How to overcome time limit exceed ...

Question

1.00/5 (1 vote)

See more:

i have a problm input n is highest 10^10;for worst case it must give TLE ...in this problem i have got probably everything ok but getting TLE for worst case may be due to using for loop...here time limit is 1 sec.there are two input n maximum 10^5,k maximum 10^9.how to overcome..thanks in advance

What I have tried:

#include <iostream>
using namespace std;

int main()
{

    long long n,k;
    while(cin>>n>>k)
    {
      long long a[n];
      for(int i=0;i<n;i++)>
      {
          cin>>a[i];
      }
        int count=0;

        int i=1;
        //count=i*(i+1)/2;
        while(k>i*(i+1)/2)
        {
            i++;
            //count=i*(i+1)/2;
        }

        int len=i*(i+1)/2;
        int j,x=0;
        for(j=len,x=i-1;j>k;j--)           //may be here is the problem i guess..
        {
            x--;
        }
        cout<<a[x]<<endl;
    }

    //cout << "Hello world!" << endl;
    return 0;
}

Posted 5-May-16 8:01am

Member 12270086

Updated 5-May-16 19:14pm

jeron1

v2

Add a Solution

Comments

Richard MacCutchan 5-May-16 14:04pm

What?

Sergey Alexandrovich Kryukov 5-May-16 14:25pm

It does not seem to make any sense. Could you explain the problem clearly?
—SA

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

leon de boer · Answer 1 · 2016-05-05T19:14:00

LOL I agree with Richard & Sergey in WTF but even funnier is reading the code

I am wondering if it ever comes out surely it just deadloops forever if value k is too big that is my guess.

These lines of code are the best ever

int i=1;
     while(k>i*(i+1)/2)

Lets give you a hint:
i) k is a LONG LONG and i is an INTEGER.
- You are going to be a little hard pressed to multiply roll an INT up to a LONG LONG

Then you go and do it all again with integer j.

I must say you got me, I can't work out if you are trying to get upperand lower longs of the long long or get the square root in a strange way. It sort of looks like you may be trying to do this

unsigned long llsqrt (unsigned long long x){
	unsigned long long op, res, one;

	op = x;
	res = 0;

	/* "one" starts at the highest power of four <= than the argument. */
	one = 0x4000000000000000;  /* second-to-top bit set  bit */
	while (one > op) one >>= 2;

	while (one != 0) {
		if (op >= res + one) {
			op -= res + one;
			res += one << 1; 
		}
		res >>= 1;
		one >>= 2;
	}
	return (unsigned long) res;
}