message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ksusha the Squirrel is standing at the beginning of a straight road, divided into n sectors. The sectors are numbered 1 to n, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector n. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector i to any of the sectors i + 1, i + 2, ..., i + k.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?
Input
The first line contains two integers n and k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105). The next line contains n characters — the description of the road: the i-th character equals ".", if the i-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".".
Output
Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes).
Examples
Input
2 1
..
Output
YES
Input
5 2
.#.#.
Output
YES
Input
7 3
.#.###.
Output
NO
Submitted Solution:
```
def squirrel(path,k):
L=[]
for i in range(0,len(path)-1):
if(path[i]==path[i+1]=="#"):
L.append(i)
else:
L=[]
if(len(L)>=k-1):
return "NO"
return "YES"
nk=input().split()
n=int(nk[0])
k=int(nk[1])
path=str(input())
s=squirrel(path,k)
print(s)
``` | instruction | 0 | 52,536 | 3 | 105,072 |
No | output | 1 | 52,536 | 3 | 105,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ksusha the Squirrel is standing at the beginning of a straight road, divided into n sectors. The sectors are numbered 1 to n, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector n. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector i to any of the sectors i + 1, i + 2, ..., i + k.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?
Input
The first line contains two integers n and k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105). The next line contains n characters — the description of the road: the i-th character equals ".", if the i-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".".
Output
Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes).
Examples
Input
2 1
..
Output
YES
Input
5 2
.#.#.
Output
YES
Input
7 3
.#.###.
Output
NO
Submitted Solution:
```
n,k = map(int,input().split())
t = input()
k1 = 0
for i in range(n):
if t[i] == "#":
k1 = k1 + 1
elif t[i] == ".":
k1 = 0
if k1 > k:
print("NO")
exit()
print("YES")
``` | instruction | 0 | 52,537 | 3 | 105,074 |
No | output | 1 | 52,537 | 3 | 105,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ksusha the Squirrel is standing at the beginning of a straight road, divided into n sectors. The sectors are numbered 1 to n, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector n. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector i to any of the sectors i + 1, i + 2, ..., i + k.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?
Input
The first line contains two integers n and k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105). The next line contains n characters — the description of the road: the i-th character equals ".", if the i-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".".
Output
Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes).
Examples
Input
2 1
..
Output
YES
Input
5 2
.#.#.
Output
YES
Input
7 3
.#.###.
Output
NO
Submitted Solution:
```
n, k = list(map(int, input().split()))
s = list(input())
x = s.count("#")
if (k >= x):
print("YES")
else:
print("NO")
``` | instruction | 0 | 52,538 | 3 | 105,076 |
No | output | 1 | 52,538 | 3 | 105,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ksusha the Squirrel is standing at the beginning of a straight road, divided into n sectors. The sectors are numbered 1 to n, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector n. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector i to any of the sectors i + 1, i + 2, ..., i + k.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?
Input
The first line contains two integers n and k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105). The next line contains n characters — the description of the road: the i-th character equals ".", if the i-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".".
Output
Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes).
Examples
Input
2 1
..
Output
YES
Input
5 2
.#.#.
Output
YES
Input
7 3
.#.###.
Output
NO
Submitted Solution:
```
road_etc = list(map(int, input().split()))
road = list(input())
perek = 0
if road_etc[0] % 2 == 0:
road.insert(0, "A")
perek = 1
step = road_etc[1]
end = False
while not end:
if len(road) <= 3 and perek == 1:
print("YES")
end = True
continue
if len(road) <= 1:
print("YES")
end = True
else:
if road[step] == ".":
if perek == 1:
del(road[1:step])
else:
del(road[:step])
elif road[step] == "#":
print("NO")
end = True
# A.########.#..#.#.........#...#..#.##..#...#...##.#..#.#..##..##.##.####.#..#..#####.....#...##..###.#
# 0123456789
# A.#.#.
# 012345
# A..
# 012
#A..
#0123456789
``` | instruction | 0 | 52,539 | 3 | 105,078 |
No | output | 1 | 52,539 | 3 | 105,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The commanding officers decided to drop a nuclear bomb on the enemy's forces. You are ordered to determine the power of the warhead that needs to be used.
The enemy has N strategically important objects. Their positions are known due to the intelligence service. The aim of the strike is to deactivate at least K important objects of the enemy. The bombing impact point is already determined and has coordinates of [X0; Y0].
The nuclear warhead is marked by the estimated impact radius R ≥ 0. All the buildings that are located closer than R to the bombing epicentre will be destroyed. All the buildings that are located further than R from the epicentre, can also be deactivated with some degree of probability. Let's assume that D is the distance between a building and the epicentre. This building's deactivation probability P(D, R) is calculated according to the following formula:
<image> We should regard <image> as ea, where e ≈ 2.7182818284590452353602874713527
If the estimated impact radius of the warhead is equal to zero, then all the buildings located in the impact point will be completely demolished and all the rest of important objects will not be damaged.
The commanding officers want the probability of failing the task to be no more than ε. Nuclear warheads are too expensive a luxury, that's why you have to minimise the estimated impact radius of the warhead.
Input
The first line contains an integer N which represents the number of the enemy's objects (1 ≤ N ≤ 100). The second line contains two integers: K is the required number of deactivated objects, and ε is the maximally permitted probability of not completing the task, given in per mils (1 ≤ K ≤ N, 1 ≤ ε ≤ 999). The third line contains X0 and Y0 which are the coordinates of the strike impact point. The next N lines contain two numbers Xi and Yi each which are the coordinates of every strategically important object. All the coordinates are integer, their absolute values do not exceed 1000.
Let us remind you that there are a thousand per mils in unity (number one).
There can be several objects in one point.
Output
Print the sought estimated impact radius of the warhead. The absolute or relative measure of the inaccuracy of your answer should not exceed 10 - 6.
Examples
Input
1
1 500
5 5
1 2
Output
3.84257761518762740
Input
5
3 100
0 0
3 4
60 70
100 100
10 10
5 12
Output
13.45126176453737600 | instruction | 0 | 52,597 | 3 | 105,194 |
Tags: binary search, dp, probabilities
Correct Solution:
```
import math
n = int(input())
k, epsilon = list(map(int, input().split(" ")))
x0, y0 = list(map(int, input().split(" ")))
epsilon /= 1000.0
l = []
for i in range(n):
l.append(list(map(int, input().split(" "))))
d = sorted([(p[0] - x0) ** 2 + (p[1] - y0) ** 2 for p in l])
rmin = 0
rmax = math.sqrt(d[k - 1])
while(1):
if(rmax - rmin < 10e-9):
print((rmin + rmax)/2)
break
r = (rmin + rmax)/2
p = [math.exp(1 - i/(r**2)) if i > r**2 else 1.0 for i in d]
dp = [[0] * (n + 1) for i in range(n + 1)]
dp[0][0] = 1
for i in range(1, n + 1):
for j in range(i + 1):
if(j > 0):
dp[i][j] = p[i - 1] * dp[i - 1][j - 1]
if(i != j):
dp[i][j] += (1 - p[i - 1]) * dp[i - 1][j]
s = 0
for j in range(k, n + 1):
s += dp[n][j]
if(s > 1 - epsilon):
rmax = r
else:
rmin = r
``` | output | 1 | 52,597 | 3 | 105,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The commanding officers decided to drop a nuclear bomb on the enemy's forces. You are ordered to determine the power of the warhead that needs to be used.
The enemy has N strategically important objects. Their positions are known due to the intelligence service. The aim of the strike is to deactivate at least K important objects of the enemy. The bombing impact point is already determined and has coordinates of [X0; Y0].
The nuclear warhead is marked by the estimated impact radius R ≥ 0. All the buildings that are located closer than R to the bombing epicentre will be destroyed. All the buildings that are located further than R from the epicentre, can also be deactivated with some degree of probability. Let's assume that D is the distance between a building and the epicentre. This building's deactivation probability P(D, R) is calculated according to the following formula:
<image> We should regard <image> as ea, where e ≈ 2.7182818284590452353602874713527
If the estimated impact radius of the warhead is equal to zero, then all the buildings located in the impact point will be completely demolished and all the rest of important objects will not be damaged.
The commanding officers want the probability of failing the task to be no more than ε. Nuclear warheads are too expensive a luxury, that's why you have to minimise the estimated impact radius of the warhead.
Input
The first line contains an integer N which represents the number of the enemy's objects (1 ≤ N ≤ 100). The second line contains two integers: K is the required number of deactivated objects, and ε is the maximally permitted probability of not completing the task, given in per mils (1 ≤ K ≤ N, 1 ≤ ε ≤ 999). The third line contains X0 and Y0 which are the coordinates of the strike impact point. The next N lines contain two numbers Xi and Yi each which are the coordinates of every strategically important object. All the coordinates are integer, their absolute values do not exceed 1000.
