poj-1146 ID codes
Description
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
Output
Sample Input
abaacb
cbbaa
#
Sample Output
ababac
No Successor
题目大意:
一组字符全排列,如果有下一个全排列则则输出 若无,则输出NO;
注意:
在STL中有个next_permutation(const *str,int length);函数。若有下一个全排列则返回true并生成下一个全排列,
反之返回false;
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#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int permutation(char str[],int length)
{ int j=0;
// sort(str,str+length); while(next_permutation(str,str+length))
{
for(int i=0;i<length;i++)
cout<<str[i]; cout<<endl;
j++;
break; }
return j;
}
int main ()
{
while(1)
{ char str[100];
cin>>str;
if(str[0]=='#')break;
int len=strlen(str);
int q=permutation(str,len);
if(!q)cout<<"No Successor"<<endl;
} return 0;
}
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