codeforces Sereja and Mugs

最新推荐文章于 2017-09-26 15:58:06 发布

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最新推荐文章于 2017-09-26 15:58:06 发布

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分类专栏： codeforce 文章标签： codeforces

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本文介绍了一种简单的游戏胜负判断算法，通过计算多个水杯的水量总和，在扣除最大水量后的值是否小于等于杯子容量来判断游戏是否存在输家。文章通过具体的输入输出样例展示了算法的应用过程。

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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
Input
3 4
1 1 1
Output
YES
Input
3 4
3 1 3
Output
YES
Input
3 4
4 4 4
Output
NO

n个数相加减去最大的一个后小于等于m就输出yes；

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int main()
{
    int n,m;
    cin>>n>>m;
    int max=0,s=0;
    for(int i=0; i<n; i++)
    {
        int d;
        cin>>d;
        s+=d;
        if(max<d)
            max=d;
    }
    s-=max;
    if(s<=m)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
    return 0;
}