Elven Postman
Elven Postman
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1108 Accepted Submission(s): 587
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.

For each test case, there is a number n(n≤1000) on a line representing the number of rooms in this tree. n integers representing the sequence written at the root follow, respectively a1,...,an where a1,...,an∈{1,...,n}.
On the next line, there is a number q representing the number of mails to be sent. After that, there will be q integers x1,...,xq indicating the destination room number of each mail.
Note that for simplicity, we assume the postman always starts from the root regardless of the room he had just visited.
#include<iostream>
#include<cstdio>
#include<cstring> using namespace std; const int maxn = 1e3+;
int a[maxn]; int main()
{
int c, n, q, x, num, d;
scanf("%d", &c);
while(c--)
{
memset(a, , sizeof(a)); scanf("%d", &n);
for(int i = ; i < n; i++)
scanf("%d", &a[i]);
scanf("%d", &q);
while(q--)
{
scanf("%d", &x);
if(x == a[])
{
puts("");
continue;
}
num = a[];
d = x - num; for(int i = ; i < n; i++)
{
if(d > )
{
while(a[i] < num) // 跳过所有与当前目标方向相反的点
i++;
printf("W");
num = a[i];
d = x - num;
}
else
{
while(a[i] > num)
i++;
printf("E");
num = a[i];
d = x - num;
}
if(a[i] == x)
break;
}
puts("");
}
}
return ;
}
写代码的时候就好好写代码……………………………………………………………………………………………………………………………………………………………………
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