Revenge of Fibonacci
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 204800/204800 K (Java/Others)
Total Submission(s): 2355 Accepted Submission(s): 587
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
题意：给你一个字符串，求最小的斐波那契数的下标，使这个字符串是这个斐波那契数的前缀。
题解：将0~99999斐波那契数存入字典树，只存前40位，在查找。
（注：加法只需保存前50位就可以保证精度了;第100000个斐波那契不能压入字典树，会wa;字典树只能压入40位数，不然会超内存）
#include<cstring>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<algorithm>
#define N 100005
using namespace std;
int n;
char s[50],a[60],b[60],c[60];
struct Trie {
int id;
struct Trie *Next[10];
Trie() {
id=-1;
for(int i=0; i<10; i++) {
Next[i]=NULL;
}
}
};
Trie *p;
void Trie_Inser(Trie *p,char s[],int id) {
int i=0;
Trie *q=p;
while(s[i]&&i<41) {
int nx=s[i]-48;
if(q->Next[nx]==NULL) {
q->Next[nx]=new Trie;
}
q=q->Next[nx];
if(q->id==-1) {
q->id=id;
}
i++;
}
}
void Add() {
int la=strlen(a);
int lb=strlen(b);
int res[100];
int l=0;
la--,lb--;
int t=0;
while(la>=0) {
res[l]=(a[la]-48)+(b[lb]-48)+t;
t=0;
if(res[l]>=10) {
res[l]-=10;
t=1;
}
la--;
lb--;
l++;
}
while(lb>=0) {
res[l]=b[lb]-48+t;
t=0;
if(res[l]>=10) {
res[l]-=10;
t=1;
}
lb--,l++;
}
if(t) {
res[l++]=1;
}
int i=0;
int k=0;
if(l>50) {
k=l-50;
}
l--;
while(l>=0&&i<50) {
c[i++]=res[l--]+48;
}
c[i]='\0';
lb=strlen(b);
int lc=strlen(c);
for(int j=0; j<lb-k; j++) {
a[j]=b[j];
}
a[lb-k]='\0';
for(int j=0; j<lc; j++) {
b[j]=c[j];
}
b[lc]='\0';
}
void Init() {
p=new Trie;
a[0]='1',a[1]='\0';
Trie_Inser(p,a,0);
b[0]='1',b[1]='\0';
for(int i=2; i<100000; i++) {
Add();
Trie_Inser(p,c,i);
}
}
int Trie_Serch(Trie *p,char s[]) {
int i=0;
Trie *q=p;
while(s[i]) {
int nx=s[i]-48;
if(q->Next[nx]==NULL)return -1;
q=q->Next[nx];
i++;
}
return q->id;
}
int main() {
//freopen("test.in","r",stdin);
Init();
int t;
cin>>t;
int ca=1;
while(t--) {
scanf("%s",s);
printf("Case #%d: %d\n",ca++,Trie_Serch(p,s));
}
return 0;
}
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