HDU-4686 Arc of Dream 构造矩阵

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4686

　　因为a_i = a_i-1*AX+AY ，b_i = b_i-1*BX+BY ，那么ai*bi=AX*BX*A*a_i-1*b_i-1+AX*BY*a_i-1+BX*AY*b_i-1+AY*BYAY。令Sn为ai*bi前n项的和，Sn=Sn-1 + an*bn，因此我们可以构造一个如下的转移矩阵：

　　然后矩阵乘法优化就可以了。。。

　　注意此题n=0的情况！

　　其实矩阵大小只要5就可以了，那几个常数项可以合并到一列。。。

 //STATUS:C++_AC_1296MS_232KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 LL n,a0,ax,ay,b0,bx,by;

 const int size=;

 struct Matrix{
     LL ma[size][size];
     Matrix friend operator * (const Matrix a,const Matrix b){
         Matrix ret;
         mem(ret.ma,);
         int i,j,k;
         for(i=;i<size;i++)
             for(j=;j<size;j++)
                 for(k=;k<size;k++)
                     ret.ma[i][j]=(ret.ma[i][j]+a.ma[i][k]*b.ma[k][j]%MOD)%MOD;
         return ret;
     }
 }A;

 Matrix mutilpow(LL k)
 {
     int i;
     Matrix ret;
     mem(ret.ma,);
     for(i=;i<size;i++)
         ret.ma[i][i]=;
     for(;k;k>>=(1LL)){
         if(k&(1LL))ret=ret*A;
         A=A*A;
     }
     return ret;
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j;
     LL ans;
     LL B[size];
     Matrix F;
     while(~scanf("%I64d",&n))
     {
         scanf("%I64d%I64d%I64d",&a0,&ax,&ay);
         scanf("%I64d%I64d%I64d",&b0,&bx,&by);
         if(n==){printf("0\n");continue;}

         a0%=MOD;ax%=MOD;ay%=MOD;
         b0%=MOD;bx%=MOD;by%=MOD;
         mem(A.ma,);
         A.ma[][]=A.ma[][]=;
         A.ma[][]=ax;A.ma[][]=ay;
         A.ma[][]=;
         A.ma[][]=bx;A.ma[][]=by;
         A.ma[][]=;
         A.ma[][]=ax*by%MOD;A.ma[][]=bx*ay%MOD;A.ma[][]=ax*bx%MOD;A.ma[][]=ay*by%MOD;
         A.ma[][]=;
         F=mutilpow(n-);
         B[]=a0*b0%MOD;
         B[]=(a0*ax%MOD+ay)%MOD;B[]=;
         B[]=(b0*bx%MOD+by)%MOD;B[]=;
         B[]=B[]*B[]%MOD;B[]=;
         ans=;
         for(i=;i<size;i++){
             ans=(ans+F.ma[][i]*B[i]%MOD)%MOD;
         }

         printf("%I64d\n",ans);
     }
     return ;
 }