杭电OJ-- 2095 find your present (依然很水)

Problem Description

In the new year party, everybody will get a "special present".Now it's your turn to get your special present, a lot of presents now putting on the desk, and only one of them will be yours.Each present has a card number on it, and your present's card number will be the one that different from all the others, and you can assume that only one number appear odd times.For example, there are 5 present, and their card numbers are 1, 2, 3, 2, 1.so your present will be the one with the card number of 3, because 3 is the number that different from all the others.

 

Input

The input file will consist of several cases. 
Each case will be presented by an integer n (1<=n<1000000, and n is odd) at first. Following that, n positive integers will be given in a line, all integers will smaller than 2^31. These numbers indicate the card numbers of the presents.n = 0 ends the input.

 

Output

For each case, output an integer in a line, which is the card number of your present.

 

Sample Input

   
   
   
   

    
    
    
    5
1 1 3 2 2
3
1 2 1
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
2


    
    
    
    
 
     
 
      Hint 
     
Hint 
    
    
    
    
     
use scanf to avoid Time Limit Exceeded
   
   
   
   
   
   
   
   


   
   
   
   

题目不难，要注意控制一下内存，内存不要超出即可。开始时准备使用multiset，结果发现超出了内存，后来直接用数组了，后来改用map来做。

map中存储<num,index>对，num表示礼品的编号，index表示该编号对应数组的下标。总之自己看代码吧，这道题也很水。

代码如下：

/*
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <set>
using namespace std;

//const int MAX_NUM = 1000000;

int main()
{
	multiset<int> num_set;
	int num;
	while (cin >> num && num != 0)
	{
		//int numOfPresent[MAX_NUM];
		//memset(numOfPresent, 0, sizeof(numOfPresent));
		int present_num;
		num_set.clear();
		for (int i = 0; i < num; ++i)
		{
			scanf("%d", &present_num);
			num_set.insert(present_num);
		}
		multiset<int>::iterator set_it = num_set.begin();

		for (; set_it != num_set.end(); ++set_it)
		{
			if (num_set.count(*set_it) % 2)
			{
				cout << *set_it << endl;
				break;
			}
		}


	}
	return 0;
}
*/
/*使用set不太现实，因为内存会超出限制*/
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <map>
using namespace std;
const int MAX_NUM = 1000000;
int num_of_present[MAX_NUM];

int main()
{
	map<int, int> num_map;
	
	int num;
	while (cin >> num && num != 0)
	{
		int present_num = 0;
		int index = 0; //指针
		num_map.clear();
		memset(num_of_present, 0, sizeof(num_of_present));
		for (int i = 0; i < num; ++i)
		{
			scanf("%d", &present_num);
			map<int, int>::iterator map_it = num_map.find(present_num);
			if (map_it != num_map.end())
			{
				++num_of_present[map_it->second];
				//cout << "要插入的数字是" << present_num << endl;
				//cout << "map_it->second = " << map_it->second << endl;
				//cout << "num_of_present[map_it->second] = " << num_of_present[map_it->second] << endl;
			}
			else
			{
				//cout << "要插入的数字是" << present_num << endl;
				//cout << "map中无数据" << endl;
				//cout << "index = " << index << endl;
				num_map.insert(make_pair(present_num, index));
				++num_of_present[index];
				++index;
			}

		}
		map<int, int>::iterator map_it = num_map.begin();
		while (map_it != num_map.end())
		{
			if (num_of_present[map_it->second] % 2 != 0)
			{
				cout << map_it->first << endl;
				break;
			}
			++map_it;
		}
		
	}
	return 0;
}