hdu 4686 Arc of Dream（矩阵快速幂乘法）

Problem Description

An Arc of Dream is a curve defined by following function:

where
a0 = A0
ai = ai-*AX+AY
b0 = B0
bi = bi-*BX+BY
What is the value of AoD(N) modulo ,,,?

 

Input

There are multiple test cases. Process to the End of File.
Each test case contains  nonnegative integers as follows:
N
A0 AX AY
B0 BX BY
N is no more than , and all the other integers are no more than ×.

Output

For each test case, output AoD(N) modulo ,,,.

 

Sample Input

 

Sample Output

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

 

因为：a[i]*b[i]=(a[i-1]*AX+AY)*(b[i-1]*BX+BY)

      =(a[i-1]*b[i-1]*AX*BX+a[i-1]*AX*BY+b[i-1]*BX*AY+AY*BY)

构造矩阵：

                                                          |  1         0    0     0      0   |

                                                          |  AX*BY   AX   0     AX*BY    0   |

           {AoD(n-1),a[i-1],b[i-1],a[i-1]*b[i-1],1}*      |  BX*AY    0   BX    BX*AY    0   |      ={AoD(n),a[i],b[i],a[i]*b[i],1}

                                                          |  AX*BX    0   0      AX*BX   0   |

                                                          |  AY*BY   AY   BY    AY*BY    1   |

 

 

另外注意：

if(n==0){//这个判断条件很重要，没有就会超时
printf("0\n");
continue;
}

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 1000000
 #define inf 1e12
 ll n;
 ll A0,Ax,Ay,B0,Bx,By;
 struct Matrix{
    ll mp[][];
 };
 Matrix Mul(Matrix a,Matrix b){
    Matrix res;
    for(ll i=;i<;i++){
       for(ll j=;j<;j++){
          res.mp[i][j]=;
          for(ll k=;k<;k++){
             res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%MOD+MOD)%MOD;
          }
       }
    }
    return res;
 }
 Matrix fastm(Matrix a,ll b){
    Matrix res;
    memset(res.mp,,sizeof(res.mp));
    for(ll i=;i<;i++){
       res.mp[i][i]=;
    }
    while(b){
       if(b&){
          res=Mul(res,a);
       }
       a=Mul(a,a);
       b>>=;
    }
    return res;
 }
 int main()
 {
    while(scanf("%I64d",&n)==){
       scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&A0,&Ax,&Ay,&B0,&Bx,&By);

       if(n==){//这个判断条件很重要，没有就会超时
          printf("0\n");
          continue;
       }

       ll a0=A0;
       ll b0=B0;

       Matrix tmp;
       memset(tmp.mp,,sizeof(tmp.mp));
       tmp.mp[][]=%MOD;
       tmp.mp[][]=Ax*By%MOD;
       tmp.mp[][]=Ax%MOD;
       tmp.mp[][]=Ax*By%MOD;
       tmp.mp[][]=Bx*Ay%MOD;
       tmp.mp[][]=Bx%MOD;
       tmp.mp[][]=Bx*Ay%MOD;
       tmp.mp[][]=Ax*Bx%MOD;
       tmp.mp[][]=Ax*Bx%MOD;
       tmp.mp[][]=Ay*By%MOD;
       tmp.mp[][]=Ay%MOD;
       tmp.mp[][]=By%MOD;
       tmp.mp[][]=Ay*By%MOD;
       tmp.mp[][]=%MOD;

       Matrix cnt=fastm(tmp,n-);

       Matrix g;
       memset(g.mp,,sizeof(g.mp));
       g.mp[][]=a0*b0%MOD;
       g.mp[][]=a0%MOD;
       g.mp[][]=b0%MOD;
       g.mp[][]=a0*b0%MOD;
       g.mp[][]=%MOD;
       Matrix ans=Mul(g,cnt);
       printf("%I64d\n",ans.mp[][]);
    }
     return ;
 }