Bear and Three Balls熊和三个小球

最新推荐文章于 2020-11-13 09:33:23 发布

原创最新推荐文章于 2020-11-13 09:33:23 发布 · 490 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#sort

编程语言专栏收录该内容

307 篇文章

订阅专栏

本文介绍了一个有趣的问题：如何从多个球中选择三个大小不同的球作为礼物，且任意两个球的大小之差不超过2。通过编程解决这个问题，使用了排序和数组处理技术。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Limak is a little polar bear. He has n balls, the i-th ball has size t_i.

Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:

No two friends can get balls of the same size.
No two friends can get balls of sizes that differ by more than 2.

For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).

Your task is to check whether Limak can choose three balls that satisfy conditions above.

Input

The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.

The second line contains n integers t₁, t₂, ..., t_n (1 ≤ t_i ≤ 1000) where t_i denotes the size of the i-th ball.

Output

Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).

Sample Input

Input

4
18 55 16 17

Output

YES

Input

6
40 41 43 44 44 44

Output

NO

Input

8
5 972 3 4 1 4 970 971

Output

YES

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
#include <iostream>  //题意：是否存在三个连续的数
#include <cstdio>  
#include <cstring>  
#include <algorithm>   
using namespace std;   
int main()  
{  
    int n,a[55],b[55];  
    while(~scanf("%d",&n))  
    {  
        int ans=1,f=0,k=0;  
        for(int i=0;i<n;i++)  
            scanf("%d",&a[i]);  
        sort(a,a+n);  
        b[0]=a[0];  
        for(int i=1;i<n;i++)  
        {  
            if(a[i]!=a[i-1]) //消除重复的 
                b[++k]=a[i];  
        }  
        for(int i=2;i<=k;i++)  
        {  
            if((b[i]-b[i-1]==1)&&(b[i]-b[i-2]==2))  //判断连续
            {  
                f=1;  
                break;  
            }  
        }  
        if(f)  
            printf("YES\n");  
        else  
            printf("NO\n");  
    }  
    return 0;  }