HDU4099 Revenge of Fibonacci（高精度+Trie）

　　　　　　　　　　　　　　Revenge of Fibonacci

　　　　　　　　　　　　　　　　　　　　　　　　　　Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 204800/204800 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Total Submission(s): 2582    Accepted Submission(s): 647

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

15 1 12 123 1234 12345 9 98 987 9876 98765 89 32 51075176167176176176 347746739 5610

 

Sample Output

Case #1: 0 Case #2: 25 Case #3: 226 Case #4: 1628 Case #5: 49516 Case #6: 15 Case #7: 15 Case #8: 15 Case #9: 43764 Case #10: 49750 Case #11: 10 Case #12: 51 Case #13: -1 Case #14: 1233 Case #15: 22374

 

Source

2011 Asia Shanghai Regional Contest

【思路】

       高精度加法+Trie。

       离线处理出F(99999)以内所有的F在长度40以内的前缀并构造一棵字典树。因为精度原因，保留60位进行加法计算就可以达到精确。

       Trie的结点维护一个val表示经过该节点的所有字符串中的最小标号，对应每一个查询用O(maxn)的时间求解。

　　 总的时间复杂度为O(99999*60+T*maxn)。

【代码】

#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 = 10;
//Trie相关 
struct Node{
    int val;
    Node* next[sigmasize];
};
struct Trie{
    Node *root;
    Trie() {
        root=new Node;
        for(int i=0;i<sigmasize;i++) root->next[i]=NULL;
        root->val=0;
    }
    int idx(char c) { return c-'0'; }
    void insert(char* s,int v) {
        int n=strlen(s);
        Node* u=root;
        for(int i=0;i<min(n,41);i++) {
            int c=idx(s[i]);
            if(u->next[c]==NULL) {
                Node* tmp=new Node;
                tmp->val=0;
                for(int i=0;i<sigmasize;i++) tmp->next[i]=NULL;
                u->next[c]=tmp;
            }
            u=u->next[c];
            if(u->val==0) u->val=v;
        }
        if(u->val==0) u->val=v;   //存储最小的F
    }
    int find(char* s) {
        int n=strlen(s);
        Node* u=root;
        for(int i=0;i<min(n,41);i++) {
            int c=idx(s[i]);
            if(u->next[c]==NULL) return 0;
            else u=u->next[c];
        }
        return u->val;
    }
    void del(Node *root) {
        for(int i=0;i<sigmasize;i++) {
            if(root->next[i]!=NULL) del(root->next[i]);
        }
        free(root);
    }
}trie;
//题目相关 
int n;
const int maxn = 100;
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=0,p=0;  
    aa=strlen(a)-1,bb=strlen(b)-1;
    while(aa>=0||bb>=0) {
        if(aa<0)i=0;   else i=a[aa]-'0';  
        if(bb<0)j=0;   else j=b[bb]-'0';  
        k=i+j+t;
        d[p++]=k%10+'0';
        t=k/10;
        aa--,bb--;
    }
    while(t) {
        d[p++]=t%10+'0';  
        t=t/10;  
    }
    for(i=0;i<p;i++) c[i]=d[p-i-1];
    c[p]='';
}

int main() {
    F1[0]='1',F1[1]='';
    F2[0]='1',F2[1]='';
    trie.insert(F1,1),trie.insert(F2,1);
    FOR(i,2,100000-1) {
        strcpy(Ftmp,F2);
        int len1=strlen(F1),len2=strlen(F2);
        if(len1>60) {
            F1[60]=F2[60]='';
        }
        Add(F1,F2,F2);
        trie.insert(F2,i+1);
        strcpy(F1,Ftmp);
    }
    int T,kase=0;
    scanf("%d",&T);
    while(T--) {
        scanf("%s",s);
        int ans=trie.find(s);
        if(!ans) ans=-1;  else ans--;
        printf("Case #%d: %d
",++kase,ans);
    }
    trie.del(trie.root);
    return 0;
}