Let us remind you that there are a thousand per mils in unity (number one).
There can be several objects in one point.
Output
Print the sought estimated impact radius of the warhead. The absolute or relative measure of the inaccuracy of your answer should not exceed 10 - 6.
Examples
Input
1
1 500
5 5
1 2
Output
3.84257761518762740
Input
5
3 100
0 0
3 4
60 70
100 100
10 10
5 12
Output
13.45126176453737600 | instruction | 0 | 52,598 | 3 | 105,196 |
Tags: binary search, dp, probabilities
Correct Solution:
```
#!/usr/bin/python3
from math import sqrt, exp
n = int(input())
k, eps = tuple(map(int, input().split()))
x0, y0 = tuple(map(int, input().split()))
val = []
for _ in range(n):
x, y = tuple(map(int, input().split()))
val.append((x - x0)**2 + (y - y0)**2)
#val.sort()
l = 0.0
r = 1e9
cnt = 0
while abs(r - l) > 1e-12 and cnt < 100:
cnt += 1
r2 = (r + l) / 2
p = [exp(1 - d2 / r2) if d2 > r2 else 1.0 for d2 in val]
dp = [[1.0 ] + [0.0] * k for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, k + 1):
dp[i][j] = p[i - 1] * dp[i - 1][j - 1] + (1.0 - p[i - 1]) * dp[i - 1][j]
if 1000 * (1.0 - dp[n][k]) > eps:
l = r2
else:
r = r2
print('%0.9f' % sqrt((r + l) / 2))
``` | output | 1 | 52,598 | 3 | 105,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The commanding officers decided to drop a nuclear bomb on the enemy's forces. You are ordered to determine the power of the warhead that needs to be used.
The enemy has N strategically important objects. Their positions are known due to the intelligence service. The aim of the strike is to deactivate at least K important objects of the enemy. The bombing impact point is already determined and has coordinates of [X0; Y0].
The nuclear warhead is marked by the estimated impact radius R ≥ 0. All the buildings that are located closer than R to the bombing epicentre will be destroyed. All the buildings that are located further than R from the epicentre, can also be deactivated with some degree of probability. Let's assume that D is the distance between a building and the epicentre. This building's deactivation probability P(D, R) is calculated according to the following formula:
<image> We should regard <image> as ea, where e ≈ 2.7182818284590452353602874713527
If the estimated impact radius of the warhead is equal to zero, then all the buildings located in the impact point will be completely demolished and all the rest of important objects will not be damaged.
The commanding officers want the probability of failing the task to be no more than ε. Nuclear warheads are too expensive a luxury, that's why you have to minimise the estimated impact radius of the warhead.
Input
The first line contains an integer N which represents the number of the enemy's objects (1 ≤ N ≤ 100). The second line contains two integers: K is the required number of deactivated objects, and ε is the maximally permitted probability of not completing the task, given in per mils (1 ≤ K ≤ N, 1 ≤ ε ≤ 999). The third line contains X0 and Y0 which are the coordinates of the strike impact point. The next N lines contain two numbers Xi and Yi each which are the coordinates of every strategically important object. All the coordinates are integer, their absolute values do not exceed 1000.
Let us remind you that there are a thousand per mils in unity (number one).
There can be several objects in one point.
Output
Print the sought estimated impact radius of the warhead. The absolute or relative measure of the inaccuracy of your answer should not exceed 10 - 6.
Examples
Input
1
1 500
5 5
1 2
Output
3.84257761518762740
Input
5
3 100
0 0
3 4
60 70
100 100
10 10
5 12
Output
13.45126176453737600
Submitted Solution:
```
#!/usr/bin/python3
from math import sqrt, exp
n = int(input())
k, eps = tuple(map(int, input().split()))
x0, y0 = tuple(map(int, input().split()))
val = []
for _ in range(n):
x, y = tuple(map(int, input().split()))
val.append((x - x0)**2 + (y - y0)**2)
val.sort()
l = 0.0
r = 2e6
while abs(r - l) > 1e-15:
r2 = (r + l) / 2
p = [exp(1 - d2 / r2) if d2 > r2 else 1.0 for d2 in val]
dp = [[1.0 ] + [0.0] * k for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, k + 1):
dp[i][j] = p[i - 1] * dp[i - 1][j - 1] + (1.0 - p[i - 1]) * dp[i - 1][j]
if 1000 * (1.0 - dp[n][k]) > eps:
l = r2
else:
r = r2
break
print('%0.9f' % sqrt((r + l) / 2))
``` | instruction | 0 | 52,599 | 3 | 105,198 |
No | output | 1 | 52,599 | 3 | 105,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The commanding officers decided to drop a nuclear bomb on the enemy's forces. You are ordered to determine the power of the warhead that needs to be used.
The enemy has N strategically important objects. Their positions are known due to the intelligence service. The aim of the strike is to deactivate at least K important objects of the enemy. The bombing impact point is already determined and has coordinates of [X0; Y0].
The nuclear warhead is marked by the estimated impact radius R ≥ 0. All the buildings that are located closer than R to the bombing epicentre will be destroyed. All the buildings that are located further than R from the epicentre, can also be deactivated with some degree of probability. Let's assume that D is the distance between a building and the epicentre. This building's deactivation probability P(D, R) is calculated according to the following formula:
<image> We should regard <image> as ea, where e ≈ 2.7182818284590452353602874713527
If the estimated impact radius of the warhead is equal to zero, then all the buildings located in the impact point will be completely demolished and all the rest of important objects will not be damaged.
The commanding officers want the probability of failing the task to be no more than ε. Nuclear warheads are too expensive a luxury, that's why you have to minimise the estimated impact radius of the warhead.
Input
The first line contains an integer N which represents the number of the enemy's objects (1 ≤ N ≤ 100). The second line contains two integers: K is the required number of deactivated objects, and ε is the maximally permitted probability of not completing the task, given in per mils (1 ≤ K ≤ N, 1 ≤ ε ≤ 999). The third line contains X0 and Y0 which are the coordinates of the strike impact point. The next N lines contain two numbers Xi and Yi each which are the coordinates of every strategically important object. All the coordinates are integer, their absolute values do not exceed 1000.
Let us remind you that there are a thousand per mils in unity (number one).
There can be several objects in one point.
Output
Print the sought estimated impact radius of the warhead. The absolute or relative measure of the inaccuracy of your answer should not exceed 10 - 6.
Examples
Input
1
1 500
5 5
1 2
Output
3.84257761518762740
Input
5
3 100
0 0
3 4
60 70
100 100
10 10
5 12
Output
13.45126176453737600
Submitted Solution:
```
#!/usr/bin/python3
from math import sqrt, exp
n = int(input())
k, eps = tuple(map(int, input().split()))
x0, y0 = tuple(map(int, input().split()))
val = []
for _ in range(n):
x, y = tuple(map(int, input().split()))
val.append((x - x0)**2 + (y - y0)**2)
val.sort()
l = 0.0
r = 1e9
while abs(r - l) > 1e-9:
r2 = (r + l) / 2
p = [exp(1 - d2 / r2) if d2 > r2 else 1.0 for d2 in val]
dp = [[1.0 ] + [0.0] * k for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, k + 1):
dp[i][j] = p[i - 1] * dp[i - 1][j - 1] + (1.0 - p[i - 1]) * dp[i - 1][j]
'''from pprint import pprint
pprint(p)
pprint(dp)
pprint(val)
print(l, r2, r, 1000 * dp[n][k], eps)'''
if 1000 * (1.0 - dp[n][k]) > eps:
l = r2
else:
r = r2
print('%0.9f' % sqrt((r + l) / 2))
``` | instruction | 0 | 52,600 | 3 | 105,200 |
No | output | 1 | 52,600 | 3 | 105,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The commanding officers decided to drop a nuclear bomb on the enemy's forces. You are ordered to determine the power of the warhead that needs to be used.
The enemy has N strategically important objects. Their positions are known due to the intelligence service. The aim of the strike is to deactivate at least K important objects of the enemy. The bombing impact point is already determined and has coordinates of [X0; Y0].
The nuclear warhead is marked by the estimated impact radius R ≥ 0. All the buildings that are located closer than R to the bombing epicentre will be destroyed. All the buildings that are located further than R from the epicentre, can also be deactivated with some degree of probability. Let's assume that D is the distance between a building and the epicentre. This building's deactivation probability P(D, R) is calculated according to the following formula:
<image> We should regard <image> as ea, where e ≈ 2.7182818284590452353602874713527
If the estimated impact radius of the warhead is equal to zero, then all the buildings located in the impact point will be completely demolished and all the rest of important objects will not be damaged.
