《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：

 // 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.
 // Note that you can only fill a jug, empty a jug or pour from one into another.
 // x and y will be positive and relatively prime.
 #include <cstdio>
 using namespace std;

 int gcd(int x, int y)
 {
     return x ==  ? y : gcd(y % x, x);
 }

 int main()
 {
     int x, y;
     int vx, vy;
     int v;

     while (scanf("%d%d%d", &x, &y, &v) ==  && (x >  && y >  && v > )) {
         if (x > y) {
             continue;
         }
         if (gcd(x, y) > ) {
             continue;
         }

         vx = ;
         vy = y;
         printf("[%d, %d]: fill y\n", vx, vy);
         while (vy != v) {
             if (vy < x) {
                 vx = vy;
                 vy = ;
                 printf("[%d, %d]: pour y into x\n", vx, vy);
                 vy = y;
                 printf("[%d, %d]: fill y\n", vx, vy);
                 vy = vy - (x - vx);
                 vx = x;
                 printf("[%d, %d]: pour y into x\n", vx, vy);
                 vx = ;
                 printf("[%d, %d]: empty x\n", vx, vy);
             } else {
                 vy -= x;
                 vx = x;
                 printf("[%d, %d]: pour y into x\n", vx, vy);
                 vx = ;
                 printf("[%d, %d]: empty x\n", vx, vy);
             }
         }
     }

     return ;
 }