Problem B : ID Codes

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 and Output

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'.

Sample input

abaacbcbbaa#

Sample output

ababacNo Successor

题意为：按字典序输出当前排列的下一个排列，如果没有就输出“No Successor”。方法不唯一，这里介绍两种1.构造法，考虑存在的情况，因为字典序比较是从前向后比较，要使俩个排列最接近，那么只需让字典序大的前移，然后后面的字典序按从小到大排列。注意，这里仅仅将最后一个字符前移是不够的，要想最接近还需要后面的字典序递增。。

#include<iostream>#include<string>#include<vector>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char ch[60];bool cmp(char a,char b){    return b<a;}int main(){    while(scanf("%s",&ch)!=EOF){        if(ch[0]=='#') break;        char c[60],cha;        strcpy(c,ch);        sort(ch,ch+strlen(ch),cmp);        if(strcmp(c,ch)==0) printf("No Successor\n");        else {            int flag=0;            for(int i=strlen(c)-1;i>=0;i--){                for(int j=i-1;j>=0;j--){                    if(c[j]<c[i]){                       cha=c[j];                       c[j]=c[i];                       c[i]=cha;                       sort(c+j+1,c+strlen(c));//后面的元素必须按增序排列，保证最接近                       flag=1;                       break;                    }                }            if(flag) break;            }            printf("%s\n",c);        }    }    return 0;}

方法2：调用next_permutation 函数，直接按字典序输出下一个排列。

#include<iostream>#include<algorithm>#include<string>using namespace std;string str;int main(){    while(cin>>str){        if(str[0]=='#') break;        if(next_permutation(str.begin(),str.end())) cout<<str<<endl;        else cout<<"No Successor"<<endl;    }    return 0;}