3070 Fibonacci 二分

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本文介绍了一种快速计算斐波那契数列中指定项最后四位数字的方法，使用矩阵快速幂运算来提高效率，并提供了完整的代码实现。

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3459		Accepted: 2426

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_n_{− 1} + F_n_{− 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

[f[n+1] f[n]]=[1 1]^n

[f[n] f[n-1]] [1 0]

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

Sample Output

#include<iostream>
#include<cstdio>
using namespace std;
int a[4],b[4];
void binary(int n)
{
   // cout<<n<<".."<<endl;
    if(n==1||n==0) 
    {
        //b[3]=1,b[2]=1,b[1]=1,b[0]=0;
        a[3]=1,a[2]=1,a[1]=1,a[0]=0;
        return ;
    }
    else if(n&1)
    {
       // for(int i=3;i>=0;i--) cout<<a[i]<<"..";cout<<endl;
        binary(n/2);
        //for(int i=3;i>=0;i--) cout<<a[i]<<"..";cout<<endl;
        b[3]=(a[3]*a[3]+a[2]*a[1])%10000;
        b[2]=(a[3]*a[2]+a[2]*a[0])%10000;
        b[1]=(a[1]*a[3]+a[0]*a[1])%10000;
        b[0]=(a[1]*a[2]+a[0]*a[0])%10000;
        for(int i=3;i>=0;i--) a[i]=b[i];
        //for(int i=3;i>=0;i--) cout<<a[i]<<"..";cout<<endl;
        b[3]=(a[3]+a[2])%10000;
        b[2]=a[3];
        b[1]=(a[1]+a[0])%10000;
        b[0]=a[1];
        for(int i=3;i>=0;i--) a[i]=b[i];
       // for(int i=3;i>=0;i--)  cout<<a[i]<<"..";cout<<endl;
    }
    else 
    {
        binary(n/2);
        b[3]=(a[3]*a[3]+a[2]*a[1])%10000;
        b[2]=(a[3]*a[2]+a[2]*a[0])%10000;
        b[1]=(a[1]*a[3]+a[0]*a[1])%10000;
        b[0]=(a[1]*a[2]+a[0]*a[0])%10000;
        for(int i=3;i>=0;i--) a[i]=b[i];
    }
}          
int main()
{
    long long n;
    while(cin>>n)
    {
        if(n==-1) break;
        if(n==0) {printf("0/n");continue;}
       // a[3]=1,a[2]=1,a[1]=1,a[0]=0;
        binary(n-1);
        printf("%d/n",a[3]);
    }
    return 0;
}