HDU - 3306 Another kind of Fibonacci 矩阵快速幂

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题目大意：A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).
And we want to Calculate S(N) , S(N) = A(0) 2 +A(1) 2+……+A(n) 2.

解题思路：将An^2化开，得x * x * A(n-1) * A(n-1) + y * y * A(n-2) * A(n-2) + 2 * x * y * A(n-1) * A(n-2)，由这个公式就可以得到矩阵了
这里写图片描述

#include<cstdio>typedef long long ll;const int N = 4;const ll mod = 10007;struct Matrix{    ll mat[N][N];}A, B, tmp;ll n, X, Y;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[1][2] = 1;    A.mat[1][0] = A.mat[1][1] = (X * X) % mod;    A.mat[2][0] = A.mat[2][1] = (Y * Y) % mod;    A.mat[3][0] = A.mat[3][1] = (2 * X * Y) % mod;    A.mat[1][3] = X % mod;    A.mat[3][3] = Y % mod;}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] += (x.mat[i][k] * y.mat[k][j] ) % 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%I64d%I64d", &n, &X, &Y) != EOF) {        if(n == 0) {            printf("1\n");            continue;        }        else if(n == 1) {            printf("2\n");            continue;        }        init();        n--;        solve();        printf("%I64d\n", (2 * B.mat[0][0] + B.mat[1][0] + B.mat[2][0] + B.mat[3][0] ) % mod);    }    return 0;}

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