The commanding officers want the probability of failing the task to be no more than ε. Nuclear warheads are too expensive a luxury, that's why you have to minimise the estimated impact radius of the warhead.
Input
The first line contains an integer N which represents the number of the enemy's objects (1 ≤ N ≤ 100). The second line contains two integers: K is the required number of deactivated objects, and ε is the maximally permitted probability of not completing the task, given in per mils (1 ≤ K ≤ N, 1 ≤ ε ≤ 999). The third line contains X0 and Y0 which are the coordinates of the strike impact point. The next N lines contain two numbers Xi and Yi each which are the coordinates of every strategically important object. All the coordinates are integer, their absolute values do not exceed 1000.
Let us remind you that there are a thousand per mils in unity (number one).
There can be several objects in one point.
Output
Print the sought estimated impact radius of the warhead. The absolute or relative measure of the inaccuracy of your answer should not exceed 10 - 6.
Examples
Input
1
1 500
5 5
1 2
Output
3.84257761518762740
Input
5
3 100
0 0
3 4
60 70
100 100
10 10
5 12
Output
13.45126176453737600
Submitted Solution:
```
#!/usr/bin/python3
from math import sqrt, exp
n = int(input())
k, eps = tuple(map(int, input().split()))
x0, y0 = tuple(map(int, input().split()))
val = []
for _ in range(n):
x, y = tuple(map(int, input().split()))
val.append((x - x0)**2 + (y - y0)**2)
val.sort()
l = 0.0
r = 1e9
while abs(r - l) > 1e-9:
r2 = (r + l) / 2
s = 0.0
v = 0
for _ in range(k):
if val[_] > r2:
s += val[_]
v += 1
p = 1000 * (1 - exp(v - s / r2))
if p > eps:
l = r2
else:
r = r2
print('%0.9f' % sqrt((r + l) / 2))
``` | instruction | 0 | 52,601 | 3 | 105,202 |
No | output | 1 | 52,601 | 3 | 105,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The commanding officers decided to drop a nuclear bomb on the enemy's forces. You are ordered to determine the power of the warhead that needs to be used.
The enemy has N strategically important objects. Their positions are known due to the intelligence service. The aim of the strike is to deactivate at least K important objects of the enemy. The bombing impact point is already determined and has coordinates of [X0; Y0].
The nuclear warhead is marked by the estimated impact radius R ≥ 0. All the buildings that are located closer than R to the bombing epicentre will be destroyed. All the buildings that are located further than R from the epicentre, can also be deactivated with some degree of probability. Let's assume that D is the distance between a building and the epicentre. This building's deactivation probability P(D, R) is calculated according to the following formula:
<image> We should regard <image> as ea, where e ≈ 2.7182818284590452353602874713527
If the estimated impact radius of the warhead is equal to zero, then all the buildings located in the impact point will be completely demolished and all the rest of important objects will not be damaged.
The commanding officers want the probability of failing the task to be no more than ε. Nuclear warheads are too expensive a luxury, that's why you have to minimise the estimated impact radius of the warhead.
Input
The first line contains an integer N which represents the number of the enemy's objects (1 ≤ N ≤ 100). The second line contains two integers: K is the required number of deactivated objects, and ε is the maximally permitted probability of not completing the task, given in per mils (1 ≤ K ≤ N, 1 ≤ ε ≤ 999). The third line contains X0 and Y0 which are the coordinates of the strike impact point. The next N lines contain two numbers Xi and Yi each which are the coordinates of every strategically important object. All the coordinates are integer, their absolute values do not exceed 1000.
Let us remind you that there are a thousand per mils in unity (number one).
There can be several objects in one point.
Output
Print the sought estimated impact radius of the warhead. The absolute or relative measure of the inaccuracy of your answer should not exceed 10 - 6.
Examples
Input
1
1 500
5 5
1 2
Output
3.84257761518762740
Input
5
3 100
0 0
3 4
60 70
100 100
10 10
5 12
Output
13.45126176453737600
Submitted Solution:
```
#!/usr/bin/python3
from math import sqrt, exp
n = int(input())
k, eps = tuple(map(int, input().split()))
x0, y0 = tuple(map(int, input().split()))
val = []
for _ in range(n):
x, y = tuple(map(int, input().split()))
val.append((x - x0)**2 + (y - y0)**2)
val.sort()
l = 0.0
r = 1e9
while abs(r - l) > 1e-9:
r2 = (r + l) / 2
s = 0.0
v = 0
for _ in range(k):
if val[_] > r2:
s += val[_]
v += 1
p = 1000 * (1 - exp(v - s / r2))
print(l, r2, r, p)
if p > eps:
l = r2
else:
r = r2
print('%0.9f' % sqrt((r + l) / 2))
``` | instruction | 0 | 52,602 | 3 | 105,204 |
No | output | 1 | 52,602 | 3 | 105,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
n = int(input())
print('YNEOS'[n & (n - 1) > 0 :: 2])
``` | instruction | 0 | 52,611 | 3 | 105,222 |
Yes | output | 1 | 52,611 | 3 | 105,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
from math import *
n=int(input())
print(['NO','YES'][int(n==2**round(log(n,2)))])
``` | instruction | 0 | 52,612 | 3 | 105,224 |
Yes | output | 1 | 52,612 | 3 | 105,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
x = int( input() )
print( "NO" if x & x-1 > 0 else "YES" )
``` | instruction | 0 | 52,613 | 3 | 105,226 |
Yes | output | 1 | 52,613 | 3 | 105,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
from math import log
a = int(input())
t = log(a, 2)
if len(str(t)) == 3:
print('YES')
else:
print('NO')
``` | instruction | 0 | 52,614 | 3 | 105,228 |
Yes | output | 1 | 52,614 | 3 | 105,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
n = int(input())
v = [0] * n
pos = 0
for i in range(n):
pos += i
pos %= n
v[pos] += 1
a = "NO"
for i in range(n):
if not v[i]:
a = "YES"
break
print(a)
``` | instruction | 0 | 52,615 | 3 | 105,230 |
No | output | 1 | 52,615 | 3 | 105,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
n=int(input())
c=0
s=set()
for i in range(1,2*n +1):
c+=i
s.add(c%n)
print("yes" if len(s)==n else "no")
``` | instruction | 0 | 52,616 | 3 | 105,232 |
No | output | 1 | 52,616 | 3 | 105,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
def f(n):
rl = [True]*n #True = nonvisited
nn = (n*(n-1))//2
for k in range(n):
kk = (k*(k+1))//2
rl[kk%n] = False
if n%2==0:
rl[(kk+nn)%n] = False
return sum(rl)==0
``` | instruction | 0 | 52,617 | 3 | 105,234 |
No | output | 1 | 52,617 | 3 | 105,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 ≤ n ≤ 1000 — number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO
Submitted Solution:
```
quant = int(input())
pos = 0
positions = []
for i in range(1,quant*2):
pos+= i-1
if pos not in positions:
positions.append(pos)
if quant == 1:
print('YES')
elif pos == len(positions):
print('YES')
else:
print('NO')
``` | instruction | 0 | 52,618 | 3 | 105,236 |
No | output | 1 | 52,618 | 3 | 105,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,801 | 3 | 105,602 |
Tags: implementation
Correct Solution:
```
n = int(input())
counter = 0
a = [[0 for j in range(n)] for i in range(n)]
for i in range(n - 1):
i, j = map(int, input().split())
a[i - 1][j - 1] = 1
a[j - 1][i - 1] = 1
for element in a:
if element.count(1) == 1:
