斐波纳契数+Trie+hdu4099

最新推荐文章于 2023-11-10 16:02:08 发布

原创最新推荐文章于 2023-11-10 16:02:08 发布 · 555 阅读

0 ·

CC 4.0 BY-SA版权

字典树|Trie树|前缀树专栏收录该内容

29 篇文章

订阅专栏

本博客探讨了一个寻找特定数字开始的最小费波那契数的问题，并提供了使用字母树进行查找的算法实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Online Judge	Online Exercise	Online Teaching	Online Contests	Exercise Author
F.A.Q Hand In Hand Online Acmers Forum \| Discuss Statistical Charts	Problem Archive Realtime Judge Status Authors Ranklist	C/C++/Java Exams ACM Steps Go to Job Contest LiveCast ICPC@China	Best Coder ^beta VIP \| STD Contests Virtual Contests DIY \| Web-DIY ^beta Recent Contests	lee Mail 0 (0) Control Panel Sign Out

[BestCoder Round #5]冠军的奖励是小米3手机一部
 恭喜福州大学杨楠获得[BestCoder Round #4]冠军（iPad Mini一部）
《BestCoder用户手册》下载

Revenge of Fibonacci

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 204800/204800 K (Java/Others)
Total Submission(s): 2014 Accepted Submission(s): 472

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

这题重点是在蛋疼的大数相加。。。

我们只需要保存前40位插入到字母树中就可以

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=100000;
char f[60];
int f1[60],f2[60],tmp[60];
struct node
{
    node *next[10];
    int val;
    node(){memset(next,NULL,sizeof(next));val=-1;}
};
struct TRIE
{
    node *root;
    void clear(){root=new node();}
    int idx(char x){return x-'0';}
    void insert(char *s,int id)
    {
        int n=strlen(s);
        node *p=root;
        for(int i=0;i<n;i++)
        {
            int c=idx(s[i]);
            if(!p->next[c])
                p->next[c]=new node();
            p=p->next[c];
            if(p->val==-1)p->val=id;
            else if(p->val>id)p->val=id;

        }
    }
    int find(char *s)
    {
        int n=strlen(s);
        node *p=root;
        for(int i=0;i<n;i++)
        {
            int c=idx(s[i]);
            if(!p->next[c])return -1;
            p=p->next[c];
        }
        return p->val;
    }
}tree;
void add(int *a,int *b)
{
    for(int i=0;i<60;i++)
        a[i]+=b[i];
    int c=0;
    for(int i=0;i<60;i++)
    {
        a[i]+=c;
        c=0;
        if(a[i]>=10)
        {
            c=a[i]/10;
            a[i]%=10;
        }
    }
    bool flag=true;
    int cnt=0;
    for(int i=59;i>=0;i--)
    {
        if(a[i]==0&&flag)continue;
        flag=false;
        f[cnt++]=a[i]+'0';
        if(cnt>40)break;
    }
    f[cnt]='\0';
}
void change(int *a,int *b)
{
    int pos=0;
    for(int i=59;i>=0;i--)
        if(a[i]!=0){pos=i;break;}
    if(pos>55)
    {
        for(int j=0;j<=59;j++)
            a[j]=a[j+1];
        for(int j=0;j<=59;j++)
            b[j]=b[j+1];
    }
    for(int i=0;i<60;i++)tmp[i]=b[i];
    for(int i=0;i<60;i++)
    {
        b[i]=a[i];
        a[i]=tmp[i];
    }
}
void init()
{
    tree.clear();
    tree.insert("1",0);
    memset(f1,0,sizeof(f1));
    memset(f2,0,sizeof(f2));
    f1[0]=f2[0]=1;
    for(int i=2;i<maxn;i++)
    {
        add(f1,f2);
        change(f1,f2);
        tree.insert(f,i);
    }
}
int main()
{
    int T,cas=1;
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",f);
        printf("Case #%d: %d\n",cas++,tree.find(f));
    }
    return 0;
}