Discussion About Compute F(N)

#1

This is Discussion thread about Compute F(N)

#2

Please explain what logic will be used to solve this problem ??
My logic is getting TLE

#3

are you mad test case is 10^18?
if we can answer in O(1) then also it is giving tle to me.

#6

package HackerBlocks;

import java.util.Scanner;

public class Series {

public static void main(String[] args) {
	int times = 0;
	Scanner scan=new Scanner(System.in);
	if(scan.hasNextInt()) {
		times=scan.nextInt();
	}
	while(times>0) {
		long n=0;
		if(scan.hasNextInt())
			n=scan.nextInt();
		 long sum=0;
		for(int i=1;i<=n;i++) {
			if(i%2==0) {
				sum+=i;
			}else {
				sum-=i;
			}
		}
		System.out.println(sum);
		times--;
	}
}

}

this is my code. It is running fine in eclipse but it is showing error here, it would be great if someone could help

#7

We can do it like this if the number is even then the sum=n/2 and if odd then it will be sum=-(n/2 + 1)
eg: for n=1 sum=-1;
n=2 sum=1; n=3 sum =-2 , n=4 sum=2 ,n=5 sum=-3 n=6 sum=3 …

#8

This is clearly a recursive relation for which we can derive an O(1) formula.
f(n) = f(n-2k) + k(-1^n)
if n is even:
let k = n/2;
then f(n) = f(0) + n/2
else (when n is odd):
let k = (n-1)/2;
then f(n) = f(1) - (n-1)/2; (negative because (-1)^odd number)

Now the base cases are: f(0) = 0, f(1) = -1;
~ O(T*1)

#9 
            if(i%2==0)
			sum+=i;
			else
			sum-=i;
#10

#include<bits/stdc++.h>
using namespace std;
int main() {
long long a,b,s,e;
cin>>a;
while(a–){
cin>>b;
s=0;
e=0;
long long f;
f=b/2;
s=((f+1)f);
if(b%2==0){
e=ff;
}
else
e=(f+1)*(f+1);

cout<<s-e<<endl;
}
return 0;
}
i think this better than other…