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  • UVA146 POJ1146 ID Codes【全排列】

    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 6932   Accepted: 4115

    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

    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

    Output will consist of one line for each code read containing the successor code or the words 'No Successor'.

    Sample Input

    abaacb
    cbbaa
    #

    Sample Output

    ababac
    No Successor

    Source


    问题链接UVA146 POJ1146 ID Codes

    题意简述:参见上文。

    问题分析求输入字符串的下一个排列,不存在的话则输出“No Successor”。

    程序说明:(略)

    题记(略)


    AC的C++语言程序如下:

    /* UVA146 POJ1146 ID Codes */
    
    #include <iostream>
    #include <algorithm>
    #include <stdio.h>
    #include <string.h>
    
    using namespace std;
    
    const int N = 50;
    char s[N+1];
    
    int main()
    {
        while(scanf("%s", s) != EOF) {
            if(s[0] == '#')
                break;
            if(next_permutation(s, s + strlen(s)))
                printf("%s
    ", s);
            else
                printf("No Successor
    ");
        }
    
        return 0;
    }



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  • 原文地址:https://www.cnblogs.com/tigerisland/p/7563623.html
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