第六次个人赛D题 Fibonacci Numbers 递归+公式

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Fibonacci Numbers

http://acm.hdu.edu.cn/showproblem.php?pid=3117

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 445 Accepted Submission(s): 216

Problem 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 10⁸ 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

0123453536373839406465

Sample Output

0112359227465149303522415781739088169632459861023...41551061...77231716...7565
Fibonacci数大家都很熟悉，也许你刚开始学递归的样例就是它，但是这里要求到10⁸ 位，数据量庞大，计算机算出来肯定超时了，而且就算
是long long类型也不能暂时保存，因此这里用到了公式！！！！！也许大家对Fibonacci数的公式还不熟悉：
斐波那契数列,有如下通项公式:
                     Fn={[(1+√5)/2]^n - [(1-√5)/2]^n} / √5
 
 
首先所求的fib数位数<=8的我们直接打表，这里略去。只考虑位数>8的：先是
前四位，这要用到我们的通项公式，公式里的后一半可以抹掉，因为它太小了，
小的看不见…这样我们取 
x=log10(F(n))=n*log10((sqrt(5.0)+1.0)/2)-0.5*(log10(5.0))
所以pow(10,x)=pow(10,[x]+{x})=F(n)   [x]表示x的整数部分，{x}表示x的小数部分
pow(10,{x})=F(n)/pow(10,[x])
 
后四位：n要分奇偶数的情况，具体情况见代码
#include "algorithm"#include "iostream"#include "cmath"#include "stdio.h"#include "string.h"#include "stdlib.h"using namespace std;#define MOD 10000int a[110000];int fun(int n){   int b,c,d;   if(n<=100000) return a[n];   else {       if(n%2==0){             //后四位           b=fun(n/2);          c=fun(n/2-1);          return (b*b%MOD+c*c%MOD)%MOD;                  }       else{           b=fun(n/2);           c=fun(n/2-1);           d=fun(n/2+1);           return (d*b%MOD+b*c%MOD)%MOD;            }        }    }int main(){   int i,n,h[1000];   double ans;   a[0]=1;  a[1]=1; a[2]=2;   for(i=3;i<=100000;i++){       a[i]=a[i-1]+a[i-2];       a[i]%=MOD;                      }   h[0]=0; h[1]=h[2]=1;    for(i=3;i<=40;i++){       h[i]=h[i-1]+h[i-2];    }   while(scanf("%d",&n)!=EOF){        if(n<=39){             printf("%d/n",h[n]); continue;         }                               ans=n*log10((sqrt(5.0)+1.0)/2)-0.5*(log10(5.0));   //前四位         ans-=(int)ans;        ans=pow(10.0,ans);        while(ans<1000)            ans*=10;        printf("%d...",(int)ans);        printf("%04d/n",fun(n-1));                          }                                                 //system("pause");return 0;    }