An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set.

For example, suppose it is decided that a code will contain exactly 3 occurrences of `a’, 2 of `b’ and 1 of `c’, then three of the allowable 60 codes under these conditions are:

      abaabc

      abaacb

      ababac

These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order.

Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message `No Successor’ if the given code is the last in the sequence for that set of characters.

Question link:http://poj.org/problem?id=1146

Title: Generate the next arrangement according to the preface of the dictionary order.

Question analysis: The next arrangement step is generated according to the preface of the dictionary order:
1) Find the last positive order: t1 = max {(j | sj-1> = sj)}
2) T1 = -1 indicates that the current is the largest, exit the output No Successor
3) Otherwise, find the last greater than ST1-1 ST2: T2 = {k | ST1-1 <sk}
4) Exchange ST1-1 and ST2
5) The number behind the anti -row ST2

#include <cStdio> 
 #include <cstring> 
 #include <algorithm> 
 using namespace std; 
 char s [55]; 
  
 BOOL GET () 
 {{ 
     int l = strlen (s), T1 = -1; 
     for (int i = l-1; i> 0; i-- 
         if (s [i -1] <s [i]) 
         {{ 
             t1 = i; // Find the last liter T1 
             Break; 
         } 
     if (t1 == -1) 
         Return false; 
     int T2 = T1; 
     for (int i = t1+1; i <l; i ++) 
     {{ 
         if (s [t1 -1]> = s [i]) 
             constinue; 
         T2 = i; // The last number greater than ST1-1 after finding T1 
     } 
     swap (s [t1 -1], s [t2]); // exchange ST1-1 and ST2 
     sort (s + t1, s + l); // The number after ST2 after the anti -row exchange 
     Return true; 
 } 
  
 int Main () 
 {{ 
     While (scanf ("%s", s)! = EOF && S [0]! = '#') 
         Printf ("%s \ n", get ()? s: "no successor"); 
 }