zoukankan      html  css  js  c++  java
  • fibonacci封闭公式及矩阵连乘

    Description

    The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one. 

    What is the numerical value of the nth Fibonacci number?
     

    Input

    For each test case, a line will contain an integer i between 0 and 10 8 inclusively, for which you must compute the ith Fibonacci number fi. Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“...”). 

    There is no special way to denote the end of the of the input, simply stop when the standard input terminates (after the EOF). 
     

    Sample Input

    0 1 2 3 4 5 35 36 37 38 39 40 64 65
     

    Sample Output

    0 1 1 2 3 5 9227465 14930352 24157817 39088169 63245986 1023...4155 1061...7723 1716...7565
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    using namespace std;
    
    typedef struct
    {
      long long m[2][2];
    }mat;
    
    mat p={0,1,1,1},I={1,0,0,1};
    
    mat calc(mat a,mat b)
    {
        int i,j,k;
        mat c;
        for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            c.m[i][j]=0;
           for(k=0;k<2;k++)
          {
            c.m[i][j]+=a.m[i][k]*b.m[k][j]%10000;
          }
          c.m[i][j]%=10000;
        }
        return c;
    }
    
    mat matirx(int n)
    {
        mat m=p,b=I;
        while(n>=1)
        {
            if(n&1) b=calc(b,m);
            n>>=1;
            m=calc(m,m);
        }
        return b;
    }
    
    int main()
    {
        int i,a[42],len;
        int n;
        double s,d;
        a[0]=0;
        a[1]=1;
        for(i=2; i<40; i++)
            a[i]=a[i-1]+a[i-2];
        while(scanf("%d",&n)!=EOF)
        {
            if(n<40)
                printf("%d
    ",a[n]);
            else
            {
                s=log10(1.0/sqrt(5))+n*log10((1+sqrt(5))/2.0);//fibonacci封闭公式求前四位
                len=(int)s;
                d=s+3-len;
                printf("%d...",(int)pow(10,d));
    
                mat tmp;
                tmp=matirx(n);                                //矩阵连乘求后四位
                printf("%04d
    ",tmp.m[0][1]);
            }
        }
        return 0;
    }
  • 相关阅读:
    springboot 实现 aop
    使用@import导入实现了ImportBeanDefinitionRegistrar接口的类,不能被注册为bean
    Spring Boot 动态数据源(多数据源自动切换)
    springboot2动态数据源的绑定
    三分钟学会@Autowired@Qualifier@Primary注解
    java复制文件的4种方式
    五款最好的免费同步软件
    springboot2多数据源完整示例
    WebSocket 结合 Nginx 实现域名及 WSS 协议访问
    Spring的注解@Qualifier注解
  • 原文地址:https://www.cnblogs.com/chen9510/p/4734709.html
Copyright © 2011-2022 走看看