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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
2024-05-07 04:17:41
Engui + Emuujin = <3
2024-05-07 04:17:41
Engui + Emuujin = <3
2024-05-07 04:17:41
Engui + Emuujin = <3
2024-05-07 04:17:41
Engui + Emuujin = <3
2024-05-07 04:17:41
Engui + Emuujin = <3
2024-05-07 04:17:41
Engui + Emuujin = <3
2024-05-07 04:17:41
Engui + Emuujin = <3
2024-03-14 08:45:31



#include <iostream>
using namespace std;

int main()
{
int a,b,c,d=0;
cin>>a>>b>>c;

if(a+b>c){
d++;
}
if(a+c>b){
d++;
}
if(b+c>a){
d++;
}
if(d==3){
cout<<"YES";
}
else{
cout<<"NO";
}
return 0;
}
2024-01-19 05:53:46
gust
2023-12-06 08:12:18
#include <bits/stdc++.h>
using namespace std;

int main() {
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
if (a + b > c && b + c > a && a + c > b){
printf("YES");
}
else printf("NO");
return 0;
}

Last edit: 2023-12-06 08:12:43
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