HDU1029——Ignatius and the Princess IV

Ignatius and the Princess IV

Problem Description： http://acm.hdu.edu.cn/showproblem.php?pid=1029

"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 
   
   
   
   

题目分析：本题其实就是要找出数组中出现次数超过一半的数，这里限制了数组的大小为奇数，这样难度就大大降低了。

同时删除不同的两个数，剩下的一定就是出现次数超过一半的那个数。至于为什么自己画一画，想一想就明白了。

#include<stdio.h>

#define M 1000000

int a[M];

void main()
{
	int i;
	int n;
	int key;
	int time;
	while(scanf("%d",&n) != EOF)
	{
		time = 0;
		for(i = 0; i < n; ++i)
		{
			scanf("%d",&a[i]);
			if(time == 0)
			{
				key = a[i];
				time = 1;
			}
			else if(key == a[i])
			{
				++time;
			}
			else
			{
				--time;
			}
		}
		printf("%d\n",key);
	}
}