• UVA


     The well-known Fibonacci sequence is defined as following:

            F(0) = F(1) = 1

            F(n) = F(n − 1) + F(n − 2) ∀n ≥ 2

     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

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 struct Node{
     4     int id;
     5     Node * next[10];
     6         Node(){
     7         id = -1;
     8         for(int i = 0; i < 10; ++i)
     9             next[i] = NULL;
    10     }
    11 };
    12 char Fib[50], In[50];
    13 int F[2][1024000];
    14 Node * const root = new Node();
    15 void add_node(char *str, int id)
    16 {
    17     Node * u = root;
    18     for(int i = 0, len = strlen(str); i < len && i <= 40; ++i){
    19         int v = str[i] - '0';
    20         if(!u->next[v])
    21             u->next[v] = new Node();
    22         u =  u->next[v];
    23         if(u->id == -1)
    24             u->id = id;
    25     }
    26 }
    27 int query(char *str)
    28 {
    29     Node * u = root;
    30     for(size_t i = 0, len = strlen(str); i < len; ++i){
    31         u = u->next[str[i]-'0'];
    32         if(!u) return -1;
    33     }
    34     return u->id;
    35 }
    36 void init()
    37 {
    38     memset(F, 0, sizeof(F));
    39     F[0][0] = F[1][0] = 1;
    40     int s = 0, e = 1;
    41     add_node((char *)"1", 0);
    42     add_node((char *)"1", 1);
    43     for(int i = 2; i < 100000; ++i){
    44         int p = i%2, q = (i+1)%2;
    45         for(int j = s; j < e; ++j)
    46             F[p][j] = F[p][j] + F[q][j];
    47         for(int j = s; j < e; ++j)
    48             if(F[p][j]>=10){
    49                  F[p][j] %= 10;
    50                  F[p][j+1] += 1;
    51             }
    52         if(F[p][e])  ++e;
    53         if(e - s > 50)  ++s;
    54         int r = e - 1, cnt = 0;
    55         memset(Fib, 0, sizeof(Fib));
    56         while(r >= 0 && cnt<45)
    57             Fib[cnt++] = F[p][r--] + '0';
    58         add_node(Fib, i);
    59     }
    60 }
    61 int main()
    62 {
    63     ios::sync_with_stdio(false);
    64     init();
    65     int T; cin >> T;
    66     for(int i = 1; i <= T; ++i){
    67         cin >> In;
    68         printf("Case #%d: %d
    ", i, query(In));
    69     }
    70     return 0;
    71 }
  • 相关阅读:
    UIProgressView的详细使用
    Android拍照上传代码样例
    UILabel的详细使用及特殊效果
    TextView属性android:ellipsize实现跑马灯效果
    Android中WebView实现Javascript调用Java类方法
    有效获取状态栏(StatusBar)高度
    详解iPhone Tableview分批显示数据
    TextView显示插入的图片
    ObjectiveC语法快速参考
    UISegmentedControl的详细使用
  • 原文地址:https://www.cnblogs.com/wydxry/p/7248983.html
Copyright © 2020-2023  润新知