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-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)

 

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 ");
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[5][5];
};
Matrix Mul(Matrix a,Matrix b){
   Matrix res;
   for(ll i=0;i<5;i++){
      for(ll j=0;j<5;j++){
         res.mp[i][j]=0;
         for(ll k=0;k<5;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,0,sizeof(res.mp));
   for(ll i=0;i<5;i++){
      res.mp[i][i]=1;
   }
   while(b){
      if(b&1){
         res=Mul(res,a);
      }
      a=Mul(a,a);
      b>>=1;
   }
   return res;
}
int main()
{
   while(scanf("%I64d",&n)==1){
      scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&A0,&Ax,&Ay,&B0,&Bx,&By);
      
      if(n==0){//这个判断条件很重要，没有就会超时
         printf("0
");
         continue;
      }
      
      
      ll a0=A0;
      ll b0=B0;

      Matrix tmp;
      memset(tmp.mp,0,sizeof(tmp.mp));
      tmp.mp[0][0]=1%MOD;
      tmp.mp[1][0]=Ax*By%MOD;
      tmp.mp[1][1]=Ax%MOD;
      tmp.mp[1][3]=Ax*By%MOD;
      tmp.mp[2][0]=Bx*Ay%MOD;
      tmp.mp[2][2]=Bx%MOD;
      tmp.mp[2][3]=Bx*Ay%MOD;
      tmp.mp[3][0]=Ax*Bx%MOD;
      tmp.mp[3][3]=Ax*Bx%MOD;
      tmp.mp[4][0]=Ay*By%MOD;
      tmp.mp[4][1]=Ay%MOD;
      tmp.mp[4][2]=By%MOD;
      tmp.mp[4][3]=Ay*By%MOD;
      tmp.mp[4][4]=1%MOD;

      Matrix cnt=fastm(tmp,n-1);

      Matrix g;
      memset(g.mp,0,sizeof(g.mp));
      g.mp[0][0]=a0*b0%MOD;
      g.mp[0][1]=a0%MOD;
      g.mp[0][2]=b0%MOD;
      g.mp[0][3]=a0*b0%MOD;
      g.mp[0][4]=1%MOD;
      Matrix ans=Mul(g,cnt);
      printf("%I64d
",ans.mp[0][0]);
   }
    return 0;
}