【hdu 1029】Ignatius and the Princess IV

Ignatius and the Princess IV

 

"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

思路：

我们很容易看出来，在一个序列中去掉两个不同的元素，

那么原序列中的多元素，在新序列中仍然是多元素，

我们从第一个数开始遍历，将t赋值给result，增加一个计数器cnt=1,

后面的数如果和其相同，则cnt++，反之cnt--，

当cnt=0时，说明这个数不是所要求的数，

则从这个数开始继续重复上述过程。

最后求出来的result则是所要求的多元素数字。

代码：

#include<cstdio>
int main()
{
	int n;
	int t,result;
	int cnt;
	while(~scanf("%d",&n))
	{
		cnt=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&t);
			if(cnt==0)
			{
				result=t;
				cnt=1;    //一定要重新赋值
			}
			else
			{
				if(t==result)
				cnt++;
				else 
				cnt--; 
			} 
		}
		printf("%d\n",result);
	}
}