Jugs
Time Limit: 2 Seconds Memory Limit: 65536 KB Special Judge
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" 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.
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
题意
有两个瓶子A，B。A，B瓶子容量已知（B的容量>=A的容量），但是每个瓶子内没有刻度，问如何操作能够量取体积为N的水
思路
因为B瓶子的容量>=A瓶子的容量，所以可以利用A，B瓶子容量的差值来进行求解：
1.如果A中没有水，将A装满
2.将A中的水全部倒入B中（此时根据A，B中的总水量来判断A倒入B后的各个瓶子里面的水量）
3.判断B中的水量是否等于N，如果等于N，停止操作，否则，将B中的水全部倒出
参考：https://blog.csdn.net/mdreamlove/article/details/46662083
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ull unsigned long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
const double E=exp(1);
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
int main(int argc, char const *argv[])
{
ios::sync_with_stdio(false);
int a,b,n;//b>=a
while(cin>>a>>b>>n)
{
int ca=0;
int cb=0;
/**
* 这里不加特判也可以，但是最后的两个if的顺序要改变一下
*/
if(n==a)
{
cout<<"fill A\nsuccess"<<endl;
break;
}
if(n==b)
{
cout<<"fill B\nsuccess"<<endl;
break;
}
while(1)
{
if(ca==0)
{
ca=a;
cout<<"fill A"<<endl;
}
else if(ca+cb<=b)
{
cb+=ca;
ca=0;
cout<<"pour A B"<<endl;
}
else
{
ca=ca-(b-cb);
cb=b;
cout<<"pour A B"<<endl;
}
if(cb==b)
{
cout<<"empty B"<<endl;
cb=0;
}
if(cb==n)
{
cout<<"success"<<endl;
break;
}
}
}
return 0;
}