蓝桥杯2013年第四届真题-买不到的数目 （Java代码）

        题目描述：

                        说白了，其实就是求a和b不能组成的最大数；

       题目解析：

                        这个题我看了网络上给的答案，可是我感觉有点投机取巧的感觉了，答案是直接给输出来的，

                直接就a*b-a-b，可是这个答案一般人都不知道的，所以我用动态规划来解决这个题，下面我来介绍

                一下我的方法，其实就是我打个比方，定义一个长度为100的array数组，并且给他们全部都赋值为0

                如下所示：

                   0 0 0 0 0 0 0 0 0 0 0 

                  1 2 3 4 5 6 7 8 9 10 11

                  打个比方就是输入的是a=3,b=5;

                  那么就把array[3]=1,array[5]=1;

                  然后我们可以遍历这个数组的，如果被遍历的数组的值是1的话，那么我们把array[i+a]和

                   array[i+b]都赋值为1，赋值为1的意思就是这个数可以被a和b给组合起来的，

                  但是为了找到最后的大数，我们还要定义两个数number,count;

                    number是用来表示数组的下标的，而count是用来表示相邻的数组的值为1的个数，

                    在上面你会发现一旦count大于或等于a,b中的大数的时候就可以了，这时候的number表示的下                           标就可以表示我们要找的数了

                    代码如下：

                                    import java.util.Arrays;
                                    import java.util.Scanner;

                                        public class Main {
                                                  public static void main(String[] args) {
                                                       Scanner m=new Scanner (System.in);
   
                                                       int a=m.nextInt();
                                                       int b=m.nextInt();
                                                       int max=a>b?a:b;
                                                       long array[]=new long[9999999];
   
                                                       Arrays.fill(array,0);
                                                       array[a]=1;
                                                       array[b]=1;
       
                                                        int number=0;
                                                        int count=0;
                                                       for(int k=1;k<array.length;k++)
                                                           {
                                                                if(array[k]==0)
                                                                    {
                                                                     number=k;
                                                                     count=0;
                                                                    }
                                                                else
                                                                {
                                                                     array[k+a]=1;
                                                                     array[k+b]=1;
                                                                     count++;
                                                                        }
                                                        if(count==max)
                                                             break;
                                                               }
                                                       System.out.println(number);
                                                          }
                                                        }