message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy finally got to work, where there is a huge amount of tasks waiting for him. Vasiliy is given a matrix consisting of n rows and m columns and q tasks. Each task is to swap two submatrices of the given matrix.
For each task Vasiliy knows six integers ai, bi, ci, di, hi, wi, where ai is the index of the row where the top-left corner of the first rectangle is located, bi is the index of its column, ci is the index of the row of the top-left corner of the second rectangle, di is the index of its column, hi is the height of the rectangle and wi is its width.
It's guaranteed that two rectangles in one query do not overlap and do not touch, that is, no cell belongs to both rectangles, and no two cells belonging to different rectangles share a side. However, rectangles are allowed to share an angle.
Vasiliy wants to know how the matrix will look like after all tasks are performed.
Input
The first line of the input contains three integers n, m and q (2 ≤ n, m ≤ 1000, 1 ≤ q ≤ 10 000) — the number of rows and columns in matrix, and the number of tasks Vasiliy has to perform.
Then follow n lines containing m integers vi, j (1 ≤ vi, j ≤ 109) each — initial values of the cells of the matrix.
Each of the following q lines contains six integers ai, bi, ci, di, hi, wi (1 ≤ ai, ci, hi ≤ n, 1 ≤ bi, di, wi ≤ m).
Output
Print n lines containing m integers each — the resulting matrix.
Examples
Input
4 4 2
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
1 1 3 3 2 2
3 1 1 3 2 2
Output
4 4 3 3
4 4 3 3
2 2 1 1
2 2 1 1
Input
4 2 1
1 1
1 1
2 2
2 2
1 1 4 1 1 2
Output
2 2
1 1
2 2
1 1
Submitted Solution:
```
def get_index(i,j):
global m
return i*(m+1)+j
if __name__=='__main__':
n,m,q = [int(i) for i in input().split()]
a = [[0,-1,-1] for i in range(m+1)]
for i in range(1,n+1):
a.append([0,-1,-1])
a.extend([[int(j),-1,-1] for j in input().split()]) #a[index][0]为值 a[index][1]为右邻的index a[index][2]为下邻的index
# print(a)
for i in range(n+1):
for j in range(m):
index = get_index(i,j)
a[index][1] = get_index(i,j+1)
for i in range(n):
for j in range(m+1):
index = get_index(i,j)
a[index][2] = get_index(i+1,j)
# print(a)
for k in range(q):
x1,y1,x2,y2,h,w = [int(j) for j in input().split()]
index1,index2 = get_index(x1,0),get_index(x2,0)
for i in range(y1-1):
index1 = a[index1][1]
for i in range(y2-1):
index2 = a[index2][1]
for i in range(h):
a[index1][1],a[index2][1] = a[index2][1],a[index1][1]
index1 = a[index1][2]
index2 = a[index2][2]
index1,index2 = get_index(x1,0),get_index(x2,0)
for i in range(y1-1+w):
index1 = a[index1][1]
for i in range(y2-1+w):
index2 = a[index2][1]
for i in range(h):
a[index1][1],a[index2][1] = a[index2][1],a[index1][1]
index1 = a[index1][2]
index2 = a[index2][2]
index1, index2 = get_index(0,y1), get_index(0,y2)
for i in range(x1-1):
index1 = a[index1][2]
for i in range(x1-1):
index2 = a[index2][2]
for i in range(w):
a[index1][2],a[index2][2] = a[index2][2],a[index1][2]
index1 = a[index1][1]
index2 = a[index2][1]
index1, index2 = get_index(0,y1), get_index(0,y2)
for i in range(x1-1+h):
index1 = a[index1][2]
for i in range(x1-1+h):
index2 = a[index2][2]
for i in range(w):
a[index1][2],a[index2][2] = a[index2][2],a[index1][2]
index1 = a[index1][1]
index2 = a[index2][1]
for i in range(1, n + 1):
index = get_index(i, 0)
for j in range(m):
index = a[index][1]
print(a[index][0], end=' ')
print()
# print(a)
``` | instruction | 0 | 8,714 | 23 | 17,428 |
No | output | 1 | 8,714 | 23 | 17,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy finally got to work, where there is a huge amount of tasks waiting for him. Vasiliy is given a matrix consisting of n rows and m columns and q tasks. Each task is to swap two submatrices of the given matrix.
For each task Vasiliy knows six integers ai, bi, ci, di, hi, wi, where ai is the index of the row where the top-left corner of the first rectangle is located, bi is the index of its column, ci is the index of the row of the top-left corner of the second rectangle, di is the index of its column, hi is the height of the rectangle and wi is its width.
It's guaranteed that two rectangles in one query do not overlap and do not touch, that is, no cell belongs to both rectangles, and no two cells belonging to different rectangles share a side. However, rectangles are allowed to share an angle.
Vasiliy wants to know how the matrix will look like after all tasks are performed.
Input
The first line of the input contains three integers n, m and q (2 ≤ n, m ≤ 1000, 1 ≤ q ≤ 10 000) — the number of rows and columns in matrix, and the number of tasks Vasiliy has to perform.
Then follow n lines containing m integers vi, j (1 ≤ vi, j ≤ 109) each — initial values of the cells of the matrix.
Each of the following q lines contains six integers ai, bi, ci, di, hi, wi (1 ≤ ai, ci, hi ≤ n, 1 ≤ bi, di, wi ≤ m).
Output
Print n lines containing m integers each — the resulting matrix.
