Sicily 1863. Elegant Fibonacci numbers again (Fibonacci + matrix power)

Question:

Fibonacci series, F [0] = 0, F [1] = 1, F [N] = f [N-2] + F [n-1], n> 1

Given N (0 <= n <2 ^ 32) and M, evaluate f [N] mod m.

 

Solution:

N is too large, so we cannot use the cyclic recursive Calculation of O (n ).

 

Conclusion: For matrices | 1 1 |, and matrices | Fn Fn-1 |, after the two are multiplied to obtain a new matrix | Fn + 1FN|, FN indicates the nth Number of fiber ACCI

         | 1 0 |     | Fn-1 Fn-2 |              | Fn  Fn-1

 

Therefore, we can use the rapid power modulo + matrix multiplication of O (logn) to calculate f [N].

Fast Power Process: A ^ B =

1, B = 0

A, B = 1

A ^ (B/2) * A ^ (B/2), B is even

A * a ^ (b-1), B is odd

#include<iostream>#include<cstdio>using namespace std;struct Node{    int arr[2][2];}o;Node MUL(Node a, Node b,const int &m){    Node tmp;    tmp.arr[0][0] = (a.arr[0][0]*b.arr[0][0] + a.arr[0][1]*b.arr[1][0]) % m;    tmp.arr[0][1] = (a.arr[0][0]*b.arr[0][1] + a.arr[0][1]*b.arr[1][1]) % m;    tmp.arr[1][0] = (a.arr[1][0]*b.arr[0][0] + a.arr[1][1]*b.arr[1][0]) % m;    tmp.arr[1][1] = (a.arr[1][0]*b.arr[0][1] + a.arr[1][1]*b.arr[1][1]) % m;    return tmp;}Node exp(Node mat, int n, const int &m){    if(n == 1) return o;        if(n & 1)    {        return MUL(o, exp(mat,n-1,m), m);    }    else    {        Node tmp = exp(mat,n>>1, m);        return MUL(tmp,tmp,m);    }}int main(){    o.arr[0][0] = o.arr[0][1] = o.arr[1][0] = 1; o.arr[1][1] = 0;    int cas;    scanf("%d",&cas);    while(cas --)    {        int n,m;        scanf("%d%d",&n,&m);        if(n == 0)        {            printf("0\n");            continue;        }                Node f = o;        f = exp(f,n,m);                printf("%d\n",f.arr[0][1] % m);    }}