HDU-1029-Ignatius and the Princess IV

最新推荐文章于 2020-02-07 11:05:28 发布

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最新推荐文章于 2020-02-07 11:05:28 发布

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分类专栏： STL_map

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本文介绍了一个算法问题，任务是找出一组数字中出现次数超过(N+1)/2次的特殊整数，并提供了一种使用C++ map实现的有效解决方案。

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Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 33930 Accepted Submission(s): 14732

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

题意：输入一个奇数，输出个数大于（n+1）/2的数

直接用map比较简单。map[t]就表示t出现的次数。如果没见过map的一定要记一下，很简单

#include<stdio.h>
#include<map>	//map头文件
using namespace std;
int main()
{
	int t,n,ans;
	while(~scanf("%d",&n))
	{

		map<int,int> map;	//建立map数组，类型为int，int。即map[t]中t是整形，次数也是整形
		map.clear();
		for(int i=0;i<n;i++)
		{
			scanf("%d",&t);
			map[t]++;	//出现一次t，次数加1
			if(map[t]>=(n+1)/2)	//比较t出现的次数
			ans=t;
		}
			printf("%d\n",ans);
		
	}
	return 0;
}