poj1146全排列问题

ID Codes

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

abaacbcbbaa#

Sample Output

ababacNo Successor

题目的意思是说字符串全排列后，这个字符串的下一个状态是什么。
步骤：
1、找到从字符串后面开始，第一个发现顺序的那个字符，记录位置为addr=3。如abaacb中的a；
2、从c这个字符位置开始，后面与它最接近的那个字符，记录位置为addr2=5。如b；
3、将这两字符调换位置，得到ababca；
4、对addr位置后的字符串进行排序，就得到ababac。

源代码：

#include <iostream>#include <string>#include <cmath>#include <algorithm>using namespace std;int main(){int i;string str;char arr[1000];while(cin>>str){string str2 = str;if(str == "#") break;else{for(i = 0; i < str.length(); i++){arr[i] = str[i];}arr[i] = '\0';int addr=-1;for(i = str.length()-1; i >= 1; i--){if(str[i-1] < str[i]) {addr = i-1;break;}}if(addr != -1)    //防止addr=0{int addr2=0;int sum = 999999;for(i = addr+1; i < str.length(); i++){int temp = abs(arr[i]-arr[addr]);    //找最接近的那个字符if(temp < sum && arr[i] != arr[addr])  //与它接近的那个字符不能与本身一样{sum = temp;addr2 = i;}}char ch = str[addr];    //调换那个字符的位置char ch1 = str[addr2];str[addr] = str[addr2];str[addr2] = ch;for(i = 0; i < str.length(); i++){arr[i] = str[i];}arr[i] = '\0';sort(arr+addr+1,arr+str.length());   //对addr后面的字符进行排序}string str1;for(i = 0; i < str.length(); i++){str1 += arr[i];}if(str1 == str2 || addr == -1)    //如果最后的结果一样或没有比较过，则已经是最后一个状态了printf("No Successor\n");else{for(i = 0; i < str.length(); i++){printf("%c",arr[i]);}printf("\n");}}}return 0;} 

之前大一的时候写的，现在从网易博客移过来~