Fibonacci

最新推荐文章于 2017-11-02 01:20:15 发布

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本文介绍了一种计算斐波那契数列中任意项最后四位数字的高效算法，利用矩阵乘法和快速幂运算优化计算过程。通过实例演示和代码实现，展示了如何在大数范围内快速准确地获取斐波那契数列的特定项。

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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

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

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

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.
练练手感：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define N 10000;
struct r{
    long long int m[2][2];
}ans, tmp;

r multi(r a,r b){
    r cnt;
    cnt.m[0][0]=(a.m[0][0]*b.m[0][0]+a.m[0][1]*b.m[1][0])%N;
    cnt.m[0][1]=(a.m[0][0]*b.m[0][1]+a.m[0][1]*b.m[1][1])%N;
    cnt.m[1][0]=(a.m[1][0]*b.m[0][0]+a.m[1][1]*b.m[1][0])%N;
    cnt.m[1][1]=(a.m[1][0]*b.m[0][1]+a.m[1][1]*b.m[1][1])%N;
    return cnt;
}

 void  Matrix_power(int n){
    tmp.m[0][0]=tmp.m[0][1]=tmp.m[1][0]=1;
    tmp.m[1][1]=0;
    ans.m[0][0]=ans.m[1][1]=1,ans.m[1][0]=ans.m[0][1]=0;
    n-=2;
    while(n){
        if(n&1)
        ans=multi(tmp,ans);
        tmp=multi(tmp,tmp);
        n>>=1;
    }
    int s=(ans.m[0][0]+ans.m[0][1])%N;
    cout<<s<<endl;
}

int main(){
    int nn;
    while(cin>>nn&&nn!=-1){
            if(nn==0){
                cout<<0<<endl;
                continue;
            } else if(nn==1||nn==2){
            cout<<1<<endl;
            continue;
            }
    Matrix_power(nn);
    }
    return 0;
}