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.
A Garden with Ponds
Mr. Gardiner is a modern garden designer who is excellent at utilizing the terrain features. His design method is unique: he first decides the location of ponds and design them with the terrain features intact.
According to his unique design procedure, all of his ponds are rectangular with simple aspect ratios. First, Mr. Gardiner draws a regular grid on the map of the garden site so that the land is divided into cells of unit square, and annotates every cell with its elevation. In his design method, a pond occupies a rectangular area consisting of a number of cells. Each of its outermost cells has to be higher than all of its inner cells. For instance, in the following grid map, in which numbers are elevations of cells, a pond can occupy the shaded area, where the outermost cells are shaded darker and the inner cells are shaded lighter. You can easily see that the elevations of the outermost cells are at least three and those of the inner ones are at most two.
<image>
A rectangular area on which a pond is built must have at least one inner cell. Therefore, both its width and depth are at least three.
When you pour water at an inner cell of a pond, the water can be kept in the pond until its level reaches that of the lowest outermost cells. If you continue pouring, the water inevitably spills over. Mr. Gardiner considers the larger capacity the pond has, the better it is. Here, the capacity of a pond is the maximum amount of water it can keep. For instance, when a pond is built on the shaded area in the above map, its capacity is (3 − 1) + (3 − 0) + (3 − 2) = 6, where 3 is the lowest elevation of the outermost cells and 1, 0, 2 are the elevations of the inner cells. Your mission is to write a computer program that, given a grid map describing the elevation of each unit square cell, calculates the largest possible capacity of a pond built in the site.
Note that neither of the following rectangular areas can be a pond. In the left one, the cell at the bottom right corner is not higher than the inner cell. In the right one, the central cell is as high as the outermost cells.
<image>
Input
The input consists of at most 100 datasets, each in the following format.
d w
e1, 1 ... e1, w
...
ed, 1 ... ed, w
The first line contains d and w, representing the depth and the width, respectively, of the garden site described in the map. They are positive integers between 3 and 10, inclusive. Each of the following d lines contains w integers between 0 and 9, inclusive, separated by a space. The x-th integer in the y-th line of the d lines is the elevation of the unit square cell with coordinates (x, y).
The end of the input is indicated by a line containing two zeros separated by a space.
Output
For each dataset, output a single line containing the largest possible capacity of a pond that can be built in the garden site described in the dataset. If no ponds can be built, output a single line containing a zero.
Sample Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output for the Sample Input
0
3
1
9
Example
Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output
0
3
1
9
Submitted Solution:
```
while 1:
d, w = map(int, input().split())
if d == w == 0:
break
E = [[*map(int, input().split())] for i in range(d)]
ans = 0
for x2 in range(d):
for y2 in range(w):
for x1 in range(x2-1):
for y1 in range(y2-1):
# [x1, x2] x [y1, y2]
co = 10**18
for x in range(x1, x2+1):
co = min(co, E[x][y1], E[x][y2])
for y in range(y1, y2+1):
co = min(co, E[x1][y], E[x2][y])
ci = 0
cnt = 0
for x in range(x1+1, x2):
for y in range(y1+1, y2):
ci = max(ci, E[x][y])
cnt += E[x][y]
if ci < co:
ans = max(ans, (x2-x1-1)*(y2-y1-1)*co - cnt)
print(ans)
``` | instruction | 0 | 63,099 | 23 | 126,198 |
No | output | 1 | 63,099 | 23 | 126,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Garden with Ponds
Mr. Gardiner is a modern garden designer who is excellent at utilizing the terrain features. His design method is unique: he first decides the location of ponds and design them with the terrain features intact.
According to his unique design procedure, all of his ponds are rectangular with simple aspect ratios. First, Mr. Gardiner draws a regular grid on the map of the garden site so that the land is divided into cells of unit square, and annotates every cell with its elevation. In his design method, a pond occupies a rectangular area consisting of a number of cells. Each of its outermost cells has to be higher than all of its inner cells. For instance, in the following grid map, in which numbers are elevations of cells, a pond can occupy the shaded area, where the outermost cells are shaded darker and the inner cells are shaded lighter. You can easily see that the elevations of the outermost cells are at least three and those of the inner ones are at most two.
<image>
A rectangular area on which a pond is built must have at least one inner cell. Therefore, both its width and depth are at least three.
When you pour water at an inner cell of a pond, the water can be kept in the pond until its level reaches that of the lowest outermost cells. If you continue pouring, the water inevitably spills over. Mr. Gardiner considers the larger capacity the pond has, the better it is. Here, the capacity of a pond is the maximum amount of water it can keep. For instance, when a pond is built on the shaded area in the above map, its capacity is (3 − 1) + (3 − 0) + (3 − 2) = 6, where 3 is the lowest elevation of the outermost cells and 1, 0, 2 are the elevations of the inner cells. Your mission is to write a computer program that, given a grid map describing the elevation of each unit square cell, calculates the largest possible capacity of a pond built in the site.
Note that neither of the following rectangular areas can be a pond. In the left one, the cell at the bottom right corner is not higher than the inner cell. In the right one, the central cell is as high as the outermost cells.
<image>
Input
The input consists of at most 100 datasets, each in the following format.
d w
e1, 1 ... e1, w
...
ed, 1 ... ed, w
The first line contains d and w, representing the depth and the width, respectively, of the garden site described in the map. They are positive integers between 3 and 10, inclusive. Each of the following d lines contains w integers between 0 and 9, inclusive, separated by a space. The x-th integer in the y-th line of the d lines is the elevation of the unit square cell with coordinates (x, y).
The end of the input is indicated by a line containing two zeros separated by a space.
Output
For each dataset, output a single line containing the largest possible capacity of a pond that can be built in the garden site described in the dataset. If no ponds can be built, output a single line containing a zero.
Sample Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output for the Sample Input
0
3
1
9
Example
Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output
0
3
1
9
Submitted Solution:
```
while True:
d, w = map(int, input().split())
if d==w==0:
break
E=[list(map(int, input().split())) for i in range(d)]
a=0
for x2 in range(d):
for y2 in range(w):
for x1 in range(x2-1):
for y1 in range(y2-1):
co=10**18
for x in range(x1, x2+1):
co=min(co, E[x][y1], E[x][y2])
for y in range(y1, y2+1):
co=min(co, E[x1][y], E[x2][y])
ci=cnt=0
for x in range(x1+1, x2):
for y in range(y1+1, y2):
ci=max(ci, E[x][y])
cnt+=E[x][y]
if ci < co:
a=max(ans, (x2-x1-1)*(y2-y1-1)*co - cnt)
print(a)
``` | instruction | 0 | 63,100 | 23 | 126,200 |
No | output | 1 | 63,100 | 23 | 126,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Garden with Ponds
Mr. Gardiner is a modern garden designer who is excellent at utilizing the terrain features. His design method is unique: he first decides the location of ponds and design them with the terrain features intact.
According to his unique design procedure, all of his ponds are rectangular with simple aspect ratios. First, Mr. Gardiner draws a regular grid on the map of the garden site so that the land is divided into cells of unit square, and annotates every cell with its elevation. In his design method, a pond occupies a rectangular area consisting of a number of cells. Each of its outermost cells has to be higher than all of its inner cells. For instance, in the following grid map, in which numbers are elevations of cells, a pond can occupy the shaded area, where the outermost cells are shaded darker and the inner cells are shaded lighter. You can easily see that the elevations of the outermost cells are at least three and those of the inner ones are at most two.
<image>
A rectangular area on which a pond is built must have at least one inner cell. Therefore, both its width and depth are at least three.