Examples
Input
4 4 2
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
1 1 3 3 2 2
3 1 1 3 2 2
Output
4 4 3 3
4 4 3 3
2 2 1 1
2 2 1 1
Input
4 2 1
1 1
1 1
2 2
2 2
1 1 4 1 1 2
Output
2 2
1 1
2 2
1 1
Submitted Solution:
```
def get_index(i,j):
global m
return i*(m+1)+j
if __name__=='__main__':
n,m,q = [int(i) for i in input().split()]
a = [[0,-1,-1] for i in range(m+1)]
for i in range(1,n+1):
a.append([0,-1,-1])
a.extend([[int(j),-1,-1] for j in input().split()]) #a[index][0]为值 a[index][1]为右邻的index a[index][2]为下邻的index
# print(a)
for i in range(n+1):
for j in range(m):
index = get_index(i,j)
a[index][1] = get_index(i,j+1)
for i in range(n):
for j in range(m+1):
index = get_index(i,j)
a[index][2] = get_index(i+1,j)
# print(a)
for k in range(q):
print(k)
x1,y1,x2,y2,h,w = [int(j) for j in input().split()]
index1,index2 = get_index(x1,0),get_index(x2,0)
for i in range(y1-1):
index1 = a[index1][1]
for i in range(y2-1):
index2 = a[index2][1]
for i in range(h):
a[index1][1],a[index2][1] = a[index2][1],a[index1][1]
index1 = a[index1][2]
index2 = a[index2][2]
# for i in range(1, n + 1):
# index = get_index(i, 0)
# for j in range(m):
# index = a[index][1]
# print(a[index][0], end=' ')
# print()
index1,index2 = get_index(x1,0),get_index(x2,0)
for i in range(y1-1+w):
index1 = a[index1][1]
for i in range(y2-1+w):
index2 = a[index2][1]
for i in range(h):
a[index1][1],a[index2][1] = a[index2][1],a[index1][1]
index1 = a[index1][2]
index2 = a[index2][2]
# for i in range(1, n + 1):
# index = get_index(i, 0)
# for j in range(m):
# index = a[index][1]
# print(a[index][0], end=' ')
# print()
index1, index2 = get_index(0,y1), get_index(0,y2)
for i in range(x1-1):
index1 = a[index1][2]
for i in range(x2-1):
index2 = a[index2][2]
for i in range(w):
a[index1][2],a[index2][2] = a[index2][2],a[index1][2]
index1 = a[index1][1]
index2 = a[index2][1]
# for i in range(1, n + 1):
# index = get_index(i, 0)
# for j in range(m):
# index = a[index][1]
# print(a[index][0], end=' ')
# print()
index1, index2 = get_index(0,y1), get_index(0,y2)
for i in range(x1-1+h):
index1 = a[index1][2]
for i in range(x2-1+h):
index2 = a[index2][2]
for i in range(w):
a[index1][2],a[index2][2] = a[index2][2],a[index1][2]
index1 = a[index1][1]
index2 = a[index2][1]
for i in range(1, n + 1):
index = get_index(i, 0)
for j in range(m):
index = a[index][1]
print(a[index][0], end=' ')
print()
# print(a)
``` | instruction | 0 | 8,715 | 23 | 17,430 |
No | output | 1 | 8,715 | 23 | 17,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy finally got to work, where there is a huge amount of tasks waiting for him. Vasiliy is given a matrix consisting of n rows and m columns and q tasks. Each task is to swap two submatrices of the given matrix.
For each task Vasiliy knows six integers ai, bi, ci, di, hi, wi, where ai is the index of the row where the top-left corner of the first rectangle is located, bi is the index of its column, ci is the index of the row of the top-left corner of the second rectangle, di is the index of its column, hi is the height of the rectangle and wi is its width.
It's guaranteed that two rectangles in one query do not overlap and do not touch, that is, no cell belongs to both rectangles, and no two cells belonging to different rectangles share a side. However, rectangles are allowed to share an angle.
Vasiliy wants to know how the matrix will look like after all tasks are performed.
Input
The first line of the input contains three integers n, m and q (2 ≤ n, m ≤ 1000, 1 ≤ q ≤ 10 000) — the number of rows and columns in matrix, and the number of tasks Vasiliy has to perform.
Then follow n lines containing m integers vi, j (1 ≤ vi, j ≤ 109) each — initial values of the cells of the matrix.
Each of the following q lines contains six integers ai, bi, ci, di, hi, wi (1 ≤ ai, ci, hi ≤ n, 1 ≤ bi, di, wi ≤ m).
Output
Print n lines containing m integers each — the resulting matrix.
Examples
Input
4 4 2
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
1 1 3 3 2 2
3 1 1 3 2 2
Output
4 4 3 3
4 4 3 3
2 2 1 1
2 2 1 1
Input
4 2 1
1 1
1 1
2 2
2 2
1 1 4 1 1 2
Output
2 2
1 1
2 2
1 1
Submitted Solution:
```
def get_index(i,j):
global m
return i*(m+1)+j
if __name__=='__main__':
n,m,q = [int(i) for i in input().split()]
a = [[0,-1,-1] for i in range(m+1)]
for i in range(1,n+1):
a.append([0,-1,-1])
a.extend([[int(j),-1,-1] for j in input().split()]) #a[index][0]为值 a[index][1]为右邻的index a[index][2]为下邻的index
# print(a)
for i in range(n+1):
for j in range(m):
index = get_index(i,j)
a[index][1] = get_index(i,j+1)
for i in range(n):
for j in range(m+1):
index = get_index(i,j)
a[index][2] = get_index(i+1,j)
# print(a)
for k in range(q):
x1,y1,x2,y2,h,w = [int(j) for j in input().split()]
index1,index2 = get_index(x1,0),get_index(x2,0)
for i in range(y1-1):
index1 = a[index1][1]
for i in range(y2-1):
index2 = a[index2][1]
for i in range(h):
a[index1][1],a[index2][1] = a[index2][1],a[index1][1]
index1 = a[index1][2]
index2 = a[index2][2]
index1,index2 = get_index(x1,0),get_index(x2,0)
for i in range(y1-1+w):
index1 = a[index1][1]
for i in range(y2-1+w):
index2 = a[index2][1]
for i in range(h):
a[index1][1],a[index2][1] = a[index2][1],a[index1][1]
index1 = a[index1][2]
index2 = a[index2][2]
index1, index2 = get_index(0,y1), get_index(0,y2)
for i in range(x1-1):
