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All problems were prepared by Alexdat2000 with the help of coauthors.
1634A - Reverse and Concatenate
Idea: sevlll777
Hint 1
What property will the string have after the first operation (regardless of the type of operation)?
Hint 2
What happens if you apply an operation to a palindrome?
Tutorial
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Solution
q = int(input())
for _ in range(q):
n, k = map(int, input().split())
s = input()
if s == s[::-1] or k == 0:
print(1)
else:
print(2)
1634B - Fortune Telling
Idea: crazyilian and antontrygubO_o
Hint 1
Can you figure out which of your friends can't get the number $$$y$$$ regardless of the order of operations? The answer to the problem would be the other person, since the jury guarantees that exactly one of your friends could get it.
Hint 2
What do the numbers $$$a \oplus b$$$ and $$$a + b$$$ have in common for any $$$a, b$$$?
Hint 3
What do all the numbers that can be obtained by all combinations of operations starting with $$$x$$$ have in common? Does the set of these numbers intersect with the set of numbers that can be obtained starting with $$$x + 3$$$?
Tutorial
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Solution
def main():
n, x, y = map(int, input().split())
a = list(map(int, input().split()))
if (sum(a) + x + y) % 2 == 0:
print('Alice')
else:
print('Bob')
for _ in range(int(input())):
main()
1634C - OKEA
Idea: sevlll777
Hint 1
If $$$k > 1$$$, for which $$$n$$$ is there no solution?
Hint 1.1
For which $$$n$$$ is there no solution when $$$k = 2$$$? Does the same rule of thumb hold for other $$$k > 1$$$?
Hint 2
Note that the array $$$[1, 3, 5, 7, 9, ...]$$$ satisfies the requirements (that is, the arithmetic mean of each subsegment is an integer) because the sum of the first $$$X$$$ elements of this array is $$$X^2$$$. Try to generalize this special case and find other arrays that satisfy the conditions.
Tutorial
Tutorial is loading...
Solution
def solve():
n, k = map(int, input().split())
if k == 1:
print("YES")
for i in range(1, n + 1):
print(i)
return
if n % 2 == 1:
print("NO")
return
print("YES")
for i in range(1, n + 1):
print(*[i for i in range(i, n * k + 1, n)])
for _ in range(int(input())):
solve()
1634D - Finding Zero
Idea: sevlll777
Hint 1
The problem can be reworded as follows: find $$$n - 2$$$ indices that definitely can't contain a zero.
Hint 2
Solve the problem for $$$n = 4$$$.
Hint 2.1
Denote by $$$\bar{i}$$$ the answer to the query for the three indices except $$$i$$$. Knowing $$$\bar{a}$$$, $$$\bar{b}$$$, $$$\bar{3}$$$, $$$\bar{4}$$$, can you find two indexes that don't contain zero for certain? Use the fact that there is exactly one zero in the array.
Hint 3
Solve the problem for even $$$n$$$ using the solution for $$$n = 4$$$.
Tutorial
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Solution
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define len(x) (int) (x).size()
using namespace std;
int get(const vector <int>& x) {
cout << "? " << x[0] + 1 << " " << x[1] + 1 << " " << x[2] + 1 << endl;
int ans;
cin >> ans;
return ans;
}
signed main() {
ios_base::sync_with_stdio();
cin.tie(nullptr);
int n;
cin >> n;
pair <int, int> possible = {0, 1};
for (int i = 2; i < n - 1; i += 2) {
vector <pair <int, int>> ch(4);
vector <int> now = {possible.first, possible.second, i, i + 1};
for (int j = 0; j < 4; j++) {
vector <int> x = now;
x.erase(x.begin() + j);
ch[j] = {get(x), now[j]};
}
sort(all(ch));
possible = {ch[0].second, ch[1].second};
}
if (n % 2 == 1) {
int other = 0;
while (possible.first == other || possible.second == other) {
other++;
}
vector <pair <int, int>> ch(4);
vector <int> now = {possible.first, possible.second, n - 1, other};
for (int j = 0; j < 4; j++) {
vector <int> x = now;
x.erase(x.begin() + j);
ch[j] = {get(x), now[j]};
}
sort(all(ch));
possible = {ch[0].second, ch[1].second};
}
cout << "! " << possible.first + 1 << " " << possible.second + 1 << endl;
return 0;
}
1634E - Fair Share
Idea: sevlll777
Hint 1
In what cases is the answer definitely NO?
