【51Nod】1126 - 求递推数列的第N项（矩阵快速幂 & C++运算符重载）

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define PI acos(-1.0)
#define LL long long
const int MOD = 7;
struct Matrix
{
int h,w;
int m[3][3];
void init(int op)
{
if (op == 0)        //初始化矩阵
CLR(this->m,0);
else if (op == 1)       //初始化为单位矩阵
{
this->h = this->w = 2;
CLR(this->m,0);
this->m[1][1] = this->m[2][2] = 1;
}
else if (op == 2)       //初始化为初始矩阵
{
this->h = 1;
this->w = 2;
CLR(this->m,0);
this->m[1][1] = this->m[1][2] = 1;
}
}
void init(int A,int B)
{
this->w = this->h = 2;
CLR(this->m,0);
this->m[1][1] = A;
this->m[2][1] = B;
this->m[1][2] = 1;
}
Matrix operator * (Matrix a)
{
Matrix t;
t.h = this->h;
t.w = a.w;
t.init(0);
for (int i = 1 ; i <= this->h ; i++)
{
for (int j = 1 ; j <= a.w ; j++)
{
if (this->m[i][j])
for (int k = 1 ; k <= this->w ; k++)
{
t.m[i][k] = (t.m[i][k] + this->m[i][j] * a.m[j][k] % MOD) % MOD;
}
}
}
return t;
}
Matrix quickMod(int n)
{
Matrix t;
t.init(1);
while (n)
{
if (n & 1)
t = t * (*this);
*this = (*this) * (*this);
n >>= 1;
}
return t;
}
};

int main()
{
int A,B,n;
scanf ("%d %d %d",&A,&B,&n);
if (n == 1 || n == 2)
puts("1");
else
{
Matrix pr,ans,ori;
pr.init(2);
ori.init(A,B);
ori = ori.quickMod(n-2);
ans = pr * ori;
printf ("%d\n",(ans.m[1][1] % MOD + MOD) % MOD);
}
return 0;
}