HDOJ2095

find your present (2)

Time Limit: 1000/2000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 666    Accepted Submission(s): 245

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

51 1 3 2 231 2 10

 

Sample Output

32
Hint
Hint
 use scanf to avoid Time Limit Exceeded

 

Author

8600

 

Source

HDU2007spring校赛__热身(1)

 

Recommend

8600

 

 

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
...{
     int n,b;
     while(scanf("%d",&n))
     ...{
                  if(n==0) break;
     int a=0;
         for(int i=0;i<n;i++)
         ...{
                   scanf("%d",&b);
                   a=(a^b);
                 }
                 cout<<(a^0)<<endl;

}
return 0;
}

PS:用scanf 输入时，用了300多MS，而cin却要比它多出1000多MS，呵呵，我正在考虑是不是输入以后都用scanf呢？

另外，这个题用了异或操作比较方便，因为只有一个odd的数，1^0^1=0，所以这个方法较好。