HDURevenge of Fibonacci --- 高精度 + 斐波那契数列 + 字典树
Total Submission(s): 2110 Accepted Submission(s): 504
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.
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.
15 1 12 123 1234 12345 9 98 987 9876 98765 89 32 51075176167176176176 347746739 5610
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
lcy | We have carefully selected several similar problems for you: 4091 4097 4096 4093 4092
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[N];
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;
}