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  • Fibonacci

    Interesting Fibonacci

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 672    Accepted Submission(s): 116

    Problem Description
    In mathematics, the Fibonacci numbers are a sequence of numbers named after Leonardo of Pisa, known as Fibonacci (a contraction of filius Bonaccio, "son of Bonaccio"). Fibonacci's 1202 book Liber Abaci introduced the sequence to Western European mathematics, although the sequence had been previously described in Indian mathematics.   The first number of the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers of the sequence itself, yielding the sequence 0, 1, 1, 2, 3, 5, 8, etc. In mathematical terms, it is defined by the following recurrence relation: That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonacci numbers (sequence A000045 in OEIS), also denoted as F[n]; F[n] can be calculate exactly by the following two expressions: A Fibonacci spiral created by drawing arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34;
    So you can see how interesting the Fibonacci number is. Now AekdyCoin denote a function G(n) Now your task is quite easy, just help AekdyCoin to calculate the value of G (n) mod C
     
    Input
    The input consists of T test cases. The number of test cases (T is given in the first line of the input. Each test case begins with a line containing A, B, N, C (10<=A, B<2^64, 2<=N<2^64, 1<=C<=300)
     
    Output
    For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the value of G(N) mod C
     
    Sample Input
    1 17 18446744073709551615 1998 139
     
    Sample Output
    Case 1: 120
     
    Author
    AekdyCoin
     
     
     
     
     
    Power of Fibonacci
    Time Limit: 5 Seconds      Memory Limit: 65536 KB

    In mathematics, Fibonacci numbers or Fibonacci series or Fibonacci sequence are the numbers of the following integer sequence:

    1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...

    By definition, the first two numbers in the Fibonacci sequence are 1 and 1, and each subsequent number is the sum of the previous two. In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation Fn = Fn - 1 + Fn - 2 with seed values F1 = 1 and F2 = 1.

    And your task is to find ΣFiK, the sum of the K-th power of the first N terms in the Fibonacci sequence. Because the answer can be very large, you should output the remainder of the answer divided by 1000000009.

    Input

    There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

    There are two integers N and K (0 <= N <= 1018, 1 <= K <= 100000).

    Output

    For each test case, output the remainder of the answer divided by 1000000009.

    Sample Input

    5
10 1
4 20
20 2
9999 99
987654321987654321 98765

    Sample Output

    143
487832952
74049690
113297124
108672406

    Hint

    The first test case, 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143.

    The second test case, 120 + 120 + 220 + 320 =3487832979, and 3487832979 = 3 * 1000000009 + 487832952, so the output is 487832952.

    Author: ZHOU, Yuchen
     
     
     
     

    2014年蓝桥杯的第九题是这样描述的：

        

        给定Fibonacci数列F[]，其中，，求表达式

           

        

        的值。其中

    E - 喵喵的遗憾

    Time Limit: 20000/10000MS (Java/Others)      Memory Limit: 512000/256000KB (Java/Others)

    Problem Description

    喵喵因为太弱，没有进Final，终身遗憾。她就挂在了这个题目上。

    已知：

    F0 = 1 , F1 = 1 , F2 = 2 , Fn = Fn-1+Fn-2

    求：

    FFFn Mod P

    ( 也就是 F[ F[ F[n] ] ]  % P ) 

    Input

    第一行一个整数 T 代表数据组数(T ≤ 20000)。                                                <del> 喵以人格担保时限肯定够！！！</del>

    以下每行两个整数 N , P (0 ≤ N ≤ 109 , 1 ≤ P ≤ 109)

