codefroce385E矩阵快速幂

时间：2023-03-09 21:09:33
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 ;

}

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