Ignatius and the Princess IV

Posted 2020-06-18

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 24410    Accepted Submission(s): 10261

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

 

本来想排序的，中间那个数肯定是要求的数， 但是我自己写了冒泡排序时间超出了，调用c++的库函数时间也超出了，所以网上搜了这个方法很好

#include<stdio.h>
int main()
{
int n;
int i,j=0;

int num,num1;
while(scanf("%d",&n)!=EOF)//输入每个测试用例中数据元素的个数
{
int a[32768]={0};
num1=(n+1)/2;//至少出现num1次

for(i=0;i<n;i++)//时间复杂度n
{
scanf("%d",&j);
a[j]++;
if(a[j]>=num1)
num=j;
}
printf("%d\n",num);
}
return 0;
}