    Output

    对于每组数据输出一个整数代表答案。

    Sample Input

    4
1 2
2 3
4 35
4 31

    Sample Output

    1
2
34
3

    Hint

    F1 = 1 , F1 = 1 , F1 = 1

    F2 = 2 , F2 = 2 , F2 = 2

    F4 = 5 , F5 = 8 , F8 = 34

      1 #include <cstdio>
  2 #include <cstring>
  3 #include <queue>
  4 #include <map>
  5 #include <cmath>
  6 #include <iostream>
  7 #include <time.h>
  8 #include <algorithm>
  9 using namespace std;
 10 #define FF(i,N) for(int i=0;i<2;i++)
 11 #define ll long long
 12 #define maxn (1<<20)
 13 struct mat{ ll m[2][2]; };
 14 ll n, s, k, t, p ,x , y, cnt, num, ans;
 15 ll gcd(ll a, ll b){ return b ? gcd(b, a%b) : a; }
 16 ll a[1005],b[105],c[105],pri[maxn],prv[maxn],GG[maxn];
 17 mat A;
 18 mat Mul(mat a, mat b, ll mod){
 19     mat res;
 20     FF(i, 2)FF(j, 2)res.m[i][j] = 0;
 21     FF(i, 2)FF(j, 2)FF(k,2)
 22         res.m[i][j] = (res.m[i][j]+a.m[i][k] * b.m[k][j]) % mod;
 23     return res;
 24 }
 25 mat Pow(mat a, ll b, ll mod){
 26     mat res;
 27     FF(i,2)FF(j,2)res.m[i][j] = (i == j);
 28     while (b){
 29         if (b & 1)res = Mul(res, a, mod);
 30         a = Mul(a, a ,mod);
 31         b >>= 1;
 32     }
 33     return res;
 34 }
 35 ll POW(ll a, ll b, ll mod){
 36     a = a%mod;
 37     ll res = 1;
 38     while (b){
 39         if (b & 1)res = res*a%mod;
 40         a = a*a%mod;
 41         b >>= 1;
 42     }
 43     return res;
 44 }
 45 ll Loop(ll p, ll mod){
 46     mat res = Pow(A, p - 1, mod);
 47     if ((res.m[0][0] + res.m[1][0])%mod == 1 && (res.m[0][1] + res.m[1][1])%mod == 0)return 1;
 48     return 0;
 49 }
 50 ll G(ll p){
 51     if (p < 1000000 && GG[p] != -1)return GG[p];
 52     ll mod = p; k = 0;
 53     if (POW(5, (p - 1)>>1, mod) == 1)p = p - 1;
 54     else p = 2 * (p + 1);
 55     int stop = sqrt(1.0*p);
 56     for (int i = 1; i <= stop; i++){
 57         if (p%i)continue;
 58         a[k++] = i;
 59         if(i*i!=p)a[k++] = p / i;
 60     }
 61     sort(a, a + k);
 62     for (int i = 0; i < k;i++)
 63     if (Loop(a[i], mod)){
 64         if (mod < 1000000)return GG[mod] = a[i];
 65         return a[i];
 66     }
 67 }
 68 void init(){
 69     A.m[0][0] = 1; A.m[0][1] = 1;
 70     A.m[1][0] = 1; A.m[1][1] = 0;
 71 }
 72 ll LOOP(ll p){
 73     s = 1; num = 0; c[0] = 0;
 74     for (int i = 0; pri[i]*pri[i] <= p; i++){
 75         c[num] = 0;
 76         if (p%pri[i] == 0){
 77             b[num] = pri[i];
 78             while (p%pri[i] == 0)p /= pri[i], c[num]++;
 79             num++;
 80         }
 81     }
 82     if (p>1)c[num] = 1, b[num++] = p;
 83     for (int i = 0; i < num; i++){
 84         if (c[i]){
 85             if (b[i] == 2)y = 3;
 86             else if (b[i] == 3)y = 8;
 87             else if (b[i] == 5)y = 20;
 88             else y = G(b[i]);
 89             for (int j = 1; j < c[i]; j++)y = y*b[i];
 90             s = s/gcd(s, y)*y;
 91         }
 92     }
 93     return s;
 94 }
 95 int main(){
 96     memset(GG, -1, sizeof GG);
 97     int stop = sqrt(1.0 * 1000000);
 98     for (int i = 2; i<=stop;i++)
 99     if (!prv[i])
100     for (int j = i*i; j <= 1000000;j+=i)
101     if (!prv[j])prv[j] = 1;
102     for (int i = 2; i <= 1000000; i++)if (!prv[i])pri[cnt++] = i;
103     int T;scanf("%d",&T);
104     while (T--){
105         init();
106         scanf("%d%d", &n, &p);
107         if (p == 1){ printf("0
"); continue; }
108         if (n == 0){ printf("1
"); continue; }
109         ll mod = p;
110         ll mod1 = LOOP(p);
111         ll mod2 = LOOP(mod1);
112         mat res;
113         res = Pow(A, n-1, mod2);
114         n = (res.m[0][0] + res.m[1][0]) % mod2;
115         res = Pow(A, (n + mod2 - 1) % mod2, mod1);
116         n = (res.m[0][0] + res.m[1][0]) % mod1;
117         res = Pow(A, (n + mod1 - 1) % mod1, mod);
118         ans = (res.m[0][0] + res.m[1][0]) % mod;
119         printf("%d
", ans);
120     }
121     return 0;
122 }
    View Code

