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.
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.
He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
Constraints
* -10^9 \leq A,B \leq 10^9
* 1 \leq V,W \leq 10^9
* 1 \leq T \leq 10^9
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A V
B W
T
Output
If "it" can catch the other child, print `YES`; otherwise, print `NO`.
Examples
Input
1 2
3 1
3
Output
YES
Input
1 2
3 2
3
Output
NO
Input
1 2
3 3
3
Output
NO
Submitted Solution:
```
A,V=map(int,input().split())
B,W=map(int,input().split())
T=int(input())
X=abs(B-A)
sp=V-W
Z=sp*T-X
if Z<0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 85,514 | 3 | 171,028 |
Yes | output | 1 | 85,514 | 3 | 171,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.
He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
Constraints
* -10^9 \leq A,B \leq 10^9
* 1 \leq V,W \leq 10^9
* 1 \leq T \leq 10^9
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A V
B W
T
Output
If "it" can catch the other child, print `YES`; otherwise, print `NO`.
Examples
Input
1 2
3 1
3
Output
YES
Input
1 2
3 2
3
Output
NO
Input
1 2
3 3
3
Output
NO
Submitted Solution:
```
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if w>v or t*abs(w-v)<abs(b-a):
print("NO")
else:
print("YES")
``` | instruction | 0 | 85,515 | 3 | 171,030 |
Yes | output | 1 | 85,515 | 3 | 171,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.
He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
Constraints
* -10^9 \leq A,B \leq 10^9
* 1 \leq V,W \leq 10^9
* 1 \leq T \leq 10^9
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A V
B W
T
Output
If "it" can catch the other child, print `YES`; otherwise, print `NO`.
Examples
Input
1 2
3 1
3
Output
YES
Input
1 2
3 2
3
Output
NO
Input
1 2
3 3
3
Output
NO
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
s_readline = sys.stdin.readline
def solve():
A, V = map(int, readline().split())
B, W = map(int, readline().split())
T = int(readline())
if A + V * T >= B + W * T:
ans = 'YES'
else:
ans = 'NO'
print(ans)
if __name__ == '__main__':
solve()
``` | instruction | 0 | 85,516 | 3 | 171,032 |
No | output | 1 | 85,516 | 3 | 171,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.
He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
Constraints
* -10^9 \leq A,B \leq 10^9
* 1 \leq V,W \leq 10^9
* 1 \leq T \leq 10^9
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A V
B W
T
Output
If "it" can catch the other child, print `YES`; otherwise, print `NO`.
Examples
Input
1 2
3 1
3
Output
YES
Input
1 2
3 2
3
Output
NO
Input
1 2
3 3
3
Output
NO
Submitted Solution:
```
A, V = map(int,input().split())
B, W = map(int,input().split())
T = int(input())
l = abs(A-B)
d = V - W
if d <= 0:
print("NO")
elif l % d != 0:
print("NO")
elif l // d <= T:
print("YES")
else :
print("NO")
#
``` | instruction | 0 | 85,517 | 3 | 171,034 |
No | output | 1 | 85,517 | 3 | 171,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.
He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
Constraints
* -10^9 \leq A,B \leq 10^9
* 1 \leq V,W \leq 10^9
* 1 \leq T \leq 10^9
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A V
B W
T
Output
If "it" can catch the other child, print `YES`; otherwise, print `NO`.
Examples
Input
1 2
3 1
3
Output
YES
Input
1 2
3 2
3
Output
NO
Input
1 2
3 3
3
Output
NO
Submitted Solution:
```
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
d = B - A
s = (V - W) * T
print(d,s)
if d <= s:
print("YES")
else:
print("NO")
``` | instruction | 0 | 85,518 | 3 | 171,036 |
No | output | 1 | 85,518 | 3 | 171,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.
He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
Constraints
* -10^9 \leq A,B \leq 10^9
* 1 \leq V,W \leq 10^9
* 1 \leq T \leq 10^9
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A V
B W
T
Output
If "it" can catch the other child, print `YES`; otherwise, print `NO`.
