HDU 3306 Another kind of Fibonacci(快速幂矩阵)

题目链接

构造矩阵 看的题解，剩下的就是模板了，好久没写过了，注意取余。

#include <cstring>

#include <cstdio>

#include <string>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

#define MOD 10007

#define LL __int64

LL p[4][4],mat[4][4];

int qmod(int n)

{

    LL c[4][4];

    int i,j,k;

    while(n)

    {

        if(n&1)

        {

            memset(c,0,sizeof(c));

            for(i = 0;i < 4;i ++)

            {

                for(j = 0;j < 4;j ++)

                {

                    for(k = 0;k < 4;k ++)

                    {

                        c[i][j] += mat[i][k]*p[k][j];

                        c[i][j] %= MOD;

                    }

                }

            }

            memcpy(mat,c,sizeof(mat));

        }

        memset(c,0,sizeof(c));

        for(i = 0;i < 4;i ++)

        {

            for(j = 0;j < 4;j ++)

            {

                for(k = 0;k < 4;k ++)

                {

                    c[i][j] += p[i][k]*p[k][j];

                    c[i][j] %= MOD;

                }

            }

        }

        memcpy(p,c,sizeof(p));

        n >>= 1;

    }

    return (mat[0][1] + mat[0][0] + mat[0][2] + mat[0][3])%MOD;

}

int main()

{

    LL x,y,n;

    int i,j;

    while(scanf("%I64d%I64d%I64d",&n,&x,&y)!=EOF)

    {

        memset(p,0,sizeof(p));

        p[0][0] = p[0][1] = 1;

        p[1][1] = x*x;

        p[1][2] = y*y;

        p[1][3] = 2*x*y;

        p[2][1] = 1;

        p[3][1] = x;

        p[3][3] = y;

        for(i = 0;i < 4;i ++)

        {

            for(j = 0;j < 4;j ++)

            {

                if(i == j)

                mat[i][j] = 1;

                else

                mat[i][j] = 0;

            }

        }

        printf("%d\n",qmod(n));

    }

    return 0;

}

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