LightOJ 1065 - Number Sequence 矩阵快速幂水题

http://www.lightoj.com/volume_showproblem.php?problem=1065

题意：给出递推式f(0) = a, f(1) = b, f(n) = f(n - 1) +f(n - 2) 求f(n)

思路：给出了递推式就是水题。

/** @Date    : 2016-12-17-15.54

  * @Author  : Lweleth ([email protected])

  * @Link    : https://github.com/

  * @Version :

  */

#include<bits/stdc++.h>

#define LL long long

#define PII pair

#define MP(x, y) make_pair((x),(y))

#define fi first

#define se second

#define PB(x) push_back((x))

#define MMG(x) memset((x), -1,sizeof(x))

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;

const int INF = 0x3f3f3f3f;

const int N = 1e5+20;

const double eps = 1e-8;

struct matrix

{

    LL mat[2][2];

    void init()

    {

        mat[0][0] = mat[1][0] = mat[0][1] = mat[1][1] = 0;

    }

};

matrix mul(matrix a, matrix b, LL mod)

{

    matrix c;

    c.init();

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

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

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

            {

                c.mat[i][j] += a.mat[i][k] * b.mat[k][j];

                c.mat[i][j] %= mod;

            }

    return c;

}

matrix fpow(matrix x, LL n, LL mod)

{

    matrix r;

    r.init();

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

        r.mat[i][i] = 1;

    while(n > 0)

    {

        if(n & 1)

            r = mul(r, x, mod);

        x = mul(x, x, mod);

        n >>= 1;

    }

    return r;

}

LL Tis(LL x, LL a, LL b, LL mod)

{

    matrix t;

    t.init();

    t.mat[0][0] = 1;

    t.mat[0][1] = 1;

    t.mat[1][0] = 1;

    matrix s;

    s = fpow(t, x - 1, mod);

    LL ans = s.mat[0][0]*b + s.mat[1][0]*a;

    return ans % mod;

}

int main()

{

    int T;

    cin >> T;

    int cnt = 0;

    while(T--)

    {

        LL n, a, b, m;

        scanf("%lld%lld%lld%lld", &a, &b, &n, &m);

        LL x = 1;

        while(m--)

        {

            x*=10;

        }

        printf("Case %d: %lld\n",++cnt, Tis(n, a, b, x));

    }

    return 0;

}