A B 都是垃圾题，不说了。

C. The Labyrinth

和之前做过的某道题相似，用 DFS 确定一下每个位置属于哪个联通块，然后确定一下每个 * 周围有哪几个联通块，去重可以直接 std::set 实现。

const int MAXN = 1000 + 10;

int n, m;

int gid(int x, int y) { return m * (x - 1) + y; }

int bel[MAXN][MAXN], siz[MAXN * MAXN], belcnt;

char mat[MAXN][MAXN];

const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, 1, -1};

int sz;
void dfs(int ux, int uy, int bl) {
    if (bel[ux][uy]) return;
    ++sz;
    bel[ux][uy] = bl;
    for (int dr = 0; dr < 4; ++dr) {
        int nx = ux + dx[dr], ny = uy + dy[dr];
        if (nx < 1 || ny < 1 || nx > n || ny > m || mat[nx][ny] == '*') continue;
        dfs(nx, ny, bl);
    }
}

int main() {
    scanf("%d %d", &n, &m);
    rep (i, 1, n) scanf("%s", mat[i] + 1);
    rep (i, 1, n) {
        rep (j, 1, m) {
            if (mat[i][j] != '.') continue;
            sz = 0;
            if (!bel[i][j]) { dfs(i, j, ++belcnt); siz[belcnt] = sz; }
        }
    }
    rep (i, 1, n) {
        rep (j, 1, m) {
            if (mat[i][j] == '.') { printf("."); continue;  }
            int base_bel = 0;
            std::set<int> bels;
            for (int dr = 0; dr < 4; ++dr) {
                int nx = i + dx[dr], ny = j + dy[dr];
                if (nx < 1 || ny < 1 || nx > n || ny > m || mat[nx][ny] == '*') continue;
                bels.insert(bel[nx][ny]);
            }
            int ans = 1;
            for (auto b : bels) ans += siz[b];
            printf("%d", ans % 10);
        }
        puts("");
    }
    return 0;
}

D. Longest k-Good Segment

可以双指针，也可以二分查找。

const int MAXN = 5e5 + 10;

int n, k;
int aa[MAXN];

int nums[MAXN * 2], cnt;

void Add(int x) {
    if (nums[aa[x]] == 0) ++cnt;
    ++nums[aa[x]];
}
void Del(int x) {
    if (nums[aa[x]] == 1) --cnt;
    --nums[aa[x]];
}

int main() {
    n = read(); k = read();
    rep (i, 1, n) {
        aa[i] = read();
    }
    int l = 1, r = 0, tl = 0, tr = -1;
    while (r < n) {
        while (r <= n && cnt <= k) Add(++r);
        Del(r--);
        if (tr - tl + 1 < r - l + 1) {
            tr = r; tl = l;
        }
        Del(l++);
    }
    printf("%d %d\n", tl, tr);
    return 0;
}

const int MAXN = 5e5 + 10;

int n, k;
int aa[MAXN];

int nums[MAXN * 2], cnt;

void Add(int x) {
    if (nums[aa[x]] == 0) ++cnt;
    ++nums[aa[x]];
}
void Del(int x) {
    if (nums[aa[x]] == 1) --cnt;
    --nums[aa[x]];
}

int tl = 0, tr = 0;
bool check(int mid) {
    tl = 0, tr = 0;
    memset(nums, 0, sizeof nums); cnt = 0;
    for (int i = 1; i <= mid - 1; ++i) Add(i);
    for (int l = 1; l + mid - 1 <= n; ++l) {
        int r = l + mid - 1;
        Add(r);
        if (cnt <= k) {
            tl = l, tr = r;
            return true;
        }
        Del(l);
    }
    return false;
}

int main() {
    n = read(); k = read();
    rep (i, 1, n) {
        aa[i] = read();
    }
    int l = 1, r = n, ans = 1;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) ans = mid, l = mid + 1;
        else r = mid - 1;
    }
    check(ans);
    printf("%d %d\n", tl, tr);
    return 0;
}

E. Sum of Remainders

\(\sum (n - \lfloor\frac{n}{i}\rfloor)i\)

然后数论分块。

const lli HA = 1e9 + 7;

lli n, m;
lli ans;

lli fastpow(lli a, lli b, lli p) {
    lli r = 1; while (b) { if (b & 1) r = r * a % p; a = a * a % p; b >>= 1; } return r;
}

int main() {
    lli inv2 = fastpow(2, HA - 2, HA);
    n = readll(); m = readll();
    ans = n % HA * (m % HA) % HA;
    for (lli l = 1, r = 0; l <= std::min(n, m); l = r + 1) {
        r = std::min(m, n / (n / l));
        ans = (ans - ((n / l) % HA) * ((l + r) % HA) % HA * ((r - l + 1) % HA) % HA * inv2 % HA + HA) % HA;
    }
    printf("%lld\n", ans);
    return 0;
}