spoj4166 Four colors

spoj4166 Four colors

矩阵乘法是显然的
把末2位的状态分成4类：
S₁：黑黑、红红、蓝蓝
S₂：黑红、黑蓝、红黑、红蓝、蓝黑、蓝红
S₃：白白、黑白、蓝白、红白
S₄：白黑、白红、白蓝

则有
S'₁=S₄
S'₂=S₂+2S₄
S'₃=S₁+S₂+S₃+S₄
S'₄=3S₃

得到矩阵A

0	0	1	0
0	1	1	0
0	0	1	3
1	2	1	0

直接做矩阵乘法居然超时，时限太BT了，所以我把要求的二进制数拆成三部分，预处理求出以下值

A ⁰,A ¹,...,A ¹⁰²³
A ⁰,A ¹⁰²⁴,...,A ^1047552
A ⁰,A ^1048576,...,A ^1072693248

然后对于每个询问，只用求三个矩阵的乘积即可

#include  < iostream >
#include  < cstdlib >
#include  < cstdio >

using   namespace  std;

const   int  mo = 1000000007 ;
const   int  c[ 4 ][ 4 ] = {{ 0 , 0 , 1 , 0 },
                   { 0 , 1 , 1 , 0 },
                   { 0 , 0 , 1 , 3 },
                   { 1 , 2 , 1 , 0 }};
const   int  e[ 4 ][ 4 ] = {{ 1 , 0 , 0 , 0 },
                   { 0 , 1 , 0 , 0 },
                   { 0 , 0 , 1 , 0 },
                   { 0 , 0 , 0 , 1 }};
const   int  s[ 4 ] = { 3 , 6 , 4 , 3 };

char  st[ 20 ];

int  f[ 32 ][ 4 ][ 4 ],res1[ 1024 ][ 4 ][ 4 ],res2[ 1024 ][ 4 ][ 4 ],res3[ 1024 ][ 4 ][ 4 ],a[ 4 ][ 4 ],b[ 4 ][ 4 ];

void  multi( int  a[ 4 ][ 4 ], int  b[ 4 ][ 4 ], int  c[ 4 ][ 4 ])
{
   int  i,j,k;
   long   long  tmp;
   for (i = 0 ;i < 4 ;i ++ )
     for (j = 0 ;j < 4 ;j ++ )
    {
      tmp = 0 ;
       for (k = 0 ;k < 4 ;k ++ )
        tmp += ( long   long )(a[i][k]) * b[k][j];
      c[i][j] = tmp % mo;
    }
}

int  getint( void )
{
  gets(st);
   int  res = 0 ;
   for ( int  i = 0 ;st[i] != ' \0 ' ;i ++ )
     if (st[i] >= ' 0 ' && st[i] <= ' 9 ' )
      res = res * 10 + st[i] - 48 ;
   return  res;
}

int  main( void )
{
   int  k,i,j,u,n,ans;
   long   long  tmp;
   for (i = 0 ;i < 4 ;i ++ )
     for (j = 0 ;j < 4 ;j ++ )
    {
      f[ 0 ][i][j] = c[i][j];
      res1[ 0 ][i][j] = res2[ 0 ][i][j] = res3[ 0 ][i][j] = e[i][j];
    }
   for (k = 1 ;k <= 20 ;k ++ )
    multi(f[k - 1 ],f[k - 1 ],f[k]);
   for (k = 1 ;k < 1024 ;k ++ )
    multi(res1[k - 1 ],f[ 0 ],res1[k]);
   for (k = 1 ;k < 1024 ;k ++ )
    multi(res2[k - 1 ],f[ 10 ],res2[k]);
   for (k = 1 ;k < 1024 ;k ++ )
    multi(res3[k - 1 ],f[ 20 ],res3[k]);
  u = getint();
   while (u -- )
  {
    n = getint();
     if  (n == 1 )
    {
      printf( " %d\n " , 4 );
       continue ;
    }
    n -= 2 ;
    multi(res1[n & 1023 ],res2[(n >> 10 ) & 1023 ],b);
    multi(b,res3[(n >> 20 ) & 1023 ],a);
    tmp = 0 ;
     for (i = 0 ;i < 4 ;i ++ )
       for (j = 0 ;j < 4 ;j ++ )
        tmp += ( long   long )(s[i]) * a[i][j];
    ans = tmp % mo;
    printf( " %d\n " ,ans);
  }
   return   0 ;
}