ZOJ-1003-Crashing-Balloon

题目描述

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 49
3599 610
62 36
Sample Output：
49
610
62

解题思路

总体思路，使用回溯法。

首先，定义一个函数int test(int a, int b, int n)

该函数返回1，
当且仅当集合{2, 3, …, n}存在两个互不相交的子集合A = { x1, x2, … , xn}, B = {y1, y2, …, yn}，
使得 x1 * x2 * … * xn = a, y1 * y2 * … * yn = b。

该函数的代码如下

int test(int a, int b, int n)
{if(a==1 && b ==1) return 1;else if(n == 1) return 0;else {int t1, t2, t3;t1 = t2 = t3 =0;//分支1if(a%n == 0) {t1 = test(a/n, b, n-1);if(t1 == 1) return 1; //剪枝}//分支2if(b%n == 0) {t2 = test(a, b/n, n-1);if(t2 == 1) return 1; //剪枝}//分支3t3 = test(a, b, n-1);return t3;}
}

然后，在main函数中运用这个函数对每一对值a，b进行测试：

假设a>b
先测试a，b是否能同时满足，如果能，说明a,b都没撒谎，a赢；
如果a，b不能同时满足，那么就有可能是因为a撒谎了，需要进行下面的测试；
单独测试b，如果b可以单独满足，就可以认为是a撒谎了，b赢；
如果b不可单独满足，则b撒谎了，根据规则，b输。

测试部分的代码：

while(scanf("%d %d", &a, &b) == 2 ){if(a < b) {int t = a; a = b; b = t;} //保证a>bint answer;if( test(a, b, 100) ) answer = a; //a,b都没撒谎，a赢else if( test(1, b, 100) ) answer = b; //b没撒谎且a撒谎了，b赢else answer = a; //b撒谎了，a赢printf("%d\n", answer);}

时间复杂度分析

递归的最大深度取决于int test(int a, int b, int n)函数的变量n。
最坏情况下:
T(n)=3∗T(n−1)T(n) = 3*T(n-1) T(n)=3∗T(n−1)T(1)=1T(1) = 1 T(1)=1
所以，最坏情况下T(n)=log(3n)T(n)=log(3^n)T(n)=log(3n)。不过test函数中加入了剪枝策略，只要找到一个结果就可以立即返回，所以平均情况下不会这么慢。

POJ Accept 截图

完整代码

#include <stdio.h>
#include <stdlib.h>//测试a，b能否同时满足
int test(int a, int b, int n)
{if(a==1 && b ==1) return 1;else if(n == 1) return 0;else {int t1, t2, t3;t1 = t2 = t3 =0;//分支1if(a%n == 0) {t1 = test(a/n, b, n-1);if(t1 == 1) return 1; //剪枝}//分支2if(b%n == 0) {t2 = test(a, b/n, n-1);if(t2 == 1) return 1; //剪枝}//分支3t3 = test(a, b, n-1);return t3;}
}int main(void)
{int a, b;while(scanf("%d %d", &a, &b) == 2 ){if(a < b) {int t = a; a = b; b = t;} //保证a>bint answer;if( test(a, b, 100) ) answer = a; //a,b都没撒谎，a赢else if( test(1, b, 100) ) answer = b; //b没撒谎且a撒谎了，b赢else answer = a; //b撒谎了，a赢printf("%d\n", answer);}return 0;
}