KETEKI2A editorial

PROBLEM LINK :

Practice
Contest

Setter: Chirag Thakur

Tester: Harsh Raj

Editorialist: Harsh Raj

DIFFICULTY:

Cakewalk

PREREQUISITES:

Basic Implementatio

EXPLAINATION:

For each turn, keep on drawing cards and store the points.
When there is no card left, print the diffrence of output.

TIME COMPLEXITY:

O(N)

SOLUTIONS:

Settler's Solution

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

	string to_string(string s) {
	  return '"' + s + '"';
	}
	 
	string to_string(const char* s) {
	  return to_string((string) s);
	}
	 
	string to_string(bool b) {
	  return (b ? "true" : "false");
	}
	 
	template <typename A, typename B>
	string to_string(pair<A, B> p) {
	  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
	}
	 
	template <typename A>
	string to_string(A v) {
	  bool first = true;
	  string res = "{";
	  for (const auto &x : v) {
	    if (!first) {
	      res += ", ";
	    }
	    first = false;
	    res += to_string(x);
	  }
	  res += "}";
	  return res;
	}
	 
	void debug_out() { cerr << endl; }
	 
	template <typename Head, typename... Tail>
	void debug_out(Head H, Tail... T) {
	  cerr << " " << to_string(H);
	  debug_out(T...);
	}
	 
	#ifdef LOCAL
	#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
	#else
	#define debug(...) 42
	#endif
	 
	void test_case() {
	  int n;
	  cin >> n;
	  vector<int> C(n + 1);
	  for (int i = 1; i <= n; ++i) {
	    cin >> C[i];
	  }
	  int k = 0, p = 0;
	  int l = 1, r = n, turn = 1;
	  while (l <= r) {
	    if (turn) {
	      p += C[l++];
	    } else {
	      k += C[r--];
	    }
	    turn ^= 1;
	  }
	  cout << k - p << "\n";
	}
	 
	int main() {
	  ios_base::sync_with_stdio(0);
	  cin.tie(0);
	  cout.tie(0);
	  test_case();
	  return 0;
	}
Tester's Solution

	#include<bits/stdc++.h>
	using namespace std;
	#define ll long long int
	int main(){
		ll i,j,n,turn=0,ketu=0,petu=0;
		cin>>n;
		vector<int> c(n);
		for(i=0;i<n;i++)
			cin>>c[i];
		i=0,j=n-1;
		while(i<=j){//while there are elements remaining
			if(turn%2==0){
				petu+=c[i];
				i++;
			}
			else{
				ketu+=c[j];
				j--;
			}
			turn+=1;
		}
		cout<<(ketu-petu)<<endl;
		return 0;
	}