Examples
Input
1 2
3 1
3
Output
YES
Input
1 2
3 2
3
Output
NO
Input
1 2
3 3
3
Output
NO
Submitted Solution:
```
A_V = input().split(" ")
B_W = input().split(" ")
T = int(input())
A = int(A_V[0])
V = int(A_V[1])
B = int(B_W[0])
W = int(B_W[1])
dA = A + V * T
dB = B + W * T
ndA = A - V * T
ndB = B - W * T
if (dA < dB):
if (ndA > ndB):
print("NO")
else:
print("YES")
else:
print("YES")
``` | instruction | 0 | 85,519 | 3 | 171,038 |
No | output | 1 | 85,519 | 3 | 171,039 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,568 | 3 | 171,136 |
"Correct Solution:
```
L, N = map(int, input().split())
X = [0]
for _ in range(N):
X.append(int(input()))
s = [0] * (N + 1)
t = [0] * (N + 1)
for i in range(1, N + 1):
s[i] = s[i-1] + X[i]
t[i] = t[i-1] + (L - X[N - i + 1])
Y = [0] + [L - X[N - i] for i in range(N)]
# i >= 1
def get_d(s, x, i):
if i == N:
return x[N]
a = i + (N - i) // 2
return (
2 * (s[a] - s[i-1] + L * (N - a) - s[N] + s[a]) -
(x[a] if (N - i) % 2 == 0 else (L - x[a + 1]))
)
ans = 0
for i in range(1, N + 1):
ans = max(ans, get_d(s, X, i), get_d(t, Y, i))
print(ans)
``` | output | 1 | 85,568 | 3 | 171,137 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,569 | 3 | 171,138 |
"Correct Solution:
```
L, N = map(int, input().split())
X = [int(input()) for i in range(N)]
Y = [L - X[-i] for i in range(1, N + 1)]
def check(T, N):
X = [0] + T
P = [0] * (N + 1)
Q = [0] * (N + 1)
for i in range(1, N + 1):
P[i] = P[i - 1] + X[i]
for i in range(N, 0, -1):
Q[i - 1] = Q[i] + L - X[i]
ans = -1
for i in range(1, N + 1):
k = N - i
if k % 2 == 0:
a = k // 2
t = 2 * (P[i + a - 1] - P[i - 1]) + 2 * (Q[i + a]) + X[i + a]
else:
a = k // 2
t = 2 * (P[i + a] - P[i - 1]) + 2 * (Q[i + a + 1]) + L - X[i + a + 1]
ans = max(ans, t)
# print(t)
return ans
print(max(check(X, N), check(Y, N)))
``` | output | 1 | 85,569 | 3 | 171,139 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,570 | 3 | 171,140 |
"Correct Solution:
```
import math
L, N = map(int, input().split())
X = [None for i in range(N)]
DistAntiClock, DistClock = [0 for i in range(N+1)], [0 for i in range(N+1)]
for i in range(N):
X[i] = int(input())
DistAntiClock[i+1] = DistAntiClock[i] + 2 * X[i]
for i in range(N):
DistClock[i+1] = DistClock[i] + (L - X[-1-i]) * 2
maxLength = max(X[N-1], L - X[0])
for i in range(1, N):
finIndex = math.ceil((i + N)/2)
if (i + N) % 2 == 0:
AntiClockLength = DistAntiClock[finIndex] - DistAntiClock[i-1] - X[finIndex-1] + DistClock[N - finIndex]
ClockLength = DistClock[finIndex] - DistClock[i-1] - (L-X[-finIndex]) + DistAntiClock[N - finIndex]
else:
AntiClockLength = DistAntiClock[finIndex - 1] - DistAntiClock[i-1] + DistClock[N - finIndex + 1] - (L - X[finIndex - 1])
ClockLength = DistClock[finIndex - 1] - DistClock[i-1] + DistAntiClock[N - finIndex + 1] - X[N - finIndex]
maxLength = max(max(AntiClockLength, ClockLength), maxLength)
print(maxLength)
``` | output | 1 | 85,570 | 3 | 171,141 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,571 | 3 | 171,142 |
"Correct Solution:
```
L,N = map(int,input().split())