    这道题卡的地方，硬是找了半天都没找到，最后发现，n=0时，有些p输不出答案，估计是n-1小于0了那些地方就卡住了。。。Pow(A,n-1,mod)这里0 0。。。一开始写的是把最外面的那层再求一个循环节，

    其实没必要，第一个n又不会超ll ，而且快速幂至少比求循环节快点，然后那时写的代码又戳，这个地方那时候不是这么写，多求了个循环节，n=n%s;Pow(A,(n-1+s)%s,mod)所以那时就没这个问题，后面优化着就

    出问题了。。。结果就卡了大半天。。。

    3286: Fibonacci矩阵

    Time Limit: 15 Sec  Memory Limit: 128 MB Submit: 131  Solved: 29 [Submit][Status]

    Description

     

    Input

    八个用空格隔开的整数n, m, a, b, c, d, e, f，其中n, m, a, b, d, e为正整数，c, f为非负整数。

    Output

    一个整数，表示Fib[n][m]对2012182013取模的值。

    Sample Input

    3 4 1 1 0 1 1 0

    Sample Output

    144

    HINT

    n,m,a,b,c,d,e,f<=10^1000000

    Source

    tangjz原创

    优美的Fibonacci数列与矩阵

    分类： 数学问题 2014-05-09 16:11 212人阅读 评论(0) 收藏 举报

    题目：http://codeforces.com/contest/392/problem/C

    题意：给定Fibonacci数列F[]，令，求的值。

        

    广义Fibonacci数列找循环节             

            分类：             数学问题              2014-05-12 15:53     221人阅读     评论(0)     收藏     举报    

    今天将来学习如何求广义Fibonacci数列的循环节。

     

    问题：给定，满足，求的循

         环节长度。

     

    来源：http://acdreamoj.sinaapp.com/ 1075题

     
    题目来源：
    基准时间限制：
    斐波那契数列Mod 一个数N会得到一个新的数列，根据同余可以得知，这个数列中的数会出现循环。例如： 
    F (mod 4) = 0 1 1 2 3 1 ... 
    F (mod 5) = 0 1 1 2 3 0 3 3 1 4 0 4 4 3 2 0 2 2 4 1 ...
     
    后面的数会出现循环，因此Fib Mod 4的循环节长度是6，Fib Mod 5的循环节长度是20。
    给出一个数N，求斐波那契数列Mod N的循环节的长度。
    Input
    Output
    Input 示例
    2
4
5
    Output 示例
    6
20


