A. Sereja and Mugs

time limit per test

1 second
memory limit per test

256 megabytes
input

standard input
output

standard output
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and

*n*water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.
As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (

*n*- 1) friends. Determine if Sereja's friends can play the game so that nobody loses.
Input

The first line contains integers

*n*and*s*(2 ≤*n*≤ 100; 1 ≤*s*≤ 1000) — the number of mugs and the volume of the cup. The next line contains*n*integers*a*1,*a*2, ...,*a**n*(1 ≤*a**i*≤ 10). Number*a**i*means the volume of the*i*-th mug.
Output

In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.

import java.util.Arrays; import java.util.Scanner; public class SerejaAndMugs { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int n = scan.nextInt(); //no of players including Sereja int c = scan.nextInt(); //capacity of the cup int mugs[] = new int[n]; for (int i = 0; i < n; ++i) { mugs[i] = scan.nextInt(); } Arrays.sort(mugs); int t = 0; for (int i = 0; i < n - 1; ++i) { t += mugs[i]; if (mugs[i] <= c && t <= c) { continue; } else { System.out.println("NO"); System.exit(0); } } System.out.println("YES"); } }

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