X = [int(input()) for i in range(N)]
if N == 1:
print(max(X[0],L-X[0]))
exit()
cum_l = [0]
for x in X:
cum_l.append(cum_l[-1] + x)
cum_r = [0]
for x in reversed(X):
cum_r.append(cum_r[-1] + L-x)
ans = 0
for st_l in range(1,N+1):
m = N - st_l
r = m - m//2
l = st_l + m//2
tmp = (cum_l[l] - cum_l[st_l-1] + cum_r[r]) * 2
if m%2:
tmp -= (L - X[-r])
else:
tmp -= X[l-1]
ans = max(ans, tmp)
for st_r in range(1,N+1):
m = N - st_r
l = m - m//2
r = st_r + m//2
tmp = (cum_r[r] - cum_r[st_r-1] + cum_l[l]) * 2
if m%2:
tmp -= X[l-1]
else:
tmp -= (L - X[-r])
ans = max(ans, tmp)
print(ans)
``` | output | 1 | 85,571 | 3 | 171,143 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,572 | 3 | 171,144 |
"Correct Solution:
```
L, N = map(int, input().split())
X = []
for i in range(N):
X.append(int(input()))
X = sorted(X)
ma = 0
for k in range(2):
if k == 1:
X = [L-X[-i] for i in range(1, N+1)]
s = X[0]
for i in range(1, N):
if i % 2 == 1:
s += L - X[-(i//2)-1] + X[i//2]
else:
s += X[i//2] + (L - X[-(i//2)])
ma = max(ma, s)
lt = (N-1) // 2 if N % 2 == 1 else -N//2
for i in range(1, N):
if lt > 0:
if N % 2 == 1:
s += X[lt] - X[i-1]*2 - (L - X[lt + 1])
else:
s += X[lt] - X[i-1] * 2 - (L - X[lt + 1])
lt = 1 + lt - N
ma = max(ma, s)
else:
if N % 2 == 1:
s += X[lt] - X[i-1] * 2 - (L - X[lt])
else:
s += X[lt]*2 - L - X[i-1] * 2
lt = N + lt
ma = max(ma, s)
print(ma)
``` | output | 1 | 85,572 | 3 | 171,145 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,573 | 3 | 171,146 |
"Correct Solution:
```
#!/usr/bin/env python3
from itertools import accumulate
L, N = map(int, input().split())
X = [int(input()) for _ in range(N)]
Y = [L - x for x in X][::-1]
sx = [0] + list(accumulate(X))
sy = [0] + list(accumulate(Y))
ans = 0
for i in range(N):
j = N - i - 1
mn = min(i, j)
now = X[i]
now += 2 * (sx[i] - sx[i - mn]) + 2 * (sy[j] - sy[j - mn - (i < j)])
ans = max(ans, now)
now = Y[j]
now += 2 * (sx[i] - sx[i - mn - (i > j)]) + 2 * (sy[j] - sy[j - mn])
ans = max(ans, now)
print(ans)
``` | output | 1 | 85,573 | 3 | 171,147 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,574 | 3 | 171,148 |
"Correct Solution:
```
L,N=map(int,input().split())
X=[int(input()) for i in range(N)]
X=[0]+X
Y=[0]
for i in range(N,0,-1):
Y.append(L-X[i])
sum_X=[0]*(N+1)
sum_Y=[0]*(N+1)
for i in range(1,N+1):
sum_X[i]=X[i]+sum_X[i-1]
sum_Y[i]=Y[i]+sum_Y[i-1]
ans=0
for i in range(N//2+1,N+1):
j=N-i
num_1=2*(sum_Y[j]+sum_X[i]-sum_X[i-j])-Y[j]
if i!=j:
num_2=2*(sum_Y[j]+sum_X[i]-sum_X[i-j-1])-X[i]
ans=max(ans,num_1,num_2)
X,Y=Y[:],X[:]
sum_X,sum_Y=sum_Y[:],sum_X[:]
for i in range(N//2+1,N+1):
j=N-i
num_1=2*(sum_Y[j]+sum_X[i]-sum_X[i-j])-Y[j]
if i!=j:
num_2=2*(sum_Y[j]+sum_X[i]-sum_X[i-j-1])-X[i]
ans=max(ans,num_1,num_2)
if N%2==0:
num_1=2*(sum_X[N//2]+sum_Y[N//2])-X[N//2]
num_2=2*(sum_X[N//2]+sum_Y[N//2])-Y[N//2]
ans=max(ans,num_1,num_2)
print(ans)
``` | output | 1 | 85,574 | 3 | 171,149 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749 | instruction | 0 | 85,575 | 3 | 171,150 |
"Correct Solution:
```
