ZOJ Problem Set - 1003
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
这题比较难，虽然做出来了，但始终认为一个 int 有 4 个字节，64 bit，最大仅为 0xffff ffff ffff ffff = 1.8 × 1019
而挑战双方是在砸 1~100 的气球，所以极端情况为一方不砸，另一方砸完，即 small = 1, big = 100! = 9.3 × 10157
所以我认为这题应该采用专门的数据结构来计算如此大的数，如100w进制等。但无奈的是即便是 int 也能 AC ……1 #include <stdio.h>
2 /* 获胜者 挑战者 胜方
3 假 真 挑战者
4 真 真 获胜者
5 真 假 获胜者
6 假 假 获奖者 */
7 #define true 1
8 #define false 0
9
10 int F_BOTH = false, F_SMALL = false;
11
12 void check(int big, int small, int n);
13
14 int main()
15 {
16 int small, big;
17
18 while(scanf("%d%d", &small, &big) == 2)
19 {
20 if(small > big)
21 {
22 small ^= big;
23 big ^= small;
24 small ^= big;
25 }
26
27 F_BOTH = false, F_SMALL = false;
28
29 check(big, small, 100);
30
31 if(F_BOTH == true || F_SMALL == false)
32 printf("%d\n", big);
33 else
34 printf("%d\n", small);
35 }
36 return 0;
37 }
38
39 void check(int big, int small, int n)
40 {
41 if(F_BOTH == true) return; /* 二者都是真的，退出所有层，直接返回到 main */
42
43 else if(big==1 && small==1) /* 二者都被分解完 */
44 {
45 F_BOTH = true;
46 F_SMALL = true;
47 return;
48 }
49
50 else if(small == 1) /* small 被某几个 n 恰好分解完，这里不需要了解 big 的状态 */
51 {
52 F_SMALL = true;
53 }
54
55 if(n<2) return; /* n 达极尽 */
56 /***************************************************/
57 if(big%n == 0) check(big/n, small, n-1); /* 遍历所有情况 ## 注意这种宏观把握的思想！ */
58
59 if(small%n == 0) check(big, small/n, n-1); /* 不用 else */
60 /***************************************************/
61 check(big, small, n-1); /* 都不能被n整除， 进入下一层 */
62 }