      1 #include <cstdio>
  2 #include <cstring>
  3 #include <queue>
  4 #include <map>
  5 #include <cmath>
  6 #include <iostream>
  7 #include <time.h>
  8 #include <algorithm>
  9 using namespace std;
 10 #define FF(i,N) for(int i=0;i<2;i++)
 11 #define ll long long
 12 #define maxn (1<<20)
 13 struct mat{ ll m[2][2]; };
 14 ll n, s, k, t, p, x, y, cnt, num, ans;
 15 ll gcd(ll a, ll b){ return b ? gcd(b, a%b) : a; }
 16 ll a[1005], b[105], c[105], pri[maxn], prv[maxn], GG[maxn];
 17 mat A;
 18 mat Mul(mat a, mat b, ll mod){
 19     mat res;
 20     FF(i, 2)FF(j, 2)res.m[i][j] = 0;
 21     FF(i, 2)FF(j, 2)FF(k, 2)
 22         res.m[i][j] = (res.m[i][j] + a.m[i][k] * b.m[k][j]) % mod;
 23     return res;
 24 }
 25 mat Pow(mat a, ll b, ll mod){
 26     mat res;
 27     FF(i, 2)FF(j, 2)res.m[i][j] = (i == j);
 28     while (b){
 29         if (b & 1)res = Mul(res, a, mod);
 30         a = Mul(a, a, mod);
 31         b >>= 1;
 32     }
 33     return res;
 34 }
 35 ll POW(ll a, ll b, ll mod){
 36     a = a%mod;
 37     ll res = 1;
 38     while (b){
 39         if (b & 1)res = res*a%mod;
 40         a = a*a%mod;
 41         b >>= 1;
 42     }
 43     return res;
 44 }
 45 ll Loop(ll p, ll mod){
 46     mat res = Pow(A, p - 1, mod);
 47     if ((res.m[0][0] + res.m[1][0]) % mod == 1 && (res.m[0][1] + res.m[1][1]) % mod == 0)return 1;
 48     return 0;
 49 }
 50 ll G(ll p){
 51     if (p < 1000000 && GG[p] != -1)return GG[p];
 52     ll mod = p; k = 0;
 53     if (POW(5, (p - 1) >> 1, mod) == 1)p = p - 1;
 54     else p = 2 * (p + 1);
 55     int stop = sqrt(1.0*p);
 56     for (int i = 1; i <= stop; i++){
 57         if (p%i)continue;
 58         a[k++] = i;
 59         if (i*i != p)a[k++] = p / i;
 60     }
 61     sort(a, a + k);
 62     for (int i = 0; i < k; i++)
 63     if (Loop(a[i], mod)){
 64         if (mod < 1000000)return GG[mod] = a[i];
 65         return a[i];
 66     }
 67 }
 68 void init(){
 69     A.m[0][0] = 1; A.m[0][1] = 1;
 70     A.m[1][0] = 1; A.m[1][1] = 0;
 71 }
 72 ll LOOP(ll p){
 73     s = 1; num = 0; c[0] = 0;
 74     for (int i = 0; pri[i] * pri[i] <= p; i++){
 75         c[num] = 0;
 76         if (p%pri[i] == 0){
 77             b[num] = pri[i];
 78             while (p%pri[i] == 0)p /= pri[i], c[num]++;
 79             num++;
 80         }
 81     }
 82     if (p>1)c[num] = 1, b[num++] = p;
 83     for (int i = 0; i < num; i++){
 84         if (c[i]){
 85             if (b[i] == 2)y = 3;
 86             else if (b[i] == 3)y = 8;
 87             else if (b[i] == 5)y = 20;
 88             else y = G(b[i]);
 89             for (int j = 1; j < c[i]; j++)y = y*b[i];
 90             s = s / gcd(s, y)*y;
 91         }
 92     }
 93     return s;
 94 }
 95 int main(){
 96     memset(GG, -1, sizeof GG);
 97     int stop = sqrt(1.0 * 1000000);
 98     for (int i = 2; i <= stop; i++)
 99     if (!prv[i])
100     for (int j = i*i; j <= 1000000; j += i)
101     if (!prv[j])prv[j] = 1;
102     for (int i = 2; i <= 1000000; i++)if (!prv[i])pri[cnt++] = i;
103     int T; scanf("%d", &T);
104     while (T--){
105         init();
106         scanf("%d", &p);
107         ans = LOOP(p);
108         printf("%d
", ans);
109     }
110     return 0;
111 }
    View Code


     
    题目来源：
    基准时间限制：
    Fib(N)表示斐波那契数列的第N项（F(0) = 0, F(1) = 1），给出N和K，求Fib(N) mod Fib(K)。
    Input
    Output
    Input 示例
    2
5 5
13 5
    Output 示例
    0
3
    据说是由

         

    这两个公式推出下面的。。。


    有了上面两个公式后，妈蛋才发现很明显嘛。。。智商低不解释！

     1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <map>
 5 #include <cmath>
 6 #include <iostream>
 7 #include <time.h>
 8 #include <algorithm>
 9 using namespace std;
10 #define FF(i,N) for(int i=0;i<2;i++)
11 #define ll long long
12 #define maxn (1<<15)
13 ll n, m, k, s, t, mod;
14 ll f[maxn];
15 int main(){
16     int T; cin >> T;
17     f[0] = 0; f[1] = 1;
18     for (int i = 2; i <= 1005; i++)f[i] = f[i - 1] + f[i - 2];
19     while (T--){
20         cin >> n >> m;
21         if ((n/m/2*m)%2 == 0)k = 1;
22         else k = -1;
23         if (n / m % 2 == 0)s = ((f[n%m] % f[m])*k+f[m])%f[m];
24         else s = (((f[n%m] % f[m])*(f[m - 1] % f[m]) % f[m])*k + f[m]) % f[m];
25         printf("%d
", s);
26     }
27     return 0;
28 }
    View Code

