HDURevenge of Fibonacci --- 高精度 + 斐波那契数列 + 字典树

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 204800/204800 K (Java/Others)

Total Submission(s): 2110 Accepted Submission(s): 504



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



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题意;



题目给出斐波那契数列的前k位，k不超过40，找出最小的正整数n，满足F(n)的前k位与给定数的前k位相同，斐波那契数列的项数不超过100000。
超过则输出 -1

解题分析：


本题可以分为两步：

第一步就是预处理出100000项斐波那契数列的前40位，插入到字典树中。

第二步就是查询匹配求最小的n。

对于第一步，我们可以把斐波那契数列精确到50多位，然后只存40位即可，这样就防止进位的误差。在斐波那契数列加法过程中，我们只把它的前50多

位进行相加，不然存不下。


代码：
#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

const int N=10;

int f1[65],f2[65],f3[65];

class Trie

{

public:

int v;

int flag;

Trie *next
;

Trie()

{

v=-1;

memset(next,NULL,sizeof(next));

}

};

Trie *root;

void Insert(char *S,int ans)

{

int len=strlen(S);

Trie *p=root;

for(int i=0;i<len;i++)

{

int id=S[i]-'0';

if(p->next[id]==NULL)

p->next[id]=new Trie();

p=p->next[id];

if(p->v<0) p->v=ans;

}

}

int Find(char *S)

{

Trie *p=root;

int count;

int len=strlen(S);

for(int i=0;i<len;i++)

{

int id=S[i]-'0';

p=p->next[id];

if(p==NULL) return -1;

else count=p->v;

}

return count;

}

void Init()

{

int h;

char str[65]="1";

memset(f1,0,sizeof(f1));

memset(f2,0,sizeof(f2));

memset(f3,0,sizeof(f3));

f1[0]=1;f2[0]=1;

Insert(str,0);

for(int i=2;i<100000;i++)

{

memset(str,0,sizeof(str));

int r=0;

for(int j=0;j<60;j++)

{

f3[j]=f1[j]+f2[j]+r;

r=f3[j]/10;

f3[j]%=10;

}

for(int j=59;j>=0;j--)

if(f3[j])

{

h=j;

break;

}

int k=0;

for(int j=h;j>=0;j--)

{

str[k++]=f3[j]+'0';

if(k>=40) break;

}

Insert(str,i);

if(h>55)

{

for(int j=1;j<59;j++)

f3[j-1]=f3[j];

for(int j=1;j<59;j++)

f2[j-1]=f2[j];

}

for(int j=0;j<60;j++)

f1[j]=f2[j];

for(int j=0;j<60;j++)

f2[j]=f3[j];

}

}

int main()

{

root=new Trie();

Init();

char str[105];

int t,i,j,k=1;

scanf("%d",&t);

while(t--)

{

scanf("%s",str);

printf("Case #%d: ",k++);

int tmp=Find(str);

printf("%d\n",tmp);

}

return 0;

}