index1 = a[index1][2]
for i in range(x2-1):
index2 = a[index2][2]
for i in range(w):
a[index1][2],a[index2][2] = a[index2][2],a[index1][2]
index1 = a[index1][1]
index2 = a[index2][1]
index1, index2 = get_index(0,y1), get_index(0,y2)
for i in range(x1-1+h):
index1 = a[index1][2]
for i in range(x2-1+h):
index2 = a[index2][2]
for i in range(w):
a[index1][2],a[index2][2] = a[index2][2],a[index1][2]
index1 = a[index1][1]
index2 = a[index2][1]
for i in range(1, n + 1):
index = get_index(i, 0)
for j in range(m):
index = a[index][1]
print(a[index][0], end=' ')
print()
# print(a)
``` | instruction | 0 | 8,716 | 23 | 17,432 |
No | output | 1 | 8,716 | 23 | 17,433 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,840 | 23 | 17,680 |
"Correct Solution:
```
[n, d], *a = [[int(i) for i in x.split()] for x in open(0)]
print(sum(x**2 + y**2 <= d**2 for x, y in a))
``` | output | 1 | 8,840 | 23 | 17,681 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,841 | 23 | 17,682 |
"Correct Solution:
```
n, d=map(int,input().split())
sm=0
d2=d**2
for i in range(n):
x,y=map(int,input().split())
if x**2+y**2<=d2:
sm+=1
print(sm)
``` | output | 1 | 8,841 | 23 | 17,683 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,842 | 23 | 17,684 |
"Correct Solution:
```
n,d = map(int,input().split())
cnt = 0
for i in range(n):
p,q = map(int,input().split())
cnt += p*p+q*q <= d*d
print(cnt)
``` | output | 1 | 8,842 | 23 | 17,685 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,843 | 23 | 17,686 |
"Correct Solution:
```
n,d = map(int,input().split())
s = 0
for i in range(n):
x,y = map(int,input().split())
if x*x+y*y<=d*d:
s += 1
print(s)
``` | output | 1 | 8,843 | 23 | 17,687 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,844 | 23 | 17,688 |
"Correct Solution:
```
n,d=map(int,input().split())
print(sum(eval("(("+input().replace(" ",")**2+(")+")**2)")**0.5<=d for _ in range(n)))
``` | output | 1 | 8,844 | 23 | 17,689 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,845 | 23 | 17,690 |
"Correct Solution:
```
n,d=map(int,input().split())
cnt=0
for i in range(n):
a,b=map(int,input().split())
if a**2+b**2<=d**2:
cnt+=1
print(cnt)
``` | output | 1 | 8,845 | 23 | 17,691 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,846 | 23 | 17,692 |
"Correct Solution:
```
n,d=map(int,input().split())
ans=0
for i in range(n):
x,y=map(int,input().split())
if(x*x+y*y<=d*d):
ans+=1
print(ans)
``` | output | 1 | 8,846 | 23 | 17,693 |
Provide a correct Python 3 solution for this coding contest problem.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6 | instruction | 0 | 8,847 | 23 | 17,694 |
"Correct Solution:
```
N,D=map(int,input().split())
ans=0
for i in range(N):
x,y=map(int,input().split())
if(D*D>=x*x+y*y):ans+=1
print(ans)
``` | output | 1 | 8,847 | 23 | 17,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
n,d = map(int, input().split())
cnt = 0
for i in range(n):
x,y = map(int, input().split())
if (x*x + y*y)**0.5 <= d: cnt += 1
print(cnt)
``` | instruction | 0 | 8,848 | 23 | 17,696 |
Yes | output | 1 | 8,848 | 23 | 17,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
N,D=map(int,input().split())
a=0
d=D**2
for i in range(N):
x,y=map(int,input().split())
if x**2+y**2<=d:
a+=1
print(a)
``` | instruction | 0 | 8,849 | 23 | 17,698 |
Yes | output | 1 | 8,849 | 23 | 17,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
n,d=map(int,input().split())
d1=d**2;c=0
for i in range(n):
x,y=map(int,input().split())
if x**2+y**2<=d1:
c+=1
print(c)
``` | instruction | 0 | 8,850 | 23 | 17,700 |
Yes | output | 1 | 8,850 | 23 | 17,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
N,M=list(map(int,input().split()))
ans=0
for u in range(N):
a,b=map(int,input().split())
S=a*a+b*b
if M*M>=S:
ans+=1
print(ans)
``` | instruction | 0 | 8,851 | 23 | 17,702 |
Yes | output | 1 | 8,851 | 23 | 17,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
import sys
a = []
for l in sys.stdin:
a.append(l)
N,D = a[0].split(' ')
N = int(N)
D = int(D)
D = D*D
x = []
y = []
count = 0
dist = 0.0
for i in range(1,N):
x, y = a[i].split(' ')
x = int(x)
y = int(y)
dist = x*x+y*y
if dist <= D:
count = count + 1
print(count)
``` | instruction | 0 | 8,852 | 23 | 17,704 |
No | output | 1 | 8,852 | 23 | 17,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
n,d=map(int,input().split())
x = [0]*n
y = [0]*n
count=0
for i in range(n):
x[i],y[i] = map(int, input().split())
if x[i]**2+y[i]**2<=d**2:
count=+1
print(count)
``` | instruction | 0 | 8,853 | 23 | 17,706 |
No | output | 1 | 8,853 | 23 | 17,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
n, d = map(int, input().split())
cnt = 0
for ti in range(n) :
x, y = map(int, input().split())
dist = (x**2 + y**2)**0.5
if(dist >= d):
cnt += 1
print(cnt)
``` | instruction | 0 | 8,854 | 23 | 17,708 |
No | output | 1 | 8,854 | 23 | 17,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N points in the two-dimensional plane. The coordinates of the i-th point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be represented as \sqrt{p^2+q^2}.
Constraints
* 1 \leq N \leq 2\times 10^5
* 0 \leq D \leq 2\times 10^5
* |X_i|,|Y_i| \leq 2\times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N
Output
Print an integer representing the number of points such that the distance from the origin is at most D.
