426B - Sereja and Mirroring

最新推荐文章于 2019-04-04 11:09:24 发布

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介绍了一款由Sereja提出的游戏，玩家轮流将水倒入一个杯子中，如果溢出则视为失败。任务是判断Sereja的朋友能否通过特定策略避免任何一方输掉游戏。

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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 test(s)

input
3 4
1 1 1

output
YES

input
3 4
3 1 3

output
YES

input
3 4
4 4 4

output
NO

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
using namespace std;
int s[200][200];
int n,m;
int main ()
{
   while (~scanf ("%d%d",&n,&m))
   {
    for (int i = 0;i < n;i++)
        for (int k = 0;k < m;k++)
            scanf ("%d",&s[i][k]);

    if (n%2)
        printf ("%d\n",n);
    else
    {
        int t = n;
        while (1)
        {
            int f = 1;
            for (int i = 0;i < t / 2;i++)
               {
                   for (int k = 0;k < m;k++)
                    {
                        if (s[i][k] != s[t-1-i][k])
                        f = 0;
                    }
               }
            if (f==1)
            {
                if (t % 2==1)
                    break;
                t /= 2;
            }
            else if(f==0)
                break;
        }
        printf ("%d\n",t);
    }
   }
    return 0;
}