When you pour water at an inner cell of a pond, the water can be kept in the pond until its level reaches that of the lowest outermost cells. If you continue pouring, the water inevitably spills over. Mr. Gardiner considers the larger capacity the pond has, the better it is. Here, the capacity of a pond is the maximum amount of water it can keep. For instance, when a pond is built on the shaded area in the above map, its capacity is (3 − 1) + (3 − 0) + (3 − 2) = 6, where 3 is the lowest elevation of the outermost cells and 1, 0, 2 are the elevations of the inner cells. Your mission is to write a computer program that, given a grid map describing the elevation of each unit square cell, calculates the largest possible capacity of a pond built in the site.
Note that neither of the following rectangular areas can be a pond. In the left one, the cell at the bottom right corner is not higher than the inner cell. In the right one, the central cell is as high as the outermost cells.
<image>
Input
The input consists of at most 100 datasets, each in the following format.
d w
e1, 1 ... e1, w
...
ed, 1 ... ed, w
The first line contains d and w, representing the depth and the width, respectively, of the garden site described in the map. They are positive integers between 3 and 10, inclusive. Each of the following d lines contains w integers between 0 and 9, inclusive, separated by a space. The x-th integer in the y-th line of the d lines is the elevation of the unit square cell with coordinates (x, y).
The end of the input is indicated by a line containing two zeros separated by a space.
Output
For each dataset, output a single line containing the largest possible capacity of a pond that can be built in the garden site described in the dataset. If no ponds can be built, output a single line containing a zero.
Sample Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output for the Sample Input
0
3
1
9
Example
Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output
0
3
1
9
Submitted Solution:
```
def area(x_s,x_e,y_s,y_e):
return int((x_e-x_s-1)*(y_e-y_s-1))
def minimum_border(A,x_s,x_e,y_s,y_e):
min=A[x_s][y_s]
for i in range(x_s,x_e+1):
for j in range(y_s,y_e+1):
if A[i][j]<min:
min=A[i][j]
return min
def maximum_inter(A,x_s,x_e,y_s,y_e):
max=A[x_s+1][y_s+1]
for i in range(x_s+1,x_e):
for j in range(y_s+1,y_e):
if A[i][j]>max:
max=A[i][j]
return max
A=[[0 for i in range(10)]for j in range(10)]
ans=[]
while True:
n,m=map(int,input().split())
if n==m==0:
break
for i in range(n):
tmp=[]
tmp=list(map(int,input().split()))
tmp.reverse()
for j in range(m):
A[i][j]=tmp.pop()
flag=0
max_s=0
for y_s in range(m-2):
for y_e in range(m-3,m):
for x_s in range(n-2):
for x_e in range(n-3,n):
if y_e-y_s<2 or x_e-x_s<2:
pass
else:
#print(area(x_s,x_e,y_s,y_e))
if minimum_border(A,x_s,x_e,y_s,y_e)>maximum_inter(A,x_s,x_e,y_s,y_e):
if flag==0:
print("=====")
max_s=area(x_s,x_e,y_s,y_e)
else:
if area(x_s,x_e,y_s,y_e)>max_s:
max_s=area(x_s,x_e,y_s,y_e)
ans.append(max_s)
ans.reverse()
for i in range(len(ans)):
print(ans.pop())
``` | instruction | 0 | 63,101 | 23 | 126,202 |
No | output | 1 | 63,101 | 23 | 126,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Garden with Ponds
Mr. Gardiner is a modern garden designer who is excellent at utilizing the terrain features. His design method is unique: he first decides the location of ponds and design them with the terrain features intact.
According to his unique design procedure, all of his ponds are rectangular with simple aspect ratios. First, Mr. Gardiner draws a regular grid on the map of the garden site so that the land is divided into cells of unit square, and annotates every cell with its elevation. In his design method, a pond occupies a rectangular area consisting of a number of cells. Each of its outermost cells has to be higher than all of its inner cells. For instance, in the following grid map, in which numbers are elevations of cells, a pond can occupy the shaded area, where the outermost cells are shaded darker and the inner cells are shaded lighter. You can easily see that the elevations of the outermost cells are at least three and those of the inner ones are at most two.
<image>
A rectangular area on which a pond is built must have at least one inner cell. Therefore, both its width and depth are at least three.
When you pour water at an inner cell of a pond, the water can be kept in the pond until its level reaches that of the lowest outermost cells. If you continue pouring, the water inevitably spills over. Mr. Gardiner considers the larger capacity the pond has, the better it is. Here, the capacity of a pond is the maximum amount of water it can keep. For instance, when a pond is built on the shaded area in the above map, its capacity is (3 − 1) + (3 − 0) + (3 − 2) = 6, where 3 is the lowest elevation of the outermost cells and 1, 0, 2 are the elevations of the inner cells. Your mission is to write a computer program that, given a grid map describing the elevation of each unit square cell, calculates the largest possible capacity of a pond built in the site.
Note that neither of the following rectangular areas can be a pond. In the left one, the cell at the bottom right corner is not higher than the inner cell. In the right one, the central cell is as high as the outermost cells.
<image>
Input
The input consists of at most 100 datasets, each in the following format.
d w
e1, 1 ... e1, w
...
ed, 1 ... ed, w
The first line contains d and w, representing the depth and the width, respectively, of the garden site described in the map. They are positive integers between 3 and 10, inclusive. Each of the following d lines contains w integers between 0 and 9, inclusive, separated by a space. The x-th integer in the y-th line of the d lines is the elevation of the unit square cell with coordinates (x, y).
The end of the input is indicated by a line containing two zeros separated by a space.
Output
For each dataset, output a single line containing the largest possible capacity of a pond that can be built in the garden site described in the dataset. If no ponds can be built, output a single line containing a zero.
Sample Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output for the Sample Input
0
3
1
9
Example
Input
3 3
2 3 2
2 1 2
2 3 1
3 5
3 3 4 3 3
3 1 0 2 3
3 3 4 3 2
7 7
1 1 1 1 1 0 0
1 0 0 0 1 0 0
1 0 1 1 1 1 1
1 0 1 0 1 0 1
1 1 1 1 1 0 1
0 0 1 0 0 0 1
0 0 1 1 1 1 1
6 6
1 1 1 1 2 2
1 0 0 2 0 2
1 0 0 2 0 2
3 3 3 9 9 9
3 0 0 9 0 9
3 3 3 9 9 9
0 0
Output
0
3
1
9
Submitted Solution:
```
while 1:
d, w = map(int, input().split())
if d == w == 0:
break
E = [[*map(int, input().split())] for i in range(d)]
ans = 0
for x2 in range(d):
for y2 in range(w):
for x1 in range(x2-1):
for y1 in range(y2-1):
co = 10**18
for x in range(x1, x2+1):
co = min(co, E[x][y1], E[x][y2])
for y in range(y1, y2+1):
co = min(co, E[x1][y], E[x2][y])
ci = 0
cnt = 0
for x in range(x1+1, x2):
for y in range(y1+1, y2):
ci = max(ci, E[x][y])
cnt += E[x][y]
if ci < co:
ans = max(ans, (x2-x1-1)*(y2-y1-1)*co - cnt)
print(ans)
``` | instruction | 0 | 63,102 | 23 | 126,204 |
No | output | 1 | 63,102 | 23 | 126,205 |
Provide a correct Python 3 solution for this coding contest problem.
A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.
Constraints
* All inputs are integers
* 3 ≤ N ≤ 50
* 0 ≤ | Xi |, | Yi | ≤ 1000000
* The input polygon is a simple convex polygon.
* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).
Input
The input is given in the following format.
> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>
Output
If there is a point that satisfies the condition of the problem statement, the coordinates of that point
> X Y
>
Output in the format of. If the point does not exist, output "NA" on one line.