Examples
Input
4 5
0 5
-2 4
3 4
4 -4
Output
3
Input
12 3
1 1
1 1
1 1
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Output
7
Input
20 100000
14309 -32939
-56855 100340
151364 25430
103789 -113141
147404 -136977
-37006 -30929
188810 -49557
13419 70401
-88280 165170
-196399 137941
-176527 -61904
46659 115261
-153551 114185
98784 -6820
94111 -86268
-30401 61477
-55056 7872
5901 -163796
138819 -185986
-69848 -96669
Output
6
Submitted Solution:
```
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, d, i, count = 0;
cin >> n;
cin >> d;
for (i = 0; i < n; i++) {
long long int x, y;
cin >> x;
cin >> y;
if (x * x + y * y <= d * d) {
count = count + 1;
}
}
cout << count;
}
``` | instruction | 0 | 8,855 | 23 | 17,710 |
No | output | 1 | 8,855 | 23 | 17,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,351 | 23 | 18,702 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
cap=[0,1]
for i in range(2,101):
k=ceil(i/2)
k+=k-1
cap.append(cap[-1]+k)
def possible(x):
quad=x//2
load=cap[quad]
if(quad%2==0):
for i in range(quad*2+2):
rem=n-i
if(rem%4==0 and rem//4<=load):
return True
else:
quad-=1
for i in range(quad*2+2):
rem=n-i
if(rem%4==0 and rem//4<=load):
return True
quad+=1
rem=n-ceil(quad/2)*4
if(rem%4==0 and rem//4<=load-4):
return True
return False
n=Int()
if(n==2):
print(3)
exit()
for mid in range(1,100,2):
if(possible(mid)):
print(mid)
break
``` | output | 1 | 9,351 | 23 | 18,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,352 | 23 | 18,704 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
#!/usr/bin/python3.5
x = int(input())
if x == 1:
print(1)
quit()
elif x == 2:
print(3)
quit()
elif x == 3:
print(5)
quit()
else:
if x % 2 == 0:
k = x * 2
else:
k = x * 2 - 1
for n in range(1, 16, 2):
if n ** 2 >= k:
print(n)
break
``` | output | 1 | 9,352 | 23 | 18,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,353 | 23 | 18,706 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
'''input
3
'''
from sys import stdin
import math
def make_dp():
dp = [0] * 1001
dp[1] = 1
dp[3] = 5
for i in range(5, 100, 2):
dp[i] = dp[i - 2] + i + i - 2
return dp
# main starts
x = int(stdin.readline().strip())
if x == 3:
print(5)
exit()
dp = make_dp()
for i in range(1, len(dp)):
if x <= dp[i]:
print(i)
break
``` | output | 1 | 9,353 | 23 | 18,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,354 | 23 | 18,708 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
from bisect import bisect_left
arr = [((2*i+1)**2)//2 + 1 for i in range(7)]
n = int(input())
if n == 3:
print(5)
else:
print(2*bisect_left(arr,n)+1)
``` | output | 1 | 9,354 | 23 | 18,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,355 | 23 | 18,710 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
import sys
import math
MAXNUM = math.inf
MINNUM = -1 * math.inf
ASCIILOWER = 97
ASCIIUPPER = 65
def getInt():
return int(sys.stdin.readline().rstrip())
def getInts():
return map(int, sys.stdin.readline().rstrip().split(" "))
def getString():
return sys.stdin.readline().rstrip()
def printOutput(ans):
sys.stdout.write()
pass
def solve(num):
if num == 1:
return 1
if num == 2:
return 3
if num == 3:
return 5
for i in range(3, 102, 2):
big = ((i + 1) // 2) ** 2
small = (i // 2) ** 2
if big + small >= num:
return i
def readinput():
num = getInt()
print(solve(num))
readinput()
``` | output | 1 | 9,355 | 23 | 18,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,356 | 23 | 18,712 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n=int(input())
if(n==3):
print(5)
else:
x=1
xx=1
while(xx<n):
x+=2
xx=x*x//2+1
print(x)
``` | output | 1 | 9,356 | 23 | 18,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,357 | 23 | 18,714 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
u = int(input())
if u == 1:
print(1)
if 2 <= u <= 5:
if u == 3:
print(5)
else:
print(3)
if 6 <= u <= 13:
print(5)
if 14 <= u <= 25:
print(7)
if 26 <= u <= 41:
print(9)
if 42 <= u <= 61:
print(11)
if 62 <= u <= 85:
print(13)
if 86 <= u <= 100:
print(15)
``` | output | 1 | 9,357 | 23 | 18,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image> | instruction | 0 | 9,358 | 23 | 18,716 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n=int(input())
if n==1:i=1
elif n!=3and n<6:i=3
else:
i=5
while i*i//2+1<n:i+=2
print(i)
# Made By Mostafa_Khaled
``` | output | 1 | 9,358 | 23 | 18,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image>
Submitted Solution:
```
def solve(n):
if n == 1:
return 1
if n == 3:
return 5
for k in range(100):
val = 4 * (((k - 1) ** 2 + 1) // 2 + (k + 1) // 2) - 3
if val >= n:
return 2 * k - 1
print(solve(int(input())))
``` | instruction | 0 | 9,359 | 23 | 18,718 |
Yes | output | 1 | 9,359 | 23 | 18,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image>
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def steps(x):
c=0
while(x%2==0 and x>2):
x//=2
c+=1
return c
cap=[1]
for i in range(2,101):
k=ceil(i/2)
k+=k-1
cap.append(cap[-1]+k)
# print(cap)
n=Int()
need=n//4
quad=bisect_left(cap,need)+1
extra=n%4
ans=2*quad+1
# print(need,extra,quad)
if(extra==0):
print(ans)
else:
if(extra>4*(quad//2)+1):
print(ans+2)
else:
print(ans)
``` | instruction | 0 | 9,360 | 23 | 18,720 |
No | output | 1 | 9,360 | 23 | 18,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image>
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def check(extra):
if(extra%2):
if(extra-1>(quad//2)*4):
return False
else:
return True
else:
if(extra>ceil(quad/2)*4):
return False
else:
return True
cap=[0,1]
for i in range(2,101):
k=ceil(i/2)
k+=k-1
cap.append(cap[-1]+k)
# print(cap)
n=Int()
for ans in range(1,100,2):
quad=cap[ans//2]
extra=n-quad*4
if(extra<0):
extra=n
# print(ans,quad,extra)
# if(extra<0):
# print(ans+2)
# exit()
if(check(extra)):
print(ans)
exit()
``` | instruction | 0 | 9,361 | 23 | 18,722 |
No | output | 1 | 9,361 | 23 | 18,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image>
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def check(extra):
if(extra%2):
if(extra-1>(quad//2)*4):
return False
else:
return True
else:
if(extra>ceil(quad//2)*4):
return False
else:
return True
cap=[0,1]
for i in range(2,101):
k=ceil(i/2)
k+=k-1
cap.append(cap[-1]+k)
# print(cap)
n=Int()
for ans in range(1,100,2):
quad=cap[ans//2]
extra=n-quad*4
# print(quad,extra)
if(check(extra)):
print(ans)
exit()
``` | instruction | 0 | 9,362 | 23 | 18,724 |
No | output | 1 | 9,362 | 23 | 18,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.
