AcWing 第136场周赛

发表于 更新于 分类于 阅读次数： 本文字数： 485 阅读时长 ≈ 2 分钟
Live beautifully, dream passionately, love completely.

A. 装苹果

思路：

  • 输出 \(\lceil \frac{n}{x} \rceil\)​ 即可

时间复杂度：\(O(1)\)

1
2
3
4
5
6
7
8
9
10
11
signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int n, x;
    std::cin >> n >> x;

    std::cout << (n + x - 1) / x << "\n";
    
    return 0;
}

B. 旋转和翻转

思路：

  • 旋转后判断是否相同
  • 旋转后然后翻转再判断是否相同

时间复杂度：\(O(n^2)\)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
int n;
struct Matrix {
    char g[15][15];

    void init() {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                std::cin >> g[i][j];
            }
        }
    }

    Matrix rotate() {
        Matrix ans;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                ans.g[i][j] = g[n - j - 1][i];
            }
        }
        return ans;
    }   

    Matrix reflex_x() {
        Matrix ans;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                ans.g[i][j] = g[n - i - 1][j];
            }
        }
        return ans;
    }

    Matrix reflex_y() {
        Matrix ans;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                ans.g[i][j] = g[i][n - j - 1];
            }
        }
        return ans;
    }

    bool operator == (Matrix b) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (g[i][j] != b.g[i][j]) return false;
            }
        }
        return true;
    }
};

signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
        
    std::cin >> n;

    Matrix a, b, c;
    a.init(), b.init();
    for (int i = 0; i < 4; i++) {//只旋转
        if (a == b) {
            std::cout << "Yes\n";
            return 0;
        }
        a = a.rotate();
    }
    for (int i = 0; i < 4; i++) {
        c = a.reflex_x();
        for (int j = 0; j < 4; j++) {
            if (c == b) {
                std::cout << "Yes\n";
                return 0;
            }
            c = c.rotate();
        }   
        a = a.reflex_x();
    }
    for (int i = 0; i < 4; i++) {
        c = a.reflex_y();
        for (int j = 0; j < 4; j++) {
            if (c == b) {
                std::cout << "Yes\n";
                return 0;
            }
            c = c.rotate();
        }   
        a = a.reflex_y();
    }
    std::cout << "No\n";

    return 0;
}

C. 三角形

思路：

  • 枚举两个顶点即可，运用初中所学的勾股定理

时间复杂度：\(O(a^2 \times b^2)\)

附：勾股定理： \(a^2+b^2=c^2\)，\(a\), \(b\) 均为直角边，\(c\)​ 为斜边

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
        
    int a, b;
    std::cin >> a >> b;

    for (int i = 1; i * i < a * a; i++) {
        for (int j = 1; j * j < b * b; j++) {
            int x1 = a * a - i * i, x2 = b * b - j * j;
            int y1 = sqrt(x1), y2 = sqrt(x2);
            if (y1 * y1 == x1 && y2 * y2 == x2) {
                if (y1 != y2 && i * j == y1 * y2) {
                    std::cout << "YES\n";
                    std::cout << "0 0\n";
                    std::cout << -i << " " << y1 << "\n";
                    std::cout << j << " " << y2 << "\n";
                    return 0;
                }
            }
        }
    }

    std::cout << "NO\n";

    return 0;
}