KY261 Jugs

题目地址

相对素数：不能被所有给定的整数（除了0）整除的数

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 arefill 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.

分析：
A B 两个罐子，无限水，求得到指定体积水的方法步骤

思路：
显然，可以倒出的水体积是：A B 的最大公因数的倍数
Step1：如果当前A满了，清空A
Step2：如果当前B是空，倒满B
Step3：把B的水倒入A（假定A体积小于等于B体积）
Step4：如果当前A有水，但是不满，尽可能把B中水往A里面倒
不断循环Step1~Step4，直到A==N || B == N 的时候，就结束

代码：

#include<iostream>
#include<string>
#include<vector>using namespace std;int main(){int ca,cb,n;while (cin>>ca>>cb>>n){//初始化罐子内水的体积int a=0,b=0;while (!(a==n||b==n||a+b==n)){if(a==ca){//A满了，倒空a=0;cout<<"empty A"<<endl;}else if(b==0){//B空了，倒满b=cb;cout<<"fill B"<<endl;}else{//A没满，B没空，持续性的将B倒入A//B中的水：要么可以一次性倒满A，要么不可以,要考虑到A中有水但不满的情况if(a+b<=ca){a=a+b;b=0;}else{b=b-(ca-a);a=ca;}cout<<"pour B A"<<endl;   }}cout<<"success"<<endl;     }}

问题：
本算法只考虑了从B往A倒水的情况，实际上还可以从A往B倒水

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