Examples
Input
4
100 100
0 100
0 0
100 0
Output
50.00000 50.00000
Input
3
100 100
0 100
0 0
Output
NA | instruction | 0 | 63,103 | 23 | 126,206 |
"Correct Solution:
```
from sys import exit
EPS = 1e-6
#外積
def OuterProduct(one, two):
tmp = one.conjugate() * two
return tmp.imag
#内積
def InnerProduct(one, two):
tmp = one.conjugate() * two
return tmp.real
#点が線分上にあるか
def IsOnSegment(point, begin, end):
if abs(OuterProduct(begin-point, end-point)) <= EPS and InnerProduct(begin-point, end-point) <= EPS:
return True
else:
return False
#直線の交点，平行の場合はNoneを返す
def Crosspoint(a, b, c, d):
if abs(OuterProduct(b-a, d-c)) <= EPS:
return None
else:
u = OuterProduct(c-a, d-a) / OuterProduct(b-a, d-c)
return (1-u)*a + u*b
n = int(input())
d = []
for _ in range(n):
x, y = map(int, input().split())
d.append(complex(x, y))
if n%2 == 1:
print("NA")
exit()
ok = True
ans = Crosspoint(d[0], d[n//2], d[1], d[n//2+1])
#print(ans)
for i in range(n//2):
if IsOnSegment(ans, d[i], d[n//2+i]) == False or IsOnSegment(ans, d[i+1], d[(n//2+i+1)%n]) == False:
ok = False
if abs(ans - d[i]) != abs(ans - d[n//2+i]):
ok = False
if abs(ans - d[i+1]) != abs(ans - d[(n//2+i+1)%n]):
ok = False
if ok:
print(ans.real, ans.imag)
else:
print("NA")
``` | output | 1 | 63,103 | 23 | 126,207 |
Provide a correct Python 3 solution for this coding contest problem.
A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.
Constraints
* All inputs are integers
* 3 ≤ N ≤ 50
* 0 ≤ | Xi |, | Yi | ≤ 1000000
* The input polygon is a simple convex polygon.
* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).
Input
The input is given in the following format.
> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>
Output
If there is a point that satisfies the condition of the problem statement, the coordinates of that point
> X Y
>
Output in the format of. If the point does not exist, output "NA" on one line.
Examples
Input
4
100 100
0 100
0 0
100 0
Output
50.00000 50.00000
Input
3
100 100
0 100
0 0
Output
NA | instruction | 0 | 63,104 | 23 | 126,208 |
"Correct Solution:
```
def solve(lis,N):
x=(lis[0][0]+lis[N//2][0])
y=(lis[0][1]+lis[N//2][1])
for i in range(1,N//2):
if (lis[i][0]+lis[i+N//2][0])!=x or (lis[i][1]+lis[i+N//2][1])!=y:
return "NA"
return str(x/2)+" "+str(y/2)
N=int(input())
if N%2==1:
print("NA")
else:
lis=[]
for i in range(N):
lis.append(tuple(map(int,input().split())))
ans=solve(lis,N)
print(ans)
``` | output | 1 | 63,104 | 23 | 126,209 |
Provide a correct Python 3 solution for this coding contest problem.
A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.
Constraints
* All inputs are integers
* 3 ≤ N ≤ 50
* 0 ≤ | Xi |, | Yi | ≤ 1000000
* The input polygon is a simple convex polygon.
* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).
Input
The input is given in the following format.
> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>
Output
If there is a point that satisfies the condition of the problem statement, the coordinates of that point
> X Y
>
Output in the format of. If the point does not exist, output "NA" on one line.
Examples
Input
4
100 100
0 100
0 0
100 0
Output
50.00000 50.00000
Input
3
100 100
0 100
0 0
Output
NA | instruction | 0 | 63,105 | 23 | 126,210 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def area(xy):
xy = sorted(xy)
x = [xy[i][0] - xy[0][0] for i in range(3)]
y = [xy[i][1] - xy[0][1] for i in range(3)]
return abs(x[1]*y[2] - x[2]*y[1]) / 2
def ccw(a, b, c):
ax = b[0] - a[0]
ay = b[1] - a[1]
bx = c[0] - a[0]
by = c[1] - a[1]
t = ax*by - ay*bx;
if t > 0:
return 1
if t < 0:
return -1
if ax*bx + ay*by < 0:
return 2
if ax*ax + ay*ay < bx*bx + by*by:
return -2
return 0
def convex_hull(ps):
n = len(ps)
k = 0
ps.sort()
ch = [None] * (n*2)
for i in range(n):
while k >= 2 and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
t = k + 1
for i in range(n-2,-1,-1):
while k >= t and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
return ch[:k-1]
def bs(f, mi, ma):
mm = -1
while ma > mi + eps:
mm = (ma+mi) / 2.0
if f(mm):
mi = mm + eps
else:
ma = mm
if f(mm):
return mm + eps
return mm
def intersection(a1, a2, b1, b2):
x1,y1 = a1
x2,y2 = a2
x3,y3 = b1
x4,y4 = b2
ksi = (y4 - y3) * (x4 - x1) - (x4 - x3) * (y4 - y1)
eta = (x2 - x1) * (y4 - y1) - (y2 - y1) * (x4 - x1)
delta = (x2 - x1) * (y4 - y3) - (y2 - y1) * (x4 - x3)
if delta == 0:
return None
ramda = ksi / delta;
mu = eta / delta;
if ramda >= 0 and ramda <= 1 and mu >= 0 and mu <= 1:
return (x1 + ramda * (x2 - x1), y1 + ramda * (y2 - y1))
return None
def main():
n = I()
a = convex_hull([LI() for _ in range(n)])
n = len(a)
a = a + a + a
s = 0
for i in range(1,n-1):
s += area([a[0],a[i],a[i+1]])
def f(i, fk):
t = 0
tj = i + 1
ra = [a[i+1][j] - a[i][j] for j in range(2)]
ap = [a[i][j] + ra[j] * fk for j in range(2)]
for j in range(i+1, i+n):
u = area([ap, a[j], a[j+1]])
if t + u >= s / 2:
tj = j
break
t += u
ts = s / 2 - t
sa = [a[tj+1][j] - a[tj][j] for j in range(2)]
def _f(k):
b = [a[tj][j] + sa[j] * k for j in range(2)]
fs = area([a[i], a[tj], b])
return fs < ts
bk = bs(_f, 0, 1)
return [ap, [a[tj][j] + sa[j] * bk for j in range(2)]]
ls = [f(i//5, random.random()) for i in range(n*5)]
ll = len(ls)
kts = []
sx = 0
sy = 0
for i in range(ll):
for j in range(i+1,ll):
t = intersection(ls[i][0], ls[i][1], ls[j][0], ls[j][1])
if t is None:
return 'NA'
kts.append(t)
sx += t[0]
sy += t[1]
mx = sx / len(kts)
my = sy / len(kts)
keps = 1.0 / 10**5
for i in range(len(kts)):
if abs(mx-t[0]) > keps or abs(my-t[1]) > keps:
return 'NA'
return '{:.9f} {:.9f}'.format(mx, my)
print(main())
``` | output | 1 | 63,105 | 23 | 126,211 |
Provide a correct Python 3 solution for this coding contest problem.
A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.
Constraints
* All inputs are integers
* 3 ≤ N ≤ 50
* 0 ≤ | Xi |, | Yi | ≤ 1000000
* The input polygon is a simple convex polygon.
* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).
Input
The input is given in the following format.
> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>
Output
If there is a point that satisfies the condition of the problem statement, the coordinates of that point
> X Y
>
Output in the format of. If the point does not exist, output "NA" on one line.
