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
a ₀ = A0
a _i = a _i-1*AX+AY
b ₀ = 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 10 ¹⁸, and all the other integers are no more than 2×10 ⁹.

 

 

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...

  1 #define LOCAL

  2 #include<cstdio>

  3 #include<cstring>

  4 #define LL __int64

  5 using namespace std;

  6 

  7 const LL mod=1000000007;

  8 

  9 struct node

 10 {

 11    LL mat[5][5];

 12    void init(int v){

 13         for(int i=0;i<5;i++){

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

 15          if(i==j)

 16             mat[i][j]=v;

 17          else

 18             mat[i][j]=0;

 19      }

 20    }

 21 };

 22 

 23 

 24 LL AO,BO,AX,AY,BX,BY,n;

 25 node ans,cc;

 26 

 27 void init(node &a)

 28 {

 29   a.mat[4][0]=a.mat[0][0]=(AX*BX)%mod;

 30   a.mat[4][1]=a.mat[0][1]=(AX*BY)%mod;

 31   a.mat[4][2]=a.mat[0][2]=(BX*AY)%mod;

 32   a.mat[4][3]=a.mat[0][3]=(AY*BY)%mod;

 33   a.mat[1][0]=a.mat[1][3]=a.mat[1][4]=a.mat[0][4]=0;

 34   a.mat[2][0]=a.mat[2][1]=a.mat[2][4]=0;

 35   a.mat[3][0]=a.mat[3][1]=a.mat[3][2]=a.mat[3][4]=0;

 36   a.mat[3][3]=a.mat[4][4]=1;

 37   a.mat[1][1]=AX;

 38   a.mat[1][3]=AY;

 39   a.mat[2][2]=BX;

 40   a.mat[2][3]=BY;

 41 }

 42 

 43 void Matrix(node &a, node &b)

 44 {

 45   node c;

 46   c.init(0);

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

 48   {

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

 50       {

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

 52       {

 53         c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod;

 54       }

 55     }

 56   }

 57 

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

 59  {

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

 61   {

 62     a.mat[i][j]=c.mat[i][j];

 63   }

 64  }

 65 }

 66 

 67 void pow(node &a,LL w )

 68 {

 69    while(w>0)

 70    {

 71          if(w&1) Matrix(ans,a);

 72          w>>=1L;

 73          if(w==0)break;

 74          Matrix(a,a);

 75    }

 76 }

 77 

 78 int main()

 79 {

 80    LL sab;

 81    #ifdef LOCAL

 82    freopen("test.in","r",stdin);

 83    #endif

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

 85   {

 86       scanf("%I64d%I64d%I64d",&AO,&AX,&AY);

 87       scanf("%I64d%I64d%I64d",&BO,&BX,&BY);

 88     if(n==0){

 89 

 90       printf("0\n");

 91       continue;

 92     }

 93     AO%=mod;

 94     BO%=mod;

 95     AX%=mod;

 96     AY%=mod;

 97     BX%=mod;

 98     BY%=mod;

 99       ans.init(1);

100       init(cc);

101       pow(cc,n-1);

102 

103       sab=(AO*BO)%mod;

104     LL res=(ans.mat[4][0]*sab)%mod+(ans.mat[4][1]*AO)%mod+(ans.mat[4][2]*BO)%mod+ans.mat[4][3]%mod+(AO*BO)%mod;

105       printf("%I64d\n",res%mod);

106   }

107  return 0;

108 }

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