hdu-1029 Ignatius and the Princess IV

题目描述

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)Total Submission(s): 28646    Accepted Submission(s): 12134

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

解题思路

​ 题目大概意思就是找出数组中出现次数超过数组元素个数一半的那个元素，既然超过一半出现次数，那排序之后，出现在数组中间位置的元素则必然是我们所要找的元素，实现也就很简单了，具体代码如下：

Code

//
//  main.cpp
//  hdu1029
//
//  Created by Morris on 2016/12/16.
//  Copyright © 2016年 Morris. All rights reserved.
//

#include <cstdio> 
#include <algorithm> 
#include <vector>

namespace {
    using std::scanf;
    using std::printf;
    using std::vector;
    using std::sort;
}

int main(int argc, const char *argv[])
{
    int n;
    vector<int> vec;
    while (~scanf("%d", &n)) {
        int i, tmp;
        for (i = 0; i < n; ++i) {
            scanf("%d", &tmp);
            vec.push_back(tmp);
        }

        sort(vec.begin(), vec.end());
        printf("%d\n", vec[n / 2]);
        vec.clear();
    }
    return 0;
}