Fibonacci sequence F (n) "N super-Large (matrix accelerated operation) template"

Hihocoder #1143: Domino Coverage problem • Time limit:10000msSingle Point time limit:1000msMemory Limit:256MBDescribe

Dominoes, an ancient toy. Today we are going to look at the problem of Domino coverage:
We have a 2xN strip board and then use the 1x2 dominoes to cover the entire board. How many different coverage methods are there for this chessboard?
For example, for a board with a length of 1 to 3, we have the following types of coverage:

Hint: Domino Overlay

Tip: How to quickly calculate results

Input

Line 1th: an integer n. Represents the board length. 1≤n≤100,000,000

Output

Line 1th: An integer representing the number of coverage scenarios MOD 19999997

Sample input

62247088

Sample output

 17748018 
  
 Analysis: n oversize, if the Fibonacci nth term is calculated recursively, the matrix of linear algebra has the effect of accelerating operation. 
  
  But some of the sequences may be: 1 1 2 3, because a little bit different requires a slight modification of the number of times the matrix is multiplicative, which is the exponent of the matrix. 
 Code:  

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include < string> #include <vector> #include <queue> #include <math.h> #define EPS 1e-8#include <algorithm >using namespace std;//Matrix fast power operation (matrix acceleration operation) struct matrix{long long a[2][2];//define 2x2 matrix};matrix mul (Matrix x, Matrix Y, L Ong MoD) {matrix ret;//The value of each element of the RET matrix is multiplied by the matrix and then returned to it ret.a[0][0]= ((x.a[0][0]%mod) *y.a[0][0]%mod + (X.a[0][1]%mod) *y.a[    1][0]%mod)%mod;    Ret.a[0][1]= ((x.a[0][0]%mod) *y.a[0][1]%mod + (x.a[0][1]%mod) *y.a[1][1]%mod)%mod;    Ret.a[1][0]= ((x.a[1][0]%mod) *y.a[0][0]%mod + (x.a[1][1]%mod) *y.a[1][0]%mod)%mod;    Ret.a[1][1]= ((x.a[1][0]%mod) *y.a[0][1]%mod + (x.a[1][1]%mod) *y.a[1][1]%mod)%mod; return ret;}    To find the power modulus of Matrix X, E is the exponential matrix Mypow (Matrix X, Long long e, long MoD)//(X^E)%mod{matrix ret, temp;        if (e==0) {ret.a[0][0]=1; ret.a[0][1]=0; ret.a[1][0]=0;        Ret.a[1][1]=1;    return ret; } if (e==1) return x; When the index is 1 o'clock, return to the originalMatrix Temp=mypow (x, e>>1, MoD); x E/2 Ret=mul (temp, temp, mod); Ret=temp*temp if (e&1) Ret=mul (ret, x, MoD); If E is odd, ret times x return ret;    Return answer}int Main () {long n, m=19999997;//m is mod matrix ans;        while (scanf ("%ld", &n)!=eof) {//matrix initialization ans.a[0][0]=1; ans.a[0][1]=1; Ans.a[1][0]=1;        ans.a[1][1]=0;        if (n) {ans = Mypow (ans, N, m);//n Here is the exponent (considering whether n is to be modified in the corresponding topic), M is take modulus printf ("%lld\n", ans.a[0][0]);        }else{printf ("0\n"); }} return 0;}

Fibonacci sequence F (n) "N super-Large (matrix accelerated operation) template"