#include <bits/stdc++.h>
using namespace std;
#define sst string
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define freeopen freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
#define mod 1000000007
#define pb push_back
#define mk make_pair
#define ll long long
#define foor(x,vec) for(auto x:vec ){cout<<x<<" ";}
#define fi first
#define se second
#define MAXN 1000069
#define lld long double
#define cha ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ffl fflush(stdout)
#define pii pair<ll,ll>
ll mvx[]={1,-1,0,0};
ll mvy[]={0,0,-1,1};
vector <ll> find_subset(ll l,ll u,vector<ll> a){
ll n=a.size();
vector<ll> ans;
ll sum=0;
for(ll i=0;i<n;i++){
if(sum+a[i]<=u){
sum+=a[i];
ans.pb(i);
}
}
if(l<=sum && sum<=u){
return ans;
}
else{
return {};
}
}
/*
---------------------------------------------------go get gold---------------------------------------------------------------------
- If u see the problem dp there's many option to iterate, u can iterate from 1- n or 1 - (possible max number)
- If u see the problem that has unexpected constraint u can divide it into 2 problem
- If u see the problem that can use binary search, then u can use binser + check condition
- in interactive problem there's many trick using binser
- kalo misal mau cari yang sama pake pernah ke visit ga bilangan itu
- kalo problem yang high itu biasanya optimisasi nya pake 2 array
- dp bisa aja kek kamu tenzing balls dimana optimisasi 2 dp
- janlup fibonacci
- kalo binser mending r = 3 * 1e18 aja
- kalo mau dibalik itu pake value nya tinggal diubah ke size
*/
Compilation message
/usr/bin/ld: /tmp/ccOhyksQ.o: in function `main':
grader.cpp:(.text.startup+0x18d): undefined reference to `find_subset(int, int, std::vector<int, std::allocator<int> >)'
collect2: error: ld returned 1 exit status
