HDU4686 Arc of Dream 矩阵快速幂

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4246    Accepted Submission(s): 1332

Problem Description

An Arc of Dream is a curve defined by following function:

where
a₀ = A0
a_i = a_i-1*AX+AY
b₀ = B0
b_i = b_i-1*BX+BY
What is the value of AoD(N) modulo 1,000,000,007?

 

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

 

题意：已知：

a₀ = A0 b₀ = B0

a_i = a_i-1*AX+AY
b_i = b_i-1*BX+BY

Sn = a0b0+a1*b1+...+a(n-1)*b(n-1);

求 Sn

 

题解：矩阵快速幂求n项和

a[i]*b[i] = (a[i-1]*Ax+Ay)(b[i-1]*Bx+By)
= Ax*Bx*a[i-1]*b[i-1]+Ay*Bx*b[i-1]+Ax*By*a[i-1]+Ay*By
s
[s[n-1],a[n-2]*b[n-2], b[n-2], a[n-2], 1]
A
[1 ,0 ,0 ,0 ,0]
[Ax*Bx ,Ax*Bx ,0 ,0 ,0]
[Ay*Bx ,Ay*Bx ,Bx ,0 ,0]
[Ax*By ,Ax*By ,0 ,Ax ,0]
[Ay*By ,Ay*By ,By ,Ay ,1]//n-1
s
[s[n], a[n-1]*b[n-1], b[n-1], a[n-1], 1]

 

#include<bits/stdc++.h>
#define N 5
#define mes(x) memset(x, 0, sizeof(x));
#define ll long long
const ll mod = 1e9+7;
const int MAX = 0x7ffffff;
using namespace std;
struct mat {
    ll a[N][N];  
    mat() {  
        memset(a, 0, sizeof(a));  
    } 
    mat operator * (mat b) {  
        mat c;  
        for (int i = 0; i < N; i++)  
            for (int j = 0; j < N; j++)  
                for (int k = 0; k < N; k++)  
                    c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j]) % mod;  
        return c;  
    }
};
mat f(mat b, ll m) {
    mat c;
    for (int i = 0; i < N; i++)  
        c.a[i][i] = 1;
    while (m) {
        if (m & 1)
            c = c * b;  
        b = b * b;  
        m >>= 1;  
    }
    return c;  
}
int main()
{
    ll n, A0,Ax, Ay, Bx,By,B0;
    mat A, s;
    while(~scanf("%lld", &n)){
        scanf("%lld%lld%lld%lld%lld%lld", &A0, &Ax, &Ay, &B0, &Bx, &By);
        if(n == 0){
            printf("0
");
            continue;
        }
        mes(A.a);
        mes(s.a);
        s.a[0][0] = s.a[0][1] = (A0%mod*B0%mod)%mod;
        s.a[0][2] = B0%mod;
        s.a[0][3] = A0%mod;
        s.a[0][4] = 1;
        A.a[0][0] = 1; 
        A.a[1][1] = A.a[1][0] = (Ax%mod*Bx%mod)%mod;A.a[2][2] = Bx%mod;
        A.a[2][1] = A.a[2][0] = (Ay%mod*Bx%mod)%mod;A.a[3][3] = Ax%mod;
        A.a[3][1] = A.a[3][0] = (Ax%mod*By%mod)%mod;
        A.a[4][0] = A.a[4][1] = (Ay%mod*By%mod)%mod;
        A.a[4][2] = By;
        A.a[4][3] = Ay;
        A.a[4][4] = 1;
        A = f(A,n-1);
        s = s*A;
        printf("%lld
", (mod+s.a[0][0])%mod);
    } 
}