HDU 5264 pog loves szh I

Description

Pog has lots of strings. And he always mixes two equal-length strings. For example, there are two strings: "abcd" and "efgh". After mixing, a new string "aebfcgdh" is coming. 

However, szh thinks it is boring, so he reverses the second string, like changing "efgh" to "hgfe". Then mix them as usual, resulting in "ahbgcfde". 

Now, here comes a problem. Pog is given a string after mixing by szh, but he wants to find the original two strings. 

Hint : In this question, it needs a linefeed at the end of line at hack time.

 

Input

The first line has an integer, $T(1 \leq T \leq 100)$, indicating the number of cases. 

Then follows $T$ lines. Every lines has a string S, whose length is even and not more than $100$, and only consists of lower-case characters('a'~'z').

 

Output

For each cases, please output two lines, indicating two original strings.

 

Sample Input

     
     
     
     

      
      
      
      1
aabbca 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
      abc
aba
     
     
     
     

非常水的题→_→ 数组分奇偶后 第二个字符串倒序输出

#include <iostream>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
using namespace std;

int main()
{
    char a[120];
    int T,len;
    scanf("%d",&T);
    for(int i=0; i<T; i++)
    {
        char b[100]= {0},c[100]= {0};
        scanf("%s",a);
        len=strlen(a);
        int x=0,y=0;
        for(int j=0; j<len; j++)
        {
            if(j%2==0)
            {
                b[x]=a[j];
                x++;
            }
            else
            {
                c[y]=a[j];
                y++;
            }
        }
        printf("%s\n",b);
        for(int j=y-1;j>=0;j--)
        {
            printf("%c",c[j]);
        }
        printf("\n");
    }
    return 0;
}