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  • LightOJ 1096

    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
    ", ++cnt, ans);
        }
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Yumesenya/p/6219604.html
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