ZOJ 1003 Crashing Balloon

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

 

思路：要根据题目的要求进行模拟。根据裁判方法，判断哪个选手是正真的赢家。定义int  a , b , p ;因为只要输入正确的分数，所以对输入数据，总是将高分赋值给a , 低分给b。要得到他们所才的气球的编号，据需要进行因式分解，然后判断是否有公共的因子。如果有的话，分数低的那个就是赢家；如果没有，分数高的就是赢家。

     气球一共只有100个，将得分与数字2~100逐一整除，如果是分解的因子就知道是踩哪些气球了。

    如果所申报的得分就在2~100之内，不进行分解，因为有相应的气球。

   如果判定高分说谎，低分说实话，则输出低分。

  如果全部说谎或者全部正确，判定高分获胜。裁判程序是由judge()函数完成的。m ，n分别对应实参a，b。形参p就是气球的编号。

 

AC代码：

#include<iostream>#include<stdio.h>using namespace std;bool atrue,btrue;int judge(int m,int n,int p){ if(atrue) return 0; if(m==1&&n==1){atrue=true;return 0;} if(n==1)btrue=true; while(p>1) {  if(m%p==0) judge(m/p,n,p-1);  if(n%p==0) judge(m,n/p,p-1);  p--; } return 0;}int main(){ int a,b; while(scanf("%d%d",&a,&b)!=EOF) {  if(a<b){int temp=a;a=b;b=temp;}  atrue=false;btrue=false;  judge(a,b,100);  if(!atrue&&btrue)printf("%d\n",b);  else printf("%d\n",a); } return 0;}

另外还有一个AC代码也不错的：

#include<iostream>#include<algorithm>using namespace std;int a,b,temp,af,bf;void judge(int a,int b,int t){if(a==1&&b==1) bf=1;if(t>100){if(a==1&&b!=1)af=1;}else{if(a%t==0) judge(a/t,b,t+1);if(b%t==0) judge(a,b/t,t+1);judge(a,b,t+1);}}int main(){while(scanf("%d%d",&a,&b)!=EOF){if(a>b)swap(a,b);af=bf=0;judge(a,b,2);if(bf)printf("%d\n",b);else if(af) printf("%d\n",a);else printf("%d\n",b);}return 0;}