Examples
Input
4
100 100
0 100
0 0
100 0
Output
50.00000 50.00000
Input
3
100 100
0 100
0 0
Output
NA | instruction | 0 | 63,106 | 23 | 126,212 |
"Correct Solution:
```
# coding: utf-8
eps=0.0001
n=int(input())
x_sum=0
y_sum=0
data=[]
for i in range(n):
x,y=map(int,input().split())
x_sum+=x
y_sum+=y
data.append((x,y))
x_sum/=n
y_sum/=n
if n%2!=0:
print('NA')
exit()
for i in range(n//2):
if abs(((data[i][0]-x_sum)**2+(data[i][1]-y_sum)**2)**0.5\
-((data[i+n//2][0]-x_sum)**2+(data[i+n//2][1]-y_sum)**2)**0.5)<=eps:
pass
else:
print('NA')
exit()
print(x_sum,y_sum)
``` | output | 1 | 63,106 | 23 | 126,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.
Constraints
* All inputs are integers
* 3 ≤ N ≤ 50
* 0 ≤ | Xi |, | Yi | ≤ 1000000
* The input polygon is a simple convex polygon.
* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).
Input
The input is given in the following format.
> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>
Output
If there is a point that satisfies the condition of the problem statement, the coordinates of that point
> X Y
>
Output in the format of. If the point does not exist, output "NA" on one line.
Examples
Input
4
100 100
0 100
0 0
100 0
Output
50.00000 50.00000
Input
3
100 100
0 100
0 0
Output
NA
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def area(xy):
xy = sorted(xy)
x = [xy[i][0] - xy[0][0] for i in range(3)]
y = [xy[i][1] - xy[0][1] for i in range(3)]
return abs(x[1]*y[2] - x[2]*y[1]) / 2
def ccw(a, b, c):
ax = b[0] - a[0]
ay = b[1] - a[1]
bx = c[0] - a[0]
by = c[1] - a[1]
t = ax*by - ay*bx;
if t > 0:
return 1
if t < 0:
return -1
if ax*bx + ay*by < 0:
return 2
if ax*ax + ay*ay < bx*bx + by*by:
return -2
return 0
def convex_hull(ps):
n = len(ps)
k = 0
ps.sort()
ch = [None] * (n*2)
for i in range(n):
while k >= 2 and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
t = k + 1
for i in range(n-2,-1,-1):
while k >= t and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
return ch[:k-1]
def bs(f, mi, ma):
mm = -1
while ma > mi + eps:
mm = (ma+mi) / 2.0
if f(mm):
mi = mm + eps
else:
ma = mm
if f(mm):
return mm + eps
return mm
def intersection(a1, a2, b1, b2):
x1,y1 = a1
x2,y2 = a2
x3,y3 = b1
x4,y4 = b2
ksi = (y4 - y3) * (x4 - x1) - (x4 - x3) * (y4 - y1)
eta = (x2 - x1) * (y4 - y1) - (y2 - y1) * (x4 - x1)
delta = (x2 - x1) * (y4 - y3) - (y2 - y1) * (x4 - x3)
if delta == 0:
return None
ramda = ksi / delta;
mu = eta / delta;
if ramda >= 0 and ramda <= 1 and mu >= 0 and mu <= 1:
return (x1 + ramda * (x2 - x1), y1 + ramda * (y2 - y1))
return None
def main():
n = I()
a = convex_hull([LI() for _ in range(n)])
n = len(a)
a = a + a + a
s = 0
for i in range(1,n-1):
s += area([a[0],a[i],a[i+1]])
def f(i, fk):
t = 0
tj = i + 1
ra = [a[i+1][j] - a[i][j] for j in range(2)]
ap = [a[i][j] + ra[j] * fk for j in range(2)]
for j in range(i+1, i+n):
u = area([ap, a[j], a[j+1]])
if t + u >= s / 2:
tj = j
break
t += u
ts = s / 2 - t
sa = [a[tj+1][j] - a[tj][j] for j in range(2)]
def _f(k):
b = [a[tj][j] + sa[j] * k for j in range(2)]
fs = area([a[i], a[tj], b])
return fs < ts
bk = bs(_f, 0, 1)
return [ap, [a[tj][j] + sa[j] * bk for j in range(2)]]
ls = [f(i//5, random.random()) for i in range(n*5)]
ll = len(ls)
kts = []
sx = 0
sy = 0
for i in range(ll):
for j in range(i+1,ll):
t = intersection(ls[i][0], ls[i][1], ls[j][0], ls[j][1])
if t is None:
return 'NA'
kts.append(t)
sx += t[0]
sy += t[1]
mx = sx / len(kts)
my = sy / len(kts)
keps = 1.0 / 10**5
for i in range(len(kts)):
if abs(mx-t[0]) > keps or abs(my-t[1]) > keps:
return 'NA'
return '{} {}'.format(mx, my)
print(main())
``` | instruction | 0 | 63,107 | 23 | 126,214 |
No | output | 1 | 63,107 | 23 | 126,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.
Constraints
* All inputs are integers
* 3 ≤ N ≤ 50
* 0 ≤ | Xi |, | Yi | ≤ 1000000
* The input polygon is a simple convex polygon.
* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).
Input
The input is given in the following format.
> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>
Output
If there is a point that satisfies the condition of the problem statement, the coordinates of that point
> X Y
>
Output in the format of. If the point does not exist, output "NA" on one line.
Examples
Input
4
100 100
0 100
0 0
100 0
Output
50.00000 50.00000
Input
3
100 100
0 100
0 0
Output
NA
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def area(xy):
xy = sorted(xy)
x = [xy[i][0] - xy[0][0] for i in range(3)]
y = [xy[i][1] - xy[0][1] for i in range(3)]
return abs(x[1]*y[2] - x[2]*y[1]) / 2
def ccw(a, b, c):
ax = b[0] - a[0]
ay = b[1] - a[1]
bx = c[0] - a[0]
by = c[1] - a[1]
t = ax*by - ay*bx;
if t > 0:
return 1
if t < 0:
return -1
if ax*bx + ay*by < 0:
return 2
if ax*ax + ay*ay < bx*bx + by*by:
return -2
return 0
def convex_hull(ps):
n = len(ps)
k = 0
ps.sort()
ch = [None] * (n*2)
for i in range(n):
while k >= 2 and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
t = k + 1
for i in range(n-2,-1,-1):
while k >= t and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
return ch[:k-1]
def bs(f, mi, ma):
mm = -1
while ma > mi + eps:
mm = (ma+mi) / 2.0
if f(mm):
mi = mm + eps
else:
ma = mm
if f(mm):
return mm + eps
return mm
def intersection(a1, a2, b1, b2):
x1,y1 = a1
x2,y2 = a2
x3,y3 = b1
x4,y4 = b2
ksi = (y4 - y3) * (x4 - x1) - (x4 - x3) * (y4 - y1)
eta = (x2 - x1) * (y4 - y1) - (y2 - y1) * (x4 - x1)
delta = (x2 - x1) * (y4 - y3) - (y2 - y1) * (x4 - x3)
if delta == 0:
return None
ramda = ksi / delta;
mu = eta / delta;
if ramda >= 0 and ramda <= 1 and mu >= 0 and mu <= 1:
return (x1 + ramda * (x2 - x1), y1 + ramda * (y2 - y1))
return None
def main():
n = I()
a = convex_hull([LI() for _ in range(n)])
n = len(a)
a = a + a + a
s = 0
for i in range(1,n-1):
s += area([a[0],a[i],a[i+1]])
def f(i, fk):
t = 0
tj = i + 1
ra = [a[i+1][j] - a[i][j] for j in range(2)]
ap = [a[i][j] + ra[j] * fk for j in range(2)]
for j in range(i+1, i+n):
u = area([ap, a[j], a[j+1]])
if t + u >= s / 2:
tj = j
break
t += u
ts = s / 2 - t
sa = [a[tj+1][j] - a[tj][j] for j in range(2)]
def _f(k):
b = [a[tj][j] + sa[j] * k for j in range(2)]
fs = area([a[i], a[tj], b])
return fs < ts
bk = bs(_f, 0, 1)
return [ap, [a[tj][j] + sa[j] * bk for j in range(2)]]
ls = [f(i//5, random.random()) for i in range(n*5)]
ll = len(ls)
kts = []
sx = 0
sy = 0
for i in range(ll):
for j in range(i+1,ll):
t = intersection(ls[i][0], ls[i][1], ls[j][0], ls[j][1])
if t is None:
return 'NA'
kts.append(t)
sx += t[0]
sy += t[1]
mx = sx / len(kts)
my = sy / len(kts)
keps = 1.0 / 10**5
for i in range(len(kts)):
if abs(mx-t[0]) > keps or abs(my-t[1]) > keps:
return 'NA'
return '{} {}'.format(mx, my)
print(main())
``` | instruction | 0 | 63,108 | 23 | 126,216 |
No | output | 1 | 63,108 | 23 | 126,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.
