uva 10655

题目连接：uva 10655 - Contemplation! Algebra

题目大意：输入非负整数，p。q，n，求an+bn的值，当中a和b满足a+b=p,ab=q,注意a和b不一定是实数。

解题思路：定义f(n)=an+bn,则有f(n)∗(a+b)=(an+bn)∗(a+b)=an+1+abn+ban+bn+1=f(n+1)+abf(n−1), 所以f(n+1)=(a+b)f(n)−abf(n−1),用矩阵高速幂求解。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxsize = 100;
typedef long long ll;
typedef long long type;

struct Mat {
    int r, l;
    type arr[maxsize][maxsize];

    Mat (int r = 0, int l = 0) { 
        set(r, l);
        memset(arr, 0, sizeof(arr));
    }

    void set (int r, int l) {
        this->r = r;
        this->l = l;
    }

    Mat operator * (const Mat& u) {
        Mat ret(r, u.l);
        for (int k = 0; k < l; k++) {
            for (int i = 0; i < r; i++)
                for (int j = 0; j < u.l; j++)
                    ret.arr[i][j] = (ret.arr[i][j] + arr[i][k] * u.arr[k][j]);
        }
        return ret;
    }
};

void put (Mat x) {
    for (int i = 0; i < x.r; i++) {
        for (int j = 0; j < x.l; j++)
            printf("%lld ", x.arr[i][j]);
        printf("
");
    }
}

Mat pow_mat (Mat ans, Mat x, ll n) {

    while (n) {
        if (n&1)
            ans = x * ans;
        x = x * x;
        n >>= 1;
    }
    return ans;
}

int main () {
    ll p, q, n;
    while (scanf("%lld%lld%lld", &p, &q, &n) == 3 && p + q + n) {
        Mat x(2, 2);
        x.arr[0][1] = 1;
        x.arr[1][0] = -q;
        x.arr[1][1] = p;

        Mat ans(2, 1);
        ans.arr[0][0] = 2;
        ans.arr[1][0] = p;

        if (n > 1) {
            ans = pow_mat(ans, x, n-1);
            printf("%lld
", ans.arr[1][0]);
        } else
            printf("%lld
", ans.arr[n][0]);
    }
    return 0;
}