ln, kn, *ki = map(int, open(0).read().split())
# 順回転
pa = ki[-1]
ky = [pa]
kigu = 1
for ys in range(2, kn + 1):
# 要素数が偶数の場合
if kigu == 1:
pa += ki[kn - ys] * 2 + ln - ki[-(ys // 2)] * 2
# 要素数が奇数の場合
else:
pa += ki[kn - ys] * 2 - ki[-(ys // 2) - 1] + ln - ki[-(ys // 2)]
ky.append(pa)
kigu *= -1
# 逆回転
pa = ln - ki[0]
ky.append(pa)
kigu = 1
for ys in range(2, kn + 1):
# 要素数が偶数の場合
if kigu == 1:
pa += (ln-ki[ys - 1]) * 2 - ln + ki[ys // 2 - 1] * 2
# 要素数が奇数の場合
else:
pa += (ln-ki[ys - 1]) * 2+ki[ys//2]-ln+ki[ys//2-1]
ky.append(pa)
kigu *= -1
print(max(ky))
``` | output | 1 | 85,575 | 3 | 171,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
l, n = [int(item) for item in input().split()]
right = []
left = []
for i in range(n):
a = int(input())
right.append(a)
left.append(l - a)
left.reverse()
rsum = [0] * (n+1)
lsum = [0] * (n+1)
for i in range(n):
rsum[i+1] += rsum[i] + right[i]
lsum[i+1] += lsum[i] + left[i]
# Take all from left or right
ans = max(right[-1], left[-1])
# Take from left first, then ping pong
for i in range(n-1):
val = left[i] * 2
rest = n - (i + 1)
val += rsum[(rest + 1) // 2] * 2
val += (lsum[rest // 2 + i + 1] - lsum[i + 1]) * 2
if rest % 2 == 0:
val -= left[rest // 2 + i]
else:
val -= right[(rest + 1) // 2 - 1]
ans = max(ans, val)
# Take from right first, then ping pong
for i in range(n-1):
val = right[i] * 2
rest = n - (i + 1)
val += lsum[(rest + 1) // 2] * 2
val += (rsum[rest // 2 + i + 1] - rsum[i + 1]) * 2
if rest % 2 == 0:
val -= right[rest // 2 + i]
else:
val -= left[(rest + 1) // 2 - 1]
ans = max(ans, val)
print(ans)
``` | instruction | 0 | 85,576 | 3 | 171,152 |
Yes | output | 1 | 85,576 | 3 | 171,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
def is_even(num):
return num % 2 == 0
def goback(M, x, Sx):
k = M // 2
if M == 0:
return L - x[0]
elif M == 1:
return 2 * (L - x[1]) + x[0]
elif is_even(M):
return 2 * Sx[k - 1] - 2 * (Sx[M] - Sx[M - k]) - x[M - k] + L * (2 * k + 1)
else:
return 2 * Sx[k - 1] + x[k] - 2 * (Sx[M] - Sx[M - k - 1]) + L * 2 * (k + 1)
L, N = (int(i) for i in input().split())
ans = 0
pl = []
for i in range(N):
pl.append(int(input()))
sumpl = [pl[0]]
for i in range(1, N):
sumpl.append(sumpl[i - 1] + pl[i])
for i in range(0, N):
ans = max(ans, goback(i, pl, sumpl))
revpl = [L - p for p in pl[::-1]]
sumrevpl = [revpl[0]]
for i in range(1, N):
sumrevpl.append(sumrevpl[i - 1] + revpl[i])
for i in range(0, N):
ans = max(ans, goback(i, revpl, sumrevpl))
print(ans)
``` | instruction | 0 | 85,577 | 3 | 171,154 |
Yes | output | 1 | 85,577 | 3 | 171,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
L, N = [int(_) for _ in input().split()]
X = [int(input()) for _ in range(N)]
def calc(X):
cumsum = [0]
for x in X:
cumsum += [cumsum[-1] + x]
res = 0
for i in range(N):
if (i + N) % 2:
end = (i + N - 1) // 2
d = 2 * (cumsum[end + 1] - cumsum[i]) + 2 * L * (
N - 1 - end) - 2 * (cumsum[N] - cumsum[end + 1]) - X[end]
else:
end = (i + N) // 2
d = 2 * (cumsum[end] - cumsum[i]) + 2 * L * (N - end) - 2 * (
cumsum[N] - cumsum[end]) - (L - X[end])