Constraints
* All inputs are integers
* 3 ≤ N ≤ 50
* 0 ≤ | Xi |, | Yi | ≤ 1000000
* The input polygon is a simple convex polygon.
* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).
Input
The input is given in the following format.
> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>
Output
If there is a point that satisfies the condition of the problem statement, the coordinates of that point
> X Y
>
Output in the format of. If the point does not exist, output "NA" on one line.
Examples
Input
4
100 100
0 100
0 0
100 0
Output
50.00000 50.00000
Input
3
100 100
0 100
0 0
Output
NA
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def area(xy):
xy = sorted(xy)
x = [xy[i][0] - xy[0][0] for i in range(3)]
y = [xy[i][1] - xy[0][1] for i in range(3)]
return abs(x[1]*y[2] - x[2]*y[1]) / 2
def ccw(a, b, c):
ax = b[0] - a[0]
ay = b[1] - a[1]
bx = c[0] - a[0]
by = c[1] - a[1]
t = ax*by - ay*bx;
if t > 0:
return 1
if t < 0:
return -1
if ax*bx + ay*by < 0:
return 2
if ax*ax + ay*ay < bx*bx + by*by:
return -2
return 0
def convex_hull(ps):
n = len(ps)
k = 0
ps.sort()
ch = [None] * (n*2)
for i in range(n):
while k >= 2 and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
t = k + 1
for i in range(n-2,-1,-1):
while k >= t and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
return ch[:k-1]
def bs(f, mi, ma):
mm = -1
while ma > mi + eps:
mm = (ma+mi) / 2.0
if f(mm):
mi = mm + eps
else:
ma = mm
if f(mm):
return mm + eps
return mm
def intersection(a1, a2, b1, b2):
x1,y1 = a1
x2,y2 = a2
x3,y3 = b1
x4,y4 = b2
ksi = (y4 - y3) * (x4 - x1) - (x4 - x3) * (y4 - y1)
eta = (x2 - x1) * (y4 - y1) - (y2 - y1) * (x4 - x1)
delta = (x2 - x1) * (y4 - y3) - (y2 - y1) * (x4 - x3)
if delta == 0:
return None
ramda = ksi / delta;
mu = eta / delta;
if ramda >= 0 and ramda <= 1 and mu >= 0 and mu <= 1:
return (x1 + ramda * (x2 - x1), y1 + ramda * (y2 - y1))
return None
def main():
n = I()
a = convex_hull([LI() for _ in range(n)])
n = len(a)
a = a + a + a
s = 0
for i in range(1,n-1):
s += area([a[0],a[i],a[i+1]])
def f(i, fk):
t = 0
tj = i + 1
ra = [a[i+1][j] - a[i][j] for j in range(2)]
ap = [a[i][j] + ra[j] * fk for j in range(2)]
for j in range(i+1, i+n):
u = area([ap, a[j], a[j+1]])
if t + u >= s / 2:
tj = j
break
t += u
ts = s / 2 - t
sa = [a[tj+1][j] - a[tj][j] for j in range(2)]
def _f(k):
b = [a[tj][j] + sa[j] * k for j in range(2)]
fs = area([a[i], a[tj], b])
return fs < ts
bk = bs(_f, 0, 1)
return [ap, [a[tj][j] + sa[j] * bk for j in range(2)]]
ls = [f(i//5, random.random()) for i in range(n*5)]
ll = len(ls)
kts = []
sx = 0
sy = 0
for i in range(ll):
for j in range(i+1,ll):
t = intersection(ls[i][0], ls[i][1], ls[j][0], ls[j][1])
if t is None:
return 'NA'
kts.append(t)
sx += t[0]
sy += t[1]
mx = sx / len(kts)
my = sy / len(kts)
keps = 1.0 / 10**5
for i in range(len(kts)):
if abs(mx-t[0]) > keps or abs(my-t[1]) > keps:
return 'NA'
return '{:.9f} {:.9f}'.format(mx, my)
print(main())
``` | instruction | 0 | 63,109 | 23 | 126,218 |
No | output | 1 | 63,109 | 23 | 126,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5 | instruction | 0 | 63,543 | 23 | 127,086 |
Tags: greedy, sortings
Correct Solution:
```
import sys
n, k = map(int, input().split())
imos = dict()
for li, ri in (map(int, line.split()) for line in sys.stdin):
imos[li*2] = imos[li*2]+1 if li*2 in imos else 1
imos[ri*2+1] = imos[ri*2+1]-1 if ri*2+1 in imos else -1
acc = 0
ans = []
append = ans.append
minf = -(10**9 * 2 + 1)
left = minf
for x in sorted(imos.keys()):
acc += imos[x]
if left != minf and acc < k:
append(str(left >> 1) + ' ' + str(x >> 1))
left = minf
elif left == minf and acc >= k:
left = x
sys.stdout.buffer.write(
(str(len(ans)) + '\n' + '\n'.join(ans)).encode('utf-8'))
``` | output | 1 | 63,543 | 23 | 127,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5 | instruction | 0 | 63,544 | 23 | 127,088 |
Tags: greedy, sortings
Correct Solution:
```
#### IMPORTANT LIBRARY ####
############################
### DO NOT USE import random --> 250ms to load the library
############################
### In case of extra libraries: https://github.com/cheran-senthil/PyRival
######################
####### IMPORT #######
######################
from functools import cmp_to_key
from collections import deque, Counter
from heapq import heappush, heappop
from math import log, ceil
######################
#### STANDARD I/O ####
######################
import sys
import os
from io import BytesIO, IOBase
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")
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def print(*args, **kwargs):
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
def inp():
return sys.stdin.readline().rstrip("\r\n") # for fast input
def ii():
return int(inp())
def si():
return str(inp())
def li(lag = 0):
l = list(map(int, inp().split()))
if lag != 0:
for i in range(len(l)):
l[i] += lag
return l
def mi(lag = 0):
matrix = list()
for i in range(n):
matrix.append(li(lag))
return matrix
def lsi(): #string list
return list(map(str, inp().split()))
def print_list(lista, space = " "):
print(space.join(map(str, lista)))
######################
### BISECT METHODS ###
######################
def bisect_left(a, x):
"""i tale che a[i] >= x e a[i-1] < x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] < x:
left = mid+1
else:
right = mid
return left
def bisect_right(a, x):
"""i tale che a[i] > x e a[i-1] <= x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] > x:
right = mid
else:
left = mid+1
return left
def bisect_elements(a, x):
"""elementi pari a x nell'árray sortato"""
return bisect_right(a, x) - bisect_left(a, x)
######################
### MOD OPERATION ####
######################
MOD = 10**9 + 7
maxN = 5
FACT = [0] * maxN
INV_FACT = [0] * maxN
def add(x, y):
return (x+y) % MOD
def multiply(x, y):
return (x*y) % MOD
def power(x, y):
if y == 0:
return 1
elif y % 2:
return multiply(x, power(x, y-1))
else:
a = power(x, y//2)
return multiply(a, a)
def inverse(x):
return power(x, MOD-2)
def divide(x, y):
return multiply(x, inverse(y))
def allFactorials():
FACT[0] = 1
for i in range(1, maxN):
FACT[i] = multiply(i, FACT[i-1])
def inverseFactorials():
n = len(INV_FACT)
INV_FACT[n-1] = inverse(FACT[n-1])
for i in range(n-2, -1, -1):
INV_FACT[i] = multiply(INV_FACT[i+1], i+1)
def coeffBinom(n, k):
if n < k:
return 0
return multiply(FACT[n], multiply(INV_FACT[k], INV_FACT[n-k]))
######################
#### GRAPH ALGOS #####
######################
# ZERO BASED GRAPH
def create_graph(n, m, undirected = 1, unweighted = 1):
graph = [[] for i in range(n)]
if unweighted:
for i in range(m):
[x, y] = li(lag = -1)
graph[x].append(y)
if undirected:
graph[y].append(x)
else:
for i in range(m):
[x, y, w] = li(lag = -1)
w += 1
graph[x].append([y,w])
if undirected:
graph[y].append([x,w])
return graph
def create_tree(n, unweighted = 1):
children = [[] for i in range(n)]
if unweighted:
for i in range(n-1):
[x, y] = li(lag = -1)
children[x].append(y)
children[y].append(x)
else:
for i in range(n-1):
[x, y, w] = li(lag = -1)
w += 1
children[x].append([y, w])
children[y].append([x, w])
return children
def dist(tree, n, A, B = -1):
s = [[A, 0]]
massimo, massimo_nodo = 0, 0
distanza = -1
v = [-1] * n
while s:
