Lucky Substrings

而在26以内且属于fibonacci数列的数为1,2,3,5,8,13,21时间限制: 10000ms

单点时限: 1000ms

内存限制: 256MB

描述

A string s is LUCKY if and only if the number of different characters in s is a fibonacci number. Given a string consisting of only lower case letters, output all its lucky non-empty substrings in lexicographical order. Same substrings should be printed once.

输入

A string consisting no more than 100 lower case letters.

输出

Output the lucky substrings in lexicographical order, one per line. Same substrings should be printed once.

样例输入

aabcd

样例输出

a

aa 

aab 

aabc 

ab 

abc 

b 

bc 

bcd 

c 

cd 

d

思路：暴力枚举。26以内斐波那契数打下表就可以了。用set去重排序。复杂度O(n*n*n);

我用前缀和，感觉并没有优化。

    
    
    
    

     
     
     
     
 
        1 #include<stdio.h> 
      
 
        2 #include<algorithm> 
      
 
        3 #include<iostream> 
      
 
        4 #include< 
      string.h> 
      
 
        5 #include<stdlib.h> 
      
 
        6 #include<math.h> 
      
 
        7 #include<cstdio> 
      
 
        8 #include<queue> 
      
 
        9 #include<stack> 
      
 
       10 #include<map> 
      
 
       11 #include< 
      set> 
      
 
       12  
      using  
      namespace std; 
      
 
       13  
      char cou[ 
      200]; 
      
 
       14  
      int fei[ 
      30]; 
      
 
       15  
      int flag[ 
      30]; 
      
 
       16  
      char dd[ 
      200]; 
      
 
       17  
      set< 
      string>my; 
      
 
       18  
      set< 
      string>::const_iterator it; 
      
 
       19 typedef  
      struct pp 
      
 
       20 { 
      
 
       21      
      int al[ 
      26]; 
      
 
       22     pp() 
      
 
       23     { 
      
 
       24         memset(al, 
      0, 
      sizeof(al)); 
      
 
       25     } 
      
 
       26 } ss; 
      
 
       27  
      
 
       28  
      int main( 
      void) 
      
 
       29 { 
      
 
       30      
      int n,i,j,k,p,q; 
      
 
       31     fei[ 
      1]= 
      1; 
      
 
       32     fei[ 
      2]= 
      1; 
      
 
       33      
      for(i= 
      3; i< 
      30; i++) 
      
 
       34     { 
      
 
       35         fei[i]=fei[i- 
      1]+fei[i- 
      2]; 
      
 
       36          
      if(fei[i]>= 
      26) 
      
 
       37         { 
      
 
       38              
      break; 
      
 
       39         } 
      
 
       40     } 
      
 
       41      
      int zz=i; 
      
 
       42      
      for(i= 
      1; i<zz; i++) 
      
 
       43         flag[fei[i]]= 
      1; 
      
 
       44      
      while(scanf( 
      " 
      %s 
      ",cou)!=EOF) 
      
 
       45     { 
      
 
       46         my.clear(); 
      
 
       47         ss ak[ 
      200]; 
      
 
       48          
      int l=strlen(cou); 
      
 
       49          
      int cnt= 
      0; 
      
 
       50         ak[cnt].al[cou[ 
      0]- 
      ' 
      a 
      ']++; 
      
 
       51          
      for(i= 
      1; i<l; i++) 
      
 
       52         { 
      
 
       53             ak[i].al[cou[i]- 
      ' 
      a 
      ']++; 
      
 
       54              
      for(j= 
      0; j< 
      26; j++) 
      
 
       55             { 
      
 
       56                 ak[i].al[j]+=ak[i- 
      1].al[j]; 
      
 
       57             } 
      
 
       58         } 
      
 
       59          
      int yy[ 
      26]; 
      
 
       60          
      for(i= 
      0; i<l; i++) 
      
 
       61         { 
      
 
       62              
      for( 
      int s= 
      0; s< 
      26; s++) 
      
 
       63             { 
      
 
       64                 yy[s]=ak[i].al[s]; 
      
 
       65             } 
      
 
       66              
      int ans= 
      0; 
      
 
       67              
      for( 
      int s= 
      0; s< 
      26; s++) 
      
 
       68             { 
      
 
       69                  
      if(yy[s]) 
      
 
       70                 { 
      
 
       71                     ans++; 
      
 
       72                 } 
      
 
       73  
      
 
       74             } 
      
 
       75              
      if(flag[ans]) 
      
 
       76             { 
      
 
       77                 memset(dd, 
      0, 
      sizeof(dd)); 
      
 
       78                  
      int z= 
      0; 
      
 
       79                  
      int s; 
      
 
       80                  
      for( s= 
      0; s<=i; s++) 
      
 
       81                 { 
      
 
       82                     dd[z++]=cou[s]; 
      
 
       83                 } 
      
 
       84                 my.insert(dd); 
      
 
       85  
      
 
       86             } 
      
 
       87         } 
      
 
       88          
      for(i= 
      0; i<l; i++) 
      
 
       89         { 
      
 
       90              
      for(j=i+ 
      1; j<l; j++) 
      
 
       91             { 
      
 
       92                  
      for( 
      int s= 
      0; s< 
      26; s++) 
      
 
       93                 { 
      
 
       94                     yy[s]=ak[j].al[s]-ak[i].al[s]; 
      
 
       95                 } 
      
 
       96                  
      int ans= 
      0; 
      
 
       97                  
      for( 
      int s= 
      0; s< 
      26; s++) 
      
 
       98                 { 
      
 
       99                      
      if(yy[s]) 
      
 
      100                     { 
      
 
      101                         ans++; 
      
 
      102                     } 
      
 
      103  
      
 
      104                 } 
      
 
      105                  
      if(flag[ans]) 
      
 
      106                 { 
      
 
      107                     memset(dd, 
      0, 
      sizeof(dd)); 
      
 
      108                      
      int z= 
      0; 
      
 
      109                      
      int s; 
      
 
      110                      
      for( s=i+ 
      1; s<=j; s++) 
      
 
      111                     { 
      
 
      112                         dd[z++]=cou[s]; 
      
 
      113                     } 
      
 
      114                     my.insert(dd); 
      
 
      115                 } 
      
 
      116             } 
      
 
      117         } 
      
 
      118          
      for(it=my.begin(); it!=my.end(); it++) 
      
 
      119         { 
      
 
      120             cout<<*it<<endl; 
      
 
      121         } 
      
 
      122     } 
      
 
      123      
      return  
      0; 
      
 
      124 }