codefroce385E矩阵快速幂

状态变化  (x,y,dx,dy,i) 表示i时刻熊站在(x,y)处速度向量(dx,dy)下一个状态是 ( 2x+y+dx+i , x+2y+dy+i , x+y+dx , x+y+dy , i+1 )

为了方便可以把平面从(1,1)平移到(0,0)  这时速度需要+2 (因为速度每次+x+y  x和y都-1则速度都+2)矩阵对应常数的地方为2

转移矩阵：{2,1,1,0,1,2},

{1,2,0,1,1,2}，

{1,1,1,0,1,2}，

          {1,1,0,1,1,2},

        {0,0,0,0,1,1},
        {0,0,0,0,0,1}

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-;
const int N=+,maxn=<<+,inf=0x3f3f3f3f;

struct Node{
   ll row,col;
   ll a[N][N];
};
ll n;
Node mul(Node x,Node y)
{
    Node ans;
    ans.row=x.row,ans.col=y.col;
    memset(ans.a,,sizeof ans.a);
    for(ll i=;i<x.row;i++)
        for(ll j=;j<x.col;j++)
            for(ll k=;k<y.col;k++)
                ans.a[i][k]=(ans.a[i][k]+x.a[i][j]*y.a[j][k]+n)%n;
    return ans;
}
Node quick_mul(Node x,ll n)
{
    Node ans;
    ans.row=x.row,ans.col=x.col;
    memset(ans.a,,sizeof ans.a);
    for(ll i=;i<ans.col;i++)ans.a[i][i]=;
    while(n){
        if(n&)ans=mul(ans,x);
        x=mul(x,x);
        n>>=;
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie();
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    ll x,y,dx,dy,t;
    cin>>n>>x>>y>>dx>>dy>>t;
    x--,y--;
    Node A;
    A.row=,A.col=;
    A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;
    A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;
    A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;
    A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;
    A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;
    A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=,A.a[][]=;
    A=quick_mul(A,t);
    Node B;
    B.row=,B.col=;
    B.a[][]=x,B.a[][]=y,B.a[][]=dx,B.a[][]=dy,B.a[][]=,B.a[][]=;
    B=mul(A,B);
    cout<<(B.a[][]+n)%n+<<" "<<(B.a[][]+n)%n+<<endl;
    return ;
}