题解 | #放苹果#

import java.util.Scanner;

// 动态规划表：n为行，m为列
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNext()) { // 注意 while 处理多个 case
            int m = in.nextInt();
            int n = in.nextInt();
            if (n < 2 || m < 2) {
                System.out.println(1);//当只有一个盘子或者一个苹果时，结果只有一种。
            } else {
                int[][] t = new int[n + 1][m + 1];
                for (int r = 0; r <= m; r++) {
                    t[0][r] = 0;
                }
                for (int r = 0; r <= m; r++) {
                    t[1][r] = 1;
                }
                for (int r = 2; r <= m; r++) {
                    t[2][r] = r / 2 + 1;//盘子数为2时，结果有规律，直接填
                }
                for (int c = 1; c <= n; c++) {
                    t[c][0] = 1;
                }
                for (int c = 1; c <= n; c++) {
                    t[c][1] = 1;
                }
                //动态规划表公式：分苹果数小于盘子数时、苹果数大于等于盘子数时两种情况
                for (int i = 3; i <= n; i++) {
                    for (int j = 2; j <= m; j++) {
                        if (j < i) {
                            t[i][j] = t[i - 1][j];
                        } else if (j >= i) {
                            t[i][j] = t[i - 1][j] + t[i][j - i];
                        }

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


        }
    }
}