POJ 3070 Fibonacci (矩阵快速幂)

题目链接：POJ 3070

Problem Description

In the Fibonacci integer sequence, \(F_0 = 0\), \(F_1 = 1\), and \(F_n = F_{n − 1} + F_{n − 2}\) for \(n \ge 2\). 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 \(F_n\).

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing \(n\) (where \(0 \le n \le 1,000,000,000\)). 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 \(F_n\). If the last four digits of \(F_n\) are all zeros, print \(‘0’\); otherwise, omit any leading zeros (i.e., print \(F_n\) mod \(10000\)).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Solution

题意

给定正整数 \(n\)，求斐波那契数列第 \(n\) 项。

题解

矩阵快速幂

矩阵快速幂模板题。

\[ \left[ \begin{matrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{matrix} \right] = \left[ \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right] ^ n \]

Code

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e2 + 5;
const ll mod = 1e4;

struct Matrix {
    int n, m;
    ll a[maxn][maxn];
    Matrix(int n = 0, int m = 0) : n(n), m(m) {}
    void input() {
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                scanf("%lld", &a[i][j]);
            }
        }
    }
    void output() {
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                printf("%lld", a[i][j]);
                printf("%s", j == m? "\n": " ");
            }
        }
    }
    void init() {
        memset(a, 0, sizeof(a));
    }
    void unit() {
        init();
        for(int i = 1; i <= n; ++i) {
            a[i][i] = 1;
        }
    }
    Matrix operator *(const Matrix b) {
        Matrix c(n, b.m);
        c.init();
        for(int i = 1; i <= c.n; ++i) {
            for(int k = 1; k <= m; ++k) {
                for(int j = 1; j <= c.m; ++j) {
                    c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j]) % mod;
                }
            }
        }
        return c;
    }
    Matrix qmod(ll b) {
        Matrix ans = Matrix(n, n);
        ans.unit();
        if(!b) return ans;
        while(b) {
            if(b & 1) ans = ans * (*this);
            *this = (*this) * (*this);
            b >>= 1;
        }
        return ans;
    }
};

int main() {
    int n;
    while(~scanf("%d", &n)) {
        if(n == -1) break;
        ll ans = 0;
        if(n) {
            Matrix m(2, 2);
            m.a[1][1] = 1;
            m.a[1][2] = 1;
            m.a[2][1] = 1;
            m.a[2][2] = 0;
            m = m.qmod(n - 1);
            ans = m.a[1][1] % mod;
        } 
        printf("%lld\n", ans);
    }
    return 0;
}

标签：matrix,int,ll,3070,POJ,Fibonacci,ans,include
来源： https://www.cnblogs.com/wulitaotao/p/11420478.html