Atcoder Beginner Contest 300 A~E

鼠鼠第一次打这个Atcoder，但一边挂视频一边写代码效率是真的低，只写了三道题

A

在一个序列中找出一个A+B的元素

# include 
# define int long long
# define endl '\n'
# define x first
# define y second
# define INF 0x3f3f3f3f
# define pii pair<int, int>
# define uf(i, j, k) for (int i = j; i <= k; i ++)
# define df(i, j, k) for (int i = j; i >= k; i --)
# define mset(x, v) memset(x, v, sizeof x)
/*
    @see https://atcoder.jp/contests/abc300/tasks/abc300_a
*/
using namespace std;

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, a, b;
    cin >> n >>  a >> b;
    int w[n + 10];
    uf (i, 1, n) 
    {
        cin >> w[i];
        if (w[i] == a + b) cout << i << endl;
    }

    return 0;
}

B

将A向下偏移H次和向左偏移W次，找出一个和B相同的偏移量(h, w)， 模拟枚举一下

# include 
# define int long long
# define endl '\n'
# define x first
# define y second
# define INF 0x3f3f3f3f
# define pii pair<int, int>
# define uf(i, j, k) for (int i = j; i <= k; i ++)
# define df(i, j, k) for (int i = j; i >= k; i --)
# define mset(x, v) memset(x, v, sizeof x)
/*
    @see https://atcoder.jp/contests/abc300/tasks/abc300_b
*/
using namespace std;
const int N = 100, M = 100;
char a[N][M], b[N][M], ga[N][M];
int n, m;
bool check()
{
    uf (i, 0, n - 1) uf (j, 0, m - 1) 
    {
        if (a[i][j] != b[i][j]) return false;
    }
    return true;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    uf (i, 0, n - 1) cin >> a[i];
    uf (i, 0, n - 1) cin >> b[i];
    bool ans = check();
    uf (h, 1, n)
    {
        //vertical
        uf (j, 0, m - 1) 
        {
            df (k, n, 1) a[k][j] = a[k - 1][j];
            a[0][j] = a[n][j];
        }
        uf (w, 0, m)
        {
            //horiztical
            uf (j, 0, n - 1)
            {
                df (k, m, 1) a[j][k] = a[j][k - 1];
                a[j][0] = a[j][m];
            }
            ans |= check();
        }
    }
    if (ans) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}

C

找各个长度的"十字架“的个数，直接枚举

# include 
# define int long long
# define endl '\n'
# define x first
# define y second
# define INF 0x3f3f3f3f
# define pii pair<int, int>
# define uf(i, j, k) for (int i = j; i <= k; i ++)
# define df(i, j, k) for (int i = j; i >= k; i --)
# define mset(x, v) memset(x, v, sizeof x)
/*
    @see https://atcoder.jp/contests/abc300/tasks/abc300_c
*/
using namespace std;
const int N = 110, M = 110;
char g[N][M];
int n, m;

int d[][2] ={ {-1, -1}, {-1, 1}, {1, 1}, {1, -1}};

int  check(int x, int y) 
{
    int ans = 0, t = 1;
    while (true) 
    {
        uf (i, 0, 3)
        {
            int dx = x + d[i][0] * t, dy = y + d[i][1] * t;
            if (dx < 1 || dx > n || dy < 1 || dy > m) return ans;
            if (g[dx][dy] == '.') return ans;
        }
        ans = t;
        t ++;
    }
    return ans;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    uf (i, 1, n) cin >> g[i] + 1;
    vector<int> ans(min(n , m) + 1, 0);
    uf (i, 1, n) 
    {
        uf (j, 0, m - 1) 
        {
            if (g[i][j] == '#')
            {
                ans[check(i, j)] ++;
            }
            uf (i, 1, n) cout << g[i] << endl;
        }
    }
    uf (i, 1, ans.size() - 1) cout << ans[i] << " ";
    return 0;
}

D

题目的数据范围是 10¹²， 当 $a=2,b=3$ 时， c的最大取值也是在 10⁶ 内

• 先用线性筛预处理出10⁶ 以内的质数
• 枚举 a 和 c，计算得出b，枚举从a到c中小于等于b的个数即可

# include 
# define int long long
# define endl '\n'
# define x first
# define y second
# define INF 0x3f3f3f3f
# define pii pair<int, int>
# define uf(i, j, k) for (int i = j; i <= k; i ++)
# define df(i, j, k) for (int i = j; i >= k; i --)
# define mset(x, v) memset(x, v, sizeof x)
/*
    @see https://atcoder.jp/contests/abc300/tasks/abc300_d
*/
using namespace std;
const int N = 1e6 + 10;
bool st[N];
vector<int> p;
void init()
{
    uf (i, 2, N - 1)
    {
        if (!st[i]) p.push_back(i);
        for (auto x : p)
        {
            if (x * i > N) break;
            st[x * i] = true;
            if (x % i == 0) break;
        }
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    init();
    cin >> n;
    int m = p.size() - 1;
    int ans = 0;
    uf (i, 0, m) 
    {
        uf (j, i + 2, m) 
        {
            int t = p[i] * p[i] * p[j] * p[j];
            if (t < n)
            {
                int b = n / t;
                uf (k, i + 1, j - 1) 
                {
                    if (p[k] <= b) ans ++;
                }
            } 
            else break;
        }
    }
    cout << ans << endl;
    return 0;
}

