I am glad to invite you to #3 Round for Beginners (Div.4). There no time for contest so you can participate any time in virtual participation. You will have 2 hours and 5 problems. The contest will be in IOI style.

Some problems can be interesting for advanced users so I advise you to view all tasks.

• abdu11a, domalak_kaktus, jasur-tairov-kentau-2025, TimDee, Esomer, Lyrically, zhasurr for testing our round and tell useful feedback.
• MikeMirzayanov for systems Codeforces and Polygon.
• You for participating.

Link for contest: click here

Link for group with previous rounds: click here

Link for blog of previous round: click here

Good luck for everyone!

As you know, recently I did a contest for beginners. Many people liked this round (there were more participants than I expected). And now meet the second round!

I happy to invite you to #2 Round for Beginners (Div.4). You can participate in virtual participation. You will have 2 hours and 5 problems. The contest will be in IOI style.

• abdu11a, domalak_kaktus, jasur-tairov-kentau-2025, TimDee, Esomer, Lyrically, zhasurr for testing our round and tell useful feedback
• MikeMirzayanov for systems Codeforces and Polygon
• You for participating

• New group where collected all contests $$$\bigstar$$$
• Russian statements of tasks $$$\bigstar$$$
• Fixed lot of errors $$$\bigstar$$$
• No bugs with checkers $$$\bigstar$$$
• More interesting tasks! $$$\bigstar$$$

LINK FOR CONTEST: click here

Good luck for everyone!

Editorial for #1 Round for Beginners (Div. 4)

$$$A.$$$ Coins for friends

Authors: zhasurr, round-for-beginners

The most optimal solution was $$$n$$$ + $$$1$$$. Perfect can give each friend $$$1$$$ coins and he himself will get $$$1$$$ coin.

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

int main() {
    long long int n;
    cin >> n;
    cout << n + 1;

    return 0;
}

n = int(input())
print(n + 1)

$$$B.$$$ Road to be Legend

Authors: zhasurr, round-for-beginners, abdu11a

First, take a separate value that will read the lessons and the value that will store the largest. Check yesterday's lesson is the same as today's. If not, then increase the value and the value that stores the longest days will be equal to the maximum of these two values, otherwise the value is equal to $$$1$$$. At the end, check the two values again and print the value that stores the largest

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

int main(){
    int n;
    cin >> n;
    string s;
    cin >> s;
    int ans = 0, k = 1;

    for(int i = 1; i < n; i++){
        if(s[i] != s[i - 1]) k++;
        else{
            ans = max(ans, k);
            k = 1;
        }
      }
    
    ans = max(ans, k);
    cout << ans;
 
    return 0;
}

n = int(input())
arr = input()
best = 0
cnt = 0
for i in range(n - 1):
    if arr[i] != arr[i + 1]:
        cnt += 1
    else:
        best = max(best, cnt)
        cnt = 0
print(max(cnt + 1, best + 1))

$$$C.$$$ Pair of numbers

Authors: zhasurr

If the sum of all numbers in the array is odd, then NO is displayed exactly. Otherwise, take a single number that is equal to $$$sum$$$/$$$2$$$. Sort the array. Now it is exactly guaranteed that the penultimate number and $$$a_i$$$ will be equal to $$$sum$$$/$$$2$$$ or when $$$a_i$$$ + $$$a_{i + 1}$$$ will be equal to $$$sum$$$/$$$2$$$

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

#define ll long long

int main() {
    ll n;
    cin >> n;
    ll a[n];
    ll sum = 0;
    
    for (ll i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
    }
    
    if (sum % 2 == 1) {
        cout << "NO";
        return 0;
    }

    sort(a, a + n);
    ll sm = sum / 2;
        
    for (ll i = 0; i < n - 1; i++) {
        if (a[i] + a[n - 1] == sm || a[i] + a[i + 1] == sm) {
            cout << "YES";
            return 0;
        }
    }

    cout << "NO";

    return 0;
}

from collections import defaultdict
n = int(input())
arr = list(map(int,input().split()))
s = sum(arr)
if s%2 != 0:
    print("NO")
    exit()
s = s//2
d = defaultdict(int)
for i in arr:
    d[i] += 1
for i in range(n):
    if s < arr[i]:
        continue
    if s - arr[i] == arr[i]:
        if d[arr[i]] > 1:
            print("YES")
            break
    else:
        if d[s - arr[i]]!=0:
            print("YES")
            break
else:
    print("NO")

$$$D.$$$ Permutation

The task was taken from the 3rd stage of the Republican Olympiad 2020 (Kazakhstan)

The first number that is odd will jump two steps forward, i.e. the 1st number will become 3 and the 3 will jump the next two numbers and will equal 5. With the even ones everything is the same. For example, the number 2 will be equal to 4, and the 4th to 6, etc.

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

#define speed ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long

int main() {
    speed;
    ll n;
    cin >> n;
    ll a[n + 1];

    for (ll i = 1; i <= n; i++) {
        if(i % 2 != 0) a[i] = i + 2;
        else a[i] = i - 2;
    }

    if (n % 2 != 0) a[n] = n - 1;
    else a[n - 1] = n;
    a[2] = 1;

    for (ll i = 1; i <= n; i++) cout << a[i] << ' ';
}

n = int(input())
arr = [i for i in range(2, n + 1)]
arr.append(1)
print(*arr)

$$$E.$$$ Joyful numbers

Author: round-for-beginners

#include <iostream>
using namespace std;

int main() {
	long long int n;
	cin >> n;
	cout << ((n - 1) * 10 + 5) * ((n - 1) * 10 + 5);
	return 0;
}

k = int(input())
print(((k - 1) * 10 + 5) ** 2)

UPD1: Searching for testers to new rounds. If you want to be tester write message to this mail: [email protected]

Hello, Codeforces!

I am glad to invite you to the round for beginner coders. There will be 6-7 problems. Try to solve all problems. The contest will be in IOI style. Thanks for testers of the round and for the good feedback.

I tried to choose and do good tasks. I hope you find them interesting. Read the conditions from all tasks. It is not guaranteed that the tasks are in order of increasing difficulty.

Would you like to participate in this contest? Follow the link. Participate in virtual participation and solve problems.

Sorry for the bad translation, LaTeX formatting or for not clear statements. I tried my best and this my first contest. If you like this round, then in the future I will try to make more such contests.

Good luck in CP :D

UPD1: Editroial here

UPD2: Searching for testers to new rounds. If you want to be tester write message to this mail: [email protected] or write to me