异或的使用一例。。。

今天在杭电ACM steps上做1.2.1，（虽然晚了点，但还是要做的）。由于题目是随机的，所以我把题目摘下；

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find your present (2)

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

Total Submission(s): 429 Accepted Submission(s): 139

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

我定义一个很大的数组，然后每次扫描，结果都是TLE，很郁闷的上百度看了下，一个人用的是异或运算符，好巧妙啊。

 具体程序是这样滴#include <stdio.h> int main() { int result,a,t,i,j; while(scanf("%d",&t)&&t!=0) { result=0; while(t--) { scanf("%d",&i); result^=i; } printf("%d/n",result); } return 0; }

      

        解释如下：

 

 

            例如  1 2 3 2 1

                     那么result=1^2^3^2^1   

                      移项  result=1^1^2^2^3 =3 (由异或的定义我们可以知道，只要偶数次的两个数异或值就是0，1^1=0  ;2^2=0,0^0^3=3)

这样子做就不会导致超内存了，，嗯嗯！今天学到这个！