HDU 2855 Fibonacci Check-up 矩阵

这题比较神，不看题解我是想不出来T_T

证明过程如下

之后直接无脑快速幂就好了。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 5;

LL n, mod;
 
struct Matrix {
    int n, m;
    LL data[maxn][maxn];
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }
    void print() {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                cout << data[i][j] << " ";
            }
            cout << endl;
        }
    }
};
 
Matrix operator * (Matrix a, Matrix b) {
    int n = a.n, m = b.m;
    Matrix ret(n, m);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            for(int k = 1; k <= a.m; k++) {
                ret.data[i][j] += a.data[i][k] * b.data[k][j];
                ret.data[i][j] %= mod;
            }
        }
    }
    return ret;
}
 
Matrix pow(Matrix mat, LL k) {
    if(k == 0) {
        Matrix ret(mat.n, mat.m);
        for(int i = 1; i <= mat.n; i++) ret.data[i][i] = i;
        return ret;
    }
    if(k == 1) return mat;
    Matrix ret = pow(mat * mat, k / 2);
    if(k & 1) ret = ret * mat;
    return ret;
}
 

int main() {
    int T; cin >> T;
    while(T--) {
        cin >> n >> mod;
        Matrix A0(2, 1), A(2, 2);
        A0.data[1][1] = 1; A0.data[2][1] = 0;
        A.data[1][1] = A.data[1][2] = A.data[2][1] = 1;
        A.data[2][2] = 0;
        A0 = pow(A, 2 * n) * A0;
        cout << A0.data[2][1] << endl;
    }
    return 0;
}