 # BLITZ3_2 - Editorial

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

CAKEWALK

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) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

int main(int argc, char** argv)
{
#ifdef HOME
if(IsDebuggerPresent())
{
freopen("../in.txt", "rb", stdin);
freopen("../out.txt", "wb", stdout);
}
#endif
forn(tc, T)
{
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:

If you have other approaches or solutions, let’s discuss in comments.