Arc of Dream
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 932 Accepted Submission(s): 322
Problem Description
An Arc of Dream is a curve defined by following function:
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
解法:
因为a0=A0，ai=a（i-1）*Ax+Ay，
b0=B0，bi=b（i-1）*Bx+By；
所以
ai*bi=a（i-1）*b（i-1)*Ax*Bx+Ay*Bx*b(i-1)+Ax*By*a(i-1)+Ay*By;
Si=S(i-1)+ai*bi;
则构造矩阵为(k=n-1)
| Sn | 1 | Ax*Bx | Ay*BX | Ax*By | Ay*By | s(n-1) | ||||
| an*bn | 0 | Ax*Bx | Ay*BX | Ax*By | Ay*By | a(n-1)*b(n-1) | ||||
| bn | 0 | 0 | Bx | 0 | By | b(n-1) | ||||
| an | 0 | 0 | 0 | Ax | Ay | a(n-1) | ||||
| 1 | 0 | 0 | 0 | 0 | 1 | 1 |
所以构造矩阵为一个5*5的矩阵如上表的中间部分：
代码：
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <math.h> 5 #include <stdlib.h> 6 #define mod 1000000007 7 #define ll long long int 8 using namespace std; 9 int N; 10 struct matrix 11 { 12 ll a[5][5]; 13 }origin,res; 14 matrix multiply(matrix x,matrix y) 15 { 16 matrix temp; 17 memset(temp.a,0,sizeof(temp.a)); 18 for(int i=0;i<N;i++) 19 { 20 for(int j=0;j<N;j++) 21 { 22 for(int k=0;k<N;k++) 23 { 24 temp.a[i][j]+=x.a[i][k]*y.a[k][j]%mod; 25 temp.a[i][j]%=mod; 26 } 27 } 28 } 29 return temp; 30 } 31 void calc(ll n) 32 { 33 memset(res.a,0,sizeof(res.a)); 34 for(int i=0;i<N;i++) 35 res.a[i][i]=1; 36 while(n) 37 { 38 if(n&1) 39 res=multiply(res,origin); 40 n>>=1; 41 origin=multiply(origin,origin); 42 } 43 } 44 int main() 45 { 46 N=5; 47 ll n; 48 ll a0,ax,ay,b0,bx,by; 49 while(cin>>n>>a0>>ax>>ay>>b0>>bx>>by){ 50 memset(origin.a,0,sizeof(origin.a)); 51 origin.a[0][0]=1;origin.a[0][1]=ax*bx%mod; 52 origin.a[0][2]=ay*bx%mod;origin.a[0][3]=ax*by%mod; 53 origin.a[0][4]=ay*by%mod; 54 origin.a[1][1]=ax*bx%mod;origin.a[1][2]=ay*bx%mod; 55 origin.a[1][3]=ax*by%mod;origin.a[1][4]=ay*by%mod; 56 origin.a[2][2]=bx%mod;origin.a[2][4]=by%mod; 57 origin.a[3][3]=ax%mod;origin.a[3][4]=ay%mod; 58 origin.a[4][4]=1; 59 if(n){ 60 calc(n-1); 61 ll sum=0; 62 sum+=res.a[0][0]*a0%mod*b0%mod; 63 sum+=res.a[0][1]*a0%mod*b0%mod; 64 sum%=mod; 65 sum+=res.a[0][2]*b0%mod; 66 sum%=mod; 67 sum+=res.a[0][3]*a0%mod; 68 sum%=mod; 69 sum+=res.a[0][4]; 70 sum%=mod; 71 cout<<sum<<endl; 72 } 73 else cout<<0<<endl; 74 } 75 }