Brute Force & STL --- UVA 146 ID Codes

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 ID Codes 

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Problem's Link:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=3&problem=82&mosmsg=Submission+received+with+ID+14418598#

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 Mean: 

 求出可重排列的下一个排列。

analyse:

 直接用STL来实现就可。自己手动写了一个，并不复杂。

Time complexity: O(n^2)

 

Source code: 

 1.STL

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#include <iostream>  #include <cstdio>  #include <cstring>  #include <algorithm>  using namespace std;  int main()  {      char s[55];      while(scanf("%s",s)!=EOF)      {          if(s[0]=='#') break;          if(next_permutation(s,s+strlen(s))) printf("%s\n",s);          else printf("No Successor\n");          memset(s,0,sizeof(s));      }      return 0;  }

2.手写

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#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int main() {     //freopen("a.txt","r",stdin);     char s[55];     while(scanf("%s",s),s[0]!='#')     {         int i,j,len(strlen(s));         for(i=len-2; i>=0; i--)             if(s[i]<s[i+1])                 break;         if(i<0) puts("No Successor");         else         {             for(j=i+1; i<len; j++)                 if(s[i]>=s[j])                 {                     char c=s[i];                     s[i]=s[j-1];                     s[j-1]=c;                     break;                 }             sort(s+i+1,s+len);             puts(s);         }     }     return 0; }