hdu----(4686)Arc of Dream(矩阵快速幂)

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2010    Accepted Submission(s): 643

Problem Description

An Arc of Dream is a curve defined by following function:

where
a₀ = A0
a_i = a_i-1*AX+AY
b₀ = B0
b_i = b_i-1*BX+BY
What is the value of AoD(N) modulo 1,000,000,007?

Input

There are multiple test cases. Process to the End of File.
Each test case contains 7 nonnegative integers as follows:
N
A0 AX AY
B0 BX BY
N is no more than 10¹⁸, and all the other integers are no more than 2×10⁹.

Output

For each test case, output AoD(N) modulo 1,000,000,007.

Sample Input

1
1 2 3
4 5 6
2
1 2 3
4 5 6
3
1 2 3
4 5 6

Sample Output

4
134
1902

Author

Zejun Wu (watashi)

Source

2013 Multi-University Training Contest 9

这道题的分析，其实应该这样分析：

  我们看到这么个式子    然后我们知道ai=ai-1*AX+AY   ;     bi=bi-1*BX+BY;

   ai*bi =AXBX*ai-1*bi-1+AXBY*ai-1+BXAY*bi-1+AY*BY;

   对于这样一个式子，我们不妨构造这样一个矩阵......

如：

  |ai*bi|     |AX*BX , AXBY , BXAY , AYBY, 0  |^n-1   | ai-1*ai-1 |

  |  ai  |     |  0       , AX    , 0        , AY   , 0  |           |  ai-1       |

  |  bi  |  = |  0       , 0     ,   BX    , BY    , 0  |   *      |  bi-1       |

  |   1  |     |  0       , 0     ,  0       ,    1   , 0  |           |     1        |

  |  sn  |     | AXBX  , AXBY,  BXAY,  AYBY, 1 |            |     sn-1   |

然后就是快速矩阵...

 #define LOCAL
 #include<cstdio>
 #include<cstring>
 #define LL __int64
 using namespace std;

 const LL mod=;

 struct node
 {
    LL mat[][];
    void init(int v){
         for(int i=;i<;i++){
          for(int j=;j<;j++)
          if(i==j)
             mat[i][j]=v;
          else
             mat[i][j]=;
      }
    }
 };

 LL AO,BO,AX,AY,BX,BY,n;
 node ans,cc;

 void init(node &a)
 {
   a.mat[][]=a.mat[][]=(AX*BX)%mod;
   a.mat[][]=a.mat[][]=(AX*BY)%mod;
   a.mat[][]=a.mat[][]=(BX*AY)%mod;
   a.mat[][]=a.mat[][]=(AY*BY)%mod;
   a.mat[][]=a.mat[][]=a.mat[][]=a.mat[][]=;
   a.mat[][]=a.mat[][]=a.mat[][]=;
   a.mat[][]=a.mat[][]=a.mat[][]=a.mat[][]=;
   a.mat[][]=a.mat[][]=;
   a.mat[][]=AX;
   a.mat[][]=AY;
   a.mat[][]=BX;
   a.mat[][]=BY;
 }

 void Matrix(node &a, node &b)
 {
   node c;
   c.init();
   for(int i=;i<;i++)
   {
       for(int j=;j<;j++)
       {
       for(int k=;k<;k++)
       {
         c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod;
       }
     }
   }

  for(int i=;i<;i++)
  {
   for(int j=;j<;j++)
   {
     a.mat[i][j]=c.mat[i][j];
   }
  }
 }

 void pow(node &a,LL w )
 {
    while(w>)
    {
          if(w&) Matrix(ans,a);
          w>>=1L;
          if(w==)break;
          Matrix(a,a);
    }
 }

 int main()
 {
    LL sab;
    #ifdef LOCAL
    freopen("test.in","r",stdin);
    #endif
   while(scanf("%I64d",&n)!=EOF)
   {
       scanf("%I64d%I64d%I64d",&AO,&AX,&AY);
       scanf("%I64d%I64d%I64d",&BO,&BX,&BY);
     if(n==){

       printf("0\n");
       continue;
     }
     AO%=mod;
     BO%=mod;
     AX%=mod;
     AY%=mod;
     BX%=mod;
     BY%=mod;
       ans.init();
       init(cc);
       pow(cc,n-);

       sab=(AO*BO)%mod;
     LL res=(ans.mat[][]*sab)%mod+(ans.mat[][]*AO)%mod+(ans.mat[][]*BO)%mod+ans.mat[][]%mod+(AO*BO)%mod;
       printf("%I64d\n",res%mod);
   }
  return ;
 }