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.
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.
For each case,
output an integer in a line, which is the card number of your
present.
1 #include <stdio.h>
2 int main()
3 {
4 int n;
5 while(scanf("%d",&n)!=EOF&&n!=0){
6 int s=0,k;
7 for(int i=0;i<n;i++){
8 scanf("%d",&k);
9 s^=k;
10 }
11 printf("%d\n",s);
12 }
13 return 0;
14 }
很巧妙的异或位运算方法！！
and you can assume that only one number appear odd times 这句话很重要。英语很重要啊。
odd 奇数
even 偶数
