Hdu2855Fibonacci Check-up矩阵

转个图。

或者打表找的规律是  g[n] = 3*g[n-1] - g[n-2];

构造个 3 -1  的矩阵然后搞下就好了。

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#include <cstdio>

#include <algorithm>

#include <iostream>

#include <string.h>

typedef long long LL;

using namespace std;



LL mod;



struct Matrix

{

    LL m[4][4];

};



Matrix Mul(Matrix a, Matrix b)

{

    Matrix ans;

    for (LL i = 0; i < 2; i++){

        for (LL j = 0; j < 2; j++){

            ans.m[i][j] = 0;

            for (LL k = 0; k < 2; k++)

                ans.m[i][j] += a.m[i][k] * b.m[k][j], ans.m[i][j] = (ans.m[i][j] + mod) % mod;

        }

    }

    return ans;

}



Matrix quick(Matrix a, LL b)

{

    Matrix ans;

    for (LL i = 0; i < 2; i++)

    for (LL j = 0; j < 2; j++)

        ans.m[i][j] = (i == j);

    Matrix c;

    c = Mul(a, a);

    while (b){

        if (b & 1) ans = Mul(a, ans);

        a = Mul(a, a);

        b >>= 1;

    }

    return ans;

}



Matrix init()

{

    Matrix ans;

    ans.m[0][0] = 3; ans.m[0][1] = -1;

    ans.m[1][0] = 1; ans.m[1][1] = 0;

    return ans;

}

int main()

{

    LL T;

    LL n;

    cin >> T;

    while (T--){

        cin >> n >> mod;

        if (n == 0){

            cout << 0 << endl; continue;

        }

        if (n == 1){

            cout << 1 << endl; continue;

        }

        Matrix ans = init();

        ans = quick(ans, n - 1);

        cout << ans.m[0][0] << endl;

    }

    return 0;

}