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HDU 2095 find your present (2)

HDU 2095 find your present (2)

作者: itbird01 | 来源:发表于2022-04-07 07:13 被阅读0次

    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

    5
    1 1 3 2 2
    3
    1 2 1
    0

    Sample Output

    3 2

    Hint
    Hint
    use scanf to avoid Time Limit Exceeded

    Author
    8600

    c code

    include <stdio.h>

    int main()
    {
    int n,x,ans;
    while(scanf("%d",&n),n)
    {
    ans = 0;
    while(n--)
    {
    scanf("%d",&x);
    ans ^= x;
    }
    printf("%d\n",ans);
    }
    return 0;
    }
    java code(java 超时,内存超过 )

    import java.util.HashMap;
    import java.util.Iterator;
    import java.util.Scanner;
    public class Main {
        public static void main(String[] args) {
            Scanner input = new Scanner(System.in);
            while (input.hasNextInt()) {
                int n = input.nextInt();
                if (n == 0) {
                    break;
                }
                HashMap<Integer, Integer> result = new HashMap<Integer, Integer>();
                for (int i = 0; i < n; i++) {
                    int a = input.nextInt();
                    if (result.containsKey(a)) {
                        int value = result.get(a) + 1;
                        result.put(a, value);
                    } else {
                        result.put(a, 1);
                    }
                }
                Iterator iterator = result.keySet().iterator();
                while (iterator.hasNext()) {
                    Integer keyInteger = (Integer) iterator.next();
                    if (result.get(keyInteger) % 2 != 0) {
                        System.out.println(keyInteger);
                        break;
                    }
                }
            }
            input.close();
        }
    
    }
    

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