algorithm - 找到整数中不同方程数的最快方法是什么？

Question

等式看起来像这样:

x1 + x2 + x3 + x4 + ... + xk = n

和(0 <= n <= 1000, 0 < k <=1000)

例子:

n=3 and k=2

3+0=3 2+1=3
0+3=3 1+2=3

output: 4 // count of equations

我想不出任何方法，即使是最糟糕的 100 循环方法。

Answer 1

n = 0 -> 1
k = 1 -> 1
k = 2 -> n + 1
k > 2 -> func(n, k - 1) + func(n - 1, k - 1) + .... + func(0, k - 1)
      // 0 + ...          1 + ...                     n + 0 + ... + 0

因此，递归地做......

int func(int n, int k) {
    assert (n >= 0);
    assert (k > 0);
    if (n == 0 || k == 1) {
        return 1;
    }
    else if (k == 2) {
        return n + 1;
    }
    else {
        int sum = 0;
        for (int i = 0; i <= n; i++) {
            sum += func(i, k - 1);
        }
        return sum;
    }
}

消除冗余计算

int result[NMAX + 1][KMAX + 1] = {0};
int func(int n, int k) {
    assert (n >= 0);
    assert (k > 0);
    if (n == 0 || k == 1) {
        return 1;
    }
    else if (k == 2) {
        return n + 1;
    }
    else if (result[n][k] != 0) {
        return result[n][k];
    }
    else {
        int sum = 0;
        for (int i = 0; i <= n; i++) {
            sum += func(i, k - 1);
        }
        result[n][k] = sum;
        return sum;
    }
}