10、Crashing Balloon

On every June 1st, the Children's Day, there will be a game named "crashing balloon" on TV.   The rule is very simple.  On the ground there are 100 labeled balloons, with the numbers 1 to 100.  After the referee shouts "Let's go!" the two players, who each starts with a score of  "1", race to crash the balloons by their feet and, at the same time, multiply their scores by the numbers written on the balloons they crash.  After a minute, the little audiences are allowed to take the remaining balloons away, and each contestant reports hisher score, the product of the numbers on the balloons heshe's crashed.  The unofficial winner is the player who announced the highest score.

Inevitably, though, disputes arise, and so the official winner is not determined until the disputes are resolved.  The player who claims the lower score is entitled to challenge hisher opponent's score.  The player with the lower score is presumed to have told the truth, because if heshe were to lie about hisher score, heshe would surely come up with a bigger better lie.  The challenge is upheld if the player with the higher score has a score that cannot be achieved with balloons not crashed by the challenging player.  So, if the challenge is successful, the player claiming the lower score wins.

So, for example, if one player claims 343 points and the other claims 49, then clearly the first player is lying; the only way to score 343 is by crashing balloons labeled 7 and 49, and the only way to score 49 is by crashing a balloon labeled 49.  Since each of two scores requires crashing the balloon labeled 49, the one claiming 343 points is presumed to be lying.

On the other hand, if one player claims 162 points and the other claims 81, it is possible for both to be telling the truth (e.g. one crashes balloons 2, 3 and 27, while the other crashes balloon 81), so the challenge would not be upheld.

By the way, if the challenger made a mistake on calculating his/her score, then the challenge would not be upheld. For example, if one player claims 10001 points and the other claims 10003, then clearly none of them are telling the truth. In this case, the challenge would not be upheld.

Unfortunately, anyone who is willing to referee a game of crashing balloon is likely to get over-excited in the hot atmosphere that heshe could not reasonably be expected to perform the intricate calculations that refereeing requires.  Hence the need for you, sober programmer, to provide a software solution.

Input

Pairs of unequal, positive numbers, with each pair on a single line, that are claimed scores from a game of crashing balloon.

Output

Numbers, one to a line, that are the winning scores, assuming that the player with the lower score always challenges the outcome.

Sample Input

343 49
3599 610
62 36

Sample Output

49
610

62

 

 

解题思路：这是我在ZOJ上做的第二道题，虽然有自己的思路，但是感觉太繁琐，步骤太多。我想知道是不是每个OJ上的第二道题都这么刁钻。在这里感谢 infohacker——51上的推荐博客，http://liucw.blog.51cto.com/6751239/1195118

简单来说，就是在1到100之间如果可以找到两组完全不同的除数（每组的除数也不能重复），使得两个被除数（也就是两个人的分数）除以除数之后得到商为1，则输出分数较高者；如果两组除数不管如何找，必有一个除数为两组共有，这样就会使较大的被除数无法得到商为1的结果。通过分析用DFS较为适合本题，通过深度搜索寻找一个使两个被除数均能得到商为1的方案。DFS简直就是神算法，在做这类在多种可能中寻找符合题意的解的题。但是还有一个比较麻烦的问题，就是Java的输入问题，在C/C++中完全不存在这样的问题，Scanner类有很多缺陷，当我们需要循环输入的时候，且输入不止一个整数，控制就显得格外让人头疼。在本题中我是用字符串输入，之后转换为整数进行运算，但这样一来就不免增加了算法运行的时间。最后虽然在编译器可以通过，但是在ZOJ上总是会出现Time Limit Exceed的问题，现在也未解决，最后改用C才通过。希望看过我博客的同仁为我指点一下。

具体代码：

import java.util.Scanner;

public class Main {
     static int flagA, flagB;
     public static void dfs(int a,int b,int k){
          if(a==1&&b==1)
              flagA=1;
          if(b==1)
              flagB=1;
          while((k<a||k<b)&&k<100){
              k++;
              if(a%k==0){
                   dfs(a/k, b, k);
                   if(flagA==1)
                        return;
              }
              if(b%k==0){
                   dfs(a, b/k, k);
                   if(flagA==1)
                        return;
              }
          }
     }

     public static void main(String[] args) {
          Scanner scanner=new Scanner(System.in);
          char[] traslate =new char[20];
          int key,numA,numB;
          while(scanner.hasNextLine()){
              String tempString=scanner.nextLine();
              traslate=tempString.toCharArray();
              int length=traslate.length;
              int[] tra=new int[20];
              key=numA=numB=0;
              flagA=flagB=0;
            for(int i=0;i<length;i++){
                   if(traslate[i]==' '){
                        tra[i]=-1;
                        key=i;
                   }
              }
              for(int i=0;i<length;i++){
                   if(i<key){
                        tra[i]=(int) ((traslate[i]-48)*Math.pow(10, key-1-i));
                        numA+=tra[i];
                   }
                   if(i>key){
                        tra[i]=(int) ((traslate[i]-48)*Math.pow(10, length-1-i));
                        numB+=tra[i];
                   }
              }
              dfs(numA, numB, 1);
              if(flagA==0&&flagB==1)
                   System.out.println(numB);
              else {
                   System.out.println(numA);
              }
          }
     }
}

感悟：DFS需要深入学习，Java的输入问题也需要找时间好好解决。