- 题目描述：
-
In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with exactly 4 gallons. This problem generalizes that puzzle.
You have two jugs, A and B, and an infinite supply of water. There are three types of actions that you can use: (1) you can fill a jug, (2) you can empty a jug, and (3) you can pour from one jug to the other. Pouring from one jug to the other stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons and B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A.
A problem is given by a triple (Ca,Cb,N), where Ca and Cb are the capacities of the jugs A and B, respectively, and N is the goal. A solution is a sequence of steps that leaves exactly N gallons in jug B. The possible steps are
fill A
fill B
empty A
empty B
pour A B
pour B A
successwhere "pour A B" means "pour the contents of jug A into jug B", and "success" means that the goal has been accomplished.
You may assume that the input you are given does have a solution.
- 输入：
-
Input to your program consists of a series of input lines each defining one puzzle. Input for each puzzle is a single line of three positive integers: Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal. You can assume 0 < Ca <= Cb and N <= Cb <=1000 and that A and B are relatively prime to one another.
- 输出：
-
Output from your program will consist of a series of instructions from the list of the potential output lines which will result in either of the jugs containing exactly N gallons of water. The last line of output for each puzzle should be the line "success". Output lines start in column 1 and there should be no empty lines nor any trailing spaces.
- 样例输入：
-
3 5 4 5 7 3
- 样例输出：
-
fill B pour B A empty A pour B A fill B pour B A success fill A pour A B fill A pour A B empty B pour A B success
思路：
给两个罐子，互相倒水得到规定的体积。此类题解法众多，各有各的思路，网上也能搜到一大片，不多解释了。
代码：
#include <stdio.h>
int ca, cb;
int a, b;
void fill(char c)
{
if (c == 'A')
a = ca;
else
b = cb;
printf("fill %c
", c);
}
void pour(char c1, char c2)
{
if (c1 == 'A')
{
if (a+b >= cb)
{
a -= (cb-b);
b = cb;
}
else
{
b += a;
a = 0;
}
}
else
{
if (a+b >= ca)
{
b -= (ca-a);
a = ca;
}
else
{
a += b;
b = 0;
}
}
printf("pour %c %c
", c1, c2);
}
void empty(char c)
{
if (c == 'A')
a = 0;
else
b = 0;
printf("empty %c
", c);
}
int main(void)
{
int n;
while (scanf("%d%d%d", &ca, &cb, &n) != EOF)
{
a = b = 0;
while (a != n && b != n)
{
fill('B');
pour('B', 'A');
while (b != 0 && a != n && b != n)
{
empty('A');
pour('B', 'A');
}
}
printf("success
");
}
return 0;
}
/**************************************************************
Problem: 1147
User: liangrx06
Language: C
Result: Accepted
Time:0 ms
Memory:912 kb
****************************************************************/