Ignatius and the Princess IV HDU - 1029（思维）

傷城~ 2021-07-24 17:54 333阅读 0赞

### 题解目录 ###

*   *   *  1.题目
         *  2.题意
         *  3.思路
         *  4.代码

### 1.题目 ###

“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

### 2.题意 ###

寻找一个大小为N的数组中至少出现(N+1)/2次的整数。

### 3.思路 ###

sort一下之后，数组正中间处的数一定是那个数。

### 4.代码 ###

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