HDU 4686 Arc of Dream(快速幂矩阵)

题目链接

再水一发，构造啊，初始化啊。。。wa很多次啊。。

#include <cstring>

#include <cstdio>

#include <string>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

#define MOD 1000000007

#define LL long long

LL mat[5][5],p[5][5];

LL a1,b1,a0,b0;

int qmod(LL n)

{

    LL c[5][5];

    int i,j,k;

    LL ans = 0;

    while(n)

    {

        if(n&1)

        {

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

            for(i = 0; i < 5; i ++)

            {

                for(j = 0; j < 5; j ++)

                {

                    for(k = 0; k < 5; 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 < 5; i ++)

        {

            for(j = 0; j < 5; j ++)

            {

                for(k = 0; k < 5; k ++)

                {

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

                    c[i][j] %= MOD;

                }

            }

        }

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

        n >>= 1;

    }

    ans = ((mat[0][0]*a0)%MOD*b0)%MOD;

    ans = (ans + (mat[0][1]*a1)%MOD*b1)%MOD;

    ans = (ans +  (a1*mat[0][2])%MOD)%MOD;

    ans = (ans + (b1*mat[0][3])%MOD)%MOD;

    ans = (ans + mat[0][4])%MOD;

    return ans;

}

int main()

{

    LL n,ax,ay,bx,by;

    int i,j;

    while(cin>>n)

    {

        cin>>a0>>ax>>ay;

        cin>>b0>>bx>>by;

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

        a0 %= MOD;

        ax %= MOD;

        ay %= MOD;

        b0 %= MOD;

        bx %= MOD;

        by %= MOD;

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

        p[1][1] = (ax*bx)%MOD;

        p[1][2] = (ax*by)%MOD;

        p[1][3] = (ay*bx)%MOD;

        p[1][4] = (ay*by)%MOD;

        p[2][2] = ax;

        p[2][4] = ay;

        p[3][3] = bx;

        p[3][4] = by;

        p[4][4] = 1;

        for(i = 0; i < 5; i ++)

        {

            for(j = 0; j < 5; j ++)

                mat[i][j] = i == j ? 1:0;

        }

        a1 = (a0*ax + ay)%MOD;

        b1 = (b0*bx + by)%MOD;

        if(n == 0)

            printf("0\n");

        else

            printf("%d\n",qmod(n-1));

    }

    return 0;

}

HDU 4686 Arc of Dream(快速幂矩阵),布布扣,bubuko.com