HDU 3117 Fibonacci Numbers

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本文介绍了一种高效计算斐波那契数列的方法，利用矩阵快速幂和通项公式来解决大数问题，并提供了完整的C++代码实现。

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题目链接 : HDU3117

Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1311 Accepted Submission(s): 549

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 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

Sample Output

  
   0
1
1
2
3
5
9227465
14930352
24157817
39088169
63245986
1023...4155
1061...7723
1716...7565

Source

IPCP 2005 Northern Preliminary for Northeast North-America

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#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
using namespace std;


int fib[45];
void mult(int a[][2],int b[][2],int c[][2])
{
    int d[2][2]={0};
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                d[i][j]=(d[i][j]+(a[i][k]*b[k][j])%10000)%10000;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            c[i][j]=d[i][j];
}
void pow_mat(int a[][2],int k)
{
    int ii[2][2]={0};
    for(int i=0;i<2;i++)
        ii[i][i]=1;
    while(k)
    {
        if(k%2) mult(a,ii,ii);
        k/=2;
        mult(a,a,a);
    }
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            a[i][j]=ii[i][j];
}
double pow_sa(double fa,int k)
{
    double s=1.0,lim=1e20;
    while(k)
    {
        if(k&1)
        {
            s=s*fa;
            while(s>lim) s/=1000000000;
        }
        k>>=1;
        fa=fa*fa;
        while(fa>lim) fa/=1000000000;
    }
    return s;
}
int main()
{
 //   freopen("1.txt","r",stdin);
    int n;
    fib[1]=fib[2]=1;
    for(int i=3;i<40;i++)
        fib[i]=fib[i-1]+fib[i-2];
    double k=sqrt(5.0);
    while(scanf("%d",&n)==1)
    {
        if(n<40) printf("%d\n",fib[n]);
        else
        {
            int s[2][2]={1,1,1,0};
            pow_mat(s,n);
            int last=s[1][0];
            long double ans=k/5.0*(pow_sa((1+k)/2,n)-pow_sa((1-k)/2,n));
            while(ans>100000.0) ans/=10.0;
            int fir=(int)ans;
            printf("%d...%04d\n",fir/10,last);
        }
    }
    return 0;
}