Is there an efficient algorithm for integer partitioning with restricted number of parts?

Question

I have to create a method that takes two integers, let them be n and m, and returns how many ways there are to sum m positive numbers

Answer 1

public static long p(final long n, final long m) {
    System.out.println("n=" + n + ", m=" + m);
    return p(n, 1, m);
}

private static long p(final long n, final long digitFrom, final long m) {
    final long digitTo = n - m + 1;
    if (digitFrom > digitTo) return 0;
    if (m == 1) return 1;
    long sum = 0;
    for (long firstDigit = digitFrom; firstDigit <= digitTo; firstDigit++)
        sum += p(n - firstDigit, firstDigit, m - 1);
    return sum;
}

Example debug:

n=6, m=3
1+1+4
1+2+3
2+2+2

From debug version:

public static long p(final long n, final long m) {
    System.out.println("n=" + n + ", m=" + m);
    return p(n, 1, m, new Stack());
}

private static long p(final long n, final long digitFrom, final long m, Stack digitStack) {
    final long digitTo = n - m + 1;
    if (digitFrom > digitTo) return 0;
    if (m == 1) {
        digitStack.push(n + "");
        printStack(digitStack);
        digitStack.pop();
        System.out.println();
        return 1;
    }
    long sum = 0;
    for (long firstDigit = digitFrom; firstDigit <= digitTo; firstDigit++) {
        digitStack.push(firstDigit + "+");
        sum += p(n - firstDigit, firstDigit, m - 1, digitStack);
        digitStack.pop();
    }
    return sum;
}