D - find your present (2)

D - find your present (2)

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

题意分析：

这道题的题意基本上就是说，给你N个卡号，这N个卡号中只有1个卡号出现的次数是奇数，最后输出这个卡号。

解题思路：

这道题最大的问题就是数据的范围太大，很容易超时，并且，也没办法打表。并且由于个人对于位运算不够熟悉，这道题一开始是没有思路的。后来查了题解，这道题本质上就是一个知识点，有关位运算的地方。a^b^b = a。如果知道这个道理，这道题就迎刃而解了。直接在输入的时候做^运算，最后直接输出结果。

编码：

#include <bits/stdc++.h> using namespace std; int main() { int n; while (scanf("%d",&n)) { if (n <= 0) { break; } else { int ans; scanf("%d",&ans); for (int i = 1; i < n; i++) { int a; scanf("%d",&a); ans = ans ^ a; } cout << ans << endl; } } return 0; }

最后：

最后还是要注意cin，cout速度的问题，也不是第一次遇到这样的问题了，而且人家提示都说了。