# SUBSCRIBE_ - Editorial

Setter: Jeevan Jyot Singh
Tester: Takuki Kurokawa, Lavish Gupta
Editorialist: Yash Kulkarni

504

Basic Math

# PROBLEM:

A new TV streaming service was recently started in Chefland called the Chef-TV.

A group of N friends in Chefland want to buy Chef-TV subscriptions for each of them. We know that 6 people can share one Chef-TV subscription. Also, the cost of one Chef-TV subscription is X rupees.
Determine the total cost that the group of N friends will incur so that everyone in the group is able to use Chef-TV.

# EXPLANATION:

The N friends will form \lfloor \frac{N}{6} \rfloor sharing groups having 6 people in each sharing group. If N is not divisible by 6 then the remaining N (mod 6) people will form an additional sharing group having less than 6 people. In order words the N friends require \lceil \frac{N}{6} \rceil number of Chef-TV subscriptions. The total cost that the group of N friends will incur is X.\lceil \frac{N}{6} \rceil.

# TIME COMPLEXITY:

O(1) for each test case.

# SOLUTION:

Setter's solution
    #include <wtsh.h>
#else
#include <bits/stdc++.h>
using namespace std;
#define dbg(...)
#endif

#define int long long
#define endl "\n"
#define sz(w) (int)(w.size())
using pii = pair<int, int>;

const long long INF = 1e18;

const int N = 1e6 + 5;

// -------------------- Input Checker Start --------------------

long long readInt(long long l, long long r, char endd)
{
long long x = 0;
int cnt = 0, 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)
{
if(is_neg)
x = -x;
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(false);
}
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 readStringSp(int l, int r) { return readString(l, r, ' '); }
void readEOF() { assert(getchar() == EOF); }

vector<int> readVectorInt(int n, long long l, long long r)
{
vector<int> a(n);
for(int i = 0; i < n - 1; i++)
a[n - 1] = readIntLn(l, r);
return a;
}

// -------------------- Input Checker End --------------------

void solve()
{
cout << (n + 5) / 6 * x << endl;
}

int32_t main()
{
ios::sync_with_stdio(0);
cin.tie(0);
for(int tc = 1; tc <= T; tc++)
{
// cout << "Case #" << tc << ": ";
solve();
}
return 0;
}


Editorialist's Solution
using namespace std;

int main() {
int T;
cin >> T;
while(T--){
int n,x;
cin >> n >> x;
int m=n/6;
if(n%6!=0)m++;
cout << m*x << endl;
}
return 0;
}