zoj 3432 与运算运用

/****************************************************

该题充分运用与运算的特点，成对消除，单的最后留下

****************************************************/

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

int main()

{

char str1[10], str2[10];

int n;

while(~scanf("%d", &n) )

{

getchar();

gets(str1);

for(int i=0; i<2*n-2; i++)

	{

gets(str2);

for(int j=0; j<7; j++)

str1[j] = str1[j] ^ str2[j];				//两个字符串各自与运算，可以判定最后剩下的不成对的一个

}

printf("%s\n", str1);

}

}

/*

Description

Alice bought a lot of pairs of socks yesterday. But when she went home, she found that she has lost one of them. Each sock has a name which contains exactly 7 charaters.

Alice wants to know which sock she has lost. Maybe you can help her.

Input

There are multiple cases. The first line containing an integer n (1 <= n <= 1000000) indicates that Alice bought n pairs of socks. For the following 2*n-1 lines, each line is a string with 7 charaters indicating the name of the socks that Alice took back.

Output

The name of the lost sock.

Sample Input

2

aabcdef

bzyxwvu

bzyxwvu

4

aqwerty

eas fgh

aqwerty

easdfgh

easdfgh

aqwerty

aqwerty

2

0x0abcd

0ABCDEF

0x0abcd

Sample Output

aabcdef

eas fgh

0ABCDEF

*/

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