Why does my code give runtime error timeout?

Question

I just want to add the sqrt of n to the divisors vector if its integer but everytime I try the code It goes Haywire. My vector.push_back(sqrt(n)) Function is giving the problem I sense.

Runtime error occurs . But when I remove those 3 Lines it works Fine.

Been trying for a long time!

INPUT - 1 100 8 23 11

I NEED HELP!

#include <bits/stdc++.h>
#include <stdio.h>
#include <iostream>

#define ll long long int
#define li long int

using namespace std;

bool isPerfectSquare(long long n) {
  long long squareRootN = (long long)round((sqrt(n)));

  if (squareRootN * squareRootN == n) {
    return true;
  } else {
    return false;
  }
}
int main() {
  ios::sync_with_stdio(0);
  ll t;
  cin >> t;
  while (t--) {
    ll n, a, b, c;
    cin >> n;
    cin >> a >> b >> c;

    vector<ll> divisors;
    for (ll i = 1; i < sqrt(n); i++) {
      if (n % i == 0) {
        divisors.push_back(i);
        if ((n / i) != i)
          divisors.push_back(n / i);
      }
    }
    // HEREEEEEEE ISS THE PROBLEMMMM
    ll y = sqrt(n);
    if (isPerfectSquare(n))
      divisors.push_back(y);

    ll z = divisors.size();
    for (ll i = 0; i < z; i++)
      cout << divisors[i] << ' ';

    sort(divisors.begin(), divisors.end());
    cout << '\n';

    ll x = divisors.size();
    for (ll i = 0; i < x; i++)
      cout << divisors[i] << ' ';

    ll endd = divisors.size();

    ll counter = 0;
    for (ll i = 0; i < endd; i++) {
      for (ll j = 0; j <= endd; j++) {
        if (n % (divisors[i] * divisors[j]) == 0 &&
            n / (divisors[i] * divisors[j]) <= c && divisors[i] <= a &&
            divisors[j] <= b) {
          cout << '\n'
               << divisors[i] << ' ' << divisors[j] << ' '
               << n / (divisors[i] * divisors[j]);
          counter++;
        }
      }
    }

    cout << '\n' << counter << '\n';
  }
  return 0;
}

Thanks A Lot!

asking someone to review code that uses `#define li long int` is not nice. Do you really thing this adds to clarity and readability? — 463035818_is_not_an_ai, Feb 21 '18 at 15:03
I don't know of any standard mechanisms that can cause a timeout error, and nothing in the provided code indicates that it could either. — François Andrieux, Feb 21 '18 at 15:03
[Why should I not #include ?](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — François Andrieux, Feb 21 '18 at 15:04
I feel it's common sense to inter-prate ll as long long int but then I didn't know. — Akhil Vaid, Feb 21 '18 at 15:04
online coding judges are the wrong tool to debug your program. I feel it is common sense that if I want a `long long int` i write `long long int` and not a cryptic 2 letter name — 463035818_is_not_an_ai, Feb 21 '18 at 15:06
@NoName did you try to execute your code without this `#include`? What exactly in your code justifies its use? This is a question you should ask yourself before adding any `#include` to your code. — Rafalon, Feb 21 '18 at 15:12
@NoName I don't know. It seems you are leaving something out. Nothing here is likely to cause a timeout error. There is no such thing in standard c++, so it must originate from your code. Perhaps it's related to the environment you are running this in. — François Andrieux, Feb 21 '18 at 15:13
I don't see how it makes a difference but still I tried it. No difference.@Rafalon — Akhil Vaid, Feb 21 '18 at 15:14
I tried on Ideone as well Error there too @FrançoisAndrieux https://ideone.com/PQ6igm — Akhil Vaid, Feb 21 '18 at 15:15
@NoName My problem is that the question mentions a **timeout** error. — François Andrieux, Feb 21 '18 at 15:15
@FrançoisAndrieux I don't think I know the difference between timeout and runtime. Sorry. Just starting out. :| — Akhil Vaid, Feb 21 '18 at 15:24
Please [edit] your code to reduce it to a [mcve] of your problem. Your current code includes much that is peripheral to your problem - a minimal sample normally looks similar to a good unit test: only performing one task, with input values specified for reproducibility. — Toby Speight, Feb 21 '18 at 18:10
"Runtime error occurs" - please edit your question and include the text of the error(s) you're seeing. Thanks. — Bob Jarvis - Слава Україні, Feb 21 '18 at 18:50

score 2 · Accepted Answer · edited Feb 21 '18 at 15:14

2

ll endd=diviors.size();
// ...
for (ll j = 0; j <= endd; j++) 
// ...            ^
    divisors[j]

When j = ennd you access out of the vector

edited Feb 21 '18 at 15:14

François Andrieux

28,148
6
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answered Feb 21 '18 at 15:14

yannick ott

76
4

Wouldn't this imply an `IndexOutOfRange` exception instead of a timeout? – Rafalon Feb 21 '18 at 15:15
1

@Rafalon That's undefined behavior and the result could be anything. `IndexOutOfRange` is something a platform might do if it's able to detect the error, but it's not something you should expect or rely on. Perhaps you are thinking of [`std::out_of_range`](http://en.cppreference.com/w/cpp/error/out_of_range) that can be thrown by the `at` functions. – François Andrieux Feb 21 '18 at 15:17
@FrançoisAndrieux well I've done too much C# and got used to well defined and explicit exceptions. Still I don't understand why the OP mentioned a **timeout**. – Rafalon Feb 22 '18 at 07:53

Why does my code give runtime error timeout?

1 Answers1