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。。。

  1 //STATUS:C++_AC_1296MS_232KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=10000010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=1000000007,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 LL n,a0,ax,ay,b0,bx,by;
 59 
 60 const int size=7;
 61 
 62 struct Matrix{
 63     LL ma[size][size];
 64     Matrix friend operator * (const Matrix a,const Matrix b){
 65         Matrix ret;
 66         mem(ret.ma,0);
 67         int i,j,k;
 68         for(i=0;i<size;i++)
 69             for(j=0;j<size;j++)
 70                 for(k=0;k<size;k++)
 71                     ret.ma[i][j]=(ret.ma[i][j]+a.ma[i][k]*b.ma[k][j]%MOD)%MOD;
 72         return ret;
 73     }
 74 }A;
 75 
 76 Matrix mutilpow(LL k)
 77 {
 78     int i;
 79     Matrix ret;
 80     mem(ret.ma,0);
 81     for(i=0;i<size;i++)
 82         ret.ma[i][i]=1;
 83     for(;k;k>>=(1LL)){
 84         if(k&(1LL))ret=ret*A;
 85         A=A*A;
 86     }
 87     return ret;
 88 }
 89 
 90 int main(){
 91  //   freopen("in.txt","r",stdin);
 92     int i,j;
 93     LL ans;
 94     LL B[size];
 95     Matrix F;
 96     while(~scanf("%I64d",&n))
 97     {
 98         scanf("%I64d%I64d%I64d",&a0,&ax,&ay);
 99         scanf("%I64d%I64d%I64d",&b0,&bx,&by);
100         if(n==0){printf("0\n");continue;}
101 
102         a0%=MOD;ax%=MOD;ay%=MOD;
103         b0%=MOD;bx%=MOD;by%=MOD;
104         mem(A.ma,0);
105         A.ma[0][0]=A.ma[0][5]=1;
106         A.ma[1][1]=ax;A.ma[1][2]=ay;
107         A.ma[2][2]=1;
108         A.ma[3][3]=bx;A.ma[3][4]=by;
109         A.ma[4][4]=1;
110         A.ma[5][1]=ax*by%MOD;A.ma[5][3]=bx*ay%MOD;A.ma[5][5]=ax*bx%MOD;A.ma[5][6]=ay*by%MOD;
111         A.ma[6][6]=1;
112         F=mutilpow(n-1);
113         B[0]=a0*b0%MOD;
114         B[1]=(a0*ax%MOD+ay)%MOD;B[2]=1;
115         B[3]=(b0*bx%MOD+by)%MOD;B[4]=1;
116         B[5]=B[1]*B[3]%MOD;B[6]=1;
117         ans=0;
118         for(i=0;i<size;i++){
119             ans=(ans+F.ma[0][i]*B[i]%MOD)%MOD;
120         }
121 
122         printf("%I64d\n",ans);
123     }
124     return 0;
125 }

来源：https://www.cnblogs.com/zhsl/p/3271284.html

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矩阵乘法