Let's call matrix A clear if no two cells containing ones have a common side.
Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.
Let's define the sharpness of matrix A as the number of ones in it.
Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.
Input
The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.
Output
Print a single number — the sought value of n.
Examples
Input
4
Output
3
Input
9
Output
5
Note
The figure below shows the matrices that correspond to the samples:
<image>
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
cap=[0,1]
for i in range(2,101):
k=ceil(i/2)
k+=k-1
cap.append(cap[-1]+k)
def possible(x):
quad=x//2
load=cap[quad]
if(quad%2==0):
for i in range(quad*2+2):
rem=n-i
if(rem%4==0 and rem//4<=load):
return True
else:
quad-=1
for i in range(quad*2+2):
rem=n-i
if(rem%4==0 and rem//4<=load):
return True
quad+=1
rem=n-ceil(quad/2)*4
if(rem%4==0 and rem//4<=load-4):
return True
return False
n=Int()
for mid in range(1,100,2):
if(possible(mid)):
print(mid)
break
``` | instruction | 0 | 9,363 | 23 | 18,726 |
No | output | 1 | 9,363 | 23 | 18,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cake shop made a lot of roll cakes of various sizes. You have been tasked with arranging this cake in a box.
The roll cake is so soft that it will collapse if another roll cake is on top. Therefore, as shown in Fig. (A), all roll cakes must be arranged so that they touch the bottom of the box. Sorting also changes the required width.
<image>
---
Figure (a)
<image>
Figure (b)
Read the radii r1, r2, ..., rn of n roll cakes and the length of the box, judge whether they fit well in the box, and devise the order of arrangement. ", Create a program that outputs" NA "if it does not fit in any order.
It is assumed that the cross section of the roll cake is a circle and the height of the wall of the box is high enough. However, the radius of the roll cake should be an integer between 3 and 10. In other words, there is no extreme difference in cake radii, and small cakes do not get stuck between large cakes as shown in Figure (b).
Input
The input consists of multiple datasets. Each dataset is given in the following format:
W r1 r2 ... rn
First, the integer W (1 ≤ W ≤ 1,000) representing the length of the box is given. It is then given the integer ri (3 ≤ ri ≤ 10), which represents the radius of each roll cake, separated by blanks. The number of cakes n is 12 or less.
The number of datasets does not exceed 50.
Output
Print OK or NA on one line for each dataset.
Example
Input
30 4 5 6
30 5 5 5
50 3 3 3 10 10
49 3 3 3 10 10
Output
OK
OK
OK
NA
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
import os
import math
def pythagoras(a, b):
return 2 * math.sqrt(a * b)
for s in sys.stdin:
lst = list(map(int, s.split()))
W = lst[0]
R = lst[1:]
R.sort()
n = len(R)
if n == 0:
print('OK')
exit()
if n == 1:
if W >= R[0]:
print('OK')
exit()
else:
print('NA')
exit()
left = []
right = []
left.append(R.pop(0))
right.append(R.pop(0))
l = left[0] + right[0]
while R:
min_R = R[0]
max_R = R[-1]
left_R = left[-1]
right_R = right[-1]
if left_R <= right_R:
if right_R - min_R >= max_R - left_R:
right.append(R.pop(0))
l += pythagoras(right_R, min_R)
else:
left.append(R.pop(-1))
l += pythagoras(max_R, left_R)
else:
if left_R - min_R >= max_R - right_R:
left.append(R.pop(0))
l += pythagoras(left_R, min_R)
else:
right.append(R.pop(-1))
l += pythagoras(max_R, right_R)
l += pythagoras(left[-1], right[-1])
if l <= W:
print('OK')
else:
print('NA')
``` | instruction | 0 | 9,797 | 23 | 19,594 |
No | output | 1 | 9,797 | 23 | 19,595 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,893 | 23 | 19,786 |
"Correct Solution:
```
N = int(input())
xs = set();ys = set()
P = []
for i in range(N):
x1,y1,x2,y2 = map(int, input().split())
xs.add(x1);xs.add(x2)
ys.add(y1);ys.add(y2)
P.append((x1,x2,y1,y2))
X = sorted(xs);Y = sorted(ys)
MX = {x: i for i, x in enumerate(X)}
MY = {y: i for i, y in enumerate(Y)}
H = len(Y);W = len(X)
S = [[0]*(W+1) for i in range(H+1)]
for x1, x2, y1, y2 in P:
p1 = MY[y1]; q1 = MX[x1]
p2 = MY[y2]; q2 = MX[x2]
S[p1][q1] += 1
S[p1][q2] -= 1
S[p2][q1] -= 1
S[p2][q2] += 1
#右側に囲まれている長方形の辺があれば正になる？
for i in range(H):
for j in range(W):
S[i][j+1] += S[i][j]
#上側に囲まれている長方形の辺があれば正になる？
for j in range(W):
for i in range(H):
S[i+1][j]+= S[i][j]
ans = 0
for i in range(H):
for j in range(W):
if S[i][j]:
ans += (Y[i+1]-Y[i])*(X[j+1]-X[j])
print(ans)
``` | output | 1 | 9,893 | 23 | 19,787 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,894 | 23 | 19,788 |
"Correct Solution:
```
N = int(input())
XS = set(); YS = set()
P = []
for i in range(N):
x1, y1, x2, y2 = map(int, input().split())
XS.add(x1); XS.add(x2)
YS.add(y1); YS.add(y2)
P.append((x1, y1, x2, y2))
X = sorted(XS); Y = sorted(YS)