el, dis = s.pop()
if dis > massimo:
massimo = dis
massimo_nodo = el
if el == B:
distanza = dis
for child in tree[el]:
if v[child] == -1:
v[child] = 1
s.append([child, dis+1])
return massimo, massimo_nodo, distanza
def diameter(tree):
_, foglia, _ = dist(tree, n, 0)
diam, _, _ = dist(tree, n, foglia)
return diam
def dfs(graph, n, A):
v = [-1] * n
s = [[A, 0]]
v[A] = 0
while s:
el, dis = s.pop()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
def bfs(graph, n, A):
v = [-1] * n
s = deque()
s.append([A, 0])
v[A] = 0
while s:
el, dis = s.popleft()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
#FROM A GIVEN ROOT, RECOVER THE STRUCTURE
def parents_children_root_unrooted_tree(tree, n, root = 0):
q = deque()
visited = [0] * n
parent = [-1] * n
children = [[] for i in range(n)]
q.append(root)
while q:
all_done = 1
visited[q[0]] = 1
for child in tree[q[0]]:
if not visited[child]:
all_done = 0
q.appendleft(child)
if all_done:
for child in tree[q[0]]:
if parent[child] == -1:
parent[q[0]] = child
children[child].append(q[0])
q.popleft()
return parent, children
# CALCULATING LONGEST PATH FOR ALL THE NODES
def all_longest_path_passing_from_node(parent, children, n):
q = deque()
visited = [len(children[i]) for i in range(n)]
downwards = [[0,0] for i in range(n)]
upward = [1] * n
longest_path = [1] * n
for i in range(n):
if not visited[i]:
q.append(i)
downwards[i] = [1,0]
while q:
node = q.popleft()
if parent[node] != -1:
visited[parent[node]] -= 1
if not visited[parent[node]]:
q.append(parent[node])
else:
root = node
for child in children[node]:
downwards[node] = sorted([downwards[node][0], downwards[node][1], downwards[child][0] + 1], reverse = True)[0:2]
s = [node]
while s:
node = s.pop()
if parent[node] != -1:
if downwards[parent[node]][0] == downwards[node][0] + 1:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][1])
else:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][0])
longest_path[node] = downwards[node][0] + downwards[node][1] + upward[node] - min([downwards[node][0], downwards[node][1], upward[node]]) - 1
for child in children[node]:
s.append(child)
return longest_path
### TBD SUCCESSOR GRAPH 7.5
### TBD TREE QUERIES 10.2 da 2 a 4
### TBD ADVANCED TREE 10.3
### TBD GRAPHS AND MATRICES 11.3.3 e 11.4.3 e 11.5.3 (ON GAMES)
######################
## END OF LIBRARIES ##
######################
n, k = li()
seg = []
for i in range(n):
x, y = li()
seg.append(2*x)
seg.append(2*y+1)
seg.sort()
c = 0
tot = []
res = None
for x in seg:
i = x % 2
x = x // 2
if i == 0:
c += 1
if c == k and res is None:
res = x
else:
c -= 1
if c == k-1 and res is not None:
tot.append([res, x])
res = None
print(len(tot))
for x, y in tot:
print_list([x, y])
``` | output | 1 | 63,544 | 23 | 127,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5 | instruction | 0 | 63,545 | 23 | 127,090 |
Tags: greedy, sortings
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
from collections import defaultdict
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2*10**9, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
# --------------------------------------------------binary-----------------------------------
n,k=map(int,input().split())
l=defaultdict(int)
for i in range(n):
a,b=map(int,input().split())
l[2*a]+=1
l[2*b+1]-=1
cur=0
ans=[]
for i in sorted(l):
cur+=l[i]
if cur>=k and cur-l[i]<k:
start=i//2
if cur<k and cur-l[i]>=k:
end=i//2
ans.append((start,end))
print(len(ans))
for i in range(len(ans)):
print(*ans[i])
``` | output | 1 | 63,545 | 23 | 127,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5 | instruction | 0 | 63,546 | 23 | 127,092 |
Tags: greedy, sortings
Correct Solution:
```
import sys
n, k = map(int, sys.stdin.buffer.readline().decode('utf-8').split())
imos = dict()
for li, ri in (map(int, line.decode('utf-8').split()) for line in sys.stdin.buffer):
imos[li*2] = imos[li*2]+1 if li*2 in imos else 1
imos[ri*2+1] = imos[ri*2+1]-1 if ri*2+1 in imos else -1
acc = 0
ans = []
append = ans.append
minf = -(10**9 * 2 + 1)
left = minf
for x in sorted(imos.keys()):
acc += imos[x]
if left != minf and acc < k:
append(str(left >> 1) + ' ' + str(x >> 1))
left = minf
elif left == minf and acc >= k:
left = x
sys.stdout.buffer.write(
(str(len(ans)) + '\n' + '\n'.join(ans)).encode('utf-8'))
``` | output | 1 | 63,546 | 23 | 127,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5
Submitted Solution:
```
from heapq import heappush,heappop
import sys
n,k=map(int,input().split())
ss=[]
ee=[]
for i in sys.stdin.readlines():
s, e = map(int,i.split())
heappush(ss, s)
heappush(ee, e)
ss.sort()
ee.sort()
m,i,j,p=0,0,0,0
s = ""
while True:
if i<n:
if j<n:
if ss[i] < ee[j]:
x = ss[i]
m += 1
i += 1
while i < n and ss[i] == x:
i += 1
m += 1
else:
x = ee[j]
m -= 1
j += 1
while j < n and ee[j] == x:
j += 1
m -= 1
else:
x = ss[i]
m += 1
i += 1
while i < n and ss[i] == x:
i += 1
m += 1
else:
if j<n:
x = ee[j]
m -= 1
j += 1
while j < n and ee[j] == x:
j += 1
m -= 1
else:
break
if p % 2 == 0 and m >= k:
s += str(x) + ' '
p += 1
elif p % 2 != 0 and m < k:
s += str(x) + "\n"
p += 1
print(p//2)
print(s)
``` | instruction | 0 | 63,547 | 23 | 127,094 |
No | output | 1 | 63,547 | 23 | 127,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5
Submitted Solution:
```
from sys import stdin, stdout
from math import sqrt
n, k = map(int, stdin.readline().split())
start, finish = [], []
for i in range(n):
s, f = map(int, stdin.readline().split())
start.append(s)
finish.append(f)
start.sort()
finish.sort()
cnt = 0
ans = []
i, j = 0, 0
while i != len(start) and j != len(start):
if start[i] < finish[j]:
cnt += 1
if cnt == k:
ans.append([start[i], -1])
i += 1
elif start[i] > finish[j]:
if cnt == k:
ans[-1][1] = finish[j]
cnt -= 1
j += 1
else:
i += 1
j += 1
continue
for i in range(j, len(finish)):
if cnt + j - i == k:
ans[-1][1] = finish[i]
for i in range(len(ans) - 1):
if ans[i + 1][0] == ans[i][1]:
ans[i + 1][0] = ans[i][0]
stdout.write(str(len(ans)) + '\n')
for v in ans:
stdout.write(str(v[0]) + ' ' + str(v[1]) + '\n')
``` | instruction | 0 | 63,548 | 23 | 127,096 |
No | output | 1 | 63,548 | 23 | 127,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5
Submitted Solution:
```
import sys
input = sys.stdin.readline
INF = 10**18
N, K = map(int, input().split())
I = []
V = []
for _ in range(N):
l, r = map(int, input().split())
V.append(l)
V.append(r)
I.append((l, r))
V = sorted(set(V))
comp = {v : k for k, v in enumerate(V)}
imos = [0 for _ in range(len(V) + 1)]
for i in range(N):
l, r = I[i]
imos[comp[l]] += 1
imos[comp[r]] -= 1
for i in range(len(V)):
imos[i + 1] += imos[i]
A = []
for i in range(len(V) + 1):
if imos[i] >= K:
A.append(i)
A.append(INF)
res = []
l = A[0]
r = A[0]
for i in range(len(A) - 1):
if A[i] + 1 == A[i + 1]:
r = A[i + 1]
else:
res.append(str(V[l]) + ' ' + str(V[r + 1]))
l = A[i + 1]
r = A[i + 1]
print(len(res))
print('\n'.join(res))
``` | instruction | 0 | 63,549 | 23 | 127,098 |
No | output | 1 | 63,549 | 23 | 127,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at least k segments. Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points and no others.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106) — the number of segments and the value of k.