E

概率DP

• 数据范围n在 10¹⁸ 内，不能开数组存

• 用dfs搜一下，因为每个状态x只能从 $\frac{x}{2},\frac{x}{3},\frac{x}{4},\frac{x}{5},\frac{x}{6}$ 转换而来，开一个map存一下搜索的状态

• 注意概率不是$\frac{1}{6}$， 而是 $\frac{1}{5}$， 因为1点数是不会改变状态，因此每个状态改变时可能穿插多个点数1，不能把点数1忽略

  抛出点数1的概率$p_1={\sum }_i^{n\to \infty }(\frac{1}{6}{)}^i=\frac{1}{1-\frac{1}{6}}=\frac{6}{5}$

  与抛出其他点数的概率相乘 $p=p_1\ast \frac{1}{6}=\frac{1}{5}$

# include 
# define int long long
# define endl '\n'
# define x first
# define y second
# define INF 0x3f3f3f3f
# define pii pair<int, int>
# define uf(i, j, k) for (int i = j; i <= k; i ++)
# define df(i, j, k) for (int i = j; i >= k; i --)
# define mset(x, v) memset(x, v, sizeof x)
/*
    @see https://atcoder.jp/contests/abc300/tasks/abc300_e
*/
using namespace std;
const int mod = 998244353;
unordered_map<int, int> mp;
int p;

int dfs (int u)
{
    if (u == 1) return 1;
    if (u <= 0) return 0;
    if (mp.count(u)) return mp[u];
    int ans = 0;
    uf (i, 2, 6)
    {
        if (u % i == 0)
        {  
            ans = (ans +  dfs(u / i) * p % mod) % mod;
        } 
    }
    mp[u] = ans;
    return ans;
}

signed main()
{

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    auto qmi = [&](int x, int y) {
        int ans = 1;
        while(y) 
        {
            if (y & 1) ans = ans * x % mod;
            x = x*x % mod;
            y >>= 1;
        }
        return ans;
    };
    p = qmi(5, mod - 2);
    int n;
    cin >> n;
    cout << dfs(n) << endl;
    return 0;
}

看到别的题解有用质因数进行dp，确定2,3,5这三个质因数的个数，然后用一个四维数组做一个dp

• $n=2^a\cdot 3^b\cdot 5^c$， a的值最多是60， 即 $a\le 60,a+b+c\le 60$
• $dp[i,j,k,h]$ 表示第i次转化到质因数个数分别是j,k,h的概率
• 最后只要统计 $dp[i,a,b,c]$ 的概率即可

# include 
# define int long long
# define endl '\n'
# define x first
# define y second
# define INF 0x3f3f3f3f
# define pii pair<int, int>
# define uf(i, j, k) for (int i = j; i <= k; i ++)
# define df(i, j, k) for (int i = j; i >= k; i --)
# define mset(x, v) memset(x, v, sizeof x)

using namespace std;
const int mod = 998244353, N = 70;
int p, n, a, b, c, ans;
int dp[N][N][N][N];

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    auto qmi = [&](int x, int y) {
        int ans = 1;
        while(y) 
        {
            if (y & 1) ans = ans * x % mod;
            x = x*x % mod;
            y >>= 1;
        }
        return ans;
    };
    p = qmi(5, mod - 2);
    dp[0][0][0][0] = 1;
    cin >> n;
    while (n % 2 == 0) a ++, n /= 2;
    while (n % 3 == 0) b ++, n /= 3;
    while (n % 5 == 0) c ++, n /= 5;
    if (n == 1)
    {
        int m = N - 3;
        uf (i, 0, m) uf (j, 0, m) uf(k, 0, m) uf (h, 0, m)
        {
            int  t = dp[i][j][k][h];
            dp[i + 1][j + 1][k][h] += t * p % mod;
            dp[i + 1][j][k + 1][h] += t * p % mod;
            dp[i + 1][j][k][h + 1] += t * p % mod;
            dp[i + 1][j + 2][k][h] += t * p % mod;
            dp[i + 1][j + 1][k + 1][h] += t * p % mod;
        }
        uf (i, 0, N - 1) ans += dp[i][a][b][c], ans %= mod;
    }
    cout << ans << endl;
    return 0;
}

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