Java实现POJ 1017：装箱问题

话不多说，直接上代码：

//miss 标注的是我提交Wrong Answer 之后发现的错误

package POJ; //NOTE: delete this line when you submit your answer!import java.util.*;public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);while (true) {int [] a = new int[6];a[0] = in.nextInt();a[1] = in.nextInt();a[2] = in.nextInt();a[3] = in.nextInt();a[4] = in.nextInt();a[5] = in.nextInt();int sum = a[0] + a[1] + a[2] + a[3] + a[4] + a[5];if (sum == 0) {break;}int result = calculate(a);System.out.println(result);}}public static int calculate(int [] a) {int p_1 = a[0]; int p_2 = a[1]; int p_3 = a[2];int p_4 = a[3]; int p_5 = a[4]; int p_6 = a[5];int sum = 0;sum = sum + p_6;//cal_5if (p_5 > 0) {sum = sum + p_5;//cal_1 to make upif (p_1 >= (p_5 * 11)) {p_1 = p_1 - (p_5 * 11);}else {p_1 = 0;}}//cal_4if (p_4 > 0) {sum = sum + p_4;//cal_2 to make upif (p_2 >= (p_4 * 5)) {p_2 = p_2 - (p_4 * 5);}else {int r_num = (20 * p_4) - (4 * p_2);p_2 = 0; //miss//cal_1 to make upif (p_1 >= r_num) {p_1 = p_1 - r_num;}else {p_1 = 0;}}}//cal_3if (p_3 > 0) {sum = sum + (p_3 / 4);if (p_3 % 4 == 1) {sum++;//cal_2 to make upif (p_2 >= 5) {p_2 = p_2 - 5;//cal_1 to make upif (p_1 >= 7) {p_1 = p_1 - 7;}else {p_1 = 0;}}else {int r_num = 27 - (4 * p_2);p_2 = 0; //miss//cal_1 to make upif (p_1 >= r_num) {p_1 = p_1 - r_num;}else {p_1 = 0;}}}else if (p_3 % 4 == 2) {sum++;//cal_2 to make upif (p_2 >= 3) {p_2 = p_2 - 3;//cal_1 to make upif (p_1 >= 6) {p_1 = p_1 -6;}else {p_1 = 0;}}else {int r_num = 18 - (4 * p_2);p_2 = 0; // miss//cal_1 to make upif (p_1 >= r_num) {p_1 = p_1 - r_num;}else {p_1 = 0;}}}else if (p_3 % 4 == 3) {sum++;//cal_2 to make upif (p_2 >= 1) {p_2 = p_2 - 1;//cal_1 to make upif (p_1 >= 5) {p_1 = p_1 -5;}else {p_1 = 0;}}else {p_2 = 0;//cal_1 to make upif (p_1 >= 9) {p_1 = p_1 - 9;}else {p_1 = 0;}}}}//cal_2if (p_2 > 0) {sum = sum + (p_2 / 9);if (p_2 % 9 != 0) { //misssum++;int r_num = 36 - (4 * (p_2 % 9));//cal_1 to make upif (p_1 >= r_num) {p_1 = p_1 - r_num;}else {p_1 = 0;}}}//cal_1if (p_1 > 0) {sum = sum + (p_1 / 36);if (p_1 % 36 != 0) {sum++;}}return sum;}}

解题思路：

装箱肯定是要从大盒子开始装，即从6*6的箱子开始到1*1的箱子结束。原则就是，用尽量大的箱子来填满空间（其实本题只涉及到用2*2或1*1的箱子来填充剩余空间）。

6*6箱子：本身占一个单位；

5*5箱子：本身占一个单位，其余用11个1*1的箱子填充；

4*4箱子：本身占一个单位，其余用5个2*2的箱子填充；

3*3箱子：4个占一个单位。故要分四种情况（设其数量为n）考虑：

若（n%4 == 0）: 理想情况，没什么要担心的；

若（n%4 == 1）: 要用5个2*2和7个1*1的箱子填充；

若（n%4 == 2）: 要用3个2*2和6个1*1的箱子填充；

若（n%4 == 3）: 要用1个2*2和5个1*1的箱子填充；

2*2箱子：9个占一个单位。其余用1*1的箱子填充；

1*1箱子：36个占一个单位。若无法整除，需要额外的一个箱子。

当然，上面的都是理想情况；即，如果用来填充的2*2的箱子数不够，则需要1*1的箱子来进行替代填充（1*1的箱子也存在用完的情况，这时以后的装箱就不再考虑填充的情况了）。