uva 571 Jugs

原題：
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
success
where pour A B" meanspour 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.

Input

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.
Output

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.
Sample Input

3 5 4
5 7 3
Sample Output

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
大意：
給你兩個壺a,b還有一個目標水量c。現在讓你用a,b和無限的水倒出目標水量。其中a,b互質。

#include<iostream>#include<algorithm>#include<map>#include<string>#include<cstring>#include<sstream>#include<cstdio>#include<vector>#include<cmath>#include<stack>#include<queue>#include<iomanip>#include<set>#include<fstream>#include <climits>using namespace std;//fstream input,output;int main(){    ios::sync_with_stdio(false);    int ja,jb,aim;    int a,b;    while(cin>>ja>>jb>>aim)    {        if(ja==aim)        {            cout<<"fill A"<<endl;            cout<<"success"<<endl;            continue;        }        if(jb==aim)        {            cout<<"fill B"<<endl;            cout<<"success"<<endl;            continue;        }        a=b=0;        while(b!=aim)        {            if(a==0)            {                cout<<"fill A"<<endl;                a=ja;            }            if(b<jb)            {                cout<<"pour A B"<<endl;                if(b+ja>jb)                {                    a=ja-(jb-b);                    b=jb;                }                else                {                    b+=a;                    a=0;                }            }            if(b==jb)            {                cout<<"empty B"<<endl;                b=0;                cout<<"pour A B"<<endl;                b=a;                a=0;            }        }        cout<<"success"<<endl;    }//  input.close();//  output.close();    return 0;}

解答：
很經典的一個問題，剛讀題的時候還以為讓我判斷能否倒出目標水量來，結果卻是讓我求操作方法。
首先先說明一下對任意給定的兩個數a,b能否實現操作出目標c。現在假設b大於a，首先讓a裡的水裝滿不斷倒進b中，直到b中的水裝滿。此時a裡剩下的水是x，再把b裡的水倒空，把x裝進b中，不斷重複就能發現這是一個經典的操作——更相減損的方法。也就是求最大公約數的方法。所以只要c的數量能夠除開a與b的最大公約數就能夠倒出目標水量，同時要求目標水量要小於b。

讀完題後糾結了好長時間如何處理最少操作步驟，設a往b的裡裝水操作x，b往a裡裝水是y。則能得到，ax+by=1這樣一個方程，再利用擴充歐幾裡得公式求出一組x,y的解，然後求得最小的x或者y….
後來想不出程式該怎麼寫了，看別人的題解居然發現任意一組解就可以….CNM的。