res = max(res, d)
return res
print(max(calc(X), calc([L - x for x in X[::-1]])))
``` | instruction | 0 | 85,578 | 3 | 171,156 |
Yes | output | 1 | 85,578 | 3 | 171,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
import sys
def main():
L, N = map(int, input().split())
lines = sys.stdin.readlines()
trees_ccw = [None] * N
for i in range(N):
trees_ccw[i] = int(lines[i])
trees_cw = [L - trees_ccw[-(i+1)] for i in range(N)]
trees_ccw_accum = [0] + [None] * N
trees_cw_accum = [0] + [None] * N
for i in range(N):
trees_ccw_accum[i+1] = trees_ccw_accum[i] + trees_ccw[i]
trees_cw_accum[i+1] = trees_cw_accum[i] + trees_cw[i]
#print(trees_ccw)
#print(trees_cw)
#print(trees_ccw_accum)
#print(trees_cw_accum)
dist_max = 0
for i in range(1,N+1):
offset_left = i
n_left = offset_left + (N-offset_left) // 2 #ccwの数
n_right = N - n_left #cwの数
'''
[print('L', end='') for _ in range(offset_left-1)]
print('L', end='_')
for i_ in range(N-offset_left):
if i_ % 2 == 0:
print('R', end='')
else:
print('L', end='')
print()
print(offset_left, n_left, n_right)
'''
dist_1 = 2 * (trees_ccw_accum[n_left] - trees_ccw_accum[i-1]) + 2 * trees_cw_accum[n_right]
dist_2 = 2 * (trees_cw_accum[n_left] - trees_cw_accum[i-1]) + 2 * trees_ccw_accum[n_right]
if (N-offset_left) % 2 == 0:
dist_1 -= trees_ccw[n_left-1]
dist_2 -= trees_cw[n_left-1]
else:
dist_1 -= trees_cw[n_right-1]
dist_2 -= trees_ccw[n_right-1]
if dist_max < dist_1:
dist_max = dist_1
if dist_max < dist_2:
dist_max = dist_2
#print(n_left, dist_1, dist_2, dist_max)
print(dist_max)
return
main()
``` | instruction | 0 | 85,579 | 3 | 171,158 |
Yes | output | 1 | 85,579 | 3 | 171,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
from collections import deque
L,N=map(int,input().split())
Y=[int(input()) for i in range(N)]
anss=[]
for k in range(2):
X=deque([i for i in Y])
ans=0
p=0
i=k
while(len(X)>0):
a=(X[0]-p)%L
b=(-X[-1]+p)%L
if i%2==0:
ans+=a
p=X.popleft()
else:
ans+=b
p=X.pop()
i+=1
anss.append(ans)
X=deque([Y[i] for i in range(N-1)])
ans=0
p=0
i=1
while(len(X)>0):
a=(X[0]-p)%L
b=(-X[-1]+p)%L
if i%2==0:
ans+=a
p=X.popleft()
else:
ans+=b
p=X.pop()
i+=1
anss.append(ans)
X=deque([Y[i] for i in range(1,N)])
ans=0
p=0
i=0
while(len(X)>0):
a=(X[0]-p)%L
b=(-X[-1]+p)%L
if i%2==0:
ans+=a
p=X.popleft()
else:
ans+=b
p=X.pop()
i+=1
anss.append(ans)
print(max(anss))
``` | instruction | 0 | 85,580 | 3 | 171,160 |
No | output | 1 | 85,580 | 3 | 171,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
L, N = map(int, input().split())
X = []
for i in range(N):
X.append(int(input()))
left = X[0]
lcount = 0
rcount = -1
right = X[-1]
now = 0
ans = 0
if left >= L - right:
ans += left
side = 0
now = left
lcount += 1
left = X[lcount]
side = 0
else:
ans += L - right
side = 1
now = right
rcount -= 1
right = X[rcount]
side = 1
for i in range(N-1):
if side == 0:
if left - now > now + (L - right):
ans += left - now
now = left
lcount += 1
left = X[lcount]
side = 0
else:
ans += now + (L - right)
now = right
rcount -= 1
right = X[rcount]
side = 1
else:
if L + left - now > now - right:
