HDOJ 2095 find your present (2) 位运算

最新推荐文章于 2023-05-11 01:02:56 发布

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#位运算

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本文介绍了一种通过位运算解决寻找特殊礼物问题的算法。在新年派对中，每个人都会收到一个独特的礼物，礼物上有一个卡片编号。特殊礼物是唯一出现奇数次的编号。通过输入礼物数量和编号，算法能够输出特殊礼物的卡片编号。

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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

题目大意：找出不同的数，输出。

思路：可以用位运算的^。 3^5=6,做法是先将两个数换成2进制，然后进行异或相同为0,不同为1.

用^可以将最后的出现奇数次的数输出，偶数次的话为0。

PS： a^=b;b^a;a^=b;可以实现两数的交换。。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
LL x,y;
LL a[1000010];
int main()
{
    LL n,m,k,i,j,l,c;
    while(~scanf("%lld",&n)&&n)
    {
        scanf("%lld",&i);
        n--;
        while(n--)
        {
            scanf("%lld",&j);
            i=i^j;
        }
        printf("%lld\n",i);
    }
    return 0;
}