寒假D3 A Find the Lost Sock

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

Hint

Because of HUGE input, scanf is recommended.

方法1：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<stdlib.h>
char a[2000005][8];
//注;若用qsort对二维数组排序  应该用strcmp  若不是-
int cmp(const void *a,const void *b)
{
    return strcmp((char*)a,(char*)b);
}
int main()
{
    int i,j,k;
    int n;
    //快排，检测相邻的字符串
    while(scanf("%d",&n)!=-1)
    {
        getchar();
        for(i=0;i<2*n-1;i++)
        gets(a[i]);
        qsort(a,2*n-1,sizeof(a[0]),cmp);

        if(strcmp(a[0],a[1])!=0) {printf("%s
",a[0]);continue;}
        for(i=0;i<2*n-1;i+=2)
        if(strcmp(a[i],a[i+1])!=0) {printf("%s
",a[i]);break;}
    }

}

 方法2：每一串都为7个字符，用数组对每一串的每一个字符计数  若最后某字符的个数为奇数  输出该字符（get）