Elven Postman

Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lightly. Also, they live on trees. However, there is something about them you may not know. Although delivering stuffs through magical teleportation is extremely convenient (much like emails). They still sometimes prefer other more “traditional” methods. 

So, as a elven postman, it is crucial to understand how to deliver the mail to the correct room of the tree. The elven tree always branches into no more than two paths upon intersection, either in the east direction or the west. It coincidentally looks awfully like a binary tree we human computer scientist know. Not only that, when numbering the rooms, they always number the room number from the east-most position to the west. For rooms in the east are usually more preferable and more expensive due to they having the privilege to see the sunrise, which matters a lot in elven culture. 

Anyways, the elves usually wrote down all the rooms in a sequence at the root of the tree so that the postman may know how to deliver the mail. The sequence is written as follows, it will go straight to visit the east-most room and write down every room it encountered along the way. After the first room is reached, it will then go to the next unvisited east-most room, writing down every unvisited room on the way as well until all rooms are visited. 

Your task is to determine how to reach a certain room given the sequence written on the root. 

For instance, the sequence 2, 1, 4, 3 would be written on the root of the following tree. 

Input

First you are given an integer T(T≤10)T(T≤10)indicating the number of test cases. 

For each test case, there is a number n(n≤1000)n(n≤1000)on a line representing the number of rooms in this tree. nn integers representing the sequence written at the root follow, respectively a1,...,ana1,...,anwhere a1,...,an∈{1,...,n}a1,...,an∈{1,...,n}. 

On the next line, there is a number qqrepresenting the number of mails to be sent. After that, there will be qq integers x1,...,xqx1,...,xqindicating the destination room number of each mail.

Output

For each query, output a sequence of move (EE or WW) the postman needs to make to deliver the mail. For that EE means that the postman should move up the eastern branch and WW the western one. If the destination is on the root, just output a blank line would suffice. 

Note that for simplicity, we assume the postman always starts from the root regardless of the room he had just visited.

Sample Input

2
4
2 1 4 3
3
1 2 3
6
6 5 4 3 2 1
1
1

Sample Output

E

WE
EEEEE

题意：给出测试数据组数T，然后接着一个n表示有n个结点，将这n个节点有序输入后开始查询，数字t表示t次查询，每次查询给出一个节点，问从根节点都这个节点怎样走？（此题只可以向东向西两个方向走）。

注意：如果要查询的节点就是根节点直接输出空行即可。

思路：做这个题的时候可以看着给出的图来做，这样的话会比较清晰，先把要查询的节点 x 与根节点相比，如果小于根节点的话向东走，找到东边的节点作为当前节点num，输出一个E，然后x继续与当前节点比，小于的话向东走否则向西走,直到找到节点x。代码如下：

#include<stdio.h>
#include<string.h>
#define maxn 1010
int a[maxn];
int main()
{
    int i,j,T,n,t,x,num,s;
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&t);
        for(i=0;i<t;i++)
        {
            scanf("%d",&x);
            if(x==a[0])
            {
                printf("\n");
                continue;
            }
            num=a[0];
            s=x-num;
            for(j = 1; j < n; j++)
            {
                if(s>0)
                {
                    while(a[j]<num)  //这都是要跳过的点,要找比num大的点。
                        j++;
                    printf("W"); //如果x比当前节点大，向西走
                    num=a[j];   //num为当前节点的值
                    s=x-num; //更新s的值
                }
                else
                {
                    while(a[j]>num)  //这是要跳过的点
                        j++;
                    printf("E");  //x比当前节点小，向东走
                    num=a[j];     //num为当前节点的值
                    s=x-num;       //s是差值
                }
                if(a[j]==x)
                    break;   //找到了就结束
            }
           printf("\n");
        }
    }
    return 0;
}