HDU2095---find your present (2)

最新推荐文章于 2021-05-01 18:50:51 发布

原创最新推荐文章于 2021-05-01 18:50:51 发布 · 191 阅读

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在一个新年派对上，每个人都会收到一份特殊的礼物。这份礼物的编号在众多礼物中是唯一的，且出现次数为奇数次。通过使用异或运算或set容器，可以有效地找到这份特殊礼物的编号。

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Find your present (2)

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

思路：

直接循环内存不够，是过不了的。这道题有两种方法：

1.第一种是用异或。首先我们先要了解异或。十进制换成二进制，相同为0，不同为1。而a^a=0,a^0=0,a^b^a=a^a^b=b。所以这道题只要把用二进制异或相加的结果就是你的礼物。

代码：

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<set>
using namespace std;
int main()
{
   int n;
   while(scanf("%d",&n)&&n)
   {
       int b=0,a;
     for(int i=1;i<=n;i++){
        scanf("%d",&a);
        b^=a;
       }
       printf("%d\n",b);
   }
    return 0;
}

2.还可以使用set容器。如果容器里有这个数，删除，如果没有就插入。最后剩的一个元素就是。

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<set>
using namespace std;、
set<int> gift;
int main()
{
    int n,a;
    while(scanf("%d",&n)!=EOF&&n)
    {
        while(n--)
        {
            scanf("%d",&a);
            if(gift.find(a)==gift.end())
            {
                gift.insert(a);
            }
            else
                gift.erase(a);

        }
        printf("%d\n",*gift.begin());
    }
    return 0;
}