HDU1029-Ignatius and the Princess IV

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面对一系列包含奇数个整数的数据集，任务是找出那个至少出现(N+1)/2次的特殊整数。本文提供两种解决方案：一种是通过排序确定特殊整数；另一种是采用线性时间复杂度的方法遍历数组，高效定位目标。

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Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 34150 Accepted Submission(s): 14867

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

Author
Ignatius.L

题目大意：有奇数个数字，输出个数大于等于(n+1)/2的数字。
解题思路：排序，或者O(n)扫一遍。

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
const int INF=0x3f3f3f3f;
const int MAXN=1e6+5;
int a[MAXN];

int main()
{
  int n;
  while(scanf("%d",&n)!=EOF)
  {
    for(int i=1;i<=n;i++)
    {
      scanf("%d",&a[i]);
    }
    sort(a+1,a+n+1);
    if(n&1)
    {
      printf("%d\n",a[(n+1)/2]);
    }else
    {
      int x=a[n/2],y=a[n/2+1];
      if(a[1]==x&&a[n]!=y)  printf("%d\n",x);
      else printf("%d\n",y);
    }
  }
}
/*
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

3
5
1
*/

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
const int INF=0x3f3f3f3f;
const int MAXN=1e6+5;
int a[MAXN];

int main()
{
  int n;
  while(scanf("%d",&n)!=EOF)
  {
    int ans;
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
      scanf("%d",&a[i]);
      if(cnt==0)
      {
        ans=a[i];cnt++;
      }else
      {
        if(a[i]==ans) cnt++;
        else cnt--;
      }
    }
    printf("%d\n",ans);
  }
  return 0;
}