This is just a implementation problem:

The idea was just check each car name is complete or not, and then get only unique one's count.

#include <bits/stdc++.h>
using namespace std;

/*
      ---===ASCII help===---
     '0' -> 48     '9' -> 57
     'A' -> 65     'Z' -> 90
     'a' -> 97     'z' -> 122
*/

const long long mod = 1e9 + 7;

void solve() {
    int n; cin >> n;
    set<string> cars;
    string s;
    while (getline(cin, s)) {
        vector<string> words;
        stringstream str(s);
        string word;
        while (str >> word) words.push_back(word);
        if (words.size() == 2 && words[0].size() >= 5 && words[1].size() >= 5) cars.insert(s);
    }
    cout << cars.size() << "\n";
}

int main() {
    int t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
}

This is a brute-force problem because of constraints. You should do a brute-force from l to r and check for current number is Armstrong number or not.

#include <bits/stdc++.h>
using namespace std;

/*
      ---===ASCII help===---
     '0' -> 48     '9' -> 57
     'A' -> 65     'Z' -> 90
     'a' -> 97     'z' -> 122
 */

const long long mod = 1e9 + 7;

bool check(int x) {
    string str = to_string(x);
    long long sum = 0;
    for (char c : str) {
        sum += pow(c - '0', str.size());
    }
    return sum == x;
}

void solve() {
    int l, r; cin >> l >> r;
    int ans = 0;
    for (int i = l; i <= r; i++) {
        if (check(i)) ans++;
    }
    cout << ans << "\n";
}

int main() {
    //freopen("factory.in", "r", stdin);
    //freopen("factory.out", "w", stdout);
    int t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
}

This one is tricky, because you can't just brute-force. Instead, we can precompute! Because there are a small amount of Armstrong numbers, we can just find all of them by sacrificing our PC, and then get all Armstrong numbers, put it in array and done! Or, if you want more math variant, you can do this:

For a fixed digit lenght d:

n = c[0] * 0^d + c[1] * 1^d.... + c[9] * 9^d, Where c[i] = how many times digit i appears. Constraints:

c[0] + c[1]... + c[9] = d.

So instead of iterating over numbers, we iterate over digit frequency distributions.

This is a classical Stars and Bars problem:

Distribution count: 17! / 9! * 8! = 24310. This is extremely fast.

Precompute pow[d][digit] = digit^d. Then, for each digit lenght d = 1...8:

Generate all distributions:

c[0] + c[1]...c[9] = d

Then compute:

num = sum(c[i] + pow[d][i]), Add to array, then output the answer for l, r.

#include <bits/stdc++.h>
using namespace std;

/*
      ---===ASCII help===---
     '0' -> 48     '9' -> 57
     'A' -> 65     'Z' -> 90
     'a' -> 97     'z' -> 122
 */

const long long mod = 1e9 + 7;

vector<long long> arm;
long long pw[10][10];
int cnt[10], cur[10];

void gen(int d, int pos, int left) {
    if (pos == 10) {
        if (left) return;
        long long s = 0;
        for (int i = 0; i < 10; i++) s += 1LL * cnt[i] * pw[d][i];
        string t = to_string(s);
        if ((int)t.size() != d) return;
        memset(cur, 0, sizeof cur);
        for (char c : t) cur[c - '0']++;
        if (memcmp(cnt, cur, sizeof cnt) == 0) arm.push_back(s);
        return;
    }
    for (int i = 0; i <= left; i++) {
        cnt[pos] = i;
        gen(d, pos + 1, left - i);
    }
}

void solve() {
    for (int d = 1; d <= 8; d++) for (int i = 0; i <= 9; i++) pw[d][i] = pow(i, d);
    for (int d = 1; d <= 8; d++) {
        memset(cnt, 0, sizeof cnt);
        gen(d, 0, d);
    }
    sort(arm.begin(), arm.end());
    arm.erase(unique(arm.begin(), arm.end()), arm.end());
    long long l, r;
    cin >> l >> r;
    if (l > r) {
        cout << 0;
        return;
    }
    cout << upper_bound(arm.begin(), arm.end(), r) - lower_bound(arm.begin(), arm.end(), l);
}

int main() {
    //freopen("speeding.in", "r", stdin);
    //freopen("speeding.out", "w", stdout);
    int t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
}

We should have a fast access to messages, their ID's, and who send that message. For that we can use std::map. We should declare:

map<int, set<int>> user; // The users messages(`std::set` for fast deleting and checking, we have any message here).

map<int, int> message; // The message ID, and who send.

ans = 0; // answer right now(for answer to third query fast)

m = 0; // The ID of the message, which we will give to the next message.

After that, the task already solved:

If we get first type query, just add that message to given person. If this message is the first for this person, increase answer to 1.

If we get second type query, just delete that message by ID. If the person don't have any messages, decrease answer by 1.

Else output the answer.

#include <bits/stdc++.h>
using namespace std;

/*
      ---===ASCII help===---
     '0' -> 48     '9' -> 57
     'A' -> 65     'Z' -> 90
     'a' -> 97     'z' -> 122
*/

const long long mod = 1e9 + 7;

void solve() {
    map<int, set<int>> user;
    map<int, int> message;
    int q, ans = 0, m = 1; cin >> q;
    while (q--) {
        int type; cin >> type;
        if (type == 1) {
            int x; cin >> x;
            user[x].insert(m);
            message[m] = x;
            m++;
            if (user[x].size() == 1) ans++;
        }
        else if (type == 2) {
            int x; cin >> x;
            user[message[x]].erase(x);
            if (user[message[x]].size() == 0) ans--;
        }
        else cout << ans << "\n";
    }
}

int main() {
    int t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
}

To the Standing will only include true Div. 4 participants, who have maximum Pupil in their rank when contest started.