矩阵基础1002 CodeForces 450B

题意:
F[i]=F[i-1]+F[i+1]
求F[n]
思路:
F[i+1]=F[i]-F[i-1]->F[i]=F[i-1]-F[i-2]
然后矩阵快速幂去做
这道题有些特殊性,简单推导一下就会发现
每6次一个循环,打个小表就行了

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<map>
#include<set>
using namespace std;
#define lowbit(x) (x&(-x))
typedef long long LL;
const int maxn = 100005;
const int inf=(1<<28)-1;
#define Matrix_Size 5
const LL MOD = 1e9+7;
int Size;
struct Matrix
{
    LL mat[Matrix_Size][Matrix_Size];
    void clear()
    {
        memset(mat,0,sizeof(mat));
    }
    void output()
    {
        for(int i = 0;i < Size;i++)
        {
            for(int j = 0;j < Size;j++)
                printf("%d ",mat[i][j]);
            printf("\n");
        }
    }
    Matrix operator *(const Matrix &b)const
    {
        Matrix ret;
        for(int i = 0;i < Size;i++)
            for(int j = 0;j < Size;j++)
            {
                ret.mat[i][j] = 0;
                for(int k = 0;k < Size;k++)
                {
                    long long tmp = (long long)mat[i][k]*b.mat[k][j]%MOD;
                    ret.mat[i][j] = (ret.mat[i][j]+tmp);
                    if(ret.mat[i][j]>=MOD)
                        ret.mat[i][j] -= MOD;
                    if(ret.mat[i][j]<0)//注意是否需要MOD 
                        ret.mat[i][j] += MOD;
                }
            }
        return ret;
    }
};
Matrix pow_M(Matrix a,long long n)
{
    Matrix ret;
    ret.clear();
    for(int i = 0;i < Size;i++)
        ret.mat[i][i] = 1;
    Matrix tmp = a;
    while(n)
    {
        if(n&1)ret = ret*tmp;
        tmp = tmp*tmp;
        n>>=1;
    }
    return ret;
}

int main()
{
    LL x,y,n;
    scanf("%lld%lld%lld",&x,&y,&n);
    if(n==1)
    {
        printf("%lld\n",(x+MOD)%MOD);
        return 0;
    }
    Size=2;
    Matrix A,B;
    A.mat[0][0]=x,A.mat[0][1]=y;
    A.mat[1][0]=0,A.mat[1][1]=0;
    B.mat[0][0]=0,B.mat[0][1]=-1;
    B.mat[1][0]=1,B.mat[1][1]=1;
    B=pow_M(B,n-2);
    A=A*B;
    //B.output();
    //printf("%lld\n",((x*B.mat[1][0])%MOD+(y*B.mat[1][1])%MOD+MOD)%MOD);
    //printf("%lld\n",A.mat[0][1]);
    printf("%lld\n",(A.mat[0][1]+MOD)%MOD);
    return 0;
}