标签：
find your present (2)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18235 Accepted Submission(s):
7021
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
use scanf to avoid Time Limit Exceeded
Author
8600
Source
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//这题巧妙的运用了位运算的方法，同一个数位运算两次，化为0；多个数的话就留下了出现次数是奇数的那个数；
1 #include <stdio.h>
2 int main()
3 {
4 int n ;
5 while(scanf("%d", &n) , n)
6 {
7 int i, num, temp = 0 ;
8 for(i=0; i<n; i++)
9 {
10 scanf("%d", &num) ;
11 temp = temp ^ num ;
12 }
13 printf("%d\n", temp) ;
14 }
15 return 0 ;
16 }
杭电2095--find your present (2) （异或）
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原文地址：http://www.cnblogs.com/fengshun/p/4641717.html
