ZOJ Problem Set - 1005 Jugs
来源:互联网 发布:阿里云主机免费 编辑:程序博客网 时间:2024/06/08 06:09
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 45 7 3
Sample Output
fill Bpour B Aempty Apour B Afill Bpour B Asuccessfill Apour A Bfill Apour A Bempty Bpour A Bsuccess
我才知道原来special judge 是可以和用例输出不一样的:
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int a,b,c;
while(cin>>a>>b>>c)
{
int d=0;
while(1)
{
cout<<"fill A"<<endl;
cout<<"pour A B"<<endl;
d+=a;
if(d>b)
{
d-=b;
cout<<"empty B"<<endl;
cout<<"pour A B"<<endl;
}
if(d==c)break;
}
cout<<"success"<<endl;
}
}
- ZOJ Problem Set - 1005 Jugs
- ZOJ Problem Set - 1005 Jugs
- ZOJ Problem Set - 1005 Jugs
- ZOJ Problem Set - 1005 Jugs
- ZOJ Problem Set - 1005
- ZOJ-1005-Jugs
- ZOJ 1005 Jugs
- ZOJ 1005 Jugs
- zoj 1005 Jugs
- ZOJ 1005 Jugs (DFS)
- zoj 1005 Jugs
- zoj 1005 Jugs
- ZOJ 1005 Jugs 【BFS】
- ZOJ 1005 Jugs
- zoj 1005 jugs
- ZOJ 1005 Jugs
- zoj 1005Jugs
- zoj 1005 Jugs (模拟)
- iOS TableViewCell的使用总结
- 【如何保护Windows XP系统帐户的安全】
- Valid Sudoku
- 系统安装小结
- Android GestureDetector 手势基础
- ZOJ Problem Set - 1005 Jugs
- QT隐藏mouse
- hdu 1232——畅通工程 (无优化)
- 实施质量保证与实施质量控制的区别与联系
- ubuntu上安装PAC Manager
- Web测试方法
- 如何以PDU格式 备份和恢复彩信(MMS)
- 图片上传前预览
- ORA-28000: the account is locked-的解决办法