counter += 1
print(counter)
``` | output | 1 | 52,801 | 3 | 105,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,802 | 3 | 105,604 |
Tags: implementation
Correct Solution:
```
import os
import sys
debug = True
if debug and os.path.exists("input.in"):
input = open("input.in", "r").readline
else:
debug = False
input = sys.stdin.readline
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return s[:len(s) - 1] # Remove line char from end
def invr():
return (map(int, input().split()))
test_count = 1
if debug:
test_count = inp()
for t in range(test_count):
if debug:
print("Test Case #", t + 1)
# Start code here
n = inp()
counts = [0] * (n + 1)
for i in range(n - 1):
a, b = invr()
counts[a] += 1
counts[b] += 1
ans = 0
for i in range(1, n + 1):
if counts[i] <= 1:
ans += 1
print(ans)
``` | output | 1 | 52,802 | 3 | 105,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,803 | 3 | 105,606 |
Tags: implementation
Correct Solution:
```
t=int(input())
dic1={}
dic2={}
remotes=[False]*t
for i in range(t-1):
a,b=input().split()
k=dic1.get(a,-1)
if k==-1:
dic1[a]=list()
dic1[a].extend([b])
k=dic2.get(b,-1)
if k==-1:
dic2[b]=list()
dic2[b].extend([a])
for i in range(1,t+1):
res=0
if str(i) in dic1:
if len(dic1[str(i)])>1:
res=1
elif str(i) in dic2 and len(dic1[str(i)])+len(dic2[str(i)])>1:
res=1
if str(i) in dic2:
if len(dic2[str(i)])>1:
res=1
elif str(i) in dic1 and len(dic1[str(i)])+len(dic2[str(i)])>1:
res=1
remotes[i-1]=res
print(t-sum(remotes))
``` | output | 1 | 52,803 | 3 | 105,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,804 | 3 | 105,608 |
Tags: implementation
Correct Solution:
```
N = int(input())
deg = [0] * N
for i in range(N - 1):
a, b = map(int, input().split())
deg[a - 1] += 1
deg[b - 1] += 1
num = 0
for i in range(N):
if deg[i] == 1:
num += 1
print(num)
``` | output | 1 | 52,804 | 3 | 105,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,805 | 3 | 105,610 |
Tags: implementation
Correct Solution:
```
__author__ = 'Esfandiar'
n = int(input())
g = [[] for i in range(n)]
for i in range(n-1):
u,v = map(int,input().split())
g[u-1].append(v-1)
g[v-1].append(u-1)
print(sum([len(g[i])==1 for i in range(n)]))
``` | output | 1 | 52,805 | 3 | 105,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,806 | 3 | 105,612 |
Tags: implementation
Correct Solution:
```
n=int(input())
a,d,f=[],0,0
for i in range(n-1):
k,l=map(int,input().split())
a.append(k)
a.append(l)
for i in range(len(a)):
d=0
for j in range(len(a)):
if i!=j and a[i]==a[j]:d=d+1
if d==0:f=f+1
print(f)
``` | output | 1 | 52,806 | 3 | 105,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,807 | 3 | 105,614 |
Tags: implementation
Correct Solution:
```
n=int(input())
arr=[]
for i in range(0,n-1):
li=list(map(int,input().split()))
u=li[0]
v=li[1]
arr.append(u)
arr.append(v)
cnt=0
for i in arr:
if arr.count(i)==1:
cnt+=1
print(cnt)
``` | output | 1 | 52,807 | 3 | 105,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium. | instruction | 0 | 52,808 | 3 | 105,616 |
Tags: implementation
Correct Solution:
```
n = int(input())
a = set()
b = set()
for _ in range(n-1):
c1,c2 = map(int,input().split())
for c in [c1,c2]:
if c in b:
pass
else:
if c in a:
b.add(c)
a.remove(c)
else:
a.add(c)
print(len(a))
``` | output | 1 | 52,808 | 3 | 105,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
z = int(input())
all_connected = []
repeated = []
while z-1:
n,k = map(int, input().rstrip().split(" "))
if n not in all_connected:
all_connected.append(n)
else:
repeated.append(n)
if k not in all_connected:
all_connected.append(k)
else:
repeated.append(k)
z = z- 1
print(len(all_connected)-len(list(set(repeated))))
``` | instruction | 0 | 52,809 | 3 | 105,618 |
Yes | output | 1 | 52,809 | 3 | 105,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
n = int(input())
ref = n * [0]
for j in range(n - 1):
a, b = [int(j) for j in input().split()]
ref[a - 1] += 1
ref[b - 1] += 1
count = 0
for j in range(n):
if ref[j] == 1:
count += 1
print(count)
``` | instruction | 0 | 52,810 | 3 | 105,620 |
Yes | output | 1 | 52,810 | 3 | 105,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
n=int(input())
k=[]
for i in range(n-1):
a,b=map(int,input().split())
k.append(a)
k.append(b)
ans=0
for i in k:
if k.count(i)==1:
ans+=1
print(ans)
``` | instruction | 0 | 52,811 | 3 | 105,622 |
Yes | output | 1 | 52,811 | 3 | 105,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
from collections import Counter
import io
import os
import sys
from atexit import register
import random
import math
import itertools
##################################### Flags #####################################
# DEBUG = True
DEBUG = False
STRESSTEST = False
##################################### IO #####################################
if not DEBUG:
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
sys.stdout = io.BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
tokens = []
tokens_next = 0
def nextStr():
global tokens, tokens_next
while tokens_next >= len(tokens):
tokens = input().split()
tokens_next = 0
tokens_next += 1
if type(tokens[tokens_next - 1]) == str:
return tokens[tokens_next - 1]
return tokens[tokens_next - 1].decode()
def nextInt():
return int(nextStr())
def nextIntArr(n):
return [nextInt() for i in range(n)]
def print(*argv, end='\n'):
for arg in argv:
sys.stdout.write((str(arg) + ' ').encode())
sys.stdout.write(end.encode())
##################################### Helper Methods #####################################
def genTestCase():
raise NotImplementedError
def bruteforce(a):
raise NotImplementedError
def doStressTest():
while True:
curTest = genTestCase()
mySoln = solve(curTest)
bruteforceSoln = bruteforce(curTest)
if mySoln != bruteforceSoln:
print('### Found case ###')
print(curTest)
print(f'{mySoln} should have been: {bruteforceSoln}')
return
def solve(edges):
c = Counter()
for e in edges:
for i in e:
c[i] += 1
return sum(1 for i in c if c[i] == 1)
##################################### Driver #####################################
if __name__ == "__main__":
if not DEBUG and STRESSTEST:
raise Exception('Wrong flags!')
if STRESSTEST:
doStressTest()
else:
### Read input here
n = nextInt()
a = []
for _ in range(n - 1):
a.append((nextInt(), nextInt()))
res = solve(a)
print(res)
sys.stdout.flush()
``` | instruction | 0 | 52,812 | 3 | 105,624 |
Yes | output | 1 | 52,812 | 3 | 105,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
a=[]
for s in [*open(0)][1:]:a+=[*map(int,s.split())]
print(sum(a.count(x)<2 for x in range(1,len(a)+2)))
``` | instruction | 0 | 52,813 | 3 | 105,626 |
No | output | 1 | 52,813 | 3 | 105,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
n=int(input())
f=[0 for i in range(n+1)]
for i in range(n-1):
x,y=map(int,input().split())
f[x]+=1
f[y]+=1
print(max(f))
``` | instruction | 0 | 52,814 | 3 | 105,628 |
No | output | 1 | 52,814 | 3 | 105,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
import sys
from collections import Counter
def main(args):
n = int(input())
counter = Counter()
for _ in range(n - 1):
u, v = map(int, input().split())
counter[u] += 1
counter[v] += 1
print(counter[1])
if __name__ == '__main__':
sys.exit(main(sys.argv))
``` | instruction | 0 | 52,815 | 3 | 105,630 |
No | output | 1 | 52,815 | 3 | 105,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output
A single integer denoting the number of remote planets.
