ID Codes
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 6229 | Accepted: 3737 |
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:
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.
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
New Zealand 1991 Division I,UVA 146
题目分析：给你一个字符串，长度不超过50. 现在要让你输出该串的下一个字典序排列。当然前提是存在就输出，否则就输出“No Successor”。
参考书籍《算法设计编程实验》(P 120)
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
char s[100];
int len;
bool get()
{
len=strlen(s);
int i=len-1;
while(i>0 && s[i-1]>=s[i] )
i--; //找到最后一个正序
if(i==0)
return false; //表示当前排列为最后一个排列，则返回false
int pos=i;
for(int j=i+1; j<len; j++)
{
if(s[j]<=s[i-1] )
continue;
if(s[j] < s[pos] )
pos=j;
}
swap(s[pos], s[i-1]);
sort(s+i, s+len);
return true;
}
int main()
{
//判断一个序列还有没有下一个全排列序列 如果存在就输出
while(scanf("%s", s)!=EOF )
{
if(s[0]=='#') break;
if( get()==true )
printf("%s
", s);
else
printf("No Successor
");
}
return 0;
}
分析：利用该算法，也可以一直输出下去，从串的当前字典位置开始，直到输出至最后一个字典序。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
char s[100];
int len;
bool get()
{
len=strlen(s);
int i=len-1;
while(i>0 && s[i-1]>=s[i] )
i--; //找到最后一个正序
if(i==0)
return false; //表示当前排列为最后一个排列，则返回false
int pos=i;
for(int j=i+1; j<len; j++)
{
if(s[j]<=s[i-1] )
continue;
if(s[j] < s[pos] )
pos=j;
}
swap(s[pos], s[i-1]);
sort(s+i, s+len);
return true;
}
int main()
{
//判断一个序列还有没有下一个全排列序列 如果存在就输出
while(scanf("%s", s)!=EOF )
{
if(s[0]=='#') break;
if(get()==true){
printf("%s
", s);
while( get()==true )
printf("%s
", s);
}
else
printf("No Successor
");
}
return 0;
}
还可以这样变体：将初始输入的字符串进行排序：然后在利用上面的代码，输出当前串的所有排列
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
char s[100];
int len;
bool get()
{
int i=len-1;
while(i>0 && s[i-1]>=s[i] )
i--; //找到最后一个正序
if(i==0)
return false; //表示当前排列为最后一个排列，则返回false
int pos=i;
for(int j=i+1; j<len; j++)
{
if(s[j]<=s[i-1] )
continue;
if(s[j] < s[pos] )
pos=j;
}
swap(s[pos], s[i-1]);
sort(s+i, s+len);
return true;
}
int main()
{
//判断一个序列还有没有下一个全排列序列 如果存在就输出
while(scanf("%s", s)!=EOF )
{
if(s[0]=='#') break;
len=strlen(s);
sort(s, s+len);
printf("%s
", s);
if(get()==true){
printf("%s
", s);
while( get()==true )
printf("%s
", s);
}
else
printf("No Successor
");
}
return 0;
}