HDU 1029 Ignatius and the Princess IV

最新推荐文章于 2020-01-27 10:07:07 发布

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最新推荐文章于 2020-01-27 10:07:07 发布

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分类专栏：训练文章标签： dp

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本文介绍了一种算法，用于从一组包含特殊整数的序列中找出该整数。特殊整数的特点是在序列中出现的次数至少为序列长度的一半加一。文章提供了具体的实现方法，并附带AC代码。

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"OK, you are not too bad, em... But you can never pass the next test." feng5166 says.

"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.

"But what is the characteristic of the special integer?" Ignatius asks.

"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.

Can you find the special integer for Ignatius?

Input

The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file.

Output

For each test case, you have to output only one line which contains the special number you have found.

Sample Input

5
1 3 2 3 3
11
1 1 1 1 1 5 5 5 5 5 5
7
1 1 1 1 1 1 1

Sample Output

3
5
1

这道题乍一看就是个暴力（事实证明直接暴力也能过），直接统计每个数出现次数，次数超过（n+1）/2直接输出。

但是细细一想，如果一个数出现超过一半，那么去掉任意两个数，它的出现次数还是超过一半的。设n个数中特殊的数为a（n），则a(n)=a(n-1)，这就有点dp的味道了。因此，可以把输入的第一个数记成t 开一个cnt=1，输入和t相等cnt++，不等cnt--，cnt为0就把t换成现在的输入，最后输出t就行了。

然后是我代码的思路，一个数出现次数超过(n+1)/2时，该数组（排列好之后）的第(n+1)/2个数一定是它，有点类似抽屉放苹果的原理吧，所以直接排序输出就好。

AC代码：

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

int a[1000005];
int main()
{
    int n;
    while(cin>>n&&n)
    {
        memset(a,0,sizeof(a));
        for(int i=0;i<n;i++)
            cin>>a[i];
        sort(a,a+n);
        cout<<a[(n+1)/2]<<endl;
    }
    return 0;
}