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 }