UVA - 146
ID Codes Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu
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Description
It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.)
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.
Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #.
Output will consist of one line for each code read containing the successor code or the words `No Successor'.
abaacb cbbaa #
ababac No Successor
Source
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 2. Data Structures and Libraries :: Data Structures With Built-in Libraries :: STL algorithmSubmit Status
思路：STL中next_permutation的使用，生成當前序列按照字典序排列的的下一個序列，對於字符串也可以，對於可重集也行，要注意：就這個題目而言，如果當前序列是最後一個序列的話，就輸出No Successor，所以用next_permutation之前還要判斷一下是否是最後一個序列。
AC代碼：
#include#include #include #include using namespace std; char str[55]; int fun(char a[]) { int len = strlen(a); for(int i = 0; i < len - 1; i++) { if(a[i] < a[i + 1]) return 1; } return 0; } int main() { while(scanf("%s", str), str[0] != '#') { int len = strlen(str); if(fun(str)) { next_permutation(str, str + len); printf("%s\n", str); } else printf("No Successor\n"); } return 0; }
更新：原來next_permutation本身可以判斷是否到了最後一個按照字典序排列的序列，所以直接用next_permutation即可
AC代碼：
#include#include #include #include using namespace std; char str[55]; int main() { while(scanf("%s", str), str[0] != '#') { int len = strlen(str); if(next_permutation(str, str + len)) printf("%s\n", str); else printf("No Successor\n"); } return 0; }
