CodeForces - 385E Bear in the Field （矩阵快速幂）

转移方程：
速度方程dx[i] dy[i]
//前一个位置的速度 + K(草莓数：(原始草莓:)sx[t-1] + sy[t-1]+(到t-1时刻增加的草莓) (t - 1))
dx[t] = dx[t-1] + sx[t-1] + sy[t-1] + t-1

dy[t] = dy[t-1] + sx[t-1] + sy[t-1] + t-1

位置方程sx[i] sy[i]
//前一个位置 + dx[t] (??)
sx[t] = sx[t-1] + dx[t-1] + sx[t-1] + sy[t-1] + t-1

sy[t] = sy[t-1] + dy[t-1] + sx[t-1] + sy[t-1] + t-1

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <vector>
#include <ctime>
#define ll long long
using namespace std;

typedef vector<ll> vec;
typedef vector<vec> mat;
ll n, sx, sy, dx, dy, t;

mat mul(mat &A, mat &B) {
    mat C(A.size(), vec(B[0].size()));
    for (int i = 0; i < A.size(); i++) {
        for (int k = 0; k < B.size(); k++) {
            for (int j = 0; j < B[0].size(); j++) {
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % n;
                if(C[i][j] == 0) C[i][j] = n;
            }
        }
    }
    return C;
}

mat pow(mat A, ll n) {
    mat B(A.size(), vec(A.size()));
    for (int i = 0; i < A.size(); i++) {
        B[i][i] = 1;
    }
    while(n > 0) {
        if(n & 1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}

void solve() {
    mat A(6, vec(6));
    for(int i = 0; i < 6; i++) {
        for(int j = 0; j < 6; j++) {
            if(i == j) A[i][j] = 1;
            else A[i][j] = 0;
        }
    }
    A[0][0] = A[1][1] = 2;
    A[0][1] = A[0][2] = A[0][4] = A[1][0] = A[1][3] = A[1][4] = A[2][0] = A[2][1] = A[2][2] = A[2][4] = 1;
    A[3][0] = A[3][1] = A[3][3] = A[3][4] = A[4][4] = A[4][5] = A[5][5] = 1;
    A = pow(A, t);
    mat B(6, vec(1));
    B[0][0] = sx, B[1][0] = sy, B[2][0] = dx, B[3][0] = dy, B[4][0] = 0, B[5][0] = 1;
    B = mul(A, B);
    if(B[0][0] < 0) B[0][0] += n;
    if(B[1][0] < 0) B[1][0] += n;
    printf("%lld %lld\n",B[0][0], B[1][0]);
}

int main() {
    scanf("%lld %lld %lld %lld %lld %lld", &n, &sx, &sy, &dx, &dy, &t);
    if(n == 1) {
        printf("1 1\n");
        return 0;
    }
    solve();
}