hdu----(4686)Arc of Dream(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): 2010 Accepted Submission(s): 643

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

Sample Output

4
134
1902

Author

Zejun Wu (watashi)

Source

2013 Multi-University Training Contest 9

这道题的分析，其实应该这样分析：

我们看到这么个式子

然后我们知道ai=ai-1*AX+AY ; bi=bi-1*BX+BY;

ai*bi =AXBX*ai-1*bi-1+AXBY*ai-1+BXAY*bi-1+AY*BY;

对于这样一个式子，我们不妨构造这样一个矩阵......

如：

| ai | | 0 , AX , 0 , AY , 0 | | ai-1 |

| bi | = | 0 , 0 , BX , BY , 0 | * | bi-1 |

| 1 | | 0 , 0 , 0 , 1 , 0 | | 1 |

| sn | | AXBX , AXBY, BXAY, AYBY, 1 | | sn-1 |

然后就是快速矩阵...

  1 #define LOCAL
  2 #include<cstdio>
  3 #include<cstring>
  4 #define LL __int64
  5 using namespace std;
  6 
  7 const LL mod=1000000007;
  8 
  9 struct node
 10 {
 11    LL mat[5][5];
 12    void init(int v){
 13         for(int i=0;i<5;i++){
 14          for(int j=0;j<5;j++)
 15          if(i==j)
 16             mat[i][j]=v;
 17          else
 18             mat[i][j]=0;
 19      }
 20    }
 21 };
 22 
 23 
 24 LL AO,BO,AX,AY,BX,BY,n;
 25 node ans,cc;
 26 
 27 void init(node &a)
 28 {
 29   a.mat[4][0]=a.mat[0][0]=(AX*BX)%mod;
 30   a.mat[4][1]=a.mat[0][1]=(AX*BY)%mod;
 31   a.mat[4][2]=a.mat[0][2]=(BX*AY)%mod;
 32   a.mat[4][3]=a.mat[0][3]=(AY*BY)%mod;
 33   a.mat[1][0]=a.mat[1][3]=a.mat[1][4]=a.mat[0][4]=0;
 34   a.mat[2][0]=a.mat[2][1]=a.mat[2][4]=0;
 35   a.mat[3][0]=a.mat[3][1]=a.mat[3][2]=a.mat[3][4]=0;
 36   a.mat[3][3]=a.mat[4][4]=1;
 37   a.mat[1][1]=AX;
 38   a.mat[1][3]=AY;
 39   a.mat[2][2]=BX;
 40   a.mat[2][3]=BY;
 41 }
 42 
 43 void Matrix(node &a, node &b)
 44 {
 45   node c;
 46   c.init(0);
 47   for(int i=0;i<5;i++)
 48   {
 49       for(int j=0;j<5;j++)
 50       {
 51       for(int k=0;k<5;k++)
 52       {
 53         c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod;
 54       }
 55     }
 56   }
 57 
 58  for(int i=0;i<5;i++)
 59  {
 60   for(int j=0;j<5;j++)
 61   {
 62     a.mat[i][j]=c.mat[i][j];
 63   }
 64  }
 65 }
 66 
 67 void pow(node &a,LL w )
 68 {
 69    while(w>0)
 70    {
 71          if(w&1) Matrix(ans,a);
 72          w>>=1L;
 73          if(w==0)break;
 74          Matrix(a,a);
 75    }
 76 }
 77 
 78 int main()
 79 {
 80    LL sab;
 81    #ifdef LOCAL
 82    freopen("test.in","r",stdin);
 83    #endif
 84   while(scanf("%I64d",&n)!=EOF)
 85   {
 86       scanf("%I64d%I64d%I64d",&AO,&AX,&AY);
 87       scanf("%I64d%I64d%I64d",&BO,&BX,&BY);
 88     if(n==0){
 89 
 90       printf("0
");
 91       continue;
 92     }
 93     AO%=mod;
 94     BO%=mod;
 95     AX%=mod;
 96     AY%=mod;
 97     BX%=mod;
 98     BY%=mod;
 99       ans.init(1);
100       init(cc);
101       pow(cc,n-1);
102 
103       sab=(AO*BO)%mod;
104     LL res=(ans.mat[4][0]*sab)%mod+(ans.mat[4][1]*AO)%mod+(ans.mat[4][2]*BO)%mod+ans.mat[4][3]%mod+(AO*BO)%mod;
105       printf("%I64d
",res%mod);
106   }
107  return 0;
108 }

View Code

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原文地址：https://www.cnblogs.com/gongxijun/p/3989185.html