zoj 3432（异或的使用）

最新推荐文章于 2017-04-12 21:12:59 发布

原创最新推荐文章于 2017-04-12 21:12:59 发布 · 901 阅读

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Alice购买了许多双袜子，但回家后发现丢失了一只。每只袜子都有一个由7个字符组成的名称。通过输入2n-1条Alice带回的袜子名称记录，找出缺失的那一只。

Find the Lost Sock

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

题目大意：题目一看就是匹配字符串的。

思路：字符串匹配。。不过老是TLE。上网一看原来是异或的使用。。这种算思路的积累吧。。实在想不到。。

//大神代码。

#include<stdio.h>

/* 

为异或量身定做的题。 
异或满足结合律和交换律，则有A^B^A = A^(B^A) = A^A^B = B 
所以所有字符串异或的结果就是答案了。 
*/

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
         getchar();                     
         int i,j;
         int a[8]={0};
         char s[10];
         for(i=0;i<n;i++)
         {
              gets(s);
              for(j=0;j<7;j++)
              a[j]=a[j]^s[j];
         }
         for(i=0;i<n-1;i++)
         {
              gets(s);
              for(j=0;j<7;j++)
              a[j]=a[j]^s[j];
         }
         for(i=0;i<7;i++)
         printf("%c",a[i]);
         printf("\n");
    }
}