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RGB7072 - Нийлбэр 3

Өгөгдсөн сондгой тоо n хүртэлх натурал сондгой тоонуудын нийлбэрийг ол. Давталт ашиглахгүй.

Input

Сондгой натурал тоо өгөгдөнө.

Output

Нийлбэр.

Example

Input:
9

Output:
25

Нэмсэн:Bataa
Огноо:2013-01-06
Хугацааны хязгаарлалт:1s
Эх кодын хэмжээний хязгаарлалт:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Програмчлалын хэлүүд:ADA95 ASM32 BASH BF C NCSHARP CSHARP C++ 4.3.2 CPP C99 CLPS LISP sbcl LISP clisp D ERL FORTRAN HASK ICON ICK JAVA JS-RHINO JULIA LUA NEM NICE OCAML PAS-GPC PAS-FPC PERL PHP PIKE PRLG-swi PYTHON PYPY3 PYTHON3 RUBY SCALA SCM guile ST TCL WHITESPACE

hide comments
2023-03-25 03:13:17
here is the formula to solve this (x+1) * (x+1)/4
2023-03-25 03:12:31
sheesh
2023-03-19 04:58:15
are u gay
2023-03-03 10:03:02
c deer bichchih hun alga uu
2023-03-02 09:56:49
using namespace std;

int main() {
int a;
cin>>a;
cout<<(a+1)*(a+1)/4;
return 0;
}
huul uilsegs
2023-01-24 05:44:41
thh

Last edit: 2023-01-24 05:48:56
2023-01-24 05:44:41
thh

Last edit: 2023-01-24 05:44:59
2022-10-30 14:24:55
hooe ter too ez getsn derese 2doh n MALA BURU BAN iostreama bits/stdc++.h BOLG
2022-06-24 13:35:26
int n, i, niilber=0;
cin >> n;
for ( i = 1 ; i<=n ; i =i+2)
niilber+=i;
cout << niilber;
mai baatarhuu
2022-03-24 08:22:15
#include <iostream>
using namespace std;
int main() {int n,s,q;
scanf("%d",&n);
if(n%2==1)
s=((1+n)/2);

if(n%2==0)
s=n/2;
q=pow(s,2);
printf("%d",q);
return 0;
}
too ez
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