The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109) each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.
Output
First line contains integer m — the smallest number of segments.
Next m lines contain two integers aj, bj (aj ≤ bj) — the ends of j-th segment in the answer. The segments should be listed in the order from left to right.
Examples
Input
3 2
0 5
-3 2
3 8
Output
2
0 2
3 5
Input
3 2
0 5
-3 3
3 8
Output
1
0 5
Submitted Solution:
```
import sys
def fast2():
import os, sys, atexit
from cStringIO import StringIO as BytesIO
# range = xrange
sys.stdout = BytesIO()
atexit.register(lambda: os.write(1, sys.stdout.getvalue()))
return BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, k = map(int, sys.stdin.buffer.readline().split())
tem, cur, out = [], 0, []
for l, r in (map(int, line.decode('utf-8').split()) for line in sys.stdin.buffer):
tem.extend([(l, 0), (r, 1)])
left = 0
while sorted(tem)[::-1]:
ele = tem.pop()
if not ele[1]:
cur += 1
if cur == k:
left = ele[0]
else:
cur -= 1
if cur == k - 1:
out.append('%d %d' % (left, ele[0]))
print('%d\n%s' % (len(out), '\n'.join(out)))
``` | instruction | 0 | 63,550 | 23 | 127,100 |
No | output | 1 | 63,550 | 23 | 127,101 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,864 | 23 | 127,728 |
"Correct Solution:
```
a,b,c = map(int, input().split())
print("YES" if a - 2*b + c == 0 else "NO")
``` | output | 1 | 63,864 | 23 | 127,729 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,865 | 23 | 127,730 |
"Correct Solution:
```
a,b,c=map(int,input().split())
if b-a==c-b:
print('YES')
else:
print(('NO'))
``` | output | 1 | 63,865 | 23 | 127,731 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,866 | 23 | 127,732 |
"Correct Solution:
```
a,b,c=map(int,input().split());print('YES' if a+c==2*b else 'NO')
``` | output | 1 | 63,866 | 23 | 127,733 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,867 | 23 | 127,734 |
"Correct Solution:
```
a, b, c = map(int,input().split())
print('YES') if b - a == c - b else print('NO')
``` | output | 1 | 63,867 | 23 | 127,735 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,868 | 23 | 127,736 |
"Correct Solution:
```
a, b, c = map(int,input().split())
if (a+c)==2*b:print('YES')
else:print('NO')
``` | output | 1 | 63,868 | 23 | 127,737 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,869 | 23 | 127,738 |
"Correct Solution:
```
a,b,c = map(int,input().split())
if b * 2 == a + c: print('YES')
else: print('NO')
``` | output | 1 | 63,869 | 23 | 127,739 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,870 | 23 | 127,740 |
"Correct Solution:
```
a,b,c = list(map(int,input().split()))
if a+c == 2*b:
print("YES")
else:
print("NO")
``` | output | 1 | 63,870 | 23 | 127,741 |
Provide a correct Python 3 solution for this coding contest problem.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES | instruction | 0 | 63,871 | 23 | 127,742 |
"Correct Solution:
```
a,b,c = list(map(int,input().split(" ")))
print("YES" if b-a==c-b else 'NO')
``` | output | 1 | 63,871 | 23 | 127,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
a,b,c=map(int,input().split());print("YNEOS"[b-a!=c-b::2])
``` | instruction | 0 | 63,872 | 23 | 127,744 |
Yes | output | 1 | 63,872 | 23 | 127,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
a = [int(i) for i in input().split()]
print("YES" if a[1]*2==a[0]+a[2] else "NO")
``` | instruction | 0 | 63,873 | 23 | 127,746 |
Yes | output | 1 | 63,873 | 23 | 127,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
a,b,c=map(int,input().split())
if a-2*b+c==0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 63,874 | 23 | 127,748 |
Yes | output | 1 | 63,874 | 23 | 127,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
a, b, c = map(int, input().split())
if c-b==b-a:
print('YES')
else:
print('NO')
``` | instruction | 0 | 63,875 | 23 | 127,750 |
Yes | output | 1 | 63,875 | 23 | 127,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
a,b,c=map(int,input().split())
print("YES" if b-a==c-b else "NO)
``` | instruction | 0 | 63,876 | 23 | 127,752 |
No | output | 1 | 63,876 | 23 | 127,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
# A
a, b, c = map(int, input().split())
if abs(a-b) == abs(b-c):
print('YES')
else:
print('NO')
``` | instruction | 0 | 63,877 | 23 | 127,754 |
No | output | 1 | 63,877 | 23 | 127,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
a, b, c = map(int, input().split())
if b - c == c - b:
print('YES')
else:
print('NO')
``` | instruction | 0 | 63,878 | 23 | 127,756 |
No | output | 1 | 63,878 | 23 | 127,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.
Determine whether the arrangement of the poles is beautiful.
Constraints
* 1 \leq a,b,c \leq 100
* a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print `YES` if the arrangement of the poles is beautiful; print `NO` otherwise.
Examples
Input
2 4 6
Output
YES
Input
2 5 6
Output
NO
Input
3 2 1
Output
YES
Submitted Solution:
```
a,b,c=map(int,input().split())
if a-b = b-c:
print("YES")
else:
print("NO")
``` | instruction | 0 | 63,879 | 23 | 127,758 |
No | output | 1 | 63,879 | 23 | 127,759 |
Provide a correct Python 3 solution for this coding contest problem.
