http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3
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 hisher score, the product of the numbers on the balloons heshe'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 hisher opponent's score. The player with the lower score is presumed to have told the truth, because if heshe were to lie about hisher score, heshe 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 heshe 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 49 3599 610 62 36
Sample Output
49 610 62
题意分析：
一群人在玩游戏，有编号1-100的气球，每个人的分数是他踩破气球的数字的乘积，有的人可能会对分数有怀疑，低分的人可以挑战高分的人，比如343和49,343只能由49*7得来，而49只能由49得来，所以49已经踩了，343就不可能再出现，这时候任务挑战成功了，输出低分。
当两个分数都有可能出现时，高分获胜；
当两个分数都不可能出现时，如101和103，高分获胜。
解题思路：
对于两个数的因子深搜，看最后能不能同时变成1，如果可以证明两个分数都有可能出现。
#include <stdio.h>
#include <algorithm>
using namespace std;
int tn, tm;
void dfs(int n, int m, int k)
{
if (n == 1 && m == 1)
{
tn = 1;
return ;
}
if (m == 1)
tm = 1;
while ((k <= m || k <= n) && k <= 100)
{
if (m%k == 0)
dfs(n, m / k, k+1);
if (n%k == 0)
dfs(n / k, m, k+1);
if (tn)
return ;
k++;
}
return ;
}
int main()
{
int n, m;
while (scanf("%d%d", &n, &m) != EOF) {
if (n < m)
swap(n, m);
tn = 0;
tm = 0;
dfs(n, m, 1);
if (tm == 1 && tn == 0)
printf("%d
", m);
else
printf("%d
", n);
}
return 0;
}