2013
12-29

# hdu 2095 find your present (2)-数学[解题报告]C++

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.

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.

5
1 1 3 2 2
3
1 2 1
0

3
2

HintHint
use scanf to avoid Time Limit Exceeded

http://acm.hdu.edu.cn/showproblem.php?pid=2095

（2）you can assume that only one number appear odd times.根据这句话可以想想更高效的算法，思考中……

（3）用异或运算符

110，即6。
几个数异或满足交换律。2^3^2=2^2^3=0^3=3.
两个相同的数异或为0，普通数都出现了偶数次，所以它们异或后都是0，而0与那个特别数异或后还是那个特殊数。

Accepted 2095 562MS 180K 226 B G++ rll

#include<stdio.h>
//异或运算的运用
int main()
{
int t,a,sum;
while(scanf("%d",&t)&&t!=0)
{
scanf("%d",&sum);
t--;
while(t--)
{
scanf("%d",&a);
sum^=a;
}
printf("%d\n",sum);
}
return 0;
}

1. 给你一组数据吧：29 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 1000。此时的数据量还是很小的，耗时却不短。这种方法确实可以，当然或许还有其他的优化方案，但是优化只能针对某些数据，不太可能在所有情况下都能在可接受的时间内求解出答案。

2. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。

3. 老实说，这种方法就是穷举，复杂度是2^n，之所以能够AC是应为题目的测试数据有问题，要么数据量很小，要么能够得到k == t，否则即使n = 30，也要很久才能得出结果，本人亲测

4. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better.