Problem hidden
|This problem was hidden by Editorial Board member probably because it has incorrect language|version or invalid test data, or description of the problem is not clear.|

RGB7120 - Гурвалжин эсэх

Өгөгдсөн 3 тоогоор талаа хийсэн гурвалжинг байгуулж болох бол YES үгүй бол NO гэж хэвлэ.

Input

Нэг мөрөнд 3 натурал тоо зайгаар тусгаарлагдан өгөгдөнө.

Output

Гурвалжин байгуулж болох бол YES үгүй бол NO.

Example

Input:
3 4 5
Output:
YES

Нэмсэн:Bataa
Огноо:2011-06-21
Хугацааны хязгаарлалт:0.203s
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
2022-08-27 04:58:28
not hi










































































































2022-08-18 10:19:00
nevermind,i forgot to account for the degenerate case

Last edit: 2022-08-18 10:23:51
2022-07-08 04:22:56
i see
ur
trying
to
look
at
the
comments
2022-06-23 11:09:31
hi




















































































2022-06-23 11:08:04
help me with #include<stdio.h>
2022-05-30 03:08:14
ene bodlogo cn buruu bn huuhde
2022-03-14 15:28:56
hatnaa suga

2021-09-10 04:07:13
#buyantugs
2020-12-04 10:10:02
#include <iostream>
#include<climits>
using namespace std;
int main() {
int a,b,c;
cin >> a >> b >> c;
if (a + b > c && b + c > a && a + c > b){
cout << " YES " << endl;
}else {
cout << " NO" << endl;
}

return 0;
}
2020-11-09 07:08:30
#include<bits/stdc++.h>
using namespace std;
int main(){

int a,b,c;
scanf("%d %d %d",&a,&b,&c);
if(a+b>c && a+c>b && c+b>a) printf("YES ");
else printf("NO");
system("pause");
return 0;
}
© Spoj.com. All Rights Reserved. Spoj uses Sphere Engine™ © by Sphere Research Labs.