find your present (2)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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, andyou 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

Source

HDU 2007-Spring Programming Contest - Warm Up （1）

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本来以为这种方法一定会超时，没想到能过，可见：位运算效率还是挺高的

#include<stdio.h>
#include<algorithm>
using namespace  std;
int a[1000010];
int main(){
    int n;
    while(scanf("%d",&n),n){
        int i,k;
        for(i=0;i<n;++i){
            scanf("%d",&a[i]);
        }
        sort(a,a+n);
        k=0;
        int ok=0;
        for(i=1;i<n;++i){
            if(a[i]^a[k]){
                //printf("%d,%d\n",a[i],a[k]);
                //printf("i-k=%d\n",i-k);
                if((i-k)&1){
                    ok=1;
                    printf("%d\n",a[k]);
                    break;
                }
                else{
                    k=i;
                }
            }
        }
        if(!ok) printf("%d\n",a[n-1]);

    }
    return 0;
}

最初没想到可以直接用异或处理， 因为其他的卡片个数都是偶数，任意两个相同的数字进行异或后都是0，而异或满足交换律，运算次序无关，结果相同

#include<stdio.h>
int main(){
    int n;
    while(scanf("%d",&n),n){
        int a,i,res=0;
        for(i=0;i<n;++i){
            scanf("%d",&a);
            res=res^a;
        }
        printf("%d\n",res);
    }
    return 0;
}

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