Create a program to determine the positional relationship between a triangle and a circle on a plane. All target figures shall include boundaries. The triangle is given the position of three vertices, and the circle is given the position of the center and the radius. The position is given by a set of two integers in a Cartesian coordinate system. The radius is also given as an integer.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
x1 y1
x2 y2
x3 y3
xc yc
r
The first to third lines are given the coordinates xi, yi of the i-th vertex of the triangle. The coordinates of the center of the circle xc, yc are given on the 4th line, and the radius r of the circle is given on the 5th line. All inputs given are integers greater than or equal to 1 and less than or equal to 10,000.
The number of datasets does not exceed 100.
Output
The judgment result is output to one line in the following format for each input data set.
If the circle is included in the triangle a
If the triangle is included in the circle b
In other cases, if there is an intersection, c
D if there is no intersection
Example
Input
1 1
3 1
3 3
3 2
3
3 12
9 3
11 12
8 7
5
15 3
17 7
22 5
7 6
4
6 11
8 2
16 9
10 8
2
0 0
Output
b
c
d
a | instruction | 0 | 63,912 | 23 | 127,824 |
"Correct Solution:
```
def contain(x, y, r):
return x**2 + y**2 <= r**2
def outer_p(x0, y0, x1, y1):
return (x0*y1 - y0*x1)
def line_segment_circle(x0, y0, x1, y1, r, border=True):
A = x0**2 + y0**2
B = x0*x1 + y0*y1
C = x1**2 + y1**2 - r**2
D = B**2 - A*C
if border:
if D < 0:
return 0
if B <= 0 and B**2 > D:
return 0
if B - A >= 0 and (B - A)**2 > D:
return 0
else:
if D <= 0:
return 0
if B <= 0 and B**2 >= D:
return 0
if B - A >= 0 and (B - A)**2 >= D:
return 0
return 1
while 1:
x1, y1 = map(int, input().split())
if x1 == y1 == 0:
break
x2, y2 = map(int, input().split())
x3, y3 = map(int, input().split())
cx, cy = map(int, input().split())
r = int(input())
C1 = contain(x1-cx, y1-cy, r)
C2 = contain(x2-cx, y2-cy, r)
C3 = contain(x3-cx, y3-cy, r)
if C1 and C2 and C3:
print("b")
continue
if C1 or C2 or C3:
print("c")
continue
p1 = outer_p(x2-x1, y2-y1, cx-x1, cy-y1)
p2 = outer_p(x3-x2, y3-y2, cx-x2, cy-y2)
p3 = outer_p(x1-x3, y1-y3, cx-x3, cy-y3)
if 1 == (p1 < 0) == (p2 < 0) == (p3 < 0) or 1 == (p1 > 0) == (p2 > 0) == (p3 > 0):
p1 = line_segment_circle(x2-x1, y2-y1, cx-x1, cy-y1, r, False)
p2 = line_segment_circle(x3-x2, y3-y2, cx-x2, cy-y2, r, False)
p3 = line_segment_circle(x1-x3, y1-y3, cx-x3, cy-y3, r, False)
if p1 or p2 or p3:
print("c")
else:
print("a")
continue
p1 = line_segment_circle(x2-x1, y2-y1, cx-x1, cy-y1, r, True)
p2 = line_segment_circle(x3-x2, y3-y2, cx-x2, cy-y2, r, True)
p3 = line_segment_circle(x1-x3, y1-y3, cx-x3, cy-y3, r, True)
if p1 or p2 or p3:
print("c")
continue
print("d")
``` | output | 1 | 63,912 | 23 | 127,825 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,965 | 23 | 127,930 |
"Correct Solution:
```
W,H,x,y,r=map(int,input().split())
if r<=(x+r)<=W and r<=(y+r)<=H:
print("Yes")
else:
print("No")
``` | output | 1 | 63,965 | 23 | 127,931 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,966 | 23 | 127,932 |
"Correct Solution:
```
(W,H,x,y,r)=map(int,input().split())
if x>=r and y>=r and (H-y)>=r and (W-x)>=r:
print('Yes')
else:
print('No')
``` | output | 1 | 63,966 | 23 | 127,933 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,967 | 23 | 127,934 |
"Correct Solution:
```
W, H, x, y, r = map(int, input().split())
if r <= x <= W - r and r <= y <= H - r:
print("Yes")
else:
print("No")
``` | output | 1 | 63,967 | 23 | 127,935 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,968 | 23 | 127,936 |
"Correct Solution:
```
[W, H, x, y, r] = list(map(int, input().split()))
if (r<=x<=W-r)and(r<=y<=H-r):
print("Yes")
else:
print("No")
``` | output | 1 | 63,968 | 23 | 127,937 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,969 | 23 | 127,938 |
"Correct Solution:
```
W, H, x, y, r = list(map(int, input().split()))
print('Yes' if r <= x <= W - r and r <= y <= H - r else 'No')
``` | output | 1 | 63,969 | 23 | 127,939 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,970 | 23 | 127,940 |
"Correct Solution:
```
w,h,x,y,r = map(int, input().split())
a = w-r
b = h-r
if r <= x <= a and r <= y <= b:
print("Yes")
else:
print("No")
``` | output | 1 | 63,970 | 23 | 127,941 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,971 | 23 | 127,942 |
"Correct Solution:
```
w, h, x, y, r = map(int, input().split())
if(min([x, y, w - x, h - y]) >= r):
print("Yes")
else:
print("No")
``` | output | 1 | 63,971 | 23 | 127,943 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No | instruction | 0 | 63,972 | 23 | 127,944 |
"Correct Solution:
```
W, H, x, y, r = map(int, input().split())
if x-r < 0 or y-r < 0 or x+r > W or y+r > H:
print('No')
else:
print('Yes')
``` | output | 1 | 63,972 | 23 | 127,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No
Submitted Solution:
```
W, H, x, y, r = map(int, input().split())
print('Yes' if x - r >= 0 and x + r <= H and y - r >= 0 and y + r <= H else 'No')
``` | instruction | 0 | 63,973 | 23 | 127,946 |
Yes | output | 1 | 63,973 | 23 | 127,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No
Submitted Solution:
```
w,h,x,y,r = map(int,input().split())
if x>=r and x+r<=w and y>=r and y+r<=h:
print("Yes")
else:
print("No")
``` | instruction | 0 | 63,974 | 23 | 127,948 |
Yes | output | 1 | 63,974 | 23 | 127,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No
Submitted Solution:
```
W,H,x,y,r=map(int,input().split())
print('Yes') if x+r<=W and y+r<=H and r<=x and r<=y else print('No')
``` | instruction | 0 | 63,975 | 23 | 127,950 |
Yes | output | 1 | 63,975 | 23 | 127,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No
Submitted Solution:
```
w,h,x,y,r = map(int,input().split())
if w < x+r or h < y+r or 0 > x-r or 0> y-r:
print("No")
else:
print("Yes")
``` | instruction | 0 | 63,976 | 23 | 127,952 |
Yes | output | 1 | 63,976 | 23 | 127,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No
Submitted Solution:
```
W,H,x,y,r = [int(i) for i in input().split()]
if x<r or y<r or W<(x+r) or H<(x+r):
print('Yes')
else:
print('No')
``` | instruction | 0 | 63,977 | 23 | 127,954 |
No | output | 1 | 63,977 | 23 | 127,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
Circle inside a rectangle
Constraints
* $ -100 \leq x, y \leq 100$
* $ 0 < W, H, r \leq 100$
Input
Five integers $W$, $H$, $x$, $y$ and $r$ separated by a single space are given in a line.
Output
Print "Yes" if the circle is placed inside the rectangle, otherwise "No" in a line.
Examples
Input
5 4 2 2 1
Output
Yes
Input
5 4 2 4 1
Output
No
Submitted Solution:
```
5 4 2 4 1
``` | instruction | 0 | 63,978 | 23 | 127,956 |
No | output | 1 | 63,978 | 23 | 127,957 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.