Examples
Input
5
4 1
4 2
1 3
1 5
Output
3
Input
4
1 2
4 3
1 4
Output
2
Note
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
Submitted Solution:
```
def dfs(edges,source,path = [],count = 0):
if source not in path:
path.append(source)
if(source not in edges):
return [path,count+1]
for n in edges[source]:
path,count = dfs(edges,n,path,count)
return [path,count]
n = int(input())
edges = {}
root =[]
for i in range(n-1):
source,point = [int(x) for x in input().split()]
root.append(point)
if(source not in edges):
edges[source] = [point]
continue
edges[source] = edges[source] + [point]
#find root
for i in range(1,n+1):
if(i not in root):
root = i
break
print(dfs(edges,root)[1])
``` | instruction | 0 | 52,816 | 3 | 105,632 |
No | output | 1 | 52,816 | 3 | 105,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image> | instruction | 0 | 53,343 | 3 | 106,686 |
Tags: brute force, geometry
Correct Solution:
```
def gcd(a,b):
if b == 0:
return a
return gcd(b, a % b)
def normalize_rational(num,den):
#associate the -ve with the num or cancel -ve sign when both are -ve
if num ^ den < 0 and num > 0 or num ^ den > 0 and num < 0:
num = -num; den = -den
#put it in its simplest form
g = gcd(abs(num),abs(den))
num //= g; den //=g
return (num,den)
from sys import *
l = stdin.readline()
(n,m) = (int(tkn) for tkn in l.split())
xs = [0] * m; ys = [0] * m
#maxhits = [set() for i in range(n+1)]
maxhits = [1] * (n + 1)
maxhits[0] = 0
for i in range(m):
l = stdin.readline()
(x,y) = (int(tkn) for tkn in l.split())
xs[i] = x; ys[i] = y
line_to_points = {}
for i in range(m):
for j in range(m):
#m = dy/dx; y = (dy/dx)x + c ==> y.dx = x.dy + c.dx
#y.dx = x.dy + c'
#c' = y.dx - x.dy
#c' = ys[i]*dx - xs[i]*dy
#Now, at y = 0, x = -c' / dy
dy = ys[i] - ys[j]; dx = xs[i] - xs[j]
if dy == 0:
continue
#not a special case anymore
# if dx == 0:
# if xs[i] > n:
# continue
# else:
# count_seen_from_x = len([x for x in xs if x == xs[i]])
# maxhits[xs[i]] = max(count_seen_from_x, maxhits[xs[i]])
else:
slope = normalize_rational(dy,dx)
c_prime = ys[i]*dx - xs[i]*dy
x_intercept = -c_prime / dy
line = (slope,x_intercept)
if line in line_to_points:
#print("line: ", line)
#print("now: ", points)
points = line_to_points[line]
points.add(i); points.add(j)
#print("after addition: ", points)
continue
#if (i == 1 and j == 2):
# print(c_prime, x_intercept, dy, dx)
if int(x_intercept) == x_intercept and x_intercept <= n and x_intercept > 0:
points = set([i,j])
line_to_points[line] = points
#maxhits[int(x_intercept)] = points
# count_on_line = 2
# for k in range(m):
# if k != i and k != j and ys[k] * dx == xs[k]*dy + c_prime:
# count_on_line += 1
# maxhits[int(x_intercept)] = max(count_on_line, maxhits[int(x_intercept)])
for line,points in line_to_points.items():
x_intercept = int(line[1])
maxhits[x_intercept] = max(maxhits[x_intercept],len(points))
#print(maxhits)
#print(sum([max(len(points),1) for points in maxhits]) - 1)
print(sum(maxhits))
``` | output | 1 | 53,343 | 3 | 106,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image> | instruction | 0 | 53,344 | 3 | 106,688 |
Tags: brute force, geometry
Correct Solution:
```
from ctypes import c_int32
def gcd(a,b):
if b == 0:
return a
return gcd(b, a % b)
def normalize_rational(num,den):
#associate the -ve with the num or cancel -ve sign when both are -ve
if num ^ den < 0 and num > 0 or num ^ den > 0 and num < 0:
num = -num; den = -den
#put it in its simplest form
g = gcd(abs(num),abs(den))
num //= g; den //=g
return (num,den)
from sys import *
l = stdin.readline()
(n,m) = (int(tkn) for tkn in l.split())
xs = [0] * m; ys = [0] * m
#maxhits = [set() for i in range(n+1)]
maxhits = [c_int32(1)] * (n + 1)
maxhits[0] = c_int32(0)
for i in range(m):
l = stdin.readline()
(x,y) = (int(tkn) for tkn in l.split())
xs[i] = x; ys[i] = y
line_to_points = {}
for i in range(m):
for j in range(m):
#m = dy/dx; y = (dy/dx)x + c ==> y.dx = x.dy + c.dx
#y.dx = x.dy + c'
#c' = y.dx - x.dy
#c' = ys[i]*dx - xs[i]*dy
#Now, at y = 0, x = -c' / dy
dy = ys[i] - ys[j]; dx = xs[i] - xs[j]
if dy == 0:
continue
#not a special case anymore
# if dx == 0:
# if xs[i] > n:
# continue
# else:
# count_seen_from_x = len([x for x in xs if x == xs[i]])
# maxhits[xs[i]] = max(count_seen_from_x, maxhits[xs[i]])
else:
slope = normalize_rational(dy,dx)
c_prime = ys[i]*dx - xs[i]*dy
x_intercept = -c_prime / dy
line = (slope,x_intercept)
if line in line_to_points:
#print("line: ", line)
#print("now: ", points)
points = line_to_points[line]
points.add(i); points.add(j)
#print("after addition: ", points)
continue
#if (i == 1 and j == 2):
# print(c_prime, x_intercept, dy, dx)
if int(x_intercept) == x_intercept and x_intercept <= n and x_intercept > 0:
points = set([i,j])
line_to_points[line] = points
#maxhits[int(x_intercept)] = points
# count_on_line = 2
# for k in range(m):
# if k != i and k != j and ys[k] * dx == xs[k]*dy + c_prime:
# count_on_line += 1
# maxhits[int(x_intercept)] = max(count_on_line, maxhits[int(x_intercept)])
for line,points in line_to_points.items():
x_intercept = int(line[1])
maxhits[x_intercept] = c_int32(max(maxhits[x_intercept].value,len(points)))
#print(maxhits)
#print(sum([max(len(points),1) for points in maxhits]) - 1)
print(sum([e.value for e in maxhits]))
# Made By Mostafa_Khaled
``` | output | 1 | 53,344 | 3 | 106,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image> | instruction | 0 | 53,345 | 3 | 106,690 |
Tags: brute force, geometry
Correct Solution:
```
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")
#######################################
n,m = map(int,input().split())
l = []
anss = [1]*(n+1)
for i in range(m):
l.append(list(map(int,input().split())))
for i in range(m-1):
for j in range(i+1,m):
sx = l[i][0] - l[j][0]
sy = l[i][1] - l[j][1]
ans = 2
if sx:
if sy:
x = (l[i][0]*sy - l[i][1]*sx)/sy
if 1<=x<=n and int(x) == x:
for k in range(j+1,m):
if l[k][1]*sx == (sy*(l[k][0]-l[i][0]) + l[i][1]*sx) :
ans+=1
anss[int(x)] = max(ans,anss[int(x)])
else:
if 1<=l[i][0]<=n:
for k in range(j+1,m):
if l[k][0] == l[i][0]:
ans+=1
anss[l[i][0]] = max(anss[l[i][0]],ans)
print(sum(anss[1::]))
``` | output | 1 | 53,345 | 3 | 106,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image> | instruction | 0 | 53,346 | 3 | 106,692 |
Tags: brute force, geometry
Correct Solution:
```
n,m = map(int,input().split())
l = []
anss = [1]*(n+1)
for i in range(m):
l.append(list(map(int,input().split())))
for i in range(m-1):
for j in range(i+1,m):
sx = l[i][0] - l[j][0]
sy = l[i][1] - l[j][1]
ans = 2
if sx:
if sy:
x = (l[i][0]*sy - l[i][1]*sx)/sy
if 1<=x<=n and int(x) == x:
for k in range(j+1,m):
if l[k][1]*sx == (sy*(l[k][0]-l[i][0]) + l[i][1]*sx) :
ans+=1
anss[int(x)] = max(ans,anss[int(x)])
else:
if 1<=l[i][0]<=n:
for k in range(j+1,m):
if l[k][0] == l[i][0]:
ans+=1
anss[l[i][0]] = max(anss[l[i][0]],ans)
print(sum(anss[1::]))
``` | output | 1 | 53,346 | 3 | 106,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image> | instruction | 0 | 53,347 | 3 | 106,694 |
Tags: brute force, geometry
Correct Solution:
```
from ctypes import c_int32
def gcd(a,b):
if b == 0:
return a
return gcd(b, a % b)
def normalize_rational(num,den):
#associate the -ve with the num or cancel -ve sign when both are -ve
if num ^ den < 0 and num > 0 or num ^ den > 0 and num < 0:
num = -num; den = -den
#put it in its simplest form
g = gcd(abs(num),abs(den))
num //= g; den //=g
return (num,den)
from sys import *
l = stdin.readline()
(n,m) = (int(tkn) for tkn in l.split())
xs = [0] * m; ys = [0] * m
#maxhits = [set() for i in range(n+1)]
maxhits = [c_int32(1)] * (n + 1)
maxhits[0] = c_int32(0)
for i in range(m):
l = stdin.readline()
(x,y) = (int(tkn) for tkn in l.split())
xs[i] = x; ys[i] = y
line_to_points = {}
for i in range(m):
for j in range(m):
#m = dy/dx; y = (dy/dx)x + c ==> y.dx = x.dy + c.dx
#y.dx = x.dy + c'
#c' = y.dx - x.dy
#c' = ys[i]*dx - xs[i]*dy
#Now, at y = 0, x = -c' / dy
dy = ys[i] - ys[j]; dx = xs[i] - xs[j]
if dy == 0:
continue
