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.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<=200, and n is odd) at first. Following that, n positive integers will be given in a line. 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
题意：找出一行数中独立的数！
按位异或的3个特点:
(1) 0^0=0,0^1=1 0异或任何数＝任何数
(2) 1^0=1,1^1=0 1异或任何数－任何数取反
(3) 任何数异或自己＝把自己置0
先说一下异或运算的运算法则：
1. a ^ b = b ^ a
2. a ^ b ^ c = a ^ (b ^ c) = (a ^ b) ^ c
3. d = a ^ b ^ c 可以推出 a = d ^ b ^ c
4. a ^ b ^ a = b
对于性质1，显而易见。
对于性质2和4，就是可以查找出一组数列中具有奇数个数的数。比如：
题目：有2n+1个数，其中有n个数出现过两次，只有一个数字出现过一次。要求是找出这个数字。
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int n =sc.nextInt();
if(n==0){
return ;
}
int m =0;
int s;
while(n-->0){
s = sc.nextInt();
m = m^s;
}
System.out.println(m);
}
}
}
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逸川先生
2017.11.14
学长，我拿你的代码去还是超时了，怎么回事
丁国华
2016.04.19
感谢博主的分享。
谙忆作者
回复
丁国华
2016.04.19
客气了(^-^)
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import java.util.Scanner; public class Main{ public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(sc.hasNext()){ int n =sc.nextInt(); if(n==0){ return ; } int m =0; int s; while(n-->0){ s = sc.nextInt(); m = m^s; } System.out.println(m); } } }
