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Spoilers are broken! Sorry for the inconvenience!

Introduction

The contest can be found here
This is an unofficial editorial and I am no way associated to Vplanet!
. Also this is my second ever blog on first code forces , and I have tried to replicate many editorial blogs as far as possible, but if I have made any mistake , please do let me know :D

Problem A Arya and Bran

We notice that the maximum amount of candies that we can give is 8 the rest of the candies are carry forward to the next day
So more formally , the tofees given on day i is min(arr[i],8)
and the number of candies carry forward are max(0,arr[i]-8).

Solution

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;
const char newl='\n';
#define ll long long
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define mp make_pair
#define debug(x) cout << (#x) << "'s value is " << (x) << newl
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0); // see /general/fast-io
	if (sz(name)) {
		freopen((name+".in").c_str(), "r", stdin); // see /general/io
		freopen((name+".out").c_str(), "w", stdout);
	}
}
// Main code starts here
void answer(int i){
    int n,k;
    cin>>n>>k;
    int arr[n];
    int sumn = 0;
    for(int i = 0;i<n;i++){
        cin>>arr[i];
    }
    for(int i = 0;i<n;i++){
        int taken = min(8,arr[i]);
        int put = max(0,arr[i]-8);
        arr[i+1]+=put;
        sumn+=taken;
        if(sumn>=k){
            cout<<i+1;
            break;
        }
    }
    if(sumn<k){
        cout<<"-1"<<newl;
    }
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    //cin>>test;
    int curr = 1;
    while(test--){
        answer(curr);
        curr++;
    }
}


/* __                             
   / /   ___  ____ ___  ____  ____ 
  / /   / _ \/ __ `__ \/ __ \/ __ \
 / /___/  __/ / / / / / /_/ / / / /    is the best!!!
/_____/\___/_/ /_/ /_/\____/_/ /_/                                                                       
*/

____________________________________________________________________________________

Problem B Love Triangle

For this problem we can directly check that there is a triangle formed , by the numbers written on the array and then print the answer accordingly

Solution

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;
const char newl='\n';
#define ll long long
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define mp make_pair
#define debug(x) cout << (#x) << "'s value is " << (x) << newl
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0); // see /general/fast-io
	if (sz(name)) {
		freopen((name+".in").c_str(), "r", stdin); // see /general/io
		freopen((name+".out").c_str(), "w", stdout);
	}
}
// Main code starts here
void answer(int i){
    int n;
    cin>>n;
    int arr[n];
    for(int i = 0;i<n;i++){
        cin>>arr[i];
    }
    for(int i = 0;i<n;i++){
        int lol = arr[i];
        int lol2 = arr[lol-1];
        int lol3 = arr[lol2-1];
        if(lol3 == i+1){
            cout<<"YES";
            return;
        }
    }
    cout<<"NO";
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    //cin>>test;
    int curr = 1;
    while(test--){
        answer(curr);
        curr++;
    }
}


/* __                             
   / /   ___  ____ ___  ____  ____ 
  / /   / _ \/ __ `__ \/ __ \/ __ \
 / /___/  __/ / / / / / /_/ / / / /    is the best!!!
/_____/\___/_/ /_/ /_/\____/_/ /_/                                                                       
*/

____________________________________________________________________________________

Problem C — Chess for three

So for this problme we need to know the following details , the person who is watching , and the players who are playing , if the winner is the spectator at any point of time then the game is not possible or else its possible.

