Describe
Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat!
Initially, Alice’s cat is located in a cell (x,y) of an infinite grid. According to Alice’s theory, cat needs to move:
exactly a steps left: from (u,v) to (u−1,v);
exactly b steps right: from (u,v) to (u+1,v);
exactly c steps down: from (u,v) to (u,v−1);
exactly d steps up: from (u,v) to (u,v+1).
Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.
Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x1,x2]×[y1,y2], i.e. for every cat’s position (u,v) of a walk x1≤u≤x2 and y1≤v≤y2 holds.
Also, note that the cat can visit the same cell multiple times.
Can you help Alice find out if there exists a walk satisfying her wishes?
Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u,v) should always satisfy the constraints: x1≤u≤x2, y1≤v≤y2. The staring point is (x,y).
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1≤t≤103) — the number of testcases.
The first line of each test case contains four integers a, b, c, d (0≤a,b,c,d≤108, a+b+c+d≥1).
The second line of the test case contains six integers x, y, x1, y1, x2, y2 (−108≤x1≤x≤x2≤108, −108≤y1≤y≤y2≤108).
Output
For each test case, output “YES” in a separate line, if there exists a walk satisfying her wishes. Otherwise, output “NO” in a separate line.
You can print each letter in any case (upper or lower).
Example
input
6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0
output
Yes
No
No
Yes
Yes
Yes
Note
In the first test case, one valid exercising walk is
(0,0)→(−1,0)→(−2,0)→(−2,1)→(−2,2)→(−1,2)→(0,2)→(0,1)→(0,0)→(−1,0)
题意
一个点（x，y），向左走a步，向右走b步，向上走c步，向下走d步，判断是否还在给定的边界内
思路实现
题目很长，花了很长时间才读懂，但是题目本身不是太难，就是走的步数左右、上下分别相互抵消，然后再和左右、上下边界分别比较即可，注意边界相等的特殊情况即可
代码
#include<iostream>
using namespace std;
int t, a, b, c, d, x, y, xl, xr, yl,yr;
int main()
{
cin >> t;
while (t--)
{
int flag = 0;
cin >> a >> b >> c >> d;
cin >> x >> y >> xl >> yl >> xr >> yr;
//特判两种特殊情况
if (xl == xr && (a >= 1 || b >= 1))
flag = 0;
else if(yl == yr && (c >= 1 || d >= 1))
flag = 0;
//判断是否在边界内
else
{
x = x + (b - a);
y = y + (d - c);
if (x >= xl && x <= xr && y >= yl && y <= yr)
flag = 1;
}
if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}
码字不易，留个赞吧~
