time limit per test
2 seconds
memory limit per test
256 megabytes
There is a 2D-coordinate plane that ranges from (0,0) to (x,y). You are located at (0,0) and want to head to (x,y).
However, there are n horizontal lasers, with the i-th laser continuously spanning (0,ai) to (x,ai). Additionally, there are also m vertical lasers, with the i-th laser continuously spanning (bi,0) to (bi,y).
You may move in any direction to reach (x,y), but your movement must be a continuous curve that lies inside the plane. Every time you cross a vertical or a horizontal laser, it counts as one crossing. Particularly, if you pass through an intersection point between two lasers, it counts as two crossings.
For example, if x=y=2, n=m=1, a=[1], b=[1], the movement can be as follows:
What is the minimum number of crossings necessary to reach (x,y)?
Input
The first line contains t (1≤t≤104) --- the number of test cases.
The first line of each test case contains four integers n, m, x, and y (1≤n,m≤2⋅105,2≤x,y≤109).
The following line contains n integers a1,a2,...,an (0<ai<y) --- the y-coordinates of the horizontal lasers. It is guaranteed that ai>ai−1 for all i>1.
The following line contains m integers b1,b2,...,bm (0<bi<x) --- the x-coordinates of the vertical lasers. It is guaranteed that bi>bi−1 for all i>1.
It is guaranteed that the sum of n and m over all test cases does not exceed 2⋅105.
Output
For each test case, output the minimum number of crossings necessary to reach (x,y).
Example
Input
Copy
2
1 1 2 2
1
1
2 1 100000 100000
42 58
32
Output
Copy
2
3解题说明：此题是一道数学题，找规律能发现直接输出m+n即可。
cpp
#include <stdio.h>
int main()
{
int t, n, m, x, y, a, ans;
scanf("%d", &t);
while (t--)
{
scanf("%d %d %d %d", &n, &m, &x, &y);
ans = n + m;
while (n--)
{
scanf("%d", &a);
}
while (m--)
{
scanf("%d", &a);
}
printf("%d\n", ans);
}
return 0;
}