hdu1029-Ignatius and the Princess IV

题目传送：hdu1029
or [kuangbin带你飞]专题十二 基础DP1 B - Ignatius and the Princess IV

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

排序方法：排序后找第n/2大的数，因为特殊数出现次数超过一半，所以排序后中位数一定是特殊数。
时间复杂度：

#include
#include
using namespace std;
const int maxn=1e6+10;
int main(){
    int n;
    int a[maxn];
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        printf("%d\n",a[(n+1)/2]);
    }
    return 0;   
}    

迭代方法：因为n为奇数，且特殊值出现次数大于一半，所以特殊值做为解时num不会小于1，所以最终的解一定就是特殊值。
时间复杂度：

#include
#include
using namespace std;
int main(){
    //N为奇数(题中的old integer;a为输入数据；ans为答案；cnt计数)
    int N,a,ans,cnt;
    while(~scanf("%d",&N)){
        cnt=0;
        for(int i=0;i