POJ 1047 Round and Round We Go 最详细的解题报告

题目链接:Round and Round We Go

解题思路：用程序实现一个乘法功能，将给定的字符串依次做旋转，然后进行比较。由于题目比较简单，所以不做过多的详解。

具体算法（java版，可以直接AC）

  1 import java.util.Scanner;
  2 
  3 public class Main {
  4 
  5     public static void main(String[] args) {
  6         Scanner scanner = new Scanner(System.in);
  7         String line;
  8         while (scanner.hasNext()) {
  9             line = scanner.next();
 10             MyNumber number = new MyNumber(line);
 11             int count = 0;
 12             for (int i = 2; i <= line.length(); i++) {
 13                 NumNode[] result = number.multiply(i);
 14                 if (number.isCycle(result)) {
 15                     count++;
 16                 }
 17             }
 18             if (count == line.length()-1) {
 19                 System.out.println(line + " is cyclic");
 20             } else {
 21                 System.out.println(line + " is not cyclic");
 22             }
 23         }
 24     }
 25 
 26 }
 27 
 28 class NumNode {
 29     int value;
 30     int carry;
 31 
 32     public NumNode() {
 33         this.value = 0;
 34         this.carry = 0;
 35     } 40 }
 41 
 42 class MyNumber {
 43     private NumNode[] data;
 44     private int length;
 45     private String[] rotation;
 46     private boolean isRotated;
 47 
 48     public MyNumber(String line) {
 49         this.length = line.length();
 50         this.isRotated=false;
 51         this.data = new NumNode[this.length];
 52         this.rotation = new String[this.length];
 53         for (int i = this.length - 1; i >= 0; i--) {
 54             this.data[i] = new NumNode();
 55             this.data[i].value = line.charAt(i) - '0';
 56         }
 57     }
 58 
 59     private void rotate() {
 60         for (int i = 0; i < this.length; i++) {
 61             StringBuffer buffer = new StringBuffer();
 62             for (int j = i; j < this.length; j++) {
 63                 buffer.append(this.data[j].value);
 64             }
 65             for (int j = 0; j < i; j++) {
 66                 buffer.append(this.data[j].value);
 67             }
 68             this.rotation[i] = buffer.toString();
 69         }
 70     }
 71 
 72     public NumNode[] multiply(int a) {
 73         NumNode[] result = new NumNode[this.length];
 74         for (int i = this.length - 1; i >= 0; i--) {
 75             int value = this.data[i].value * a;
 76             result[i] = new NumNode();
 77             if (i + 1 < this.length) {
 78                 value += result[i + 1].carry;
 79             }
 80             result[i].value = value % 10;
 81             result[i].carry = value / 10;
 82         }
 83         return result;
 84     }
 85 
 86     public boolean equals(String s) {
 87         for (String str : this.rotation) {
 88             if (str.equals(s))
 89                 return true;
 90         }
 91         return false;
 92     }
 93 
 94     public boolean isCycle(NumNode[] num) {
 95         if (num[0].carry > 0)
 96             return false;
 97         if(!this.isRotated){
 98             this.rotate();
 99             this.isRotated=true;
100         }
101         StringBuffer buffer = new StringBuffer();
102         for (int i = 0; i < num.length; i++) {
103             buffer.append(num[i].value);
104         }
105         return this.equals(buffer.toString());
106     }
107 
108     public String toString() {
109         StringBuffer buffer = new StringBuffer();
110         for (int i = 0; i < this.length; i++) {
111             buffer.append(this.data[i].value);
112         }
113         return buffer.toString();
114     }
115 }

 

posted @ 2014-10-19 09:27  PinXiong  阅读(306)  评论(0)  编辑  收藏  举报