HDU-4686 Arc of Dream 构造矩阵
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：
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 }
