LightOJ 1070 - Algebraic Problem 推导+矩阵快速幂

http://www.lightoj.com/volume_showproblem.php?problem=1070

思路：\({(a+b)}^n =(a+b){(a+b)}^{n-1} \) \((ab)C_{n}^{r}a^{n-r}b{r} = C_{n+2}^{r}a^{n-r+2}b{r} - a^{n+2} - b^{n+2} \)

综上\( f(n) = (a+b)f(n-1)-(ab)f(n-2) \)

/** @Date    : 2016-12-19-19.53

  * @Author  : Lweleth (SoungEarlf@gmail.com)

  * @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;

typedef unsigned long long ull;

struct matrix

{

    ull mt[2][2];

    void init()

    {

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

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

                mt[i][j] = 0;

    }

    void cig()

    {

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

            mt[i][i] = 1;

    }

};

matrix mul(matrix a, matrix b)

{

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

            }

    return c;

}

matrix fpow(matrix a, LL n)

{

    matrix r;

    r.init();

    r.cig();

    while(n > 0)

    {

        if(n & 1)

            r = mul(r, a);

        a = mul(a, a);

        n >>= 1;

    }

    return r;

}

ull fun(LL p, LL q, LL n)

{

    if(n < 1)

    {

        return 2;

    }

    matrix base;

    base.mt[0][0] = p;

    base.mt[0][1] = -q;

    base.mt[1][0] = 1;

    base.mt[1][1] = 0;

    base = fpow(base, n - 1);

    ull ans = base.mt[0][0] * p + base.mt[0][1] * 2;

    return ans;

}

int main()

{

    int T;

    int cnt = 0;

    cin >> T;

    while(T--)

    {

        LL n, p , q;

        scanf("%lld%lld%lld", &p, &q, &n);

        ull ans = fun(p, q, n);

        printf("Case %d: %llu\n", ++cnt, ans);

    }

    return 0;

}

//f(n) = (a+b)*f(n-1) - (ab)*f(n-2)

巴特西

LightOJ 1070 - Algebraic Problem 推导+矩阵快速幂

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