zoukankan      html  css  js  c++  java
  • HDU 1104 Remainder (BFS(广度优先搜索))

    Remainder

    Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2260 Accepted Submission(s): 481


    Problem Description
    Coco is a clever boy, who is good at mathematics. However, he is puzzled by a difficult mathematics problem. The problem is: Given three integers N, K and M, N may adds (‘+’) M, subtract (‘-‘) M, multiples (‘*’) M or modulus (‘%’) M (The definition of ‘%’ is given below), and the result will be restored in N. Continue the process above, can you make a situation that “[(the initial value of N) + 1] % K” is equal to “(the current value of N) % K”? If you can, find the minimum steps and what you should do in each step. Please help poor Coco to solve this problem.

    You should know that if a = b * q + r (q > 0 and 0 <= r < q), then we have a % q = r.
     
    Input
    There are multiple cases. Each case contains three integers N, K and M (-1000 <= N <= 1000, 1 < K <= 1000, 0 < M <= 1000) in a single line.

    The input is terminated with three 0s. This test case is not to be processed.
     
    Output
    For each case, if there is no solution, just print 0. Otherwise, on the first line of the output print the minimum number of steps to make “[(the initial value of N) + 1] % K” is equal to “(the final value of N) % K”. The second line print the operations to do in each step, which consist of ‘+’, ‘-‘, ‘*’ and ‘%’. If there are more than one solution, print the minimum one. (Here we define ‘+’ < ‘-‘ < ‘*’ < ‘%’. And if A = a1a2...ak and B = b1b2...bk are both solutions, we say A < B, if and only if there exists a P such that for i = 1, ..., P-1, ai = bi, and for i = P, ai < bi)
     
    Sample Input
    2 2 2 -1 12 10 0 0 0
     
    Sample Output
    0 2 *+

    BFS(广度优先搜索)

     
    import java.io.*;
    import java.util.*;
    
    public class Main {
    	public String str="+-*%";
    	public int n,m,k,sum,km;
    	public boolean boo[]=new boolean[1000*1000*10+1];
    	public Queue<Node1> list=new LinkedList<Node1>();
    	public static void main(String[] args) {
    		
    		new Main().work();
    	}
    	public void work(){
    		Scanner sc=new Scanner(new BufferedInputStream(System.in));
    		while(sc.hasNext()){
    			list.clear();
    			Arrays.fill(boo,false);
    			n=sc.nextInt();
    			k=sc.nextInt();
    			m=sc.nextInt();
    			if(n==0&&k==0&&m==0)
    				System.exit(0);
    			Node1 node=new Node1();
    			node.n=n;
    			node.s="";
    			sum=getMode(n+1,k);
    			km=m*k;
    			boo[getMode(n,km)]=true;
    			list.add(node);
    			BFS();
    		}
    	}
    	public void BFS(){
    		while(!list.isEmpty()){
    			Node1 node=list.poll();
    			if(getMode(node.n,k)==sum){
    				System.out.println(node.s.length());
    				System.out.println(node.s);
    				return;
    			}
    			for(int i=0;i<str.length();i++){
    				int temp=0;
    				if(str.charAt(i)=='+'){
    					temp=getMode(node.n+m,km);
    				}
    				else if(str.charAt(i)=='-'){
    					temp=getMode(node.n-m,km);
    				}
    				else if(str.charAt(i)=='*'){
    					temp=getMode(node.n*m,km);
    				}
    				else if(str.charAt(i)=='%'){
    					temp=getMode(getMode(node.n,m),km);
    				}
    				if(!boo[temp]){
    					boo[temp]=true;
    					Node1 t=node.getNode();
    					t.n=temp;
    					t.s=t.s+str.charAt(i);
    					list.add(t);
    				}
    			}
    		}
    		System.out.println(0);
    	}
    	public  int getMode(int a,int b){
    		return (a%b+b)%b;
    	}
    }
    class Node1{
    	int n;
    	String s;
    	Node1(){
    		n=0;
    		s="";
    	}
    	public Node1 getNode(){
    		Node1 node=new Node1();
    		node.n=0;
    		node.s=s;
    		return node;
    	}
    }


     
  • 相关阅读:
    【Henu ACM Round#15 F】Arthur and Questions
    【Henu ACM Round#16 F】Om Nom and Necklace
    【Henu ACM Round#16 E】Paths and Trees
    JS制作的简单的三级及联
    .Net实现的批量删除(使用了repeater控件)
    setInterval和setTimeout调用方法小知识科普
    AJAX制作JSON格式的实时更新数据的方法
    关于获取网站域名的写法杂谈
    JS初识(着重讲解Date函数)
    Guid函数
  • 原文地址:https://www.cnblogs.com/james1207/p/3260634.html
Copyright © 2011-2022 走看看