2013
11-12

# Surprising Strings

The D-pairs of a string of letters are the ordered pairs of letters that are distance D from each other. A string is D-unique if all of its D-pairs are different. A string is surprising if it is D-unique for every possible distance D.

Consider the string ZGBG. Its 0-pairs are ZG, GB, and BG. Since these three pairs are all different, ZGBG is 0-unique. Similarly, the 1-pairs of ZGBG are ZB and GG, and since these two pairs are different, ZGBG is 1-unique. Finally, the only 2-pair of ZGBG is ZG, so ZGBG is 2-unique. Thus ZGBG is surprising. (Note that the fact that ZG is both a 0-pair and a 2-pair of ZGBG is irrelevant, because 0 and 2 are different distances.)

Acknowledgement: This problem is inspired by the "Puzzling Adventures" column in the December 2003 issue of Scientific American.

The input consists of one or more nonempty strings of at most 79 uppercase letters, each string on a line by itself, followed by a line containing only an asterisk that signals the end of the input.

For each string of letters, output whether or not it is surprising using the exact output format shown below.

ZGBG
X
EE
AAB
AABA
AABB
BCBABCC
*

ZGBG is surprising.
X is surprising.
EE is surprising.
AAB is surprising.
AABA is surprising.
AABB is NOT surprising.
BCBABCC is NOT surprising.

//* @author popop0p0popo
import java.util.*;
import java.io.*;

public class Main{
public static void main(String[] args){
boolean flag;
int index;
String s;
String[] pair;
while (true){
s=scanner.next();
if (s.equals("*")){
break;
}
if (s.length()==1){
System.out.println(s+" is surprising.");
continue;
}
index=0;
flag=true;
while (index< s.length()-1){
pair=new String[s.length()-index-1];
for (int i=0;i< pair.length ;i++ ){
pair[i]=""+s.charAt(i)+s.charAt(i+index+1);
}
if (!check(pair)){
flag=false;
break;
}
index++;
}
if (flag){
System.out.println(s+" is surprising.");
}
else{
System.out.println(s+" is NOT surprising.");
}
}
}

public static boolean check(String[] s){
for (int i=0;i< s.length-1 ;i++ ){
for (int j=i+1;j< s.length ;j++ ){
if (s[i].equals(s[j])){
return false;
}
}
}
return true;
}
}

1. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？

2. 学算法中的数据结构学到一定程度会乐此不疲的，比如其中的2－3树，类似的红黑树，我甚至可以自己写个逻辑文件系统结构来。