POJ 3070 Fibonacci [矩阵快速幂模板]

最新推荐文章于 2024-10-30 23:49:49 发布

Vace___yun

最新推荐文章于 2024-10-30 23:49:49 发布

阅读量164

点赞数

CC 4.0 BY-SA版权

分类专栏：矩阵快速幂文章标签：矩阵快速幂

本文链接：https://blog.youkuaiyun.com/Vace___yun/article/details/89003493

矩阵快速幂专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种利用矩阵快速幂算法高效计算斐波那契数列中任一项最后四位数字的方法，并提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

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:

题意：根据题意套用矩阵快速幂模板，f[n]=f[n-1]+f[n-2];

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
struct node
{
    int m[2][2];
};
node mul(node a,node b)
{
    node c;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
    {
        c.m[i][j]=0;
        for(int k=0;k<2;k++)
            c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%10000;
    }
    return c;
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n!=-1)
    {
        if(n==0)
        {
            printf("0\n");
            continue;
        }
        node a,b;
        a.m[0][0]=1,a.m[0][1]=1,a.m[1][0]=1,a.m[1][1]=0;
        b.m[0][0]=1,b.m[0][1]=0,b.m[1][0]=0,b.m[1][1]=1;
        while(n)
        {
            if(n&1)
                b=mul(a,b);
            a=mul(a,a);
            n=n>>1;
        }
        printf("%d\n",b.m[1][0]);
    }
    return 0;
}