    不会弄大数，这里给出小数据类似代码。。。用JAVA很方便，我也不会JAVA- -！

     
    题目来源：
    基准时间限制：
    F(n)表示斐波那契数列的第N项（F(0) = 0, F(1) = 1），给出一个数N，输出F(n)质因数分解的表示。例如：
     
    F(0)= 0
    F(1)= 1
    F(2)= 1
    F(10)= 5 * 11
    F(30)= 2^3 * 5 * 11 * 31 * 61
    Input
    Output
    Input 示例
    100
    Output 示例
    F(100)= 3 * 5^2 * 11 * 41 * 101 * 151 * 401 * 3001 * 570601


    据说是论文题。。。
    看了一下，吓尿。。。
    是不是那个什么Lucas大数分解有关的。。。反正不明白。。。
    http://mersennus.net/fibonacci/
    http://mersennus.net/fibonacci/f1000.txt



     
    基准时间限制：
    斐波那契字符串的定义如下：
     
    f(1) = a
    f(2) = b
    f(n) = f(n-1) + f(n-2)，(n > 2)
     
    所以前6个斐波那契字符串是："a", "b", "ba", "bab", "babba",  "babbabab"
    现在给出一个编号n，再给出1个字符串s，求f(n)包含多多少个s。
    例如：n = 6，s = "ba"，f(6) = "babbabab"，包含了3个"ba"。输出3。由于数量巨大，只要输出Mod 10^9 + 7的结果。
    Input
    Output
    Input 示例
    6
ba
    Output 示例
    3

    8001. Fibonacci Sum

    Problem code: FIBOSUM


    The fibonacci sequence is defined by the following relation:

    F(0) = 0

    F(1) = 1

    F(N) = F(N - 1) + F(N - 2), N >= 2

    Your task is very simple. Given two non-negative integers N and M, you have to calculate the sum (F(N) + F(N + 1) + ... + F(M)) mod 1000000007.

    Input

    The rst line contains an integer T (the number of test cases). Then, T lines follow. Each test
    case consists of a single line with two non-negative integers N and M.

    The first line contains an integer T (the number of test cases). Then, T lines follow. Each test case consists of a single line with two non-negative integers N and M.

    Output

    For each test case you have to output a single line containing the answer for the task.

    Example

    Input:
3
0 3
3 5
10 19

Output:
4
10
10857

    Constraints

    • T <= 1000
    • 0 <= N <= M <= 109


     

    1021 Fibonacci Again Leojay   (17212/35677)48.24%
    1250 Hat's Fibonacci 戴帽子的   (2278/6923)32.90%
    1568 Fibonacci daringQQ Happy 2007 (1484/3238)45.83%
    1588 Gauss Fibonacci DYGG HDU “Valentines Day” Open Programming Contest 2007-02-14 (885/2030)43.60%
    1708 Fibonacci String linle HDU 2007-Spring Programming Contest (1122/3263)34.39%
    1848 Fibonacci again and again lcy ACM Short Term Exam_2007/12/13 (1907/4528)42.12%
    2814 Interesting Fibonacci AekdyCoin HDU 1st “Old-Vegetable-Birds Cup” Programming Open Contest (116/672)17.26%
    2855 Fibonacci Check-up   2009 Multi-University Training Contest 5 - Host by NUDT (626/1120)55.89%
    3054 Fibonacci   2009 Multi-University Training Contest 15 - Host by BUAA (83/212)39.15%
    3117 Fibonacci Numbers   IPCP 2005 Northern Preliminary for Northeast North-America (676/1709)39.56%
    3306 Another kind of Fibonacci wyb HDOJ Monthly Contest – 2010.02.06 (561/1464)38.32%
    3509 Buge's Fibonacci Number Problem   2010 ACM-ICPC Multi-University Training Contest（8）——Host by ECNU (179/643)27.84%
    4099 Revenge of Fibonacci   2011 Asia Shanghai Regional Contest (456/1963)23.23%
    4786 Fibonacci Tree   2013 Asia Chengdu Regional Contest (308/1043)29.53%
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  • 原文地址：https://www.cnblogs.com/HaibaraAi/p/3825349.html
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