A - Fibonacci

最新推荐文章于 2023-06-19 22:02:25 发布

philophobia

最新推荐文章于 2023-06-19 22:02:25 发布

阅读量725

点赞数

CC 4.0 BY-SA版权

分类专栏：快速幂

本文链接：https://blog.youkuaiyun.com/philophobia/article/details/77099042

快速幂专栏收录该内容

4 篇文章

订阅专栏

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，利用矩阵快速幂技巧实现，并提供了完整的C++代码示例。

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

/*直接套快速幂模板，求矩阵的n次幂
int ans=1;
while(n){
    if(n&1)
        ans*=x;
    x*=x;
    n>>=1;
就是此模板啦，将相应的ans 和x换成矩阵就行*/

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,a[2][2],ans[2][2];
void mul(int a[2][2],int b[2][2]){
    int temp[2][2];
    memset(temp,0,sizeof temp);
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){
            for(int k=0;k<2;k++){
                temp[i][j]=(temp[i][j]+a[i][k]*b[k][j])%10000;
            }
        }
    }
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){
            b[i][j]=temp[i][j];
        }
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        if(n==-1)
           return 0;
        a[0][0]=a[0][1]=a[1][0]=1;
        a[1][1]=0;
        ans[0][0]=ans[1][1]=1;
        ans[0][1]=ans[1][0]=0;//相当于ans=1,初始化为单位矩阵
        while(n){
            if(n&1)
                mul(a,ans);//此过程相当于求ans=ans*a;
            mul(a,a);//相当于求a=a*a;
            n>>=1;
        }
        printf("%d\n",ans[0][1]);
    }
}