Solution

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;
const char newl='\n';
#define ll long long
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define mp make_pair
#define debug(x) cout << (#x) << "'s value is " << (x) << newl
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0); // see /general/fast-io
	if (sz(name)) {
		freopen((name+".in").c_str(), "r", stdin); // see /general/io
		freopen((name+".out").c_str(), "w", stdout);
	}
}
// Main code starts here
void answer(int i){
    int n;
    cin>>n;
    int player1 = 1;
    int player2 = 2;
    int coming_up = 3;
    for(int i = 0;i<n;i++){
        int a;
        cin>>a;
        if(player1==a || player2==a){
            if (player2==a){
                swap(player1,coming_up);
            }
            else{
                swap(player2,coming_up);
            }
        }
        else{
            cout<<"NO";
            return;
        }
    }
    cout<<"YES";
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    //cin>>test;
    int curr = 1;
    while(test--){
        answer(curr);
        curr++;
    }
}


/* __                             
   / /   ___  ____ ___  ____  ____ 
  / /   / _ \/ __ `__ \/ __ \/ __ \
 / /___/  __/ / / / / / /_/ / / / /    is the best!!!
/_____/\___/_/ /_/ /_/\____/_/ /_/                                                                       
*/

____________________________________________________________________________________

D. Fafa and his Company

We can run a loop from the 1 to the number given and check if (number-i)%i == 0 as it is the only way that all the leaders have same number of people working under them.

Solution

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;
const char newl='\n';
#define ll long long
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define mp make_pair
#define debug(x) cout << (#x) << "'s value is " << (x) << newl
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0); // see /general/fast-io
	if (sz(name)) {
		freopen((name+".in").c_str(), "r", stdin); // see /general/io
		freopen((name+".out").c_str(), "w", stdout);
	}
}
// Main code starts here
void answer(int test){
    int n;
    cin>>n;
    int answer = 0;
    for(int i = 1;i<n;i++){
        if((n-i)%i==0){
            answer++;
        }
    }
    cout<<answer;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    //cin>>test;
    int curr = 1;
    while(test--){
        answer(curr);
        curr++;
    }
}

____________________________________________________________________________________

E. Restaurant Tables

In this problem we can take in input and check if the person is 2 seater or one seater , then if a person is two seater then we check if we have a two seater left , else we disappoint 2 customers. If its a one seater and we donot have a 1 seater left but we have a 2 seater left ,then we can put the one person there are increase the count of one seater as now anathor one person can sit with him

Solution

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;
const char newl='\n';
#define ll long long
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define mp make_pair
#define debug(x) cout << (#x) << "'s value is " << (x) << newl
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0); // see /general/fast-io
	if (sz(name)) {
		freopen((name+".in").c_str(), "r", stdin); // see /general/io
		freopen((name+".out").c_str(), "w", stdout);
	}
}
// Main code starts here
void answer(int test){
    int n;
    cin>>n;
    int answer = 0;
    for(int i = 1;i<n;i++){
        if((n-i)%i==0){
            answer++;
        }
    }
    cout<<answer;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    //cin>>test;
    int curr = 1;
    while(test--){
        answer(curr);
        curr++;
    }
}

____________________________________________________________________________________

F. Between the Offices

We can see the number of times the person moves from S to F and from F to S and then depending depending on the cound check if we have to print no or yes

Solution

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;
const char newl='\n';
#define ll long long
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define mp make_pair
#define debug(x) cout << (#x) << "'s value is " << (x) << newl
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0); // see /general/fast-io
	if (sz(name)) {
		freopen((name+".in").c_str(), "r", stdin); // see /general/io
		freopen((name+".out").c_str(), "w", stdout);
	}
}
// Main code starts here
void answer(int i){
  int a;
  cin>>a;
  string b;
  cin>>b;
  int noooo = 0;
  int yees = 0;
  char curr = b[0];
  for(int i = 1;i<a;i++){
    char lol = b[i];
    if(curr=='F' && lol=='S'){
      noooo++;
    }
    if(curr=='S' && lol=='F'){
      yees++;
    }
    curr = lol;
  }
  if(yees>noooo){
    cout<<"YES"<<newl;
  }
  else{
    cout<<"NO"<<newl;
  }
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    //cin>>test;
    int curr = 1;
    while(test--){
        answer(curr);
        curr++;
    }
}

LemonMonster's blog

Spoilers are broken! Sorry for the inconvenience!

Introduction

Problem A Arya and Bran

Problem B Love Triangle

Problem C — Chess for three

D. Fafa and his Company

E. Restaurant Tables

F. Between the Offices