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back to boardDiscussion of Problem 1068. SumI got AC with an 28K program!! Posted by Drizzt 15 Mar 2003 08:08 program Sum (input, output);
var lI, lN : longint;
begin
readln (lN);
if lN > 0 then lI := (lN + 1) * lN div 2;
if lN = 0 then lI := 1;
if lN < 0 then lI := (abs (lN) + 1) * lN div 2 + 1;
writeln (lI);
end. it isn`t too intresting is it?!!!! i used less memory than you :D (+) Var
n :integer;
sum :longint;
begin
readln(n);
if n=0 then n:=1;
sum:=abs(n)*(abs(n)+1) div 2;
if n<0 then sum:=1-sum;
writeln(sum);
readln;
end.
~~~~~~~~~~~~~~
u used 2 Longint (2*4b) and mine 1longint (4) + a Integer (2)
so 4+2<2*4 !!! :P
Please Stop sending AC dear,
Best Regards
Aidin_n7 Wow! You are both the best programmers in the world! I couldn't ever dream about such good use of memory! ^_^ woo, so many AC codes are posted. doesn't matter my spoiling one more. #include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
int main (void) {
int n;
scanf("%d", &n);
printf("%d\n", ((1 + n) * (abs(1 - n) + 1)) >> 1);
return 0;
}
Re: it isn`t too intresting is it?!!!! i used less memory than you :D (+) How about my programme.
var
n:longint;
begin
readln(n);
writeln((1+n)*(abs(n-1)+1)div 2);
end.
Edited by author 18.02.2006 15:12
Edited by author 18.02.2006 15:12 |
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