MX = {x: i for i, x in enumerate(X)}
MY = {y: i for i, y in enumerate(Y)}
H = len(Y); W = len(X)
S = [[0]*(W+1) for i in range(H+1)]
for x1, y1, x2, y2 in P:
p1 = MY[y1]; q1 = MX[x1]
p2 = MY[y2]; q2 = MX[x2]
S[p1][q1] += 1
S[p1][q2] -= 1
S[p2][q1] -= 1
S[p2][q2] += 1
for i in range(H):
for j in range(W):
S[i][j+1] += S[i][j]
for j in range(W):
for i in range(H):
S[i+1][j] += S[i][j]
ans = 0
for i in range(H):
for j in range(W):
if S[i][j]:
ans += (Y[i+1] - Y[i]) * (X[j+1] - X[j])
print(ans)
``` | output | 1 | 9,894 | 23 | 19,789 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,895 | 23 | 19,790 |
"Correct Solution:
```
#import bisect
#from collections import product
n = int(input())
co_x = set()
co_y = set()
Rec = []
for _ in range(n):
x1,y1,x2,y2 = map(int, input().split( ))
co_x = co_x | {x1,x2}
co_y = co_y | {y1,y2}
Rec.append([x1,y1,x2,y2])
Rec.sort()
co_x = list(co_x)
co_y = list(co_y)
co_x.sort()
co_y.sort()
r = len(co_x)
l = len(co_y)
#Rec_press = []
"""
for p in Rec:
ix0 = bisect.bisect_left(co_x,p[0])
iy0 = bisect.bisect_left(co_x,p[1])
ix1 = bisect.bisect_left(co_x,p[2])
iy1 = bisect.bisect_left(co_x,p[3])
Rec_press.append([ix0,iy0,ix1,iy1])
#以上圧縮
"""
#3161838 By JAKENU0X5E参照
mx = {x:i for i,x in enumerate(co_x)}
my = {y:i for i,y in enumerate(co_y)}
h = len(mx)
w = len(my)
#print(mx,my)
s = [[0]*(w+1) for i in range(h+1)]
for p in Rec:
p1 = mx[p[0]]
q1 = my[p[1]]
p2 = mx[p[2]]
q2 = my[p[3]]
s[p1][q1] += 1
s[p1][q2] -= 1
s[p2][q1] -= 1
s[p2][q2] += 1
"""
この時点でのs[p][q]は被覆開始のフラグ
次のforで直前のs[i][j]に応じてs[i][j+1],s[i+1][j+1]を+1することで
sが被覆されているか否かのフラグになる
"""
for i in range(h):
for j in range(w):
s[i][j+1] += s[i][j]
for j in range(w):
for i in range(h):
s[i+1][j] += s[i][j]
ans = 0
for i in range(h):
for j in range(w):
if s[i][j]:
#print(co_x[i],co_y[j],co_x[i+1],co_y[j+1],(co_x[i+1]-co_x[i])*(co_y[j+1]-co_y[j]))
ans += (co_x[i+1]-co_x[i])*(co_y[j+1]-co_y[j])
print(ans)
``` | output | 1 | 9,895 | 23 | 19,791 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,896 | 23 | 19,792 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import io, sys
import bisect
def main():
N = int( sys.stdin.readline() )
x1y1x2y2_list = [ list(map(int, sys.stdin.readline().split())) for _ in range(N) ]
X1 = [a for a,b,c,d in x1y1x2y2_list]
X2 = [c for a,b,c,d in x1y1x2y2_list]
Y1 = [b for a,b,c,d in x1y1x2y2_list]
Y2 = [d for a,b,c,d in x1y1x2y2_list]
all_X = compress(X1,X2)
all_Y = compress(Y1,Y2)
matrix =[ [0]*len(all_X) for _ in range(len(all_Y)) ]
for i in range(N):
matrix[ Y1[i] ][ X1[i] ] += 1
matrix[ Y2[i] ][ X2[i] ] += 1
matrix[ Y2[i] ][ X1[i] ] -= 1
matrix[ Y1[i] ][ X2[i] ] -= 1
for row in range(len(matrix)):
for col in range(1, len(matrix[0])):
matrix[row][col] += matrix[row][col-1]
for row in range(1, len(matrix)):
for col in range(len(matrix[0])):
matrix[row][col] += matrix[row-1][col]
ans = 0
for row in range(len(matrix)):
for col in range(len(matrix[0])):
if matrix[row][col] > 0:
area = (all_X[col+1] - all_X[col]) * (all_Y[row+1] - all_Y[row])
ans += area
print(ans)
def compress(A1 :list, A2 :list):
all_A = []
#delta = [-1, 0, 1]
delta = [0]
for a in (A1 + A2):
for d in delta:
all_A.append(a + d)
all_A = sorted(set(all_A))
for i in range(len(A1)):
A1[i] = bisect.bisect_left(all_A, A1[i])
A2[i] = bisect.bisect_left(all_A, A2[i])
return all_A
if __name__ == "__main__":
main()
``` | output | 1 | 9,896 | 23 | 19,793 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,897 | 23 | 19,794 |
"Correct Solution:
```
from typing import List, Iterator, Tuple
class Rectangle(object):
def __init__(self, xy1: Tuple[int, int], xy2: Tuple[int, int]) -> None:
self.topleft = xy1
self.bottomright = xy2
def area(self) -> int:
x1, y1 = self.topleft
x2, y2 = self.bottomright
return (x2 - x1) * (y2 - y1)
class MergetdRectangles(object):
def __init__(self) -> None:
self.rects: List[Rectangle] = []
def _intersect(self, base_rect: Rectangle, new_rect: Rectangle) -> bool:
x1, y1 = base_rect.topleft
x2, y2 = base_rect.bottomright
xn1, yn1 = new_rect.topleft
xn2, yn2 = new_rect.bottomright
if (xn2 <= x1 or x2 <= xn1):
return False
if (yn2 <= y1 or y2 <= yn1):
return False
return True
def _sub(self, base_rect: Rectangle, new_rect: Rectangle) -> Iterator[Rectangle]:
x1, y1 = base_rect.topleft
x2, y2 = base_rect.bottomright
xn1, yn1 = new_rect.topleft
xn2, yn2 = new_rect.bottomright
if (x1 < xn1):
yield Rectangle((x1, y1), (xn1, y2))
if (xn2 < x2):
yield Rectangle((xn2, y1), (x2, y2))
if (y1 < yn1):
yield Rectangle((max(x1, xn1), y1), (min(x2, xn2), yn1))
if (yn2 < y2):
yield Rectangle((max(x1, xn1), yn2), (min(x2, xn2), y2))
def add(self, new_rect: Rectangle) -> None:
rects: List[Rectangle] = []
for r in self.rects:
if self._intersect(r, new_rect):
rects.extend(self._sub(r, new_rect))
else:
rects.append(r)
