ZJU 1003 解题报告

Crashing Balloon

Time Limit: 1 Seconds Memory Limit: 32768 KB

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 his\her score, the product of the numbers on the balloons he\she'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 his\her opponent's score. The player with the lower score is presumed to have told the truth, because if he\she were to lie about his\her score, he\she 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 he\she 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

Output for Sample Input

49
610
62

Source:Zhejiang University Local Contest 2001

题目的意思是说,地面上有100个标记为1-100的气球,参赛的两位选手的起始分数都是1,每踩到一个气球就将气球上的分数乘到自己已有的分数上,最终两位选手报上自己的分数,如果两个参赛的说的都是真话,则分数多的为胜利者,但有时选手会把分数弄错.所以需要写一个程序判定比赛结果.

先假设分数多的为胜利者,分数低的为挑战者

判定过程如下:

胜利者 挑战者 获胜者
正确 正确 胜利者
正确 错误 胜利者
错误 正确 挑战者
错误 错误 挑战者

由表格的分析可以看出来,挑战者胜利的唯一条件就是自己分数计算正确而胜利者算错了.使用深搜将两者的分数分解,如果挑战者的分数最后可以除到1,则挑战者说的是真话,那么只要分析胜利者的分数能否除到1就可以判定比赛结果了.

CODE

#include<stdio.h>
#include<stdbool.h>
#define swap(a,b) {int t=a;a=b;b=t;}
 
_Bool Ta,Tb;
 
void dfs(int a,int b,int c)
{
  if(Ta&&Tb) return;
  if(a==1&&b==1) Ta=Tb=true;
  if(b==1) Tb=true;
  if(c==1) return;
  if(a%c==0) dfs(a/c,b,c-1);
  if(b%c==0) dfs(a,b/c,c-1);
  dfs(a,b,c-1);
}
 
int main()
{
  int a,b;
  while(scanf("%d%d",&a,&b)!=EOF)
    {
      if(a<b) swap(a,b);
      Ta=Tb=false;
      dfs(a,b,100);
      if(!Ta&&Tb)
        printf("%d\n",b);
      else
        printf("%d\n",a);
    }
  return 0;
}

不知道为什么,如果先判定胜利者是否错误,就会WA.

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