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：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <math.h>
 5 #include <stdlib.h>
 6 #define mod 1000000007
 7 #define ll long long int
 8 using namespace std;
 9 int N;
10 struct matrix
11 {
12        ll a[5][5];
13 }origin,res;
14 matrix multiply(matrix x,matrix y)
15 {
16        matrix temp;
17        memset(temp.a,0,sizeof(temp.a));
18        for(int i=0;i<N;i++)
19        {
20                for(int j=0;j<N;j++)
21                {
22                        for(int k=0;k<N;k++)
23                        {
24                                temp.a[i][j]+=x.a[i][k]*y.a[k][j]%mod;
25                                temp.a[i][j]%=mod;
26                        }
27                }
28        }
29        return temp;
30 }
31 void calc(ll n)
32 {
33      memset(res.a,0,sizeof(res.a));
34      for(int i=0;i<N;i++)
35      res.a[i][i]=1;
36      while(n)
37      {
38              if(n&1)
39                     res=multiply(res,origin);
40              n>>=1;
41              origin=multiply(origin,origin);
42      }
43 }
44 int main()
45 {
46     N=5;
47     ll n;
48     ll a0,ax,ay,b0,bx,by;
49     while(cin>>n>>a0>>ax>>ay>>b0>>bx>>by){
50     memset(origin.a,0,sizeof(origin.a));
51     origin.a[0][0]=1;origin.a[0][1]=ax*bx%mod;
52     origin.a[0][2]=ay*bx%mod;origin.a[0][3]=ax*by%mod;
53     origin.a[0][4]=ay*by%mod;
54     origin.a[1][1]=ax*bx%mod;origin.a[1][2]=ay*bx%mod;
55     origin.a[1][3]=ax*by%mod;origin.a[1][4]=ay*by%mod;
56     origin.a[2][2]=bx%mod;origin.a[2][4]=by%mod;
57     origin.a[3][3]=ax%mod;origin.a[3][4]=ay%mod;
58     origin.a[4][4]=1;
59     if(n){
60     calc(n-1);
61     ll  sum=0;
62     sum+=res.a[0][0]*a0%mod*b0%mod;
63     sum+=res.a[0][1]*a0%mod*b0%mod;
64     sum%=mod;
65     sum+=res.a[0][2]*b0%mod;
66     sum%=mod;
67     sum+=res.a[0][3]*a0%mod;
68     sum%=mod;
69     sum+=res.a[0][4];
70     sum%=mod;
71     cout<<sum<<endl;
72     }
73     else cout<<0<<endl;
74     }
75 }

View Code

posted on 2013-08-21 17:26 ERKE 阅读(302) 评论(0) 编辑收藏举报