HDU5950（矩阵快速幂）

来源：互联网发布：如何设置子域名编辑：程序博客网时间：2024/06/04 21:00

转载：自己看方便

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5950

题意：f(n) = f(n-1) + 2*f(n-2) + n^4，f(1) = a , f(2) = b，求f(n)

思路：对矩阵快速幂的了解仅仅停留在fib上，重现赛自己随便乱推还一直算错，快两个小时才a还wa了好几次....

主要就是构造矩阵：(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1

|1 2 1 4 6 4 1| | f(n+1) | | f(n+2) |

|1 0 0 0 0 0 0| | f(n) | | f(n+1) |

|0 0 1 4 6 4 1| | (n+1)^4 | | (n+2)^4 |

|0 0 0 1 3 3 1| * | (n+1)^3 | = | (n+2)^3 |

|0 0 0 0 1 2 1| | (n+1)^2 | | (n+2)^2 |

|0 0 0 0 0 1 1| | n+1 | | n+2 |

|0 0 0 0 0 0 1| | 1 | | 1 |

#include<cstdio>using namespace std;typedef long long ll;const ll mod = 2147493647;ll n,a,b;struct Matrix{    ll m[7][7];    void init1()    {        m[0][0] = b,m[0][1] = 0,m[0][2] = 0,m[0][3] = 0,m[0][4] = 0,m[0][5] = 0,m[0][6] = 0;        m[1][0] = a,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0;        m[2][0] = 16,m[2][1] = 0,m[2][2] = 0,m[2][3] = 0,m[2][4] = 0,m[2][5] = 0,m[2][6] = 0;        m[3][0] = 8,m[3][1] = 0,m[3][2] = 0,m[3][3] = 0,m[3][4] = 0,m[3][5] = 0,m[3][6] = 0;        m[4][0] = 4,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 0,m[4][5] = 0,m[4][6] = 0;        m[5][0] = 2,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 0,m[5][6] = 0;        m[6][0] = 1,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 0;    }    void init2()    {        m[0][0] = 1,m[0][1] = 2,m[0][2] = 1,m[0][3] = 4,m[0][4] = 6,m[0][5] = 4,m[0][6] = 1;        m[1][0] = 1,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0;        m[2][0] = 0,m[2][1] = 0,m[2][2] = 1,m[2][3] = 4,m[2][4] = 6,m[2][5] = 4,m[2][6] = 1;        m[3][0] = 0,m[3][1] = 0,m[3][2] = 0,m[3][3] = 1,m[3][4] = 3,m[3][5] = 3,m[3][6] = 1;        m[4][0] = 0,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 1,m[4][5] = 2,m[4][6] = 1;        m[5][0] = 0,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 1,m[5][6] = 1;        m[6][0] = 0,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 1;    }    Matrix operator * (Matrix t)    {        Matrix res;        for (int i = 0; i < 7; i++)        {            for (int j = 0; j < 7; j++)            {                res.m[i][j] = 0;                for (int k = 0;k < 7; k++)                    res.m[i][j] = (res.m[i][j] + (m[i][k] % mod) * (t.m[k][j] % mod) % mod) % mod;            }        }        return res;    }    Matrix operator ^ (int k)    {        Matrix res,s;        res.init2();        s.init2();        while(k)        {            if(k & 1)                res = res * s;            k >>= 1;            s = s * s;        }        return res;    }};int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%lld %lld %lld",&n,&a,&b);        if(n == 1)        {            printf("%lld\n",a % mod);            continue;        }        if(n == 2)        {            printf("%lld\n",b % mod);            continue;        }        Matrix ans,t;        ans.init1();        t.init2();        ans = (t^(n-3)) * ans;        printf("%lld\n",ans.m[0][0]);    }    return 0;}

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