(HDUSTEP 2) hdu 2095 find your present (2)(找到出现奇数次的那个数)

本文链接：https://blog.youkuaiyun.com/hjd_love_zzt/article/details/43196201

本文探讨了在新年派对中获取唯一特殊礼物的问题，并通过使用异或运算符解决此问题，提供了有效的算法实现和代码解析。

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题目如下：

find your present (2)

Time Limit: 1000/2000 MS (Java/Others) Memory Limit: 32768/1024 K (Java/Others)

Total Submission(s): 6275 Accepted Submission(s): 1639

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

Author

8600

Source

HDU 2007-Spring Programming Contest - Warm Up （1）

Recommend

8600

题目分析：

主要是异或运算符的使用。

1. a ⊕ a = 0

2. a ⊕ b = b ⊕ a

3. a ⊕b ⊕ c = a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c;

4. d = a ⊕ b ⊕ c 可以推出 a = d ⊕ b ⊕ c.

5. a ⊕ b ⊕ a = b.

6.若x是二进制数0101，y是二进制数1011

则x⊕y=1110

只有在两个比较的位不同时其结果是1，否则结果为0

即“相同为0，不同为1”！

输入	运算符	输入	结果
1	⊕	0	1
1	⊕	1	0
0	⊕	0	0
0	⊕	1	1

代码如下：

#include<stdio.h>
int main()
{
    int n,x,y;
    while(scanf("%d",&n)!=EOF&&n)
    {
           x=0;
           while(n--)
           {
                scanf("%d",&y);
                x^=y;
           }
           printf("%d\n",x);
    }
    return 0;
}