MOVIE2X - Editorial

editorial
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PROBLEM LINK:

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Daanish Mahajan
Tester: Harris Leung
Editorialist: Aman Dwivedi

DIFFICULTY:

Simple

PROBLEM:

The chef started watching a movie that runs for a total of XX minutes.

Chef has decided to watch the first Y minutes of the movie at twice the usual speed as he was warned by his friends that the movie gets interesting only after the first Y minutes.

How long will Chef spend watching the movie in total?

EXPLANATION:

Let’s watch Y minutes of the movie at twice the usual speed. Hence, it will require Y/2 minutes to watch the first Y minutes of the movie.

Now, the remaining part of the movie is of (X-Y) minutes and since we are watching this part with normal speed, hence it requires (X-Y) minutes in total.

Hence total time required to watch the entire movie is

\frac{Y}{2}+(X-Y)

This is our final answer, hence simply print this.

TIME COMPLEXITY:

O(1) per test case

SOLUTIONS:

Setter 
#include<bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
#define LL long long
LL seed = chrono::steady_clock::now().time_since_epoch().count();
mt19937_64 rng(seed);
#define rand(l, r) uniform_int_distribution<LL>(l, r)(rng)
clock_t start = clock();

#define getchar getchar_unlocked

namespace IO {
long long readInt(char endd) {
    long long ret = 0;
    char c = getchar();
    while (c != endd) {
        ret = (ret * 10) + c - '0';
        c = getchar();
    }
    return ret;
}
long long readInt(long long L, long long R, char endd) {
    long long ret = readInt(endd);
    assert(ret >= L && ret <= 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 readString(int l, int r) {
    string ret = "";
    char c = getchar();
    while (c == '0' || c == '?' || c == '1') {
        ret += c;
        c = getchar();
    }
    assert((int)ret.size() >= l && (int)ret.size() <= r);
    return ret;
}
}
using namespace IO;

const int TMAX = 1'00'000;

void solve() {

}

int main() {
    int x = readIntSp(1, 1000);
    int y = readIntLn(1, 1000);
    cout << x - y / 2 << '\n';
    assert(getchar() == EOF);
    cerr << fixed << setprecision(10);
    cerr << (clock() - start) / ((long double)CLOCKS_PER_SEC) << " secs\n";
    return 0;
}
Tester
Editorialist 
#include<bits/stdc++.h>
using namespace std;

void solve()
{
  int x,y;
  cin>>x>>y;

  cout<<(x-y)+y/2<<"\n";
}

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int t;
  t=1;

  while(t--)
    solve();

return 0;
}