HDU 1104 Remainder (BFS（广度优先搜索）)-优快云博客

解决一个数学挑战，通过加、减、乘、取模操作，使初始值N经过最少步骤后，其对K的余数等于初始值加1对K的余数。使用广度优先搜索算法寻找最优解。

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

Sample Output

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;
	}
}

转载于:https://www.cnblogs.com/james1207/p/3260634.html