HDU 2095 find your present (2)

Find your present (2)

Time Limit: 1000/2000 MS (Java/others) memory limit: 32768/1024 K (Java/Others)

Total submission (s): 5186 accepted submission (s): 1513

Problem descriptionin 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.

Inputthe 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.

Outputfor 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 32HintHint use scanf to avoid Time Limit Exceeded Really pitfall... If the data volume given by the question is too large, the memory limit exceeded can only be used with an exception or

Integers are either converted into binary values and then bitwise. For example, 3 ^ 5, 3 = 101, 5 = 110, the two numbers are equal to or later than, that is, 6. A few numbers are different or meet the exchange law. 2 ^ 3 ^ 2 = 2 ^ 2 ^ 3 = 0 ^ 3 = 3. two identical numbers are equal to or equal to 0, and the common numbers are even, so they are both equal to or equal to 0, while the difference between 0 and that special number is also the special number.

#include<stdio.h>int main(){    int n,i,x,y;    while(scanf("%d",&n)!=EOF&&n)    {           x=0;           while(n--)           {                scanf("%d",&y);                x^=y;                }                    printf("%d\n",x);                 }    return 0;}