Problem Description

The well-known Fibonacci sequence is defined as following:

Here we regard n as the index of the Fibonacci number F(n).
  This
sequence has been studied since the publication of Fibonacci's book
Liber Abaci. So far, many properties of this sequence have been
introduced.
  You had been interested in this sequence, while after
reading lots of papers about it. You think there’s no need to research
in it anymore because of the lack of its unrevealed properties.
Yesterday, you decided to study some other sequences like Lucas sequence
instead.
  Fibonacci came into your dream last night. “Stupid human
beings. Lots of important properties of Fibonacci sequence have not been
studied by anyone, for example, from the Fibonacci number 347746739…”
  You
woke up and couldn’t remember the whole number except the first few
digits Fibonacci told you. You decided to write a program to find this
number out in order to continue your research on Fibonacci sequence.

 

Input

  There
are multiple test cases. The first line of input contains a single
integer T denoting the number of test cases (T<=50000).
  For each
test case, there is a single line containing one non-empty string made
up of at most 40 digits. And there won’t be any unnecessary leading
zeroes.

 

Output

  For
each test case, output the smallest index of the smallest Fibonacci
number whose decimal notation begins with the given digits. If no
Fibonacci number with index smaller than 100000 satisfy that condition,
output -1 instead – you think what Fibonacci wants to told you beyonds
your ability.

 

Sample Input

 

Sample Output

 

Source

2011 Asia Shanghai Regional Contest

 #include<cstdio>

 #include<cstring>

 #include<iostream>

 #include<algorithm>

 #define FOR(a,b,c) for(int a=(b);a<=(c);a++)

 using namespace std;

 const int sigmasize = ;

 //Trie相关

 struct Node{

     int val;

     Node* next[sigmasize];

 };

 struct Trie{

     Node *root;

     Trie() {

         root=new Node;

         for(int i=;i<sigmasize;i++) root->next[i]=NULL;

         root->val=;

     }

     int idx(char c) { return c-''; }

     void insert(char* s,int v) {

         int n=strlen(s);

         Node* u=root;

         for(int i=;i<min(n,);i++) {

             int c=idx(s[i]);

             if(u->next[c]==NULL) {

                 Node* tmp=new Node;

                 tmp->val=;

                 for(int i=;i<sigmasize;i++) tmp->next[i]=NULL;

                 u->next[c]=tmp;

             }

             u=u->next[c];

             if(u->val==) u->val=v;

         }

         if(u->val==) u->val=v;   //存储最小的F

     }

     int find(char* s) {

         int n=strlen(s);

         Node* u=root;

         for(int i=;i<min(n,);i++) {

             int c=idx(s[i]);

             if(u->next[c]==NULL) return ;

             else u=u->next[c];

         }

         return u->val;

     }

     void del(Node *root) {

         for(int i=;i<sigmasize;i++) {

             if(root->next[i]!=NULL) del(root->next[i]);

         }

         free(root);

     }

 }trie;

 //题目相关

 int n;

 const int maxn = ;

 char F1[maxn],F2[maxn],Ftmp[maxn],s[maxn];

 char d[maxn];

 void Add(char *a,char *b,char *c)

 {

     int i,j,k,aa,bb,t=,p=;

     aa=strlen(a)-,bb=strlen(b)-;

     while(aa>=||bb>=) {

         if(aa<)i=;   else i=a[aa]-'';

         if(bb<)j=;   else j=b[bb]-'';

         k=i+j+t;

         d[p++]=k%+'';

         t=k/;

         aa--,bb--;

     }

     while(t) {

         d[p++]=t%+'';

         t=t/;

     }

     for(i=;i<p;i++) c[i]=d[p-i-];

     c[p]='\0';

 }

 int main() {

     F1[]='',F1[]='\0';

     F2[]='',F2[]='\0';

     trie.insert(F1,),trie.insert(F2,);

     FOR(i,,-) {

         strcpy(Ftmp,F2);

         int len1=strlen(F1),len2=strlen(F2);

         if(len1>) {

             F1[]=F2[]='\0';

         }

         Add(F1,F2,F2);

         trie.insert(F2,i+);

         strcpy(F1,Ftmp);

     }

     int T,kase=;

     scanf("%d",&T);

     while(T--) {

         scanf("%s",s);

         int ans=trie.find(s);

         if(!ans) ans=-;  else ans--;

         printf("Case #%d: %d\n",++kase,ans);

     }

     trie.del(trie.root);

     return ;

 }