rects.append(new_rect)
self.rects = rects
def total_area(self) -> int:
return sum([r.area() for r in self.rects])
if __name__ == "__main__":
N = int(input())
rects = MergetdRectangles()
for _ in range(N):
x1, y1, x2, y2 = map(lambda x: int(x), input().split())
rects.add(Rectangle((x1, y1), (x2, y2)))
print(rects.total_area())
``` | output | 1 | 9,897 | 23 | 19,795 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,898 | 23 | 19,796 |
"Correct Solution:
```
#!/usr/bin/env python3
# DSL_4_A: Union of Rectangles
# O(N^2)
class Rect:
def __init__(self, p1, p2):
self.top_left = p1
self.bottom_right = p2
def intersect(self, other):
xs1, ys1 = self.top_left
xs2, ys2 = self.bottom_right
xo1, yo1 = other.top_left
xo2, yo2 = other.bottom_right
if xs1 >= xo2:
return False
elif xs2 <= xo1:
return False
if ys1 >= yo2:
return False
elif ys2 <= yo1:
return False
return True
def sub(self, other):
# assert self.intersect(other), "{} {}".format(self, other)
xs1, ys1 = self.top_left
xs2, ys2 = self.bottom_right
xo1, yo1 = other.top_left
xo2, yo2 = other.bottom_right
if xs1 < xo1:
yield Rect((xs1, ys1), (xo1, ys2))
if xs2 > xo2:
yield Rect((xo2, ys1), (xs2, ys2))
if ys1 < yo1:
yield Rect((max(xs1, xo1), ys1), (min(xs2, xo2), yo1))
if ys2 > yo2:
yield Rect((max(xs1, xo1), yo2), (min(xs2, xo2), ys2))
def area(self):
x1, y1 = self.top_left
x2, y2 = self.bottom_right
return (x2 - x1) * (y2 - y1)
def __str__(self):
return '<Rect({}, {})>'.format(self.top_left, self.bottom_right)
class Rects:
def __init__(self):
self.rects = []
def add(self, rect):
rects = []
for r in self.rects:
if rect.intersect(r):
rects.extend(r.sub(rect))
else:
rects.append(r)
rects.append(rect)
self.rects = rects
# assert self.not_intersect(), self
def area(self):
return sum([r.area() for r in self.rects])
def __str__(self):
s = '<Rects('
s += '\n '.join(str(r) for r in self.rects)
s += ')>'
return s
def not_intersect(self):
for i in range(len(self.rects)-1):
for j in range(i+1, len(self.rects)):
if self.rects[i].intersect(self.rects[j]):
return False
return True
def run():
n = int(input())
rects = Rects()
for _ in range(n):
x1, y1, x2, y2 = [int(i) for i in input().split()]
rect = Rect((x1, y1), (x2, y2))
rects.add(rect)
print(rects.area())
if __name__ == '__main__':
run()
``` | output | 1 | 9,898 | 23 | 19,797 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,899 | 23 | 19,798 |
"Correct Solution:
```
from itertools import accumulate
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
def main():
N = int(input())
xs = set()
ys = set()
rect = []
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
xs.add(x1)
xs.add(x2)
ys.add(y1)
ys.add(y2)
rect.append((x1, y1, x2, y2))
X = sorted(xs)
Y = sorted(ys)
dx = {x: i for i, x in enumerate(X)}
dy = {y: i for i, y in enumerate(Y)}
#print(dx, dy, dx[1])
H = len(xs)
W = len(ys)
grid = [[0] * (W + 1) for _ in range(H + 1)]
for x1, y1, x2, y2 in rect:
grid[dx[x1]][dy[y1]] += 1
grid[dx[x2]][dy[y1]] -= 1
grid[dx[x1]][dy[y2]] -= 1
grid[dx[x2]][dy[y2]] += 1
#print(grid)
for i in range(H):
grid[i] = list(accumulate(grid[i]))
#print(grid)
for i in range(H):
for j in range(W):
grid[i + 1][j] += grid[i][j]
#print(grid)
ans = 0
for i in range(H):
for j in range(W):
if grid[i][j]:
ans += (X[i + 1] - X[i]) * (Y[j + 1] - Y[j])
return ans
if __name__ == "__main__":
print(main())
``` | output | 1 | 9,899 | 23 | 19,799 |
Provide a correct Python 3 solution for this coding contest problem.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8 | instruction | 0 | 9,900 | 23 | 19,800 |
"Correct Solution:
```
from itertools import accumulate
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
def main():
N = int(input())
xs = set()
ys = set()
rect = []
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
xs.add(x1)
xs.add(x2)
ys.add(y1)
ys.add(y2)
rect.append((x1, y1, x2, y2))
X = sorted(xs)
Y = sorted(ys)
dx = {x: i for i, x in enumerate(X)}
dy = {y: i for i, y in enumerate(Y)}
H = len(xs)
W = len(ys)
grid = [[0] * (W + 1) for _ in range(H + 1)]
for x1, y1, x2, y2 in rect:
grid[dx[x1]][dy[y1]] += 1
grid[dx[x2]][dy[y1]] -= 1
grid[dx[x1]][dy[y2]] -= 1
grid[dx[x2]][dy[y2]] += 1
for i in range(H):
grid[i] = list(accumulate(grid[i]))
for i in range(H):
for j in range(W):
grid[i + 1][j] += grid[i][j]
ans = 0
for i in range(H):
for j in range(W):
if grid[i][j]:
ans += (X[i + 1] - X[i]) * (Y[j + 1] - Y[j])
return ans
if __name__ == "__main__":
print(main())
``` | output | 1 | 9,900 | 23 | 19,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.
Constraints
* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $
Input
The input is given in the following format.
$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$
($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.
Output
Print the area of the regions.
