HDU 1029

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 13878    Accepted Submission(s): 5634

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

【题目大意】

题目大意就是在一组数中寻找一个出现超过（或等于）一半的数。

【解题报告】

因此，我们可以将该数组分为两部分，一部分是等于要寻找的数x，一部分是不等于x的数，因此有：x出现的次数>=不等于它的出现的数的次数，从而可以假设该数为x，当后边出现的数等于x时，将其计数值times（初始值为0）加一，否则减一，当times==0时，则表明肯定不是该数，从而又假设x等于当前读入的数t，按照这种方法可以得到结果//https://blog.youkuaiyun.com/u012492780/article/details/17115799以上说法是这位大佬的，看了也是膜的不行

附上自己的理解：如果x没有过半，times一定会被减成负数，如果x过半，那么一定不会有足够的其他数使times减成负数，所以times只有两种可能，负数和正数，如果是负数，则没有数量过半的数，如果是正数，那么这个x就是答案

//觉得说的对的别忘了点赞

#include <stdio.h>

int main() {
	int n, k, charactor, times;
	while (EOF != scanf("%d", &n)) {
	    times = 0;
	    while(n--) {
	        scanf("%d", &k);
	        if (times<=0) charactor=k;
	        (k==charactor) ? (++times) : (--times);
	    }
	    printf("%d\n", charactor);
	}
	return 0;
}