UVA 146 - 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

 

abaacb
cbbaa
#

 

Sample output

 

ababac
No Successor

 

从后向前，找到不再递增的元素x。再从后向前，找到第一个比x大的元素y。

然后把x与y交换。把x+1到末尾的元素按从小到大的顺序排序。

 

 

#define RUN
#ifdef RUN

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <list>
#include <cctype> 
#include <algorithm>
#include <utility>
#include <math.h>
#include <ctime>

using namespace std;

#define MAXN 110

char buf[MAXN];


int cmp( const void *a , const void *b )  
{
	return *(char *)a - *(char *)b;  
} 


void play(){

	int i, j;
	for(i=strlen(buf)-1; i>0; i--){
		//buf[i-1]是要被调到后面的元素
		if(buf[i-1] < buf[i]){
			//printf("%d\n", i);
			break;
		}
	}

	for(j=strlen(buf)-1; j>0; j--){
		if(buf[j] > buf[i-1]){
			break;
		}
	}

	if(i==0 || j==0){
		printf("No Successor\n");
	}
	else{
		char c = buf[i-1];
		buf[i-1] = buf[j];
		buf[j] = c;

		//从i到末尾做从小到大的排序
		//cout << "buf[i]:" << buf[i] << endl;
		//cout << "strlen(buf)-i:" << strlen(buf)-i << endl;
		qsort(&buf[i], strlen(buf)-i, sizeof(buf[0]), cmp);
		printf("%s\n", buf);

	}

}


int main(){

#ifndef ONLINE_JUDGE
	freopen("146.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif

	memset(buf, 0, sizeof(buf));

	while(scanf("%s", buf) != EOF){
		if(buf[0] == '#'){
			break;
		}

		play();

		//printf("%s\n", buf);

		memset(buf, 0, sizeof(buf));
	}

}


#endif