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 |
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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 |
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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. |
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Output For each case, output an integer in a line, which is the card number of your present. |
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Sample Input 5 1 1 3 2 2 3 1 2 1 0 |
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Sample Output 3 2 Hint Hint use scanf to avoid Time Limit Exceeded 真坑。。。题目给的数据量过大开数组会 Memory Limit Exceeded 只能用异或 |
整数的异或是先把它们化成二进制，再按位异或。
比如3^5,3=011,5=101,两数按位异或后为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。
#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;
}
hdu 2095 find your present（2）