【水hash】#22 A. Second Order Statistics

最新推荐文章于 2021-05-27 15:41:40 发布

原创最新推荐文章于 2021-05-27 15:41:40 发布 · 871 阅读

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本文探讨了如何在一组整数中找到第二阶统计值，即排序后位于第二位的元素，提供了有效的解决方法。

A. Second Order Statistics

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.

Input

The first input line contains integer n (1 ≤ n ≤ 100) — amount of numbers in the sequence. The second line contains nspace-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.

Output

If the given sequence has the second order statistics, output this order statistics, otherwise output NO.

Sample test(s)

input
4
1 2 2 -4

output
1

input
5
1 2 3 1 1

output
2

5min内秒掉的题呢^_^~~~

感觉良好~但是学不到东西……

用pair似乎比较省内存，但是才100，我就直接上hash表了

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

int mark[201]; 

int main()
{
	int n=0;
	scanf("%d",&n);
	memset(mark,0,sizeof(mark));
	for(int i=1;i<=n;i++)
	{
		int now=0;
		scanf("%d",&now);
		mark[now+100]=1;
	}
	int flag=0;
	for(int j=0;j<=200;j++)
	{
		if(flag==1&&mark[j]==1)
		{
			printf("%d",j-100); 
			return 0;
		}
		if(flag==0&&mark[j]==1) flag=1;
	}
	printf("NO");
	return 0;
}