• Excellent:
• Good:
• Average:
• Bad:
• Terrible:

• A:
• B:
• C:
• D:
• E:
• F:
• G:

• A:
• B:
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Prove that the number $$$2$$$ is not an ideal generator. What other numbers are not ideal generators for the same reason?

Prove that the number $$$3$$$ is an ideal generator. What is the difference from the previous case?

#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
    int t;
    cin >> t;
 
    while (t--) {
        int n;
        cin >> n;
        if (n % 2 == 1) {
            cout << "YES\n";
        } else {
            cout << "NO\n";
        }
    }
 
    return 0;
}

What is the minimum cost that can be achieved?

How many non-zero digits should be in the remaining number for the cost to be minimal?

How can we ensure that the number has as many leading zeros as possible?

#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
    int t;
    cin >> t;
    while (t--) {
        string s;
        cin >> s;
        int n = s.size();
        bool met_positive = false;
        int cnt_zero = 0;
        
        for (auto i = n - 1; i >= 0; --i) {
            if (s[i] != '0') {
                met_positive = true;
            } else if (met_positive) {
                cnt_zero++;
            }
        }
        
        cout << n - (cnt_zero + 1) << '\n';
    }
    return 0;
}

$$$y$$$ is always divisible by $$$x$$$.

What happens if $$$k \geq 2$$$?

If $$$k \geq 2$$$ and $$$x \geq 2$$$, then $$$y$$$ will never be prime. Are there any edge cases with $$$k = 1$$$ or $$$x = 1$$$?

#include <bits/stdc++.h>
 
using namespace std;
 
bool is_prime(int x) {
    if (x <= 1) {
        return false;
    }
    for (int i = 2; i * i <= x; i++) {
        if (x % i == 0) {
            return false;
        }
    }
    return true;
}
 
void solve() {
    int x, k;
    cin >> x >> k;
    if (k > 1 && x > 1) {
        cout << "NO";
    } else if (k == 1) {
        cout << (is_prime(x) ? "YES" : "NO");
    } else {
        cout << ((k == 2) ? "YES" : "NO");
    }
}
 
int main() {
    int tests;
    cin >> tests;
    while (tests--) {
        solve();
        cout << '\n';
    }
}

How does the answer change when $$$n$$$ is decreased?

#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
    int t;
    cin >> t;
    
    while (t--) {
        int n, q;
        cin >> n >> q;
        
        while (q--) {
            string type;
            cin >> type;
            
            if (type == "->") {
                int x, y;
                cin >> x >> y;
                x--, y--;
                
                long long num = 0;
                for (int i = n - 1; i >= 0; --i) {
                    int cur = 1 << i;
                    if (!(x & cur) && !(y & cur))
                        num ^= 0ll << (2 * i);
                    if ((x & cur) && (y & cur))
                        num ^= 1ll << (2 * i);
                    if ((x & cur) && !(y & cur))
                        num ^= 2ll << (2 * i);
                    if (!(x & cur) && (y & cur))
                        num ^= 3ll << (2 * i);
                }
                cout << num + 1 << '\n';
            }
            else {
                long long num;
                cin >> num;
                num--;
                
                int x = 0, y = 0;
                for (int i = n - 1; i >= 0; --i) {
                    long long cur = 3ll << (2 * i);
                    if ((num & cur) >> (2 * i) == 0)
                        x ^= 0 << i, y ^= 0 << i;
                    if ((num & cur) >> (2 * i) == 1)
                        x ^= 1 << i, y ^= 1 << i;
                    if ((num & cur) >> (2 * i) == 2)
                        x ^= 1 << i, y ^= 0 << i;
                    if ((num & cur) >> (2 * i) == 3)
                        x ^= 0 << i, y ^= 1 << i;
                }
                cout << x + 1 << ' ' << y + 1 << '\n';
            }
        }
    }
    
    return 0;
}

How to check if a solution exists for a fixed $$$x$$$?

We can use binary search.

