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;elseinit.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;elseres.m[i][j] = 0;tp = matrix_mi(b, n);cout << tp.m[0][1] <<'\n';return 0;
}

转载于:https://www.cnblogs.com/RootVount/p/11469372.html