1126 求递推序列的第N项（矩阵快速幂）

思路：很普通的矩阵快速幂，直接构造就可以了。但是这题贼坑，他所谓的取膜和我们平时说的不一样，负数取膜要变成正的，浪费了我2个多小时一直以为是模板问题。总之垃圾题一道。

#include <iostream>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <sstream>
#include <time.h>
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define lson l,m,rt*2
#define rson m+1,r,rt*2+1
#define mt(A,B) memset(A,B,sizeof(A))
#define mod 7
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const int N=2;
const int inf = 0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3fLL;
LL one,two,three;
struct Mat
{
    LL mat[N][N];
};
Mat operator *(Mat a,Mat b)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    int i,j,k;
    for(i=0; i<N; i++)
    {
        for(j=0; j<N; j++)
        {
            for(k=0; k<N; k++)
            {
                c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%7;
            }
        }
    }
    return c;
}
Mat powermod(Mat a,int n)
{
    Mat ans={1,0,1,0};//初始矩阵
    int i,j;
    if(n<0)return ans;
    while(n)
    {
        if(n&1)
        {
            ans=ans*a;//矩阵a是我们所要构造的矩阵即a^n。
        }
        a=a*a;
        n/=2;
    }
    return ans;
}
int main()
{
#ifdef Local
    freopen("data.txt","r",stdin);
#endif
    int n;
    LL o;
    cin>>one>>two>>n;
    Mat p={0,(two%mod+mod)%mod,1,(one%mod+mod)%mod},q;
    q=powermod(p,n-2);
    cout<<q.mat[0][1]<<endl;
#ifdef Local
    cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
#endif
}