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)