#not a special case anymore
# if dx == 0:
# if xs[i] > n:
# continue
# else:
# count_seen_from_x = len([x for x in xs if x == xs[i]])
# maxhits[xs[i]] = max(count_seen_from_x, maxhits[xs[i]])
else:
slope = normalize_rational(dy,dx)
c_prime = ys[i]*dx - xs[i]*dy
x_intercept = -c_prime / dy
line = (slope,x_intercept)
if line in line_to_points:
#print("line: ", line)
#print("now: ", points)
points = line_to_points[line]
points.add(i); points.add(j)
#print("after addition: ", points)
continue
#if (i == 1 and j == 2):
# print(c_prime, x_intercept, dy, dx)
if int(x_intercept) == x_intercept and x_intercept <= n and x_intercept > 0:
points = set([i,j])
line_to_points[line] = points
#maxhits[int(x_intercept)] = points
# count_on_line = 2
# for k in range(m):
# if k != i and k != j and ys[k] * dx == xs[k]*dy + c_prime:
# count_on_line += 1
# maxhits[int(x_intercept)] = max(count_on_line, maxhits[int(x_intercept)])
for line,points in line_to_points.items():
x_intercept = int(line[1])
maxhits[x_intercept] = c_int32(max(maxhits[x_intercept].value,len(points)))
#print(maxhits)
#print(sum([max(len(points),1) for points in maxhits]) - 1)
print(sum([e.value for e in maxhits]))
``` | output | 1 | 53,347 | 3 | 106,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image>
Submitted Solution:
```
n,m=list(map(int,input().split(" ")))
liste=[]
for _ in range(m):
x,y=list(map(int,input().split(" ")))
liste.append((x,y))
visited=[0 for i in range(1,n+1)]
count=0
def nearest(x):
if abs(int(x)-x)<abs(int(x)+1-x):
return int(x)
else:
return int(x)+1
for i in range(m-1):
for j in range(i+1,m):
x1,y1=liste[i]
x2,y2=liste[j]
if x1==x2:
for k in range(m):
if liste[k][0]==x1:
count+=1
visited[x1-1]=max(visited[x1-1],count)
elif y1!=y2:
slope=(y2-y1)/(x2-x1)
ord_ori=y2-slope*x2
number=-ord_ori/slope
if number%1<1e-3:
number=int(number)
if 1<=number<=n:
for k in range(m):
if liste[k][1]-slope*liste[k][0]-ord_ori<1e-3:
count+=1
visited[number-1]=max(visited[number-1],count)
count=0
sum_v=sum(visited)+visited.count(0)
print(sum_v)
``` | instruction | 0 | 53,348 | 3 | 106,696 |
No | output | 1 | 53,348 | 3 | 106,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image>
Submitted Solution:
```
n,m=list(map(int,input().split(" ")))
liste=[]
for _ in range(m):
x,y=list(map(int,input().split(" ")))
liste.append((x,y))
visited=[0 for i in range(1,n+1)]
count=0
def nearest(x):
if abs(int(x)-x)<abs(int(x)+1-x):
return int(x)
else:
return int(x)+1
for i in range(m-1):
for j in range(i+1,m):
x1,y1=liste[i]
x2,y2=liste[j]
if x1==x2:
for k in range(m):
if liste[k][0]==x1:
count+=1
visited[x1-1]=max(visited[x1-1],count)
elif y1!=y2:
slope=(y2-y1)/(x2-x1)
ord_ori=y2-slope*x2
number=-ord_ori/slope
if number%1<1e-4:
number=nearest(number)
if 1<=number<=n:
for k in range(m):
if abs(liste[k][1]-slope*liste[k][0]-ord_ori)<1e-4:
count+=1
visited[number-1]=max(visited[number-1],count)
count=0
sum_v=sum(visited)+visited.count(0)
print(sum_v)
``` | instruction | 0 | 53,349 | 3 | 106,698 |
No | output | 1 | 53,349 | 3 | 106,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image>
Submitted Solution:
```
n,m=list(map(int,input().split(" ")))
liste=[]
for _ in range(m):
x,y=list(map(int,input().split(" ")))
liste.append((x,y))
visited=[0 for i in range(1,n+1)]
count=0
def nearest(x):
if abs(int(x)-x)<abs(int(x)+1-x):
return int(x)
else:
return int(x)+1
for i in range(m-1):
for j in range(i+1,m):
x1,y1=liste[i]
x2,y2=liste[j]
count=0
if x1==x2:
for k in range(m):
if liste[k][0]==x1:
count+=1
visited[x1-1]=max(visited[x1-1],count)
elif y1!=y2:
slope=(y2-y1)/(x2-x1)
ord_ori=y2-slope*x2
number=-ord_ori/slope
print(slope,ord_ori,i,j)
if abs(nearest(number)-number)%1<1e-4:
number=nearest(number)
if 1<=number<=n:
for k in range(m):
if abs(liste[k][1]-slope*liste[k][0]-ord_ori)<1e-4:
count+=1
print(count)
visited[number-1]=max(visited[number-1],count)
sum_v=sum(visited)+visited.count(0)
print(sum_v)
``` | instruction | 0 | 53,350 | 3 | 106,700 |
No | output | 1 | 53,350 | 3 | 106,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Zoo in the Grid Kingdom is represented by an infinite grid. The Zoo has n observation binoculars located at the OX axis. For each i between 1 and n, inclusive, there exists a single binocular located at the point with coordinates (i, 0). There are m flamingos in the Zoo, located at points with positive coordinates. The flamingos are currently sleeping and you can assume that they don't move.
In order to get a good view over the flamingos, each of the binoculars can be independently rotated to face any angle (not necessarily integer). Then, the binocular can be used to observe all flamingos that is located at the straight line passing through the binocular at the angle it is set. In other words, you can assign each binocular a direction corresponding to any straight line passing through the binocular, and the binocular will be able to see all flamingos located on that line.
Today, some kids from the prestigious Codeforces kindergarten went on a Field Study to the Zoo. Their teacher would like to set each binocular an angle to maximize the number of flamingos that can be seen by the binocular. The teacher is very interested in the sum of these values over all binoculars. Please help him find this sum.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 250), denoting the number of binoculars and the number of flamingos, respectively.
Then m lines follow, the i-th line will contain two space-separated integers xi and yi (1 ≤ xi, yi ≤ 109), which means that the i-th flamingo is located at point (xi, yi).
All flamingos will be located at distinct points.
Output
Print a single integer denoting the maximum total number of flamingos that can be seen by all the binoculars.
Examples
Input
5 5
2 1
4 1
3 2
4 3
4 4
Output
11
Note
This picture shows the answer to the example test case.
<image>
Submitted Solution:
```
from sys import *
l = stdin.readline()
(n,m) = (int(tkn) for tkn in l.split())
xs = [0] * m; ys = [0] * m
maxhits = [1] * (n + 1)
maxhits[0] = 0
for i in range(m):
l = stdin.readline()
(x,y) = (int(tkn) for tkn in l.split())
xs[i] = x; ys[i] = y
for i in range(m):
for j in range(m):
dy = ys[i] - ys[j]; dx = xs[i] - xs[j]
if dy == 0:
continue
if dx == 0:
if xs[i] > n:
continue
else:
count_seen_from_x = len([x for x in xs if x == xs[i]])
maxhits[xs[i]] = max(count_seen_from_x, maxhits[xs[i]])
else:
c_prime = ys[i]*dx - xs[i]*dy
x_intercept = -c_prime / dy
if int(x_intercept) == x_intercept and x_intercept <= n:
count_on_line = 2
for k in range(m):
if k != i and k != j and ys[k] * dx == xs[k]*dy + c_prime:
count_on_line += 1
maxhits[int(x_intercept)] = max(count_on_line, maxhits[int(x_intercept)])
#print(maxhits)
print(sum(maxhits))
``` | instruction | 0 | 53,351 | 3 | 106,702 |
No | output | 1 | 53,351 | 3 | 106,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number line there are n balls. At time moment 0 for each ball the following data is known: its coordinate xi, speed vi (possibly, negative) and weight mi. The radius of the balls can be ignored.
The balls collide elastically, i.e. if two balls weighing m1 and m2 and with speeds v1 and v2 collide, their new speeds will be:
<image>.
Your task is to find out, where each ball will be t seconds after.
Input
The first line contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 100) — amount of balls and duration of the process. Then follow n lines, each containing three integers: xi, vi, mi (1 ≤ |vi|, mi ≤ 100, |xi| ≤ 100) — coordinate, speed and weight of the ball with index i at time moment 0.
It is guaranteed that no two balls have the same coordinate initially. Also each collision will be a collision of not more than two balls (that is, three or more balls never collide at the same point in all times from segment [0;t]).
Output
Output n numbers — coordinates of the balls t seconds after. Output the numbers accurate to at least 4 digits after the decimal point.
