HDU1029 Ignatius and the Princess IV

Ignatius and the Princess IV

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

Total Submission(s): 27552    Accepted Submission(s): 11728
[/b]

[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

 

        这道题目求的是出现次数超过总次数一半的元素我们可以总结

发现一个规律，就是在一个多数元素为n的数列中，去掉两个不相同

的元素，剩下数列的多数元素还是n进一步推广，去掉的元素中的多

数元素如果不是n，那么剩下数列的多数元素任然是n，这个可以通

过一个出现频率的概念来理解。假设n的出现频率为0.5，而前半段

的出现频率小于0.5那么后半段的出现频率必然大于0.5。

        想清楚这点以后这道题目就可以解决了，值得注意的是没有必要

通过数组来存储每一个数据，直接在每次输入数据的时候进行操

作，可以节省内存空间的使用。

 

代码：

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n,t,res;
while(~scanf("%d",&n))
{
int cnt=0;
for(int i=0;i<n;i++)
{
scanf("%d",&t);
if(cnt==0)
{
cnt=1;
res=t;
}
else
{
if(t==res) cnt++;
else
{
cnt--;
}
}
}
printf("%d\n",res);
}
return 0;
}