( a^(n-1) + b^(n-1) ) * ( a + b ) = a^n + a^(n-1) * b + a * b^(n-1) + b^n

a^n + b^n = p * ( a^(n-1) + b^(n-1) ) - q * ( a^(n-2) + b^(n-2) )

F(n) = p * F(n-1) - q * F(n-2)

Windows或C++用%lld和%llu

Linux或G++用%I64d和%I64u

#include <iostream>
#include <cstdio>
#include <cstring>
#include 
 
<algorithm>
using namespace std;

typedef long long ll;
const int N = 4;
int msize;

struct Mat
{
    ll mat[N][N];
};

Mat operator *(Mat a, Mat b)
{
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    for(int k = 0; k < msize; ++k)
        for(int i = 0; i < msize; ++i)
            
 
if(a.mat[i][k])
                for(int j = 0; j < msize; ++j)
                    if(b.mat[k][j])
                        c.mat[i][j] = (a.mat[i][k] * b.mat[k][j] + c.mat[i][j]);
    return c;
}

Mat operator ^(Mat a, ll k)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(int i = 0
 
; i < msize; ++i)
        c.mat[i][i]=1;
    for(; k; k >>= 1)
    {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}

int main()
{
    ll n,p,q;
    msize = 2;
    Mat A;
    while(scanf("%lld%lld%lld", &p, &q, &n) == 3)
    {
        if(n == 0)
        {
            puts("2");
            continue;
        }
        A.mat[0][0] = p, A.mat[0][1] = -q;
        A.mat[1][0] = 1, A.mat[1][1] = 0;
        A = A^(n-1);
        printf("%lld\n", A.mat[0][0]*p + A.mat[0][1]*2);
    }
    return 0;
}

UVa 10655 Contemplation! Algebra 矩陣快速冪