HDU 4686 Arc of Dream （2013多校9 1001 题，矩阵）

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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)

很明显是要构造矩阵，然后用矩阵快速幂求解。

|   AX   0   AXBY   AXBY  0  |

|   0   BX  AYBX    AYBX  0  |

{a[i-1]   b[i-1]   a[i-1]*b[i-1]  AoD[i-1]  1}* |   0   0   AXBX    AXBX   0  |  = {a[i]   b[i]   a[i]*b[i]  AoD[i]  1}

|    0   0     0          1     0    |

|  AY  BY   AYBY   AYBY   1   |

然后就可以搞了

注意n==0的时候，输出0

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/8/20 12:21:51
 File Name     :F:\2013ACM练习\2013多校9\1001.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 const int MOD = 1e9+;
 struct Matrix
 {
     int mat[][];
     void clear()
     {
         memset(mat,,sizeof(mat));
     }
     void output()
     {
         for(int i = ;i < ;i++)
         {
             for(int j = ;j < ;j++)
                 printf("%d ",mat[i][j]);
             printf("\n");
         }
     }
     Matrix operator *(const Matrix &b)const
     {
         Matrix ret;
         for(int i = ;i < ;i++)
             for(int j = ;j < ;j++)
             {
                 ret.mat[i][j] = ;
                 for(int k = ;k < ;k++)
                 {
                     long long tmp = (long long)mat[i][k]*b.mat[k][j]%MOD;
                     ret.mat[i][j] = (ret.mat[i][j]+tmp);
                     if(ret.mat[i][j]>MOD)
                         ret.mat[i][j] -= MOD;
                 }
             }
         return ret;
     }
 };
 Matrix pow_M(Matrix a,long long n)
 {
     Matrix ret;
     ret.clear();
     for(int i = ;i < ;i++)
         ret.mat[i][i] = ;
     Matrix tmp = a;
     while(n)
     {
         if(n&)ret = ret*tmp;
         tmp = tmp*tmp;
         n>>=;
     }
     return ret;
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     long long n;
     int A0,AX,AY;
     int B0,BX,BY;
     while(scanf("%I64d",&n) == )
     {
         scanf("%d%d%d",&A0,&AX,&AY);
         scanf("%d%d%d",&B0,&BX,&BY);
         if(n == )
         {
             printf("0\n");
             continue;
         }
         Matrix a;
         a.clear();
         a.mat[][] = AX%MOD;
         a.mat[][] = (long long)AX*BY%MOD;
         a.mat[][] = BX%MOD;
         a.mat[][] = (long long)AY*BX%MOD;
         a.mat[][] = (long long)AX*BX%MOD;
         a.mat[][] = ;
         a.mat[][] = AY%MOD;
         a.mat[][] = BY%MOD;
         a.mat[][] = (long long)AY*BY%MOD;
         a.mat[][] = ;
         a.mat[][] = a.mat[][];
         a.mat[][] = a.mat[][];
         a.mat[][] = a.mat[][];
         a.mat[][] = a.mat[][];
         //a.output();
         a = pow_M(a,n-);
         //a.output();
         long long t1 = (long long)A0*B0%MOD;
         long long ans = t1*a.mat[][]%MOD + t1*a.mat[][]%MOD;
         if(ans > MOD)ans -= MOD;
         ans += (long long)A0*a.mat[][];
         ans %= MOD;
         ans += (long long)B0*a.mat[][];
         ans %= MOD;
         ans += (long long)a.mat[][];
         ans %= MOD;
         printf("%d\n",(int)ans);
     }
     return ;
 }