Examples
Input
2 9
3 4 5
0 7 8
Output
68.538461538
44.538461538
Input
3 10
1 2 3
4 -5 6
7 -8 9
Output
-93.666666667
-74.666666667
-15.666666667 | instruction | 0 | 53,399 | 3 | 106,798 |
Tags: brute force, implementation, math
Correct Solution:
```
class Ball:
def __init__(self, x, v, m):
self.v = v
self.x = x
self.m = m
def move(self, time):
self.x += self.v * time
def collisionTime(self, other):
if self.v == other.v:
return float("inf")
t = - (self.x - other.x) / (self.v - other.v)
if t < 1e-9:
return float("inf")
return t
def __str__(self):
return "Ball(x={:.2f} v={:.2f} m={:.2f})".format(self.x, self.v, self.m)
def findFirst():
global nBalls, balls
minTime = float("inf")
minPairs = []
for i in range(nBalls):
for j in range(i + 1, nBalls):
time = balls[i].collisionTime(balls[j])
if time < minTime:
minTime = time
minPairs = [(i, j)]
elif abs(time - minTime) < 1e-9:
minPairs.append((i, j))
return minTime, minPairs
def collidePair(i, j):
global balls
v1 = balls[i].v
v2 = balls[j].v
m1 = balls[i].m
m2 = balls[j].m
balls[i].v = ((m1 - m2) * v1 + 2 * m2 * v2) / (m1 + m2)
balls[j].v = ((m2 - m1) * v2 + 2 * m1 * v1) / (m1 + m2)
nBalls, maxTime = map(int, input().split())
balls = []
for i in range(nBalls):
x, v, m = map(int, input().split())
balls.append(Ball(x, v, m))
while True:
time, pairs = findFirst()
if time > maxTime:
break
for i in range(nBalls):
balls[i].move(time)
# print(time, maxTime)
for i, j in pairs:
collidePair(i, j)
# print(i, balls[i], j, balls[j])
# print()
maxTime -= time
for ball in balls:
ball.move(maxTime)
print("{:.6f}".format(ball.x))
``` | output | 1 | 53,399 | 3 | 106,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number line there are n balls. At time moment 0 for each ball the following data is known: its coordinate xi, speed vi (possibly, negative) and weight mi. The radius of the balls can be ignored.
The balls collide elastically, i.e. if two balls weighing m1 and m2 and with speeds v1 and v2 collide, their new speeds will be:
<image>.
Your task is to find out, where each ball will be t seconds after.
Input
The first line contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 100) — amount of balls and duration of the process. Then follow n lines, each containing three integers: xi, vi, mi (1 ≤ |vi|, mi ≤ 100, |xi| ≤ 100) — coordinate, speed and weight of the ball with index i at time moment 0.
It is guaranteed that no two balls have the same coordinate initially. Also each collision will be a collision of not more than two balls (that is, three or more balls never collide at the same point in all times from segment [0;t]).
Output
Output n numbers — coordinates of the balls t seconds after. Output the numbers accurate to at least 4 digits after the decimal point.
Examples
Input
2 9
3 4 5
0 7 8
Output
68.538461538
44.538461538
Input
3 10
1 2 3
4 -5 6
7 -8 9
Output
-93.666666667
-74.666666667
-15.666666667 | instruction | 0 | 53,400 | 3 | 106,800 |
Tags: brute force, implementation, math
Correct Solution:
```
n,t=map(int,input().split())
ar=[]
for _ in range(n):ar.append(list(map(int,input().split())))
cur=0
while True:
T=float('inf')
nums=[]
for i in range(n):
for j in range(i):
vd=ar[i][1]-ar[j][1]
xd=ar[i][0]-ar[j][0]
xd=-xd
if(vd==0):
continue
if(xd/vd<=1.0e-6):
continue
if(cur+xd/vd>=t):
continue
if(xd/vd<T and abs(xd/vd-T)>=1.0e-6):
T=xd/vd
nums=[[i,j]]
elif(abs(xd/vd-T)<1.0e-6):
nums.append([i,j])
if(T!=float('inf')):
cur+=T
for i in range(n):
ar[i][0]+=ar[i][1]*T
for el in nums:
a,b=el[0],el[1]
m1,m2=ar[a][2],ar[b][2]
v1,v2=ar[a][1],ar[b][1]
#ar[a][0]+=v1*T
#ar[b][0]+=v2*T
nv1=((m1-m2)*v1+2*m2*v2)/(m1+m2)
nv2=((m2-m1)*v2+2*m1*v1)/(m1+m2)
ar[a][1]=nv1
ar[b][1]=nv2
else:
break
#print(T,nums)
for i in range(n):
ar[i][0]+=ar[i][1]*(t-cur)
for i in range(n):
print(ar[i][0])
``` | output | 1 | 53,400 | 3 | 106,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number line there are n balls. At time moment 0 for each ball the following data is known: its coordinate xi, speed vi (possibly, negative) and weight mi. The radius of the balls can be ignored.
The balls collide elastically, i.e. if two balls weighing m1 and m2 and with speeds v1 and v2 collide, their new speeds will be:
<image>.
Your task is to find out, where each ball will be t seconds after.
Input
The first line contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 100) — amount of balls and duration of the process. Then follow n lines, each containing three integers: xi, vi, mi (1 ≤ |vi|, mi ≤ 100, |xi| ≤ 100) — coordinate, speed and weight of the ball with index i at time moment 0.
It is guaranteed that no two balls have the same coordinate initially. Also each collision will be a collision of not more than two balls (that is, three or more balls never collide at the same point in all times from segment [0;t]).
Output
Output n numbers — coordinates of the balls t seconds after. Output the numbers accurate to at least 4 digits after the decimal point.
Examples
Input
2 9
3 4 5
0 7 8
Output
68.538461538
44.538461538
Input
3 10
1 2 3
4 -5 6
7 -8 9
Output
-93.666666667
-74.666666667
-15.666666667 | instruction | 0 | 53,401 | 3 | 106,802 |
Tags: brute force, implementation, math
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, t = map(int, input().split())
t = float(t)
balls = sorted(list(map(float, input().split())) + [i] for i in range(n))
eps = 1e-9
def calc(i, j):
ok, ng = 0.0, t + 1.0
for _ in range(50):
mid = (ok + ng) / 2
if balls[i][0] + balls[i][1] * mid - eps > balls[j][0] + balls[j][1] * mid:
ng = mid
else:
ok = mid
return mid
def adv(i, delta):
balls[i][0] += balls[i][1] * delta
def collide(i, j):
balls[i][1], balls[j][1] = (
((balls[i][2] - balls[j][2]) * balls[i][1] + 2 * balls[j][2] * balls[j][1]) / (balls[i][2] + balls[j][2]),
((balls[j][2] - balls[i][2]) * balls[j][1] + 2 * balls[i][2] * balls[i][1]) / (balls[i][2] + balls[j][2])
)
while t > 0:
x, y, d = 0, 0, t + 1.0
for i in range(n - 1):
res = calc(i, i + 1)
if d > res:
d = res
x, y = i, i + 1
if d >= t:
for i in range(n):
adv(i, t)
break
for i in range(n):
adv(i, d)
collide(x, y)
t -= d
ans = [0.0] * n
for i in range(n):
ans[balls[i][-1]] = balls[i][0]
print(*ans, sep='\n')
``` | output | 1 | 53,401 | 3 | 106,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number line there are n balls. At time moment 0 for each ball the following data is known: its coordinate xi, speed vi (possibly, negative) and weight mi. The radius of the balls can be ignored.
The balls collide elastically, i.e. if two balls weighing m1 and m2 and with speeds v1 and v2 collide, their new speeds will be:
<image>.
Your task is to find out, where each ball will be t seconds after.
Input
The first line contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 100) — amount of balls and duration of the process. Then follow n lines, each containing three integers: xi, vi, mi (1 ≤ |vi|, mi ≤ 100, |xi| ≤ 100) — coordinate, speed and weight of the ball with index i at time moment 0.
It is guaranteed that no two balls have the same coordinate initially. Also each collision will be a collision of not more than two balls (that is, three or more balls never collide at the same point in all times from segment [0;t]).
Output
Output n numbers — coordinates of the balls t seconds after. Output the numbers accurate to at least 4 digits after the decimal point.
