题目如下：
find your present (2) |
| Time Limit: 1000/2000 MS (Java/Others) Memory Limit: 32768/1024 K (Java/Others) |
| Total Submission(s): 6275 Accepted Submission(s): 1639 |
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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 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. |
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Input The input file will consist of several cases. |
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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 |
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Author 8600 |
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Source HDU 2007-Spring Programming Contest - Warm Up （1） |
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Recommend 8600 |
题目分析：
主要是异或运算符的使用。
1. a ⊕ a = 0
2. a ⊕ b = b ⊕ a
3. a ⊕b ⊕ c = a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c;
4. d = a ⊕ b ⊕ c 可以推出 a = d ⊕ b ⊕ c.
5. a ⊕ b ⊕ a = b.
6.若x是二进制数0101，y是二进制数1011
则x⊕y=1110
只有在两个比较的位不同时其结果是1，否则结果为0
即“相同为0，不同为1”！
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输入 |
运算符 |
输入 |
结果 |
|
1 |
⊕ |
0 |
1 |
|
1 |
⊕ |
1 |
0 |
|
0 |
⊕ |
0 |
0 |
|
0 |
⊕ |
1 |
1 |
代码如下：
#include<stdio.h>
int main()
{
int n,x,y;
while(scanf("%d",&n)!=EOF&&n)
{
x=0;
while(n--)
{
scanf("%d",&y);
x^=y;
}
printf("%d\n",x);
}
return 0;
}