Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2.
Input
a single line containing n (where 0 ≤ n ≤ 100,000,000,000)
Output
print Fn 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:
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; }