Fibonacci（模板）【矩阵快速幂】

Fibonacci

题目链接（点击）

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 20989 Accepted: 14381

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:

.

思路：

由于输入的n数值特别大，所以如果先打表然后随机访问再输出肯定不行就用到了矩阵快速幂

(补充)矩阵相乘：

矩阵AB的第i行第j列为 A矩阵第i行与B矩阵第j列对应元素分别相乘再求和

不用字母直接上例子：

已知：(n)=f(n-1)+f(n-2) f(n-1)=f(n-1)+0*f(n-2)

可以构造下面的递推式：

可以化简为：

根据矩阵快速幂就可以先求常数矩阵的（n-1）次幂然后输出a[0][0] 即可：

a[0][0]表示的是 f(n) 但要注意考虑当n=0是无法求出 f(-1) 所以要特殊考虑n=0的情况

AC代码:

(通过结构体定义数组方便传参) 感谢@鸡冠花12138

#include<stdio.h>

#include<string.h>

const int mod=1e4;

struct node{

    int a[5][5];

};

node mat_mul(node x,node y)  //该函数返回值为node型 作用为计算两个矩阵x和y的乘积

{

    node res;

    memset(res.a,0,sizeof(res.a));

    for(int i=0;i<2;i++){

        for(int j=0;j<2;j++){

            res.a[i][j]=0;

            for(int k=0;k<2;k++){

                res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%mod;

            }

        }

    }

    return res;

}

node mat_qpow(int n)  //返回值仍然是node型  作用是求常数矩阵的n次幂

{

    node res,c;

    c.a[0][0]=1; c.a[0][1]=1;  //c.a表示的是常数矩阵

    c.a[1][0]=1; c.a[1][1]=0;

    memset(res.a,0,sizeof(res.a));

    for(int i=0;i<2;i++){

        res.a[i][i]=1;

    }

    while(n){  //快速幂

        if(n%2){

            res=mat_mul(res,c);

        }

        c=mat_mul(c,c);

        n/=2;

    }

    printf("%d\n",res.a[0][0]);

}

int main()

{

    int n;

    while(~scanf("%d",&n)&&n!=-1){

        if(n==0){  //特殊考虑n=0的情况

            printf("0\n");

        }

        else{

            mat_qpow(n-1);

        }

    }

    return 0;

}

巴特西

Fibonacci（模板）【矩阵快速幂】

Fibonacci

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20989		Accepted: 14381