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  • Bear and Three Balls

    Description

    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 45 and 3, or balls with sizes 9091 and 92. But he can't choose balls with sizes 55and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 3031 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).

    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
    

    Hint

    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 sizes1816 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:

    1. Choose balls with sizes 34 and 5.

    1. Choose balls with sizes 972970971.
    简单的说就是如果序列中能够找到连续的三个自然数就输出 YES否则NO  先排序再去重后查找
    unique为去重函数
    比如  12 3 3  5 4 4  4
    排序后为1 2 3 5 3 4 4  函数返回的是4也就是第二个3的坐标(从0开始)
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<queue>
    #include<algorithm>
    using namespace std;
    int s[10001];
    int main()
    {
    	int n;
    	scanf("%d",&n);
    	for(int i=0;i<n;i++)
    	{
    		scanf("%d",&s[i]);
    	
    	}
    	sort(s,s+n);
    	int len=unique(s,s+n)-s;//去重   
    	int flog=0;
    	for(int i=0;i<len;i++)
    	{
    		if(s[i+1]==s[i]+1&&s[i+2]==s[i+1]+1)
    		{
    			flog=1;
    			break;
    		}
    	}
    	if(flog==1)
    	printf("YES\n");
    	else
    	printf("NO\n");
    	return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027039.html
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