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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
2019-12-15 07:17:49
#include <iostream>
using namespace std;

int main() {

int a;
cin >> a >> b >> c;
if (a + b > c && b + c > a && a + c > b)
{
cout << "YES" << endl;
}
else
{
cout << "NO" << endl;
}

return 0;
}
2019-12-10 15:12:12
help me with a java
2019-11-29 08:09:03
#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;
}
2019-11-21 08:45:58
#include <bits/stdc++.h>

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

Last edit: 2019-11-21 08:46:51
2019-11-11 05:48:36
heeeeelp
2019-10-09 13:17:42
help me

Last edit: 2019-10-09 13:18:07
2019-02-27 05:40:14
@miigaa
2018-11-20 07:22:31
#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;
}
2018-10-12 10:07:04

2018-10-12 10:07:04

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