ZOJ water jugs problem

最新推荐文章于 2020-07-09 19:41:27 发布

weixin_33691817

最新推荐文章于 2020-07-09 19:41:27 发布

阅读量117

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/songwanzi/p/3507542.html

本文介绍了一种使用递归方法解决水壶问题的方案，并通过C++实现。该问题要求利用两个容量不同的水壶准确量出指定水量。文章首先尝试了循环方法但因超时未被接受，最终采用递归策略成功解决了问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Solution with loop:

#include <iostream>
using namespace std;
int main(){
	int m,n,k,r;
	while(std::cin>>m>>n>>k){
		r=n-m;
		cout<<"fill B"<<endl;
		cout<<"pour B A"<<endl;
		while(r!=k){
			if(r<m){
			  r=n-(m-r);
			  cout<<"empty A"<<endl<<"pour B A"<<endl<<"fill B"<<endl<<"pour B A"<<endl;
			}else{
				r=r-m;
				cout<<"empty A"<<endl<<"pour B A"<<endl;
			}
		}
		cout<<"success"<<endl;
	}
}

It seems to be the right solution, but ZOJ gave "Time limit Exceeded" error; I searched on the interenet about this problem, someone suggests that it may result from "cout/cin" in c++, so I tried printf in C, but it still generates error. Then I tried Iteration method:

#include <iostream>
using namespace std;
void operation(int m,int n,int mnow,int nnow,int k){
	if(nnow==k)
	return;
    if(nnow>(m-mnow)){
    	nnow=nnow-(m-mnow);
    	cout<<"pour B A"<<endl;
	    if(nnow==k)
	    return;
    	cout<<"empty A"<<endl;
    	mnow=0;
    }else{
    	mnow=nnow;
    	nnow=n;
    	cout<<"pour B A"<<endl<<"fill B"<<endl;
    }
    operation(m,n,mnow,nnow,k);
}
int main(){
	int m,n,k,mnow,nnow;
	while(cin>>m>>n>>k){
		mnow=0;
		nnow=n;
		cout<<"fill B"<<endl;
		operation(m,n,mnow,nnow,k);
		cout<<"success"<<endl;
	}
}

　　Finally, it is accepted!!!!!

转载于:https://www.cnblogs.com/songwanzi/p/3507542.html