踩气球

A - Crashing Balloon

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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Description

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：一共有100个标号为1-100的气球，两个小孩同时踩，分数是气球编号的乘积，游戏结束后分数低的选手可以进行质疑，如果分数高的选手说了假话，则分数低的选手获胜，如果两个人都说了假话，则分数高的人获胜，输出获胜人的分数。

例如：

①125和25

125=25*5，而25=25*1  高分选手想获得125 必须踩25和5号气球，而25号气球被低分踩了，所以高分说谎，低分赢。

②162和81

81只能踩81 但是162=81*2=27*3*2，所以高分没说谎，所以高分赢。

算法分析

1、输赢判断

高分说谎低分说谎 高分赢

高分实话低分实话 高分赢

高分实话低分说谎 高分赢

高分说谎低分实话 低分赢

2、那么如何判断两人是否说谎呢？

首先得分的因数必须是1-100，若两个得分为1-100之间的数，说明二人都没说谎。

超过100的数不可能是质数，否则说谎。

俩个和数看代码（不易表达）。

 代码

#include<cstdio>using namespace std;int ature=0,bture=0;int f(int a,int b,int p){    if(ature) return 0;    if(b==1&&a==1) {ature=1;return 0;}//a==1并不能说明什么如例中的162 和81，只有b把较大的因数拿完了变成1了之后 a还有除了b之外的因数才行    if(b==1){bture=1;}//没有return    while(p>1)    {        if(a%p==0) f(a/p, b, p-1);        if(b%p==0) f(a, b/p, p-1);        p--;    }    return 0;}int main(){    int a,b,t;    while(scanf("%d %d",&a,&b)!=EOF)    {        if(b>a)//既然输出的是分数，那么无所谓是哪个人，让a为高分b为低分        {            t=a;a=b;b=t;        }        ature=0;bture=0;//每次都要初始化        f(a,b,100);//从100开始给因数        if(ature==0&&bture) printf("%d\n",b);//不要忘记换行        else printf("%d\n",a);    }    return 0;}