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toshirohitsugaya    Created at    7 likes

`#include <bits/stdc++.h> using namespace std; typedef long long ll; int main(){ ll l,r; cin >> l >> r; for(int i = l ;i <= r;i++){ cout << i << " "; } return 0; }`

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omegumunot    Created at    5 likes

# Loop ## Idea We can use a for loop to iterate over all numbers from $l$ to $r$ and print it in $O(r - l)$. ## Code `#include <bits/stdc++.h> using namespace std; using ll = long long; const int MOD = 1E9 + 7; const int INF = 1E9; const ll INFLL = 1E18; int l; int r; int main() { ios_base::sync_with_stdio(false); cin.tie(0); cin >> l >> r; for(int i = l; i <= r; i++) { cout << i << "\n"; } }`