HDU-4686 Arc of Dream 构造矩阵

时间：2021-09-14 07:27:51

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

　　因为ai = ai-1*AX+AY ，bi = bi-1*BX+BY ，那么ai*bi=AX*BX*A*ai-1*bi-1+AX*BY*ai-1+BX*AY*bi-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：

HDU-4686 Arc of Dream 构造矩阵

　　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 ;

 }

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