Factory Machines
Key Idea: Binary search on time, checking whether all machines together can produce enough units by that time.
Solution
#include <bits/stdc++.h>
using namespace std;
//author: von_Braun
#define ll long long
#define lli long long int
#define pb push_back
#define rep(var, start, num) for(ulli var = start; var <start + num; var++)
#define all(x) x.begin(), x.end()
#define ulli unsigned long long int
#define ull unsigned long long
bool sortbysec(const pair<ll,ll> &a,const pair<ll,ll> &b) { return (a.second < b.second); }
void solve() {
ull n,m;
cin>>n>>m;
vector<ull> a(n);
rep(i,0,n) {cin>>a[i];}
ull l = 0;
ull r = INT64_MAX;
auto checkbruh = [&](ull time){
ull s=0;
rep(i,0,n) {s+=(time/a[i]);}
return (s>=m);
};
while(l!=r) {
if (r-l==1) {break;}
ull m = (l+r)/2;
if (checkbruh(m)) {
r=m;
} else {
l=m+1;
}
}
if (checkbruh(l)) {cout<<l;} else {cout<<r;}
}
int main() {
//add quotes incase input output file
//freopen(input.txt,r,stdin);
//freopen(output.txt,w,stdout);
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int tc = 1;
// cin >> tc;
for (int t = 1; t <= tc; t++) {
solve();
}
}