解題代碼：
粗暴而簡單。大一的水平。
package com.kyc.liantong;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
char[] input = in.next().toCharArray();
int n = input.length;
switch (n) {
case 1:
StringBuilder sb = new StringBuilder();
for (int i = 1; i <= 3; i++) {
sb.append(input[0]).append(i).append(" ");
}
System.out.println(sb.substring(0, sb.length() - 1));
break;
case 2:
StringBuilder sb2 = new StringBuilder();
for (int i = 1; i <= 3; i++) {
for (int i1 = 1; i1 <= 3; i1++) {
sb2.append(input[0]).append(i).append(input[1]).append(i1).append(" ");
}
}
System.out.println(sb2.substring(0, sb2.length() - 1));
break;
case 3:
StringBuilder sb3 = new StringBuilder();
for (int i = 1; i <= 3; i++) {
for (int i1 = 1; i1 <= 3; i1++) {
for (int i2 = 1; i2 <= 3; i2++) {
sb3.append(input[0]).append(i)
.append(input[1]).append(i1)
.append(input[2]).append(i2).append(" ");
}
}
}
System.out.println(sb3.substring(0, sb3.length() - 1));
break;
case 4:
StringBuilder sb4 = new StringBuilder();
for (int i = 1; i <= 3; i++) {
for (int i1 = 1; i1 <= 3; i1++) {
for (int i2 = 1; i2 <= 3; i2++) {
for (int i3 = 1; i3 <= 3; i3++) {
sb4.append(input[0]).append(i)
.append(input[1]).append(i1)
.append(input[2]).append(i2)
.append(input[3]).append(i3).append(" ");
}
}
}
}
System.out.println(sb4.substring(0, sb4.length() - 1));
break;
case 5:
StringBuilder sb5 = new StringBuilder();
for (int i = 1; i <= 3; i++) {
for (int i1 = 1; i1 <= 3; i1++) {
for (int i2 = 1; i2 <= 3; i2++) {
for (int i3 = 1; i3 <= 3; i3++) {
for (int i4 = 1; i4 <= 3; i4++) {
sb5.append(input[0]).append(i)
.append(input[1]).append(i1)
.append(input[2]).append(i2)
.append(input[3]).append(i3)
.append(input[4]).append(i4).append(" ");
}
}
}
}
}
System.out.println(sb5.substring(0, sb5.length() - 1));
break;
default:
}
}
}