#include <bits/stdc++.h>

using namespace std;

vector<int> nums(2e5 + 5, 0);

bool check(vector<int>& v, int k, int m)
{
    int cnt = 0;
    int cur_mex = 0;
    for (int i = 0; i < v.size(); i++) {
        if (v[i] <= v.size() + 1) {
            nums[v[i]] = 1;
        }
        while (nums[cur_mex]) {
            cur_mex++;
        }
        if (cur_mex >= m) {
            cnt++;
            for (int j = 0; j < min(m + 1, (int)v.size() + 2); j++) {
                nums[j] = 0;
            }
            cur_mex = 0;
        }
    }
    for (int j = 0; j < v.size() + 2; j++) {
        nums[j] = 0;
    }
    return cnt >= k;
}

void solve()
{
    int n, k;
    cin >> n >> k;

    vector<int> v(n);
    for (int i = 0; i < n; i++) {
        cin >> v[i];
    }

    int l = 0;
    int r = 1e9;
    while (r - l > 1) {
        int m = (r + l) / 2;
        if (check(v, k, m)) {
            l = m;
        } else {
            r = m;
        }
    }

    cout << l << '\n';
}

signed main()
{
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
        solve();
    }
    return 0;
}

We need to ensure that we construct the array $$$a$$$. Therefore, when interacting with the neural network, we should consider the worst-case scenario.

Read the previous hint. How many operations of the first type applied to the $$$i$$$-th neural network do we need to obtain at least one correct word (i.e., one such that $$$b_{i, j} = a_j$$$)?

Read the previous hint and its answer. Which neural network is optimal to start with? How many blanks will there be in the array $$$c$$$ after interacting with this neural network?

There are currently no blanks in the array $$$c$$$, but there are incorrect words in some positions. How many operations are needed to replace one incorrect word with a correct one?

#include <bits/stdc++.h>
 
using namespace std;
 
void solve() {
    int n, m;
    cin >> n >> m;
    vector<string> a(n);
    for (auto &i: a) {
        cin >> i;
    }
    vector<vector<string>> b(m, vector<string>(n));
    for (auto &ar: b) {
        for (auto &s: ar) {
            cin >> s;
        }
    }
    vector<bool> exists(n);
    int x = 0;
    for (int i = 0; i < m; i++) {
        int cnt = 0;
        for (int j = 0; j < n; j++) {
            if (a[j] == b[i][j]) {
                cnt++;
                exists[j] = true;
            }
        }
        x = max(x, cnt);
    }
    if (all_of(exists.begin(), exists.end(), identity())) {
        cout << n + 2 * (n - x) << "\n";
    } else {
        cout << "-1\n";
    }
}
 
int main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

Let the shortest segment have boundaries $$$l$$$ and $$$r$$$. What can be said about the indices $$$i$$$ and $$$j$$$ for which the maximum value of the expression $$$a_i \oplus a_j$$$ is achieved?

The operation $$$x \oplus y$$$ on int32 numbers can be represented as the XOR of two binary strings of length $$$32$$$. What can be said about the comparison of numbers? In what case does the expression $$$x \oplus y = z$$$ hold? And $$$x \oplus y \gt z$$$?

How can we use a trie on binary strings to find the nearest index $$$j$$$ such that $$$a_i \oplus a_j = k$$$?

#include <bits/stdc++.h>
 
using namespace std;
 
const int LOG_X = 29;
 
struct node {
    int children[2] { -1, -1 };
    int last = -1;
};
 
int find(const vector<node>& trie, int value, int border) {
    int res = -1;
    int current = 0;
    bool ok = true;
    for (int position = LOG_X; ok && position >= 0; position--) {
        int x_bit = (value >> position) & 1;
        int k_bit = (border >> position) & 1;
        auto& children = trie[current].children;
        if (k_bit == 1) {
            if (children[x_bit ^ 1] != -1) {
                current = children[x_bit ^ 1];
            } else {
                ok = false;
            }
        } else {
            if (children[x_bit ^ 1] != -1) {
                res = max(res, trie[children[x_bit ^ 1]].last);
            }
            if (children[x_bit] != -1) {
                current = children[x_bit];
            } else {
                ok = false;
            }
        }
    }
    if (ok) {
        res = max(res, trie[current].last);
    }
    return res;
}
 
void add(vector<node>& trie, int value, int index) {
    int current = 0;
    trie[current].last = max(trie[current].last, index);
    for (int position = LOG_X; position >= 0; position--) {
        int x_bit = (value >> position) & 1;
        if (trie[current].children[x_bit] == -1) {
            trie[current].children[x_bit] = trie.size();
            trie.push_back(node());
        }
        current = trie[current].children[x_bit];
        trie[current].last = max(trie[current].last, index);
    }
}
 
void solve() {
    int n, k;
    cin >> n >> k;
 
    int ans = n + 1;
    vector<node> trie(1);
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        add(trie, x, i);
        int y = find(trie, x, k);
        if (y != -1) {
            ans = min(ans, i - y + 1);
        }
    }
 
    if (ans == n + 1) {
        cout << "-1\n";
    } else {
        cout << ans << '\n';
    }
}
 
int main() {
    int t;
    cin >> t;
    
    while (t--) {
        solve();
    }
 
    return 0;
}