HDU 2095--Find your present (2) 【set】

最新推荐文章于 2020-07-19 12:26:10 发布

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最新推荐文章于 2020-07-19 12:26:10 发布

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分类专栏： STL HDOJ

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本文介绍了一个算法问题，即从一系列数字中找到唯一出现奇数次的数字。通过使用C++ set容器，实现了一种高效查找解决方案，并附带了完整的代码示例。

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find your present (2)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17493 Accepted Submission(s): 6684

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

Sample Output

  
  
   
   3
2


   
   
    
    
     
     Hint
    
    Hint
   
   
   
    
use scanf to avoid Time Limit Exceeded

题意：

给你n个数字，已知只有一个数字出现了奇数次，其他数字都出现了偶数次，要求你找出这个特别的数字。

用的set ，想学习set的朋友可以看一下这篇博客，写的非常好：http://www.cnblogs.com/BeyondAnyTime/archive/2012/08/13/2636375.html

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;

int main(){
    int n,m;
    set<int>s;
    while(scanf("%d",&n),n){
        s.clear();
        while(n--){
            scanf("%d",&m);
            if(s.find(m)!=s.end())
                s.erase(m);
            else
                s.insert(m);
        }
        printf("%d\n",*s.begin());
    }
    return 0;
}