zoukankan      html  css  js  c++  java
  • HDU 4099 大数+Trie

    Revenge of Fibonacci

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 204800/204800 K (Java/Others)
    Total Submission(s): 3218    Accepted Submission(s): 821


    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
    debug三天竟是数据范围看错的惨案,将<=10w改成<10w后AC   = =
    还有一点关于大数模拟,一开始string一直错我以为是这个问题就改成手打的模拟大数加法,
    关于这个char数组并不会自动初始化   假如  char s[35];  for(int i=0;i<10;++i) s[i]=i+'0'; 就会出现乱码导致我一直RE
    真是醉了
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    #define ql(a) memset(a,0,sizeof(a))
    #define LL long long
    const int UP=50;
    const int N=100000-1;
    struct node
    {
        int val;
        node *child[10];
        node(){val=-1;for(int i=0;i<10;++i) child[i]=NULL;}
    }*root;
    void ins(char *s,int num)
    {
     node *p=root;
     int minn=min(40,(int)strlen(s));
     for(int i=0;i<minn;++i){
        int t=s[i]-'0';
        if(p->child[t]==NULL){
            p->child[t]=new node();
        }
        p=p->child[t];
        if(p->val<0) p->val=num;
     }
    }
    void init()
    {
        int f1[65],f2[65],f3[65],r=0;
        ql(f1),ql(f2),ql(f3);
        ins("1",0);
        f1[0]=f1[1]=f2[0]=f2[1]=1;
        for(int i=2;i<=N;++i){ql(f3);r=0;
        int ml=max(f1[0],f2[0]);
            for(int j=1;j<=ml;j++){
                f3[j]=f1[j]+f2[j]+r;
                r=f3[j]/10;
                f3[j]%=10;
                if(j==ml&&r) ml++;
            }f3[0]=ml;
            char s[65]; ql(s);int l=0;
            for(int j=f3[0];j>=1;j--) s[l++]=f3[j]+'0';
            for(int j=41;j<=f3[0];j++) s[j-1]='';
            ins(s,i);
            ql(f1); for(int j=0;j<=f2[0];j++) f1[j]=f2[j];
            ql(f2); for(int j=0;j<=f3[0];j++) f2[j]=f3[j];
            if(ml>55){
            for(int j=1;j<f1[0];j++) f1[j]=f1[j+1]; f1[f1[0]--]=0;
            for(int j=1;j<f2[0];j++) f2[j]=f2[j+1]; f2[f2[0]--]=0;
            }
        }
    }
    int Find(char *s)
    {
        int len=strlen(s);
    if(!strcmp(s,"1")) {return 0;}
    node *p=root;
    for(int i=0;i<len;++i){
        int t=s[i]-'0';
        if(p->child[t]==NULL) return -1;
        p=p->child[t];
    }
    return p->val;
    }
    int main()
    {
        int k,cas=0;
        char p[55];
        root=new node();
        init();
        cin>>k;
        while(k--){
            scanf("%s",p);
            printf("Case #%d: %d
    ",++cas,Find(p));
        }
        return 0;
    }
  • 相关阅读:
    CentOS7安装mysql
    centos 7 firewall(防火墙)开放端口/删除端口/查看端口
    CentOS7 FTP安装与配置
    处理nuget包太占用C盘
    windows下使用nginx
    SQL Server 设置新用户只能查看并访问特定数据库
    RPC框架
    RPC与REST
    Windows 环境下 Docker 使用及配置
    “远程调试监视器(MSVSMON.EXE)似乎没有在远程计算机上运行“的 解决方法
  • 原文地址:https://www.cnblogs.com/zzqc/p/7231697.html
Copyright © 2011-2022 走看看