Brute Force & STL --- UVA 146 ID Codes

  ID Codes 

 

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

#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

#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;}