acm Bear and Three Balls

最新推荐文章于 2019-07-21 18:47:27 发布

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本文介绍了一个有趣的编程问题——礼物球问题。小北极熊Limak需要从他的球中挑选三个符合特定条件的球作为礼物送给朋友。文章详细解释了选择球的规则，并提供了完整的代码实现，帮助读者理解如何通过编程解决这一问题。


 
  Limak is a little polar bear. He has n balls, the i-th ball has size ti.
  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, ..., tn (1 ≤ ti ≤ 1000) where ti 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).
 
 
  
   Examples
  
  
   
    
     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

Note

In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes18, 16 and 17.

In the second sample, there is no way to give gifts to three friends without breaking the rules.

In the third sample, there is even more than one way to choose balls:

Choose balls with sizes 3, 4 and 5.
Choose balls with sizes 972, 970, 971.

参考资料：点击打开链接

 捂脸，想到是连续的数，但是第一次忘了查重，一直都不是用的连续数判断

是我太蠢了QAQ

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
	int n,t[1002],i;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
		scanf("%d",&t[i]);
		sort(t,t+n);
		int l=1;
		for(i=0;i<n-1;i++)
		if(t[i]==t[i+1])
		t[i]=0;
		sort(t,t+n);
		for(i=0;i<n-2;i++)
		if(t[i]==t[i+1]-1&&t[i]==t[i+2]-2&&t[i]!=0&&t[i+1]!=0)
		l++;
		 if(l>=2)
		 printf("YES");
		 else
		 printf("NO");
		
		
	}return 0;
}