Examples
Input
2 9
3 4 5
0 7 8
Output
68.538461538
44.538461538
Input
3 10
1 2 3
4 -5 6
7 -8 9
Output
-93.666666667
-74.666666667
-15.666666667
Submitted Solution:
```
class Ball:
def __init__(self, x, v, m):
self.v = v
self.x = x
self.m = m
def move(self, time):
self.x += self.v * time
def collisionTime(self, other):
if self.v == other.v or self.x == other.x:
return float("inf")
t = - (self.x - other.x) / (self.v - other.v)
if t < 0:
return float("inf")
return t
def __str__(self):
return "Ball(x={:.2f} v={:.2f} m={:.2f})".format(self.x, self.v, self.m)
def findFirst():
global nBalls, balls
minTime = float("inf")
minPairs = []
for i in range(nBalls):
for j in range(i + 1, nBalls):
time = balls[i].collisionTime(balls[j])
if time < minTime:
minTime = time
minPairs = [(i, j)]
elif abs(time - minTime) < 1e-9:
minPairs.append((i, j))
return minTime, minPairs
def collidePair(i, j):
global balls
v1 = balls[i].v
v2 = balls[j].v
m1 = balls[i].m
m2 = balls[j].m
balls[i].v = ((m1 - m2) * v1 + 2 * m2 * v2) / (m1 + m2)
balls[j].v = ((m2 - m1) * v2 + 2 * m1 * v1) / (m1 + m2)
nBalls, maxTime = map(int, input().split())
balls = []
for i in range(nBalls):
x, v, m = map(int, input().split())
balls.append(Ball(x, v, m))
while True:
time, pairs = findFirst()
if time > maxTime:
break
for i in range(nBalls):
balls[i].move(time)
for i, j in pairs:
collidePair(i, j)
# print(balls[i], balls[j])
maxTime -= time
for ball in balls:
ball.move(maxTime)
print("{:.6f}".format(ball.x))
``` | instruction | 0 | 53,402 | 3 | 106,804 |
No | output | 1 | 53,402 | 3 | 106,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number line there are n balls. At time moment 0 for each ball the following data is known: its coordinate xi, speed vi (possibly, negative) and weight mi. The radius of the balls can be ignored.
The balls collide elastically, i.e. if two balls weighing m1 and m2 and with speeds v1 and v2 collide, their new speeds will be:
<image>.
Your task is to find out, where each ball will be t seconds after.
Input
The first line contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 100) — amount of balls and duration of the process. Then follow n lines, each containing three integers: xi, vi, mi (1 ≤ |vi|, mi ≤ 100, |xi| ≤ 100) — coordinate, speed and weight of the ball with index i at time moment 0.
It is guaranteed that no two balls have the same coordinate initially. Also each collision will be a collision of not more than two balls (that is, three or more balls never collide at the same point in all times from segment [0;t]).
Output
Output n numbers — coordinates of the balls t seconds after. Output the numbers accurate to at least 4 digits after the decimal point.
Examples
Input
2 9
3 4 5
0 7 8
Output
68.538461538
44.538461538
Input
3 10
1 2 3
4 -5 6
7 -8 9
Output
-93.666666667
-74.666666667
-15.666666667
Submitted Solution:
```
n,t=map(int,input().split())
ar=[]
for _ in range(n):ar.append(list(map(int,input().split())))
cur=0
while True:
T,a,b=float('inf'),0,1
for i in range(n):
for j in range(n):
if(i==j):
continue
vd=ar[i][1]-ar[j][1]
xd=ar[i][0]-ar[j][0]
xd=-xd
if(vd==0):
continue
if(xd/vd<=1.0e-6):
continue
if(cur+xd/vd>=t):
continue
if(xd/vd<T):
T=xd/vd
a=i
b=j
if(T!=float('inf')):
cur+=T
m1,m2=ar[a][2],ar[b][2]
v1,v2=ar[a][1],ar[b][1]
#ar[a][0]+=v1*T
#ar[b][0]+=v2*T
for i in range(n):
ar[i][0]+=ar[i][1]*T
nv1=((m1-m2)*v1+2*m2*v2)/(m1+m2)
nv2=((m2-m1)*v2+2*m1*v1)/(m1+m2)
ar[a][1]=nv1
ar[b][1]=nv2
else:
break
#print(T,a,b)
for i in range(n):
ar[i][0]+=ar[i][1]*(t-cur)
for i in range(n):
print(ar[i][0])
``` | instruction | 0 | 53,403 | 3 | 106,806 |
No | output | 1 | 53,403 | 3 | 106,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number line there are n balls. At time moment 0 for each ball the following data is known: its coordinate xi, speed vi (possibly, negative) and weight mi. The radius of the balls can be ignored.
The balls collide elastically, i.e. if two balls weighing m1 and m2 and with speeds v1 and v2 collide, their new speeds will be:
<image>.
Your task is to find out, where each ball will be t seconds after.
Input
The first line contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 100) — amount of balls and duration of the process. Then follow n lines, each containing three integers: xi, vi, mi (1 ≤ |vi|, mi ≤ 100, |xi| ≤ 100) — coordinate, speed and weight of the ball with index i at time moment 0.
It is guaranteed that no two balls have the same coordinate initially. Also each collision will be a collision of not more than two balls (that is, three or more balls never collide at the same point in all times from segment [0;t]).
Output
Output n numbers — coordinates of the balls t seconds after. Output the numbers accurate to at least 4 digits after the decimal point.
Examples
Input
2 9
3 4 5
0 7 8
Output
68.538461538
44.538461538
Input
3 10
1 2 3
4 -5 6
7 -8 9
Output
-93.666666667
-74.666666667
-15.666666667
Submitted Solution:
```
n,t=map(int,input().split())
ar=[]
for _ in range(n):ar.append(list(map(int,input().split())))
cur=0
while True:
T,a,b=float('inf'),0,1
for i in range(n):
for j in range(n):
if(i==j):
continue
vd=ar[i][1]-ar[j][1]
xd=ar[i][0]-ar[j][0]
xd=-xd
if(vd==0):
continue
if(xd/vd<=1.0e-8):
continue
if(cur+xd/vd>=t):
continue
if(xd/vd<T):
T=xd/vd
a=i
b=j
if(T!=float('inf')):
cur+=T
m1,m2=ar[a][2],ar[b][2]
v1,v2=ar[a][1],ar[b][1]
#ar[a][0]+=v1*T
#ar[b][0]+=v2*T
for i in range(n):
ar[i][0]+=ar[i][1]*T
nv1=((m1-m2)*v1+2*m2*v2)/(m1+m2)
nv2=((m2-m1)*v2+2*m1*v1)/(m1+m2)
ar[a][1]=nv1
ar[b][1]=nv2
else:
break
#print(T,a,b)
for i in range(n):
ar[i][0]+=ar[i][1]*(t-cur)
for i in range(n):
print(ar[i][0])
``` | instruction | 0 | 53,404 | 3 | 106,808 |
No | output | 1 | 53,404 | 3 | 106,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number line there are n balls. At time moment 0 for each ball the following data is known: its coordinate xi, speed vi (possibly, negative) and weight mi. The radius of the balls can be ignored.
The balls collide elastically, i.e. if two balls weighing m1 and m2 and with speeds v1 and v2 collide, their new speeds will be:
<image>.
Your task is to find out, where each ball will be t seconds after.
Input
The first line contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 100) — amount of balls and duration of the process. Then follow n lines, each containing three integers: xi, vi, mi (1 ≤ |vi|, mi ≤ 100, |xi| ≤ 100) — coordinate, speed and weight of the ball with index i at time moment 0.
It is guaranteed that no two balls have the same coordinate initially. Also each collision will be a collision of not more than two balls (that is, three or more balls never collide at the same point in all times from segment [0;t]).
Output
Output n numbers — coordinates of the balls t seconds after. Output the numbers accurate to at least 4 digits after the decimal point.
Examples
Input
2 9
3 4 5
0 7 8
Output
68.538461538
44.538461538
Input
3 10
1 2 3
4 -5 6
7 -8 9
Output
-93.666666667
-74.666666667
-15.666666667
Submitted Solution:
```
n,t=map(int,input().split())
ar=[]
for _ in range(n):ar.append(list(map(int,input().split())))
cur=0
while True:
T,a,b=float('inf'),0,1
for i in range(n):
for j in range(i):
vd=ar[i][1]-ar[j][1]
xd=ar[i][0]-ar[j][0]
xd=-xd
if(vd==0):
continue
if(xd/vd<=0):
continue
if(cur+xd/vd>=t):
continue
if(xd/vd<T):
T=xd/vd
a=i
b=j
if(T!=float('inf')):
cur+=T
m1,m2=ar[a][2],ar[b][2]
v1,v2=ar[a][1],ar[b][1]
#ar[a][0]+=v1*T
#ar[b][0]+=v2*T
for i in range(n):
ar[i][0]+=ar[i][1]*T
nv1=((m1-m2)*v1+2*m2*v2)/(m1+m2)
nv2=((m2-m1)*v2+2*m1*v1)/(m1+m2)
ar[a][1]=nv1
ar[b][1]=nv2
else:
break
#print(T,a,b)
for i in range(n):
ar[i][0]+=ar[i][1]*(t-cur)
for i in range(n):
print(ar[i][0])
``` | instruction | 0 | 53,405 | 3 | 106,810 |
No | output | 1 | 53,405 | 3 | 106,811 |
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