HDU4686Arc of Dream 矩阵快速幂

   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?

InputThere 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.OutputFor 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

构造矩阵 过程爆longlong WA 了很多次，很蓝瘦
 1 #include <bits/stdc++.h>
 2 #define LL long long
 3 using namespace std;
 4 const int N = 5;
 5 int mod = 1e9+7;
 6 
 7 LL k;
 8 LL A0,Ax,Ay,B0,Bx,By;
 9 
10 struct Matrix
11 {
12     LL mat[N][N];
13     Matrix operator*(const Matrix &b)const
14     {
15         Matrix temp;
16         for(int i = 0; i < N; i++)
17         {
18             for(int j = 0; j < N; j++)
19             {
20                 temp.mat[i][j] = 0;
21                 for(int k = 0; k < N; k++)
22                 {
23                     temp.mat[i][j]+=mat[i][k]*b.mat[k][j]%mod;
24                     temp.mat[i][j]%=mod;
25                 }
26             }
27         }
28         return temp;
29     }
30 };
31 void Init(Matrix &m)
32 {
33     memset(m.mat,0,sizeof(m.mat));
34     m.mat[0][0]=Ax*Bx%mod;m.mat[0][1]=Ax*By%mod;m.mat[0][2]=Ay*Bx%mod;m.mat[0][3]=Ay*By%mod;
35     m.mat[1][1]=Ax%mod;m.mat[1][3]=Ay%mod;
36     m.mat[2][2]=Bx%mod;m.mat[2][3]=By%mod;
37     m.mat[3][3]=1;
38     m.mat[4][0]=1;m.mat[4][4]=1;
39 }
40 LL qpow(Matrix &m,LL k)
41 {
42     Matrix ans;
43     memset(ans.mat,0,sizeof(ans.mat));
44     for(int i = 0; i < N; i++)
45         ans.mat[i][i] = 1;
46     while(k)
47     {
48         if(k&1)ans = ans*m;
49         m = m*m;
50         k>>=1;
51     }
52     LL sum = 0;
53     sum = (sum+ans.mat[4][0] *A0%mod*B0%mod)%mod;
54     sum = (sum+ans.mat[4][1] *A0%mod)%mod;
55     sum = (sum+ans.mat[4][2] *B0%mod)%mod;
56     sum = (sum+ans.mat[4][3]%mod)%mod;
57     //sum = (sum+ans.mat[4][4] *0%mod)%mod;
58     return sum;
59 }
60 int main()
61 {
62     while(cin>>k)
63     {
64         cin>>A0>>Ax>>Ay>>B0>>Bx>>By;
65         //scanf("%d%d%d%d%d%d",&A0,&Ax,&Ay,&B0,&Bx,&By);
66         if(k==0){
67             printf("0\n");
68             continue;
69         }
70         Matrix m;
71         Init(m);
72         cout<<qpow(m,k)<<endl;;
73     }
74     return 0;
75 }

 

posted @ 2018-05-03 20:59  TTTCoder  阅读(218)  评论(0)  编辑  收藏  举报