Fibonacci (矩阵快速幂)

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#矩阵快速幂

本文介绍了一种高效计算斐波那契数列第n项最后四位数字的方法，利用矩阵乘法和快速幂运算，解决了大规模数值计算的问题，并提供了一个具体的C++实现示例。

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:

AC代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<time.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
using namespace std;
const int mod=10000;
struct Matrix{
    int n,m; //大小
    int d[5][5];
    Matrix(int a=0,int b=0){
        n=a,m=b;
        memset(d,0,sizeof(d));
    }
    void copy(int *tmp){
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++){
                d[i][j]=(*tmp)%mod;
                ++tmp;  //指针 指向要传入数组的某个元素
            }
    }
    friend Matrix operator * (const Matrix &a,const Matrix &b){ //矩阵乘法的重载
        Matrix c(a.n,b.m);
        for (int i=0; i<a.n; ++i)
            for (int j=0; j<b.m; ++j){
                long long tmp=0;
                for (int k=0; k<a.m; ++k){
                    tmp+=(long long)a.d[i][k]*b.d[k][j] % mod;
                    tmp%=mod;
                }
                c.d[i][j]=tmp % mod;
            }
        return c;
    }
};
int solve(int n){
    if (n==0) return 0;
    if (n==1) return 1;
    Matrix A(1,2),B(2,2);
    int aa[2]={0,1};
    int bb[2*2]={
        0,1,
        1,1,
    };
    A.copy(aa);//矩阵的构造函数copy
    B.copy(bb);
    for (int k=n; k; k>>=1){
        if (k&1)
            A=A*B;
        B=B*B;
    }
    return A.d[0][0];
}

int main()
{
    int n=0;
    while(scanf("%d",&n)!=EOF){
        if (n==-1) break;
        printf("%d\n",solve(n));
    }
    return 0;
}