    /// <summary>
    /// 与えられた数の逆元を計算します．
    /// </summary>
    /// <param name="v">逆元を取る対象となる数</param>
    /// <returns>逆元となるような値</returns>
    /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
    public static ModInt Inverse(ModInt v) { return Pow(v, Mod - 2); }
}
#endregion


class Scanner
{
    public Scanner()
    {
        _stream = new StreamReader(Console.OpenStandardInput());
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private readonly StreamReader _stream;
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = _stream.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = _stream.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}