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;
}
}

 

HDU 1104 Remainder (BFS(广度优先搜索))的更多相关文章

  1. HDU 1104 Remainder(BFS 同余定理)

    题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1104 在做这道题目一定要对同余定理有足够的了解,所以对这道题目对同余定理进行总结 首先要明白计算机里的 ...

  2. hdu - 1104 Remainder (bfs + 数论)

    http://acm.hdu.edu.cn/showproblem.php?pid=1104 注意这里定义的取模运算和计算机的%是不一样的,这里的取模只会得到非负数. 而%可以得到正数和负数. 所以需 ...

  3. BFS广度优先搜索 poj1915

    Knight Moves Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 25909 Accepted: 12244 Descri ...

  4. 0算法基础学算法 搜索篇第二讲 BFS广度优先搜索的思想

    dfs前置知识: 递归链接:0基础算法基础学算法 第六弹 递归 - 球君 - 博客园 (cnblogs.com) dfs深度优先搜索:0基础学算法 搜索篇第一讲 深度优先搜索 - 球君 - 博客园 ( ...

  5. 图的遍历BFS广度优先搜索

    图的遍历BFS广度优先搜索 1. 简介 BFS(Breadth First Search,广度优先搜索,又名宽度优先搜索),与深度优先算法在一个结点"死磕到底"的思维不同,广度优先 ...

  6. 算法竞赛——BFS广度优先搜索

    BFS 广度优先搜索:一层一层的搜索(类似于树的层次遍历) BFS基本框架 基本步骤: 初始状态(起点)加到队列里 while(队列不为空) 队头弹出 扩展队头元素(邻接节点入队) 最后队为空,结束 ...

  7. 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 Sub ...

  8. 步步为营(十六)搜索(二)BFS 广度优先搜索

    上一篇讲了DFS,那么与之相应的就是BFS.也就是 宽度优先遍历,又称广度优先搜索算法. 首先,让我们回顾一下什么是"深度": 更学术点的说法,能够看做"单位距离下,离起 ...

  9. GraphMatrix::BFS广度优先搜索

    查找某一结点的邻居: virtual int firstNbr(int i) { return nextNbr(i, n); } //首个邻接顶点 virtual int nextNbr(int i, ...

随机推荐

  1. windows系统中,创建临时环境变量

    以servlet-api.jar 包为例 set classpath=%classpath%;C:\apache-tomcat-6.0.37\lib\servlet-api.jar

  2. 删除元素(LintCode)

    删除元素 给定一个数组和一个值,在原地删除与值相同的数字,返回新数组的长度. 元素的顺序可以改变,并且对新的数组不会有影响. 样例 给出一个数组 [0,4,4,0,0,2,4,4],和值 4 返回 4 ...

  3. Flask实战第48天:首页轮播图实现

    首页的布局如下 因为以后所有的内容都是在main-container里面,所以这里我们修改front_base.html,把{% block body%}{% endblock%}放到里面去 < ...

  4. python的reduce()函数(转)

    reduce()函数也是Python内置的一个高阶函数. reduce()函数接收的参数和 map()类似,一个函数 f,一个list,但行为和 map()不同,reduce()传入的函数 f 必须接 ...

  5. [ARC103F]Distance Sums

    题意:有一棵树,对于每个点$i$,给出了它到其他点的距离和$i$,现在要还原这棵树,保证$d_i$两两不同 一个点从$u$移到相邻节点$v$时,若删掉$(u,v)$后$u$这边的连通块大小为$siz_ ...

  6. Android Studio自动化快速实现Parcelable接口序列化

    1.在线安装 然后打开File -> Settings -> Pugins -> Browse Repositories 如下,输入android parcelable code g ...

  7. [bzoj1008](HNOI2008)越狱(矩阵快速幂加速递推)

    Description 监狱有连续编号为1...N的N个房间,每个房间关押一个犯人,有M种宗教,每个犯人可能信仰其中一种.如果相邻房间的犯人的宗教相同,就可能发生越狱,求有多少种状态可能发生越狱 In ...

  8. #iOS问题记录#UITableView加载后直接滑动倒最底部

    类似QQ的聊天框,当进入聊天框,直接滑动倒最底部: 需要先将以他变了view滚动倒底部,再来移动NSIndexPath, 代码如下: -(void) doForceScrollToBottom { d ...

  9. Linux下Apache、PHP、MySQL默认安装路径

    Apache: 如果采用RPM包安装,安装路径应在 /etc/httpd 目录下 Apache配置文件:/etc/httpd/conf/httpd.conf Apache模块路径:/usr/sbin/ ...

  10. minGW cygwin gnuwin32

    首先,三个的官方网站分别是: minGW:http://www.mingw.org cygwin:  http://www.cygwin.com gnuwin32:  https://sourcefo ...