而在26以內且屬于fibonacci數列的數為1,2,3,5,8,13,21時間限制:10000ms單點時限:1000ms內存限制: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.

輸入

A string consisting no more than 100 lower case letters.

輸出

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

樣例輸入aabcd樣例輸出aaa?aab?aabc?ab?abc?b?bc?bcd?c?cd?d思路：暴力枚舉。26以內斐波那契數打下表就可以了。用set去重排序。復雜度O(n*n*n);我用前綴和，感覺并沒有優化。??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?}

轉載于:https://www.cnblogs.com/zzuli2sjy/p/5186440.html

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