算法 g(n,m)最为准确的时间复杂度分析结果为

#include <algorithm>
#include <iostream>
#include <limits>
using namespace std;

const int MAXN = 105;

const int MAXK = 105;
int h[MAXN][MAXK];

int f(int n, int m) {
    if (m == 1) return n;
    if (n == 0) return 0;
    int ret = numeric_limits<int>::max();
    for (int i = 1; i <= n; i++)
        ret = min(ret, max(f(n - i, m), f(i - 1, m - 1)) + 1);
    return ret;
}

int g(int n, int m) {
    for (int i = 1; i <= n; i++)
        h[i][1] = i;
    for (int j = 1; j <= m; j++)
        h[0][j] = 0;
    h[0][0] = 0;

    for (int i = 1; i <= n; i++) {
        for (int j = 2; j <= m; j++) {
           h[i][j] = numeric_limits<int>::max();
           for (int k = 1; k <= i; k++)
               h[i][j] = min(
                   h[i][j],
                    max(h[i - k][j], h[k - 1][j - 1]) + 1);
       }
    }
    return h[n][m];
}

int main() {
    int n, m;
    cin >> n >> m;
    cout << f(n, m) << endl << g(n, m) << endl;
    return 0;
}

假设输入的n、m均是不超过100的正整数，完成下面的判断题和单选题：