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
ai = ai-1*AX+AY
b0 = B0
bi = bi-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

 

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013/8/20 12:21:51
  4 File Name     :F:\2013ACM练习\2013多校9\1001.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int MOD = 1e9+7;
 21 struct Matrix
 22 {
 23     int mat[5][5];
 24     void clear()
 25     {
 26         memset(mat,0,sizeof(mat));
 27     }
 28     void output()
 29     {
 30         for(int i = 0;i < 5;i++)
 31         {
 32             for(int j = 0;j < 5;j++)
 33                 printf("%d ",mat[i][j]);
 34             printf("\n");
 35         }
 36     }
 37     Matrix operator *(const Matrix &b)const
 38     {
 39         Matrix ret;
 40         for(int i = 0;i < 5;i++)
 41             for(int j = 0;j < 5;j++)
 42             {
 43                 ret.mat[i][j] = 0;
 44                 for(int k = 0;k < 5;k++)
 45                 {
 46                     long long tmp = (long long)mat[i][k]*b.mat[k][j]%MOD;
 47                     ret.mat[i][j] = (ret.mat[i][j]+tmp);
 48                     if(ret.mat[i][j]>MOD)
 49                         ret.mat[i][j] -= MOD;
 50                 }
 51             }
 52         return ret;
 53     }
 54 };
 55 Matrix pow_M(Matrix a,long long n)
 56 {
 57     Matrix ret;
 58     ret.clear();
 59     for(int i = 0;i < 5;i++)
 60         ret.mat[i][i] = 1;
 61     Matrix tmp = a;
 62     while(n)
 63     {
 64         if(n&1)ret = ret*tmp;
 65         tmp = tmp*tmp;
 66         n>>=1;
 67     }
 68     return ret;
 69 }
 70 int main()
 71 {
 72     //freopen("in.txt","r",stdin);
 73     //freopen("out.txt","w",stdout);
 74     long long n;
 75     int A0,AX,AY;
 76     int B0,BX,BY;
 77     while(scanf("%I64d",&n) == 1)
 78     {
 79         scanf("%d%d%d",&A0,&AX,&AY);
 80         scanf("%d%d%d",&B0,&BX,&BY);
 81         if(n == 0)
 82         {
 83             printf("0\n");
 84             continue;
 85         }
 86         Matrix a;
 87         a.clear();
 88         a.mat[0][0] = AX%MOD;
 89         a.mat[0][2] = (long long)AX*BY%MOD;
 90         a.mat[1][1] = BX%MOD;
 91         a.mat[1][2] = (long long)AY*BX%MOD;
 92         a.mat[2][2] = (long long)AX*BX%MOD;
 93         a.mat[3][3] = 1;
 94         a.mat[4][0] = AY%MOD;
 95         a.mat[4][1] = BY%MOD;
 96         a.mat[4][2] = (long long)AY*BY%MOD;
 97         a.mat[4][4] = 1;
 98         a.mat[0][3] = a.mat[0][2];
 99         a.mat[1][3] = a.mat[1][2];
100         a.mat[2][3] = a.mat[2][2];
101         a.mat[4][3] = a.mat[4][2];
102         //a.output();
103         a = pow_M(a,n-1);
104         //a.output();
105         long long t1 = (long long)A0*B0%MOD;
106         long long ans = t1*a.mat[2][3]%MOD + t1*a.mat[3][3]%MOD;
107         if(ans > MOD)ans -= MOD;
108         ans += (long long)A0*a.mat[0][3];
109         ans %= MOD;
110         ans += (long long)B0*a.mat[1][3];
111         ans %= MOD;
112         ans += (long long)a.mat[4][3];
113         ans %= MOD;
114         printf("%d\n",(int)ans);
115     }
116     return 0;
117 }

 

 

 

 

 

 

 

 

 

posted on 2013-08-20 19:28  kuangbin  阅读(1434)  评论(1)  编辑  收藏  举报

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