Contemplation! Algebra 矩阵快速幂

Given the value of a+b and ab you will have to find the value of a n + b n Input The input file contains several lines of inputs. Each line except the last line contains 3 non-negative integers p, q and n. Here p denotes the value of a+b and q denotes the value of ab. Input is terminated by a line containing only two zeroes. This line should not be processed. Each number in the input file fits in a signed 32-bit integer. There will be no such input so that you have to find the value of 00 . Output For each line of input except the last one produce one line of output. This line contains the value of a n + b n. You can always assume that a n + b n fits in a signed 64-bit integer.

Sample Input

10 16 2 7 12 3 0 0

Sample Output

68 91

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<string>
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
#define MAXN 49
#define MOD 10000007
#define INF 1000000009
const double eps = 1e-;
/*
矩阵快速幂！
an = p an - q an-1
*/
LL p, q, n;
struct Mat
{
    Mat()
    {
        memset(a, , sizeof(a));
    }
    LL a[][];
    Mat operator * (const Mat& rhs)
    {
        Mat ret;
        for (int i = ; i < ; i++)
        {
            for (int j = ; j < ; j++)
            {
                if (a[i][j])
                {
                    for (int k = ; k < ; k++)
                        ret.a[i][k] += a[i][j] * rhs.a[j][k];
                }
            }
        }
        return ret;
    }
};
Mat fpow(Mat m, LL b)
{
    Mat tmp = m, ans;
    ans.a[][] = ans.a[][] = ;
    while (b != )
    {
        if (b & )
            ans = tmp * ans;
        tmp = tmp * tmp;
        b /= ;
    }
    return ans;
}
int main()
{
    while (scanf("%lld%lld%lld", &p, &q, &n) == )
    {

        LL a1 = p, a2 = p*p -  * q;
        if (n == )
            printf("2\n");
        else if (n == )
            printf("%lld\n", a1);
        else if (n == )
            printf("%lld\n", a2);
        else
        {
            n = n - ;
            Mat M;
            M.a[][] = p, M.a[][] = -q, M.a[][] = ;
            M = fpow(M, n);
            printf("%lld\n", M.a[][] * a2 + M.a[][] * a1);
        }
    }
}