Examples
Input
2
0 0 3 4
1 2 4 3
Output
13
Input
3
1 1 2 5
2 1 5 2
1 2 2 5
Output
7
Input
4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3
Output
8
Submitted Solution:
```
#!usr/bin/env python3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
res = list(sys.stdin.readline())
if res[-1] == "\n":
return res[:-1]
return res
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
bl = bisect.bisect_left
br = bisect.bisect_right
def solve():
n = I()
p = LIR(n)
fx = {x for (x, _, _, _) in p} | {x for (_, _, x, _) in p}
fx = list(fx)
fx.sort()
fy = {y for (_, y, _, _) in p} | {y for (_, _, _, y) in p}
fy = list(fy)
fy.sort()
h,w = len(fy),len(fx)
a = [[0]*w for i in range(h)]
for x1,y1,x2,y2 in p:
l,u,r,d = bl(fx,x1),bl(fy,y1),bl(fx,x2),bl(fy,y2)
a[u][l] += 1
a[u][r] -= 1
a[d][l] -= 1
a[d][r] += 1
s = [list(accumulate(i)) for i in a]
for j in range(w):
for i in range(h-1):
s[i+1][j] += s[i][j]
ans = 0
for i in range(h-1):
y = fy[i+1]-fy[i]
for j in range(w-1):
if s[i][j]:
x = fx[j+1]-fx[j]
ans += x*y
print(ans)
return
#Solve
if __name__ == "__main__":
solve()
``` | instruction | 0 | 9,901 | 23 | 19,802 |
Yes | output | 1 | 9,901 | 23 | 19,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,175 | 23 | 20,350 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
import sys,math
input=sys.stdin.readline
from collections import defaultdict
for _ in range(int(input())):
n,w=map(int,input().split())
ans=0
v=list(map(int,input().split()))
d=defaultdict(int)
vis=[]
for i in v:
d[i]+=1
if d[i]==1:
vis.append(i)
vis.sort(reverse=True)
cnt=0
while cnt<n:
cur=0
i=0
while i<len(vis):
if d[vis[i]]>0:
if cur+vis[i]<=w:
cnt+=1
cur+=vis[i]
d[vis[i]]-=1
i-=1
i+=1
ans+=1
print(ans)
``` | output | 1 | 10,175 | 23 | 20,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,176 | 23 | 20,352 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
from heapq import heappop, heappush
T = int(input())
for _ in range(T):
N, W = map(int, input().split())
A = list(map(int, input().split()))
A.sort(reverse=True)
lst = [0]
idx = 0
for a in A:
flag = False
v = heappop(lst)
if v + a <= W:
heappush(lst, v + a)
else:
heappush(lst, v)
heappush(lst, a)
# for idx in range(len(lst)):
# if flag:
# break
# if lst[idx] + a <= W:
# lst[idx] += a
# flag = True
# if not flag:
# lst.append(a)
# print(lst)
print(len(lst))
``` | output | 1 | 10,176 | 23 | 20,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,177 | 23 | 20,354 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
def main():
from sys import stdin, stdout
from collections import Counter
rl = stdin.readline
wl = stdout.write
for _ in range(int(rl())):
n, w = map(int, rl().split())
a = Counter(int(x) for x in rl().split())
a = dict(a.most_common())
keys = sorted(a.keys(), reverse=True)
h = -1
while True:
cur = 0
h += 1
for key in keys:
while a[key] != 0 and cur + key <= w:
cur += key
a[key] -= 1
if cur == 0:
break
wl(str(h) + '\n')
main()
``` | output | 1 | 10,177 | 23 | 20,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,178 | 23 | 20,356 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
import sys
from os import path
if(path.exists("inp.txt")):
sys.stdin = open("inp.txt",'r')
sys.stdout = open("out.txt",'w')
from collections import Counter
t=int(input())
while(t):
t-=1
n,w=map(int,input().split())
li=list(map(int,input().split()))
li.sort()
omp=Counter(li)
r=0
while(omp!={}):
r+=1
h=w
for j in sorted(omp.keys(),reverse=1):
while(omp[j]>0 and h>=j):
h-=j
omp[j]-=1
if omp[j]==0:
del(omp[j])
break
print(r)
``` | output | 1 | 10,178 | 23 | 20,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,179 | 23 | 20,358 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
import sys
import heapq
input = sys.stdin.readline
for _ in range(int(input())):
n, w = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
heap = []
ans = 1
heapq.heappush(heap, -(w-a[n-1]))
for i in range(n-2, -1, -1):
tmp = -heapq.heappop(heap)
if tmp>=a[i]:
heapq.heappush(heap, -(tmp-a[i]))
else:
ans+=1
heapq.heappush(heap, -tmp)
heapq.heappush(heap, -(w-a[i]))
print(ans)
``` | output | 1 | 10,179 | 23 | 20,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,180 | 23 | 20,360 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
t=int(input())
for _ in range(t):
n,y=map(int,input().split())
l=list(map(int,input().split()))
d={}
ans=0
for i in l:
d[i]=d.get(i,0)+1
while d:
w=y
for j in reversed(sorted(d.keys())):
#print(x)
#print(d.keys())
x=min(w//j,d[j])
d[j]-=x
w-=x*j
if d[j]==0:del d[j]
if w==0:break
ans+=1
print(ans)
``` | output | 1 | 10,180 | 23 | 20,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,181 | 23 | 20,362 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
from collections import *
for _ in range(int(input())):
n, w = map(int, input().split())
l = list(map(int, input().split()))
C = Counter(l)
height = 0
while len(C) > 0:
height += 1
rest = w
for i in range(30, -1, -1):
while (1 << i) in C and rest >= (1 << i):
rest -= (1 << i)
C[(1 << i)] -= 1
if C[1 << i] == 0:
del C[1 << i]
print(height)
``` | output | 1 | 10,181 | 23 | 20,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels. | instruction | 0 | 10,182 | 23 | 20,364 |
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
# cp template by varun
from math import ceil
from math import floor
from random import random
from math import gcd
def ar ():
return [int(x) for x in input().split()]
# -----
T = 1
T = int(input())
size = 2**20 + 7
dp = [0]*size
for t in range (T):
N,W = ar()
A = ar()
for i in A:
dp[i]+=1
ans = 0
while N > 0:
curr = W
i = 2**20
while i>0:
while dp[i]>0 and curr>=i:
dp[i]-=1
curr-=i
N-=1
i//=2
ans+=1
print (ans)
``` | output | 1 | 10,182 | 23 | 20,365 |
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