ZOJ-1003-Crashing Balloon(搜索)

 

Crashing Balloon

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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 493599 61062 36

Sample Output

4961062

就是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

最后的结果判断麻烦点，题目要求较小者发起挑战，若较大者被证明说谎，较小者胜，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

有这样的逻辑

if(min ok)

{

    if(max ok)

        cout<<max;

    else cout<<min;

}

else cout<<max;

于是，先在因子中拼较小的数，若成功，则再拼较大的数

#include <iostream>using namespace std;int factor[100];int used[101];int a,b,j,mmax,mmin,pa;bool flag;bool dfs(int i,int cur,int d){ if(i==j) { if(!flag&&cur==mmin)//flag用于控制搜索拼哪一个数 { pa=1; flag=true; if(dfs(0,1,mmax))return true; flag=false; } else if(flag&&cur==mmax) { pa=2; return true; } } else { int mul=factor[i]*cur; if(mul<=d&&used[factor[i]]==0) { used[factor[i]]=1; if(dfs(i+1,mul,d)) return true; used[factor[i]]=0; } if(dfs(i+1,cur,d))return true; } return false;}int main(){ int a,b; while(cin>>a>>b) { j=0;pa=0;flag=false; if(a>b) { mmin=b;mmax=a; } else { mmin=a;mmax=b; } //若两数都在100以内，直接出结果 if(a<=100&&b<=100&&a!=b) cout<<mmax<<endl; else { memset(used,0,sizeof(used)); //找出两数所有的各不相同的且在100以内的因数 for(int i=2;i<=mmax/2;++i) if(mmax%i==0) if(i<=100&&used[i]==0) { used[i]=1; factor[j++]=i; } for(int i=2;i<=mmin/2;++i) if(mmin%i==0) if(i<=100&&used[i]==0) { used[i]=1; factor[j++]=i; } //较小数若在100内，则其本身可能是大数的因数 if(mmin<=100&&used[mmin]==0)factor[j++]=mmin; memset(used,0,sizeof(used)); dfs(0,1,mmin); if(pa==0||pa==2) cout<<mmax<<endl; else if(pa==1) cout<<mmin<<endl; } } return 0;}