Poj - 3070 Fibonacci (矩阵快速幂

最新推荐文章于 2019-11-03 16:21:23 发布

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本文介绍了一种高效计算Fibonacci数列第n项后四位数字的方法，利用矩阵快速幂技术来减少计算复杂度。通过构建特定的矩阵进行幂运算，可以显著提高计算效率，特别适用于大规模数据计算。

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Fibonacci

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

这里写图片描述

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

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:

题意

就是求Fib的第n项后四位

题解：

矩阵快速幂板子递推构造矩阵需要学习一下

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int mod = 1e4;
const int N = 2;
struct node {
    LL arr[N][N];
}p;
LL n, k; 

node mul(node x,node y)
{
    node ans;;
    memset(ans.arr,0,sizeof(ans.arr));
    for(int i = 0;i < 2; i++) {
        for(int j = 0;j < 2; j++) {
            for(int k = 0;k < 2; k++) {
                ans.arr[i][j] = (ans.arr[i][j]+(x.arr[i][k]*y.arr[k][j])%mod)%mod;
            }
        }
    }
    return ans;
}
node powMod(LL u, node x) 
{
    node ans;
    for(int i = 0;i < 2; i++) {
        for(int j = 0;j < 2; j++) {
            if(i==j) ans.arr[i][j] = 1;
            else ans.arr[i][j] = 0;
        }
    }
    while(u) {
        if(u&1) ans = mul(ans,x);
        x = mul(x,x);
        u >>= 1;
    }
    return ans;
}
int main()
{
    int n;
    while(~scanf("%d",&n),n!=-1) {
        p.arr[0][0] = 1; p.arr[0][1] = 1;
        p.arr[1][0] = 1; p.arr[1][1] = 0;
        node x = powMod(n,p);
        printf("%d\n",x.arr[0][1]);

    }
return 0;
}