SDNU 1062.Fibonacci（矩阵快速幂）

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2.

Input

a single line containing n (where 0 ≤ n ≤ 100,000,000,000)

Output

print F_n mod 1000000007 in a single line.

Sample Input

99999999999

Sample Output

669753982

Hint

An alternative formula for the Fibonacci sequence is


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:

Source

Unknown

思路：一开始想着要暴力的办法，或者用python，但还是tle了，然后发现这玩意儿可以用矩阵快速幂，就来一波骚操作了。

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define eps 1e-9
#define pi acos(-1)

const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int maxn = 1000 + 8;

ll n;

struct matrix
{
    ll m[2][2];
}b, tp, res, init;

matrix mul(matrix a, matrix b)
{
    matrix 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] += (a.m[i][k] * b.m[k][j]) % mod;
                c.m[i][j] %= mod;
            }
        }
    }
    return c;
}

matrix matrix_mi(matrix p, ll k)
{
    matrix t = res;
    while(k)
    {
        if(k & 1)
            t = mul(t, p);
        k >>= 1;
        p = mul(p, p);
    }
    return t;
}

int main()
{
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
        {
            if(i == 1 && j == 1)
                init.m[i][j] = 0;
            else
                init.m[i][j] = 1;
        }
    cin >> n;
    b = init;
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            if(i == j)
                res.m[i][j] = 1;
            else
                res.m[i][j] = 0;
    tp = matrix_mi(b, n);
    cout << tp.m[0][1] <<'
';
    return 0;
}