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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?

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3 Answers 3

Reset to default
1
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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.)

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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.
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0
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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!

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2
  • 1
    \$\begingroup\$ 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 \$\endgroup\$ Commented Sep 11, 2020 at 22:31
  • 1
    \$\begingroup\$ how about Result := (n shr 1) - n * (n and 1);? \$\endgroup\$ Commented Sep 22, 2020 at 17:41

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