ZOJ1003 Crashing Balloon

Crashing Balloon

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Time Limit: 2 Seconds      Memory Limit: 65536 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

题目大意就是有两个人报数字，一个人报大，一个人报小，所报的都不知道真假，但是小的那个人可以质疑大的，如果质疑成功了，且小的是真的则小的赢，否则就大的赢。

规则：有编号为1-100的小球，每个人的分数即为所取得的小球编号相乘（小球编号不重复），最后各自报出自己的分数。

题目规定每次小的都会质疑大的。

质疑的过程：如果大的数的因子（即气球）在小的数的因子（即气球）中出现过的话，则可辨别出大的说谎（因为气球只有一个，小的优先）。如果大的数的因子都没有在小的数的因子中出现的话，即大的赢（无论所报的数是真的还是假的），否则小的赢。

总结：小赢的情况：小的数是真的且大的数是假的。否则都是大的赢。

本题要用到DFS遍历所有因子。

#include <iostream>#include <cstring>#include <algorithm>using namespace std;bool flag1,flag2;//大的数flag1，小的flag2void dfs(int big,int small,int k){        if(small==1)//小数的气球都找完了    {        flag2=true;//小数的是真的        if(big==1)//如果在小数的因子都找完的情况下，大数的也能到1，则大数也是真的            flag1=true;    }    if(k<2||(flag1&&flag2))//如果都是是真的话就没必要在往下搜索了        return;    //下面相当于建立分支，小的用，大的用，大小都不用    if(small%k==0)//先让小的用气球        dfs(big,small/k,k-1);    if(big%k==0)//再让大的用        dfs(big/k,small,k-1);    dfs(big,small,k-1);//都用不到就看下一个气球}int main(){    int a,b;    while(cin>>a>>b){    if(a<b)        swap(a,b);//交换数值        flag1=false;        flag2=false;        dfs(a,b,100);        if(!flag1&&flag2)cout<<b<<endl;        else cout<<a<<endl;    }    return 0;}