A. Petya and Origami
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Petya is having a party soon, and he has decided to invite his nn friends.
He wants to make invitations in the form of origami. For each invitation, he needs two red sheets, five green sheets, and eight blue sheets. The store sells an infinite number of notebooks of each color, but each notebook consists of only one color with kk sheets. That is, each notebook contains kk sheets of either red, green, or blue.
Find the minimum number of notebooks that Petya needs to buy to invite all nn of his friends.
Input
The first line contains two integers nn and kk (1≤n,k≤1081≤n,k≤108) — the number of Petya's friends and the number of sheets in each notebook respectively.
Output
Print one number — the minimum number of notebooks that Petya needs to buy.
Examples
input
Copy
3 5
output
Copy
10
input
Copy
15 6
output
Copy
38
Note
In the first example, we need 22 red notebooks, 33 green notebooks, and 55 blue notebooks.
In the second example, we need 55 red notebooks, 1313 green notebooks, and 2020 blue notebooks.
#include<cstdio>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<functional>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<numeric>
#include<cctype>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
using namespace std;
#define N 100000+5
#define rep(i,n) for(int i=0;i<n;i++)
#define sd(n) scanf("%d",&n)
#define sll(n) scanf("%I64d",&n)
#define pd(n) scanf("%d\n",n)
#define pll(n) scanf("%I64d\n",n)
#define MAX 26
typedef long long ll;
const ll mod=1e9+7;
ll n,m;
ll a[N];
//ll b[N];
int main()
{
//string s;
//cin>>s;
ll ans=0;
//ll sum=0;
scanf("%I64d",&n);
scanf("%I64d",&m);
a[1]=n*2;
a[2]=n*5;
a[3]=n*8;
for(int i=1;i<=3;i++)
{
int k=a[i]/m;
if(k*m<a[i])
k++;
//cout<<k<<endl;
ans+=k;
}
//for(int i=1;i<=n;i++)
// printf("%I64d",a[i]);
printf("%I64d\n",ans);
return 0;
}