《Cracking the Coding Interview》——第6章：智力题——题目3

2014-03-20 00:48

题目：有3升的瓶子和5升的瓶子，只允许倒满、倒到满为止、或是泼光三种操作，怎么搞出4升水呢？

解法：如果A和B是互质的两个正整数，且A<B，令X=B-A，则(X%B, 2X%B, 3X%B, ..., BX%B)正好就是0~n-1的一个排列。也就是说，两瓶子的容积如果互质，就可以配出0~B之间的任意容积。至于怎么配，请看代码。

代码：

 1 // 6.3 Given two jugs with size x and y, and a volume v between 0 and y inclusive. Show how you're gonna get that value.

 2 // Note that you can only fill a jug, empty a jug or pour from one into another.

 3 // x and y will be positive and relatively prime.

 4 #include <cstdio>

 5 using namespace std;

 6 

 7 int gcd(int x, int y)

 8 {

 9     return x == 0 ? y : gcd(y % x, x);

10 }

11 

12 int main()

13 {

14     int x, y;

15     int vx, vy;

16     int v;

17     

18     while (scanf("%d%d%d", &x, &y, &v) == 3 && (x > 0 && y > 0 && v > 0)) {

19         if (x > y) {

20             continue;

21         }

22         if (gcd(x, y) > 1) {

23             continue;

24         }

25         

26         vx = 0;

27         vy = y;

28         printf("[%d, %d]: fill y\n", vx, vy);

29         while (vy != v) {

30             if (vy < x) {

31                 vx = vy;

32                 vy = 0;

33                 printf("[%d, %d]: pour y into x\n", vx, vy);

34                 vy = y;

35                 printf("[%d, %d]: fill y\n", vx, vy);

36                 vy = vy - (x - vx);

37                 vx = x;

38                 printf("[%d, %d]: pour y into x\n", vx, vy);

39                 vx = 0;

40                 printf("[%d, %d]: empty x\n", vx, vy);

41             } else {

42                 vy -= x;

43                 vx = x;

44                 printf("[%d, %d]: pour y into x\n", vx, vy);

45                 vx = 0;

46                 printf("[%d, %d]: empty x\n", vx, vy);

47             }

48         }

49     }

50     

51     return 0;

52 }