AtCoder Beginner Contest 326

2UP3DOWN

public static void solve() {
    int x = io.nextInt(), y = io.nextInt();
    if (y - x >= -3 && y - x <= 2) {
        io.println("Yes");
    } else {
        io.println("No");
    }
}

326-like Numbers

public static void solve() {
    int n = io.nextInt();
    for (; ; n++) {
        if (n / 100 * (n / 10 % 10) == n % 10) {
            io.println(n);
            return;
        }
    }
}

Peak

public static void solve() {
    int n = io.nextInt(), m = io.nextInt();
    int[] a = new int[n];
    for (int i = 0; i < n; i++) {
        a[i] = io.nextInt();
    }
    Arrays.sort(a);

    int ans = 0;
    for (int i = 0, j = 0; j < n; j++) {
        while (a[j] - a[i] >= m) {
            i++;
        }
        ans = Math.max(ans, j - i + 1);
    }
    io.println(ans);
}

ABC Puzzle

public static void solve() {
    int n = io.nextInt();
    char[] r = io.next().toCharArray();
    char[] c = io.next().toCharArray();

    int[] row = new int[n];
    int[] col = new int[n];
    char[][] ans = new char[n][n];
    for (int i = 0; i < n; i++) {
        Arrays.fill(ans[i], '.');
    }
    boolean ok = dfs(0, 0, r, c, ans, row, col, 0);

    if (!ok) {
        io.println("No");
        return;
    }
    io.println("Yes");
    for (int i = 0; i < n; i++) {
        io.println(new String(ans[i]));
    }
}

private static boolean dfs(int x, int y, char[] r, char[] c, char[][] ans, int[] row, int[] col, int cnt) {
    int n = r.length;
    if (x == n) {
        return cnt == 3 * n;
    }
    if (n - 1 - y >= 3 - Integer.bitCount(row[x])) {
        if (dfs(x + (y + 1) / n, (y + 1) % n, r, c, ans, row, col, cnt)) {
            return true;
        }
    }
    for (int i = 0; i < 3; i++) {
        if ((row[x] >> i & 1) == 1 || (col[y] >> i & 1) == 1) {
            continue;
        }
        if (row[x] == 0 && r[x] != 'A' + i) {
            continue;
        }
        if (col[y] == 0 && c[y] != 'A' + i) {
            continue;
        }
        row[x] |= 1 << i;
        col[y] |= 1 << i;
        ans[x][y] = (char) ('A' + i);
        if (dfs(x + (y + 1) / n, (y + 1) % n, r, c, ans, row, col, cnt + 1)) {
            return true;
        }
        row[x] ^= 1 << i;
        col[y] ^= 1 << i;
        ans[x][y] = '.';
    }
    return false;
}

Revenge of “The Salary of AtCoder Inc.”

概率 DP，答案为 \(\sum_{i=1}^{n}{A_{i}P_{i}}\)，而 \(P_{i}=\frac{1}{N}\sum_{j=0}^{i-1}{P_{j}}=P_{i-1}+\frac{1}{N}P_{i-1}\)。（这么简单，真没想到）

private static final int MOD = 998244353;

public static void solve() {
    int n = io.nextInt();
    long invn = pow(n, MOD - 2);
    long ans = 0L, p = invn;
    for (int i = 0; i < n; i++) {
        int a = io.nextInt();
        ans = (ans + p * a) % MOD;
        p = (p + invn * p) % MOD;
    }
    io.println(ans);
}

private static long pow(long x, long n) {
    long res = 1L;
    for (; n != 0; x = x * x % MOD, n >>= 1) {
        if ((n & 1) == 1) {
            res = res * x % MOD;
        }
    }
    return res;
}