HDU 1029 Ignatius and the Princess IV（排序取数）

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others)

Total Submission(s): 25416 Accepted Submission(s): 10716



[align=left]Problem Description[/align]
"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?

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you have to output only one line which contains the special number you have found.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

3
5
1

[align=left]题解：[/align]

[align=left]求出出现次数为n+1/2的数。。。。sort大法。。。[/align]

[align=left]AC代码：[/align]

#include<cstdio>
#include<algorithm>
using namespace std;
int a[999999];
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+1)/2]);

}
return 0;
}