参考：

算法辅导课

DP1 时间复杂度：$O(n^2 m)$

import java.util.*;

public class Main {
    public static void main(String [] args) {
        int [][] f = new int[101][11];
        int n , m;
        Scanner sc = new Scanner(System.in);

        while(sc.hasNext()) {
            n = sc.nextInt();
            m = sc.nextInt();

            for(int i = 1 ;i <= n ;i ++) f[i][1] = i; // 测量长度为i，有1个鸡蛋。最坏情况的最小值i
            for(int i = 1 ;i <= m ;i ++) f[1][i] = 1; // 测量长度为1，有i个鸡蛋。最坏情况的最小值1

            for(int i = 2;i <= n ;i ++)
                for(int j = 2 ;j <= m ;j ++) {
                    f[i][j] = f[i][j - 1];
                    for(int k = 2 ;k <= i ;k ++) 
                        f[i][j] = Math.min(f[i][j],Math.max(f[k - 1][j - 1], f[i - k][j]) + 1);
                }

            System.out.println(f[n][m]);
        }
    }
}

DP2 时间复杂度：$O(nm)$

import java.util.*;

public class Main {
    public static void main(String [] args) {
        Scanner sc = new Scanner(System.in);

        int [][] f = new int[101][11];
        int n , m;
        while(sc.hasNext()) {
            n = sc.nextInt();
            m = sc.nextInt();

            for(int i = 1 ;i <= n ;i ++) {
                // f[i - 1][j] 没碎 
                // f[i - 1][j - 1] 碎了
                for(int j = 1 ;j <= m ;j ++) f[i][j] = f[i - 1][j] + f[i - 1][j - 1] + 1;

                if(f[i][m] >= n) {
                    System.out.println(i);
                    break;
                }
            }
        }
    }
}