【Acm】算法之美—Crashing Balloon

题目概述：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 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.

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

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简单描述

　　这也是一个很有意思的题,不认真读题,可能还不太明白.我来简单翻译一下

有一个游戏,规则是这样,有一堆气球100个,标号1->100,有两个人参与.一但开始,两个人就疯狂的踩气球,时间到就结束了(也许就10s),把他们各自踩破的球上的编号乘起来,分别是M,N,那么排名自然揭晓了

　　可是分数低的人不服气,想申诉.现在问题来了,怎么申诉呢?因为每个标号的球只有一个,所以加入B踩破的话,A就没办法踩了,申诉想要产生的矛盾就在这儿.现在假如分数是 343 49 ,343可以是踩了 7和49,49只能是踩49,他们两都同时必须要踩这个49,那么就产生矛盾了.所以49赢了.还有要是有人的分数,不能由1->100的数的成绩的出,算说假话,如果两个人都说假话,还是分高的赢.

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题目分析

　　有三种情况:

　　(1)A,B没有矛盾,那么A赢

　　(2)A,B怎么都会有矛盾,而且B说的是真话,那么B赢

　　(3)A,B怎么都会有矛盾,而且B说的是假话,那么A赢

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解题算法

　　下面的源代码中，我对主要的代码都作了详细的注释，如题

#include < stdio.h>
int flagA,flagB;
int result ;
void dfs(int m,int n,int kk)    //运用了深度优先 的搜索策略
{
    int k =kk;
    if(m==1 && n==1)    //在两个数的所有各不相同的因子中，有因子能重新乘出给出的两个数，则A说了真话
    {
       flagA=1; //A说了真话
       return ;
    }
    if(n==1) //在两个数的所有各不相同的因子中，没有任何因子能重新乘出给出的两个数，则B说了真话
        flagB =1;   //B说了真话
    while( (k <  m || k <  n) && (k< 100) )
    {
       k++;
       /*
       *依次找出两个数所有各不相同的因子，如24和12的所有因子为 2，3，4，6，8，12 ，
        再在这些因子中搜索，看是否能重新乘出给出的两个数
       */
       if(m%k ==0)
       {
            dfs(m/k,n,k);
            if(flagA)
                return ;
       }
       if(n%k ==0 )
       {
            dfs(m,n/k,k);
            if(flagA)
                return ;
       }
    }
}
int main()
{
    int A,B,t;
    while(scanf("%d%d",&A,&B)!=EOF )
    {
        if(A <  B )  //保证A大B小
        {
            t=A;
            A=B;
            B=t;
        }
        flagA =0;   //先假定AB都说假话
        flagB =0;
        dfs(A,B,1); //判断AB矛盾
        /*
        *要求:
        *较小者发起挑战，若较大者被证明说谎，较小者胜（较小者说真话，同时较大者说了假话）;
        *若较大者可以成立，则较大者胜;
        *若较小者对自己的结果计算错误，也就是较小者不能成立，如因子中包含一个大于100的质数，则挑战不会举行，较大者胜
        */
        result =A;
        if(flagA ==0 && flagB ==1)  //只有证明A说了假话，并且B说了真话，才算B赢
            result =B;
        printf("%d\n",result);
    }
    return 0;
}

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