import java.util.Arrays; //第二题:照着大神改的小demo,加了点注释. //dp[i][j] 行,当前歌曲,列,所求和,值,选择数  //dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - A]) //和为k的选择数 = 放x前的 和为k的选择数 + 放x前的 和为(k-x) 的选择数.  //i-1代表放x前,即上一次. j-a 代表和为k-x public class Main2_2 {     public static void main(String[] args) {         int K = 5;         int A = 2;         int X = 3;// X首长度为A的         int B = 3;         int Y = 3; // Y首长度为B的         // - 0 1 2 3 4 5 (k)         // 0         // 2         // 2         // 2         // 3         // 3         // (X+Y)         int dp[][] = new int[X + Y + 1][K + 1];         // 初始化         for (int i = 0; i < X + Y + 1; i++) {             dp[i][0] = 1;         }         for (int i = 1; i < X + Y + 1; i++) {             for (int j = 1; j < K + 1; j++) {                 if (i <= X) {                     if (j < A) {                         // 在不能放A的时候:J-A越界了怎么办                         dp[i][j] = dp[i - 1][j];                     } else {                         // i<=X 就要-A                         dp[i][j] = dp[i - 1][j] + dp[i - 1][j - A];                     }                 } else {                     if (j < B) {                         // 在不能放B的时候:J-B越界了怎么办                         dp[i][j] = dp[i - 1][j];                     } else {                         // X<=i<=X+Y+1 就要-B                         dp[i][j] = dp[i - 1][j] + dp[i - 1][j - B];                     }                 }             }         }         for (int i = 0; i < X + Y + 1; i++) {             System.out.println(Arrays.toString(dp[i]));         }     } }