TheForces Round #18 (JuneIsApril-Forces) Editorial

#	User	Rating
1	tourist	3985
2	jiangly	3814
3	jqdai0815	3682
4	Benq	3529
5	orzdevinwang	3526
6	ksun48	3517
7	Radewoosh	3410
8	hos.lyric	3399
9	ecnerwala	3392
9	Um_nik	3392

#	User	Contrib.
1	cry	169
2	maomao90	162
2	Um_nik	162
4	atcoder_official	161
5	djm03178	158
6	-is-this-fft-	157
7	adamant	155
8	awoo	154
8	Dominater069	154
10	luogu_official	150

Thanks you for participating in TheForces Round #18! Here is editorial:

Editorial

Solution is to print "TheForces rounds!".

Solution

#include <bits/stdc++.h>
using namespace std;
 
int32_t main() {
	cout << "TheForces rounds!";
}

104443B - Smaller than 100

Editorial

The answer is the number of TheForces Rounds with less than 100 participants (7).

Solution

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

int32_t main() {
	cout << 7;
}

104443C - Morco-Feely Palindromes

Editorial

In this problem, you need to determine if the string is a palindrome when translated into Morse.

Solution

#include <bits/stdc++.h>
using namespace std;
 
int32_t main() {
	ios::sync_with_stdio(0); cin.tie(0);
	string s; cin >> s; int n = s.size();
	bool bad = false;
	for (int i = 0; i < n; i++)
		if ((s[i] + s[n - 1 - i] - '0' - '0') % 10 != 0)
			bad = true;
	if (bad) cout << "NO";
	else cout << "YES";
}

104443D - Missing Characters

Editorial

In this problem, you need to print concatenation of missing characters in statements B, A, and D. The answer is "mty".

Solution

#include <bits/stdc++.h>
using namespace std;
 
int32_t main() {
    cout << "mty";
}

104443E - Cringemeter

Editorial

You need to find how cringe is the given string. To do it you should calculate the cost of the string and divide it by the cost of "cringe". The cost of the string is equal to $$$\sum (a_i\mod 10)$$$ where $$$a_i$$$ is the index (0-based) of $$$s_i$$$ in the English alphabet.

Solution

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

void test_case() {
	string s; cin >> s;
	int cringe = 30, sum = 0;
	for (char c : s)
		sum += (c - 'a') % 10;
	cout << sum / cringe << '\n';
}

int32_t main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int t; cin >> t;
	while (t--) test_case();
}

104443F - Rt Dg

Editorial

You have to find the degree of the root in directed tree.

Solution

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

#define N 1100

int in[N], out[N];

int main() {
	int n; cin >> n;
	for (int i = 1; i < n; i++) {
		int u, v; cin >> u >> v;
		in[v]++; out[u]++;
	}
	for (int i = 1; i < n; i++) if(in[i] == 0)
		cout << out[i];
}

104443G - Qpert pg yep

Editorial

All characters in the statement and the name of the problem are encoded in the next way: letter changes to the letter to the right of it. The name of the problem is actually "Power of two". It is not hard to guess that the answer is the last digit of $$$2^m$$$.

Solution

#include <bits/stdc++.h>
using namespace std;
 
int32_t main() {
	int m; cin >> m;
	if (m == 1) cout << 1;
	else cout << "2486"[(m - 2) % 4];
}

104443H - Random Generator

Editorial

The answer to this problem is 1. There is no reason why is it 1.

Solution

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

int32_t main() {
	cout << 1;
}

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