Hint 2
The lengths of all arrays are even numbers, and each number appears an even number of times in total. There are suspiciously many even numbers...
Hint 3
An Euler circuit is a circuit that passes through all edges of a graph exactly once (or a union of several circuits if the graph is disconnected). It turns out that an Euler circuit exists if and only if the degree of each vertex is even.
Hint 4
Construct some graph such that the degree of each of its vertices is even, find a directed Eulerian circuit in it, and reconstruct the answer to the problem using this circuit. What do the edges of such graph correspond to?
Tutorial
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Solution
#include <bits/stdc++.h>
#define len(x) (int) (x).size()
#define all(x) (x).begin(), (x).end()
#define endl "\n"
using namespace std;
const int N = 4e5 + 100, H = 2e5 + 50;
vector <pair <int, int>> g[N];
string ans[N];
int pos[N];
void dfs(int v) {
if (pos[v] == 0) {
ans[v] = string(len(g[v]), 'L');
}
while (pos[v] < len(g[v])) {
auto [i, ind] = g[v][pos[v]];
if (i == -1) {
pos[v]++;
continue;
}
g[i][ind].first = -1, g[v][pos[v]].first = -1;
if (v < H) {
ans[v][pos[v]] = 'R';
}
pos[v]++;
dfs(i);
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int m;
cin >> m;
map <int, int> ind, cnt;
int fre_ind = 0;
vector <vector <int>> nums(m);
for (int i = 0; i < m; i++) {
int n;
cin >> n;
for (int _ = 0; _ < n; _++) {
int x;
cin >> x;
if (!ind.count(x)) {
ind[x] = fre_ind++;
}
x = ind[x];
cnt[x]++;
nums[i].push_back(x);
g[i].emplace_back(x + H, len(g[x + H]));
g[x + H].emplace_back(i, len(g[i]) - 1);
}
}
for (auto [num, cn] : cnt) {
if (cn % 2 == 1) {
cout << "NO" << endl;
return 0;
}
}
for (int i = 0; i < N; i++) {
dfs(i);
}
cout << "YES" << endl;
for (int i = 0; i < m; i++) {
cout << ans[i] << endl;
}
return 0;
}
1634F - Fibonacci Additions
Idea: Mangooste
Hint 1
Let $$$C_i = A_i - B_i$$$. Perform all operations on the array $$$C$$$.
Hint 2
Nevermind Fibonacci and suppose the operation was to increase the elements of a given subsegment by a given constant. Can you solve this problem without using heavy data structures like segment tree?
Hint 3
Generalize the idea from hint 2 using the characteristic property of Fibonacci numbers ($$$F_i = F_{i-1} + F_{i-2}$$$).
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define len(x) (int) (x).size()
#define endl "\n"
#define int long long
using namespace std;
const int N = 3e5 + 100;
int MOD;
int fib[N];
vector <int> unfib;
int notzero = 0;
void upd(int ind, int delta) {
notzero -= (unfib[ind] != 0);
unfib[ind] += delta;
if (unfib[ind] >= MOD) {
unfib[ind] -= MOD;
};
notzero += (unfib[ind] != 0);
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int n, q;
cin >> n >> q >> MOD;
fib[0] = fib[1] = 1;
for (int i = 2; i < N; i++) {
fib[i] = (fib[i - 1] + fib[i - 2]) % MOD;
}
vector <int> delta(n);
for (int& i : delta) {
cin >> i;
}
for (int i = 0; i < n; i++) {
int x;
cin >> x;
delta[i] = (delta[i] - x + MOD) % MOD;
}
unfib.resize(n);
unfib[0] = delta[0];
if (len(unfib) >= 2) {
unfib[1] = (delta[1] - delta[0] + MOD) % MOD;
}
for (int i = 2; i < n; i++) {
unfib[i] = ((long long) delta[i] - delta[i - 1] - delta[i - 2] + MOD * 2) % MOD;
}
for (int i = 0; i < n; i++) {
notzero += (unfib[i] != 0);
}
while (q--) {
char c;
int l, r;
cin >> c >> l >> r;
if (c == 'A') {
upd(l - 1, 1);
if (r != n) {
upd(r, MOD - fib[r - l + 1]);
}
if (r < n - 1) {
upd(r + 1, MOD - fib[r - l]);
}
} else {
upd(l - 1, MOD - 1);
if (r != n) {
upd(r, fib[r - l + 1]);
}
if (r < n - 1) {
upd(r + 1, fib[r - l]);
}
}
if (!notzero) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
}
return 0;
}
