HDU 1029 Ignatius and the Princess IV （思维题，排序?）

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

3
5
1

Http

https://vjudge.net/problem/HDU-1029

2017.8.26 Update

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
int n;
while (cin>>n)
{
int cnt=0,Ans=0;
for (int i=1;i<=n;i++)
{
int number;
scanf("%d",&number);
if (cnt==0)
{
cnt=1;
Ans=number;
continue;
}
if (number==Ans)
cnt++;
else
cnt--;
}
printf("%d\n",Ans);
}
return 0;
}   

#include<cstdio>

int main()
{
int n;
scanf("%d",&n);
int cnt=0,Ans=0;
for (int i=1;i<=n;i++)
{
int number;
scanf("%d",&number);
if (cnt==0)
{
cnt=1;
Ans=number;
continue;
}
if (number==Ans)
cnt++;
else
cnt--;
}
printf("%d\n",Ans);
return 0;
}   

sort暴力方法：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxN=1000000;
const int inf=2147483647;

int n;
int Arr[maxN];

int main()
{
while (cin>>n)
{
for (int i=1;i<=n;i++)
scanf("%d",&Arr[i]);
sort(&Arr[1],&Arr[n+1]);
printf("%d\n",Arr[n/2+1]);
}
return 0;
}

posted @ 2017-08-23 20:15 SYCstudio 阅读(...) 评论(...) 编辑 收藏