LightOJ 1096 - nth Term 矩阵快速幂

http://www.lightoj.com/volume_showproblem.php?problem=1096

 

题意：\(f(n)  = a * f(n-1) + b * f(n-3) + c, if(n > 2) f(n)= 0, if(n ≤ 2) \)

思路：给出了递推式，构造下4X4矩阵就好。

 

 

/** @Date    : 2016-12-19-18.55
  * @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;
const LL mod = 1e4 + 7;

struct matrix
{
    LL mt[4][4];
    void init()
    {
        for(int i = 0; i < 4; i++)
            for(int j = 0; j < 4; j ++)
                mt[i][j] = 0;
    }
    void cig()
    {
        for(int i = 0; i < 4; i++)
            mt[i][i] = 1;
    }
};

matrix mul(matrix a, matrix b)
{
    matrix c;
    c.init();
    for(int i = 0; i < 4; i++)
        for(int j = 0; j < 4; j++)
            for(int k = 0; k < 4; k++)
            {
                c.mt[i][j] += a.mt[i][k] * b.mt[k][j];
                c.mt[i][j] %= mod;
            }
    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;
}

LL fun(LL a, LL b, LL c, LL n)
{
    if(n < 3)
    {
        return 0;
    }
    matrix base;
    base.init();
    base.mt[0][0] = a;
    base.mt[0][2] = b;
    base.mt[0][3] = c;
    base.mt[1][0] = 1;
    base.mt[2][1] = 1;
    base.mt[3][3] = 1;
    base = fpow(base, n - 3);
    LL ans = (base.mt[0][0] * c + base.mt[0][3]) % mod;
    return ans;
}

int main()
{
    int T;
    cin >> T;
    int cnt = 0;
    while(T--)
    {
        LL n, a, b, c;
        scanf("%lld%lld%lld%lld", &n, &a, &b, &c);
        LL ans = fun(a, b, c, n);
        printf("Case %d: %lld\n", ++cnt, ans);
    }
    return 0;
}