BLITZ3_2 - Editorial

PROBLEM LINK:

Practice
Contest: Division 3
Contest: Division 2
Contest: Division 1

Author: Daanish Mahajan
Tester: Istvan Nagy
Editorialist: Vichitr Gandas

DIFFICULTY:

CAKEWALK

PREREQUISITES:

None

PROBLEM:

Given each Chess player has a starting time of 3 minute and after each move, 2 seconds are added to the person’s clock.

After N moves, the game ends, and the clocks of the two players have A and B seconds left. Find the total duration of the game.

Note: One move is counted as a move by a single player.

EXPLANATION:

Each player has starting time of 3 minutes or 180 seconds and there are total of N moves in the game. Each move takes 2 seconds. Hence the players spent N*2 seconds time on the moves.

We are also given the time remaining for each player which is A and B seconds respectively. So first player used 180-A seconds and second player used 180-B seconds. Hence total duration of game was N*2 + 180 - A + 180 - B = 360+N*2-A-B.

TIME COMPLEXITY:

O(1) per test case

SOLUTIONS:

Setter's Solution
#include<bits/stdc++.h>
# define pb push_back 
#define pii pair<int, int>
#define mp make_pair
# define ll long long int

using namespace std;
  
const int maxt = 1e5;
const string newln = "\n", space = " ";
int main()
{   
    int t, n, A, B; cin >> t; 
    while(t--){
        cin >> n;
        int tot = 360 + 2 * n;
        cin >> A >> B;
        if(n % 2)assert(A >= 2);
        else assert(B >= 2);
        int ans = tot - A - B;
        cout << ans << endl;            
    }
} 
Tester's Solution
#include <iostream>
#include <cassert>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <random>

#ifdef HOME
	#include <windows.h>
#endif

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

template<class T> bool umin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool umax(T &a, T b) { return a < b ? (a = b, true) : false; }

using namespace std;

long long readInt(long long l, long long r, char endd) {
	long long x = 0;
	int cnt = 0;
	int fi = -1;
	bool is_neg = false;
	while (true) {
		char g = getchar();
		if (g == '-') {
			assert(fi == -1);
			is_neg = true;
			continue;
		}
		if ('0' <= g && g <= '9') {
			x *= 10;
			x += g - '0';
			if (cnt == 0) {
				fi = g - '0';
			}
			cnt++;
			assert(fi != 0 || cnt == 1);
			assert(fi != 0 || is_neg == false);

			assert(!(cnt > 19 || (cnt == 19 && fi > 1)));
		}
		else if (g == endd) {
			assert(cnt > 0);
			if (is_neg) {
				x = -x;
			}
			assert(l <= x && x <= r);
			return x;
		}
		else {
			//assert(false);
		}
	}
}

string readString(int l, int r, char endd) {
	string ret = "";
	int cnt = 0;
	while (true) {
		char g = getchar();
		assert(g != -1);
		if (g == endd) {
			break;
		}
		cnt++;
		ret += g;
	}
	assert(l <= cnt && cnt <= r);
	return ret;
}
long long readIntSp(long long l, long long r) {
	return readInt(l, r, ' ');
}
long long readIntLn(long long l, long long r) {
	return readInt(l, r, '\n');
}
string readStringLn(int l, int r) {
	return readString(l, r, '\n');
}
string readStringSp(int l, int r) {
	return readString(l, r, ' ');
}


int main(int argc, char** argv) 
{
#ifdef HOME
	if(IsDebuggerPresent())
	{
		freopen("../in.txt", "rb", stdin);
		freopen("../out.txt", "wb", stdout);
	}
#endif
	int T = readIntLn(1, 100'000);
	forn(tc, T)
	{
		int N = readIntSp(10, 100);
		int A = readIntSp(0, 180+2*((N+1)/2));
		int B = readIntLn(0, 180 + 2 * (N / 2));
		assert(A + B > 0);
		if (N & 1)
			assert(A >= 2);
		else
			assert(B >= 2);
		int sum = 2 * (180 + N);
		int res = sum - A - B;
		printf("%d\n", res);
	}
	assert(getchar() == -1);
	return 0;
}

Editorialist's Solution
/*
 * @author: vichitr
 * @date: 26th June 2021
 */

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

void solve() {
	int n, a, b;
	cin >> n >> a >> b;

	int totalTime = 180 + 180 + n * 2;
	int duration = totalTime - a - b;
	cout << duration << '\n';
}


int main() {

#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif

	int t = 1;
	cin >> t;
	while (t--)
		solve();
	return 0;
}

VIDEO EDITORIAL:

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