URAL_1036

    用f[i][j]表示递推到第i位时数字和为j的方案数，最后用下乘法原理f[N][S/2]*f[N][S/2]就是最后结果。

    由于结果比较大，所以需要高精度。

import java.math.BigInteger;import java.util.Scanner;public class Main {    static int N, S;    static BigInteger[][] f = new BigInteger[60][510];    public static void main(String[] args) {        Scanner cin = new Scanner(System.in);        while(cin.hasNext())        {            N = cin.nextInt();            S = cin.nextInt();            if(S % 2 == 1)                System.out.println("0");            else                solve();        }    }    static void solve()    {        int i, j, k;        BigInteger sum;        S /= 2;        for(i = 1; i <= S; i ++)            f[0][i] = new BigInteger("0");        f[0][0] = new BigInteger("1");        for(i = 1; i <= N; i ++)        {            sum = new BigInteger("0");            k = 0;            for(j = 0; j <= S; j ++)            {                if(j - k > 9)                {                    sum = sum.add(f[i - 1][k].negate());                    ++ k;                }                sum = sum.add(f[i - 1][j]);                f[i][j] = sum;            }        }        System.out.println(f[N][S].multiply(f[N][S]));    }}