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| Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
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 a1, a2, ..., an(1 ≤ ai ≤ 10). Number ai 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.
Sample Input
3 4
1 1 1
YES
3 4
3 1 3
YES
3 4
4 4 4
NO
Source
题意：一个空杯，有n杯水，n-1个人每个人选一杯水往一个瓶子里倒水，不溢出就YES。
题解：扔掉水最多的那一杯，剩下的和如果小于等于瓶子容积就YES。
#include <iostream> #include <algorithm> #include <string.h> #include <stdlib.h> #include <math.h> #include <stdio.h> using namespace std; int a[200050]; int main() { int i,n,t,c,sum=0; scanf("%d%d",&n,&c); for(i=0;i<n;i++) { cin>>a[i]; sum+=a[i]; } sort(a,a+n); sum-=a[n-1]; if(sum<=c) cout<<"YES"<<endl; else cout<<"NO"<<endl; return 0; }
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原文地址：http://www.cnblogs.com/Ritchie/p/5425255.html
