題目傳送：hdu1029
or [kuangbin帶你飛]專題十二 基礎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大的數(shù)，因為特殊數(shù)出現(xiàn)次數(shù)超過一半，所以排序后中位數(shù)一定是特殊數(shù)。
時間復雜度：

#include<bits/stdc++.h>
#include<cstdio>
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：因為n為奇數(shù)，且特殊值出現(xiàn)次數(shù)大于一半，所以特殊值做為解時num不會小于1，所以最終的解一定就是特殊值。
時間復雜度：

#include<bits/stdc++.h>
#include<cstdio>
using namespace std;
int main(){
    //N為奇數(shù)(題中的old integer;a為輸入數(shù)據(jù)；ans為答案；cnt計數(shù))
    int N,a,ans,cnt;
    while(~scanf("%d",&N)){
        cnt=0;
        for(int i=0;i<N;i++){
            scanf("%d",&a);
            if(cnt==0){
                ans=a;
                cnt++;
            }
            else{
                if(a==ans)cnt++;
                else cnt--;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}