Efficiently calculate the value of an arithmetic sequence

Question

I need to calculate the value for the following function:

$$f(n) = -1 + 2 - 3 + ... + -1^n$$

The time limit to calculate a given value of \$n\$ is less than 1 second. My approach does not meet that requirement at large sizes, for example \$f(1000000000)\$.

My code:

program A486;
uses wincrt, sysutils;
var
 n: LongInt;
function f(n : LongInt): LongInt;
var
 i : LongInt;
 //Result : LongInt;
begin
 Result:= 0;
 for i:= 1 to n do 
 if i mod 2 = 0 then
 Result += i
 else
 Result -= i;
 //f:= Result; 
end;
begin
 readln(n);
 writeln(f(n));
 //readkey()
end. 

Is there a better way I can do this?

Answer 1

Seems trivial. Pair them, each pair is worth -1.

f(1000) = 1 - 2 + 3 - 4 + ... + 999 - 1000
 = (1 - 2) + (3 - 4) + ... + (999 - 1000)
 = -1 -1 + ... + -1
 = -500

Odd n left as exercise for the reader.

(I wrote this when the question title still said "1 − 2 + 3 − 4 + · · ·", can't be bothered to switch all signs now.)

Answer 2

According to the help of Mr Andreas, i have come up with this code, which was helpful, but it gave a runtime error, in this test : 1000000000000000, I have to disapprove the answer as it wasn't the best performance. But i still greatly thank you for your attention.

code :

program A486;
uses wincrt, sysutils;
var
 n: LongInt;
function f(n : LongInt): LongInt;
var
 {Result,} SumOdd, SumEven : LongInt;
begin
 Result:=0; SumOdd:=0; SumEven:=0; 
 if n mod 2 = 0 then
 begin 
 SumOdd:=( n*(n div 2) ) div 2;
 SumEven:=((2+n)*(n div 2) ) div 2;
 Result:= SumOdd - SumEven;
 end else 
 begin
 SumOdd:=( (n+1)*(n div 2 + 1) ) div 2;
 SumEven:=((n+1)*(n div 2) ) div 2;
 Result:= SumOdd - SumEven; 
 end; 
 //f:= Result; 
end;
begin
 readln(n);
 writeln(f(n));
 //readkey()
end.

Answer 3

As far as we discussed it, this is the most efficient answer found:

program A486;
uses wincrt, sysutils;
var
 n: LongInt;
function f(n : LongInt): LongInt;
//var
 //Result : LongInt;
begin
 Result:=0; 
 if n mod 2 = 0 then
 Result:= n div 2
 else 
 Result:= (n div 2) - n;
 //f:= Result; 
end;
begin
 readln(n);
 writeln(f(n));
 //readkey()
end. 

thanks for everyone helped!

• Coupcoup: It might not speed up your code but you could rewrite that whole if statement as -1^(n % 2) * ceiling(n / 2) The pattern to your problem is that results go -1, 1, -2, 2, -3, 3, ... -roundUp(n / 2), roudUp(n / 2). Odd numbers mod 2 are always 1 and even numbers mod 2 are 0 so -1^(n mod 2) will keep alternating the negative sign for you and roundUp(n / 2) gives you the int value at each step
• W. Chang: how about Result := (n shr 1) - n * (n and 1);?