hdu4686 Arc of Dream 2013 Multi-University Training Contest 9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): 932    Accepted Submission(s): 322

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

解法:

因为a0=A0，ai=a（i-1）*Ax+Ay，

b0=B0，bi=b（i-1）*Bx+By；

所以

ai*bi=a（i-1）*b（i-1)*Ax*Bx+Ay*Bx*b(i-1)+Ax*By*a(i-1)+Ay*By;

Si=S(i-1)+ai*bi;

则构造矩阵为(k=n-1)

Sn			1	Ax*Bx	Ay*BX	Ax*By	Ay*By			s(n-1)
an*bn			0	Ax*Bx	Ay*BX	Ax*By	Ay*By			a(n-1)*b(n-1)
bn			0	0	Bx	0	By			b(n-1)
an			0	0	0	Ax	Ay			a(n-1)
1			0	0	0	0	1			1

所以构造矩阵为一个5*5的矩阵如上表的中间部分：

代码：

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <stdlib.h>
 #define mod 1000000007
 #define ll long long int
 using namespace std;
 int N;
 struct matrix
 {
        ll a[][];
 }origin,res;
 matrix multiply(matrix x,matrix y)
 {
        matrix temp;
        memset(temp.a,,sizeof(temp.a));
        for(int i=;i<N;i++)
        {
                for(int j=;j<N;j++)
                {
                        for(int k=;k<N;k++)
                        {
                                temp.a[i][j]+=x.a[i][k]*y.a[k][j]%mod;
                                temp.a[i][j]%=mod;
                        }
                }
        }
        return temp;
 }
 void calc(ll n)
 {
      memset(res.a,,sizeof(res.a));
      for(int i=;i<N;i++)
      res.a[i][i]=;
      while(n)
      {
              if(n&)
                     res=multiply(res,origin);
              n>>=;
              origin=multiply(origin,origin);
      }
 }
 int main()
 {
     N=;
     ll n;
     ll a0,ax,ay,b0,bx,by;
     while(cin>>n>>a0>>ax>>ay>>b0>>bx>>by){
     memset(origin.a,,sizeof(origin.a));
     origin.a[][]=;origin.a[][]=ax*bx%mod;
     origin.a[][]=ay*bx%mod;origin.a[][]=ax*by%mod;
     origin.a[][]=ay*by%mod;
     origin.a[][]=ax*bx%mod;origin.a[][]=ay*bx%mod;
     origin.a[][]=ax*by%mod;origin.a[][]=ay*by%mod;
     origin.a[][]=bx%mod;origin.a[][]=by%mod;
     origin.a[][]=ax%mod;origin.a[][]=ay%mod;
     origin.a[][]=;
     if(n){
     calc(n-);
     ll  sum=;
     sum+=res.a[][]*a0%mod*b0%mod;
     sum+=res.a[][]*a0%mod*b0%mod;
     sum%=mod;
     sum+=res.a[][]*b0%mod;
     sum%=mod;
     sum+=res.a[][]*a0%mod;
     sum%=mod;
     sum+=res.a[][];
     sum%=mod;
     cout<<sum<<endl;
     }
     else cout<<<<endl;
     }
 }