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
",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
");  
        }
        else{
            mat_qpow(n-1);
        }
    }
    return 0;
}

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