Codeforces

Bear in the Field

题目链接

分类：math matrices

1.题意概述

• 有一片 n×n 的草莓地，每个位置的初始草莓量为横坐标和纵坐标的和，然后每过一秒增长一个草莓。然后给出熊的初始位置(sx,sy)，以及移动的速度(dx,dy)，每一秒发生的事：
  
  1. 速度增加k（k为该位置的草莓数）
  2. 熊的位置发生移动
  3. 每个位置上草莓数+1

2.解题思路

• 容易推出的转移方程是：
  

  ⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪sx[t]=sx[t−1]+dx[t−1]+sx[t−1]+sy[t−1]+t−1+2sy[t]=sy[t−1]+dy[t−1]+sx[t−1]+sy[t−1]+t−1+2dx[t]=dx[t−1]+sx[t−1]+sy[t−1]+t−1+2dy[t]=dy[t−1]+sx[t−1]+sy[t−1]+t−1+2t=t−1+1
  
  但是看数据范围：
  • 1 ≤n≤ 109;1 ≤sx, sy≤ n; − 100 ≤ dx, dy ≤ 100;0 ≤ t ≤ 1018

  正常的转移时间复杂度O(t)是很难接受的，根据递推的线性，我们可以考虑构造矩阵加速转移过程：
  

  ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢sx[t]sy[t]dx[t]dy[t]t1⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢211100121100101000010100111110222211⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥×⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢sx[t−1]sy[t−1]dx[t−1]dy[t−1]t−11⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥

3.AC代码

#include <bits/stdc++.h>using namespace std;#define eps 1e-6#define e exp(1.0)#define pi acos(-1.0)#define fi first#define se second#define pb push_back#define mp make_pair#define SZ(x) ((int)(x).size())#define All(x) (x).begin(),(x).end()#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define Close() ios::sync_with_stdio(0),cin.tie(0)#define INF 0x3f3f3f3f#define maxn 100010#define N 7typedef vector<int> VI;typedef pair<int, int> PII;typedef long long ll;typedef unsigned long long ull;ll mod = 1e9 + 7;/* head */class Matrix {public:    ll a[N][N];    int n;    void Init(int key) {        memset(a, 0, sizeof a);        if (key)            rep(i, 0, n) a[i][i] = 1;    }    Matrix operator+ (const Matrix &b) {        Matrix c;        c.n = n;        rep(i, 0, n) rep(j, 0, n) c.a[i][j] = (a[i][j] + b.a[i][j]) % mod;        return c;    }    Matrix operator+ (int x) {        Matrix p = *this;        rep(i, 0, n) p.a[i][i] = (p.a[i][i] + x % mod) % mod;        return p;    }    Matrix operator- (const Matrix &b) {        Matrix c;        c.n = n;        rep(i, 0, n) rep(j, 0, n) c.a[i][j] = (a[i][j] - b.a[i][j] + mod) % mod;        return c;    }    Matrix operator* (const Matrix &b) {        Matrix c;        c.n = n;        c.Init(0);        rep(i, 0, n) rep(j, 0, n) rep(k, 0, n)        c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j] % mod + mod) % mod;        return c;    }    Matrix Power(ll t) {        Matrix ans, p = *this;        ans.n = p.n;        ans.Init(1);        while (t) {            if (t & 1) ans = ans * p;            p = p * p;            t >>= 1;        }        return ans;    }    void Print() {        rep(i, 0, n) {            rep(j, 0, n) {                if (j == 0) printf("%I64d", a[i][j]);                else printf(" %I64d", a[i][j]);            }            puts("");        }    }};int main() {#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);    long _begin_time = clock();#endif    Matrix A, f;    ll n, sx, sy, dx, dy, t;    scanf("%I64d%I64d%I64d%I64d%I64d%I64d", &n, &sx, &sy, &dx, &dy, &t);    mod = n;    f.Init(0); f.n = 6;    f.a[0][0] = f.a[0][5] = f.a[1][1] = f.a[1][5] = f.a[2][5] = f.a[3][5] = 2;    f.a[0][1] = f.a[0][2] = f.a[0][4] = f.a[1][0] = f.a[1][3] = f.a[1][4] = f.a[2][0] = f.a[2][1] = f.a[2][2] = f.a[2][4] = 1;    f.a[3][0] = f.a[3][1] = f.a[3][3] = f.a[3][4] = f.a[4][4] = f.a[4][5] = f.a[5][5] = 1;    //  f.Print();    A.Init(0); A.n = 6;    A.a[0][0] = sx - 1;    A.a[1][0] = sy - 1;    A.a[2][0] = dx;    A.a[3][0] = dy;    A.a[4][0] = 0;    A.a[5][0] = 1;    //  A.Print();    A = f.Power(t) * A;    printf("%I64d %I64d\n", A.a[0][0] + 1, A.a[1][0] + 1);#ifndef ONLINE_JUDGE    long _end_time = clock();    printf("time = %ld ms.", _end_time - _begin_time);#endif    return 0;}