hdu 1029 Ignatius and the Princess IV（水题）

最新推荐文章于 2018-04-08 20:46:54 发布

浪漫些许潦草

最新推荐文章于 2018-04-08 20:46:54 发布

阅读量524

点赞数 1

CC 4.0 BY-SA版权

分类专栏： # 水题 # 思路题 ACM 文章标签： hdu-1029

本文链接：https://blog.youkuaiyun.com/blessLZH0108/article/details/72587923

ACM 同时被 3 个专栏收录

329 篇文章

订阅专栏

思路题

68 篇文章

订阅专栏

水题

60 篇文章

订阅专栏

本文介绍了一个寻找特殊整数的问题及两种解法。特殊整数在给定的整数列表中出现次数至少为(N+1)/2次。文章提供了O(1e6)复杂度的解法，使用额外数组记录每个整数出现的频率；同时也给出了一种O(n)复杂度的高效解法，利用特殊整数出现次数的特点进行求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Ignatius and the Princess IV

Problem Description

“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

ps：水题，

O（1e6）的解法：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;

const int maxn=1000010;
int vis[maxn];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof(vis));
        int x;
        int ans=-1;
        for(int i=0;i<n;++i)
        {
            scanf("%d",&x);
            ++vis[x];
        }
        int k=(n+1)/2;
        for(int i=0;i<=1000000;++i)
        {
            if(vis[i]>=k)
            {
                ans=i;
                break;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

O（n）的解法，一种思想吧
由于最终结果出现次数≥（n+1）/2这个特性，所以最终结果的那个cnt最后会>0
代码：

#include<stdio.h>

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int x;
        int ans,cnt=0;
        for(int i=0;i<n;++i)
        {
            scanf("%d",&x);
            if(!cnt)
            {
                ans=x;
                ++cnt;
            }
            else
            {
                if(ans!=x)
                    --cnt;
                else
                    ++cnt;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}