HDU - 4686 Arc of Dream 矩阵快速幂

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

Aod(n) = a0 * b0 + a1 * b1 + a2 * b2 … a(n-1) * b(n-1)

where

a 0 = A0

a i = a i-1*AX+AY

b 0 = B0

b i = b i-1*BX+BY

What is the value of AoD(N) modulo 1,000,000,007?

解题思路：这一题的矩阵构造还是挺麻烦的

由递推公式的an*bn = (an-1*Ax+Ay)*(bn-1*Bx+By) => an-1*bn-1*Ax*Bx+an-1*Ax*By+bn-1*Ay*Bx+Ay*By




坑点：注意long long 可能会溢出，乘的时候注意

[code]#include<cstdio>
typedef long long ll;
const int N = 5;
const ll mod = 1e9 + 7;
ll n, A0, Ax, Ay, B0, Bx, By;

struct Matrix{
    ll mat

;
}A, B, tmp;

void init() {
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++) {
            A.mat[i][j] = B.mat[i][j] = 0;
            if(i == j)
                B.mat[i][j] = 1;
        }
    A.mat[0][0] = A.mat[4][4] = 1;
    A.mat[1][0] = A.mat[1][1] = (Ax * Bx) % mod;
    A.mat[2][0] = A.mat[2][1] = (Ax * By) % mod;
    A.mat[3][0] = A.mat[3][1] = (Ay * Bx) % mod;
    A.mat[4][0] = A.mat[4][1] = (Ay * By) % mod;
    A.mat[2][2] = Ax;
    A.mat[4][2] = Ay;
    A.mat[3][3] = Bx;
    A.mat[4][3] = By;
}

Matrix matMul(Matrix x, Matrix y) {
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++) {
            tmp.mat[i][j] = 0;
            for(int k = 0; k < N; k++)
                tmp.mat[i][j] = (tmp.mat[i][j] + (x.mat[i][k] * y.mat[k][j]) % mod) % mod; 
        }
    return tmp;
}

void solve() {
    while(n) {
        if(n & 1)
            B = matMul(B,A);
        A = matMul(A,A);
        n >>= 1;
    }
}

int main() {
    while(scanf("%I64d", &n) != EOF) {
        scanf("%I64d%I64d%I64d", &A0, &Ax, &Ay);
        scanf("%I64d%I64d%I64d", &B0, &Bx, &By);
        if(n == 0) {
            printf("0\n");
            continue;
        }
        if(n == 1) {
            printf("%I64d\n", (A0 * B0) % mod);
            continue;
        }
        n--;
        init();
        solve();
        ll ans = 0;
        ans += (((A0 * B0) % mod) * B.mat[0][0] % mod);
        ans += (((A0 * B0) % mod) * B.mat[1][0] % mod);
        ans += (A0 * B.mat[2][0]) % mod;
        ans += (B0 * B.mat[3][0]) % mod;
        ans += (B.mat[4][0]) % mod;
        printf("%I64d\n", ans % mod);
    }
    return 0;
}