洛谷P2233 [HNOI2002]公交车路线 题解 矩阵快速幂

题目链接：​​https://www.luogu.com.cn/problem/P2233​​

解题思路：

矩阵快速幂。

思路完全来自 ​​ghj1222大佬的博客​​

示例代码：

#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000;
struct Matrix {
    int a[8][8];
    Matrix operator * (Matrix b) const {
        Matrix c;
        memset(c.a, 0, sizeof(c.a));
        for (int i = 0; i < 8; i ++)
            for (int j = 0; j < 8; j ++)
                for (int k = 0; k < 8; k ++)
                    c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j]) % MOD;
        return c;
    }
} a, c;
int r[8][8] = {
    0, 1, 0, 0, 0, 0, 0, 1,
    1, 0, 1, 0, 0, 0, 0, 0,
    0, 1, 0, 1, 0, 0, 0, 0,
    0, 0, 1, 0, 1, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0, // E没有出边
    0, 0, 0, 0, 1, 0, 1, 0,
    0, 0, 0, 0, 0, 1, 0, 1,
    1, 0, 0, 0, 0, 0, 1, 0,
};
int n;
int main() {
    cin >> n;
    memset(c.a, 0, sizeof(c.a));
    for (int i = 0; i < 8; i ++) c.a[i][i] = 1;
    memcpy(a.a, r, sizeof(r));
    while (n > 0) {
        if (n % 2) c = c * a;
        a = a * a;
        n /= 2;
    }
    cout << c.a[0][4] << endl;
    return 0;
}