ans += L + left - now
now = left
lcount += 1
left = X[lcount]
side = 0
else:
ans += now - right
now = right
rcount -= 1
right = X[rcount]
side = 1
print(ans)
``` | instruction | 0 | 85,581 | 3 | 171,162 |
No | output | 1 | 85,581 | 3 | 171,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
l, n = [int(i) for i in input().split()]
X = [0] + [int(input()) for i in range(n)]
XJ = X[1:] + [l]
dp = [[[0] * (n+1) for i in range(n+1)] for j in range(2)]
for j in range(n, 0, -1):
for i in range(n):
dp[0][i+1][j] = max(dp[0][i][j] + X[i+1] - X[i], dp[1][i][j] + (l - XJ[j]) + X[i+1])
dp[1][i][j-1] = max(dp[0][i][j] + X[i] + (l - XJ[j-1]), dp[1][i][j] + XJ[j] - XJ[j-1])
ans = 0
for i in range(n+1):
ans = max(ans, dp[1][i][i], dp[0][i][i])
print(ans)
``` | instruction | 0 | 85,582 | 3 | 171,164 |
No | output | 1 | 85,582 | 3 | 171,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi Lake has a perimeter of L. On the circumference of the lake, there is a residence of the lake's owner, Takahashi. Each point on the circumference of the lake has a coordinate between 0 and L (including 0 but not L), which is the distance from the Takahashi's residence, measured counter-clockwise.
There are N trees around the lake; the coordinate of the i-th tree is X_i. There is no tree at coordinate 0, the location of Takahashi's residence.
Starting at his residence, Takahashi will repeat the following action:
* If all trees are burnt, terminate the process.
* Specify a direction: clockwise or counter-clockwise.
* Walk around the lake in the specified direction, until the coordinate of a tree that is not yet burnt is reached for the first time.
* When the coordinate with the tree is reached, burn that tree, stay at the position and go back to the first step.
Find the longest possible total distance Takahashi walks during the process.
Constraints
* 2 \leq L \leq 10^9
* 1 \leq N \leq 2\times 10^5
* 1 \leq X_1 < ... < X_N \leq L-1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
L N
X_1
:
X_N
Output
Print the longest possible total distance Takahashi walks during the process.
Examples
Input
10 3
2
7
9
Output
15
Input
10 6
1
2
3
6
7
9
Output
27
Input
314159265 7
21662711
77271666
89022761
156626166
160332356
166902656
298992265
Output
1204124749
Submitted Solution:
```
L,N = list(map(int,input().split()))
X=[]
Y=[]
SX=[0]
SY=[0]
sx=0
sy=0
for i in range(N):
x=int(input())
X.append(x)
sx+=x
SX.append(sx)
Y.append(L-x)
sy+=L-x
SY.append(sy)
max_score=0
for i in range(N):
if i>N-i-1:
max_score=max(2*(SX[i]-SX[2*i-N])+2*(SY[N]-SY[i+1])+Y[i],max_score)
elif i<N-i-1:
max_score=max(2*(SX[i]-SX[0])+2*(SY[2*i+2]-SY[i+1])+X[i],max_score)
ind = min(i,N-i-1)
max_score=max(2*(SX[i]-SX[i-ind])+2*(SY[i+ind]-SY[i+1])+X[i],max_score)
max_score=max(2*(SX[i]-SX[i-ind])+2*(SY[i+ind]-SY[i+1])+Y[i],max_score)
print(max_score)
``` | instruction | 0 | 85,583 | 3 | 171,166 |
No | output | 1 | 85,583 | 3 | 171,167 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,620 | 3 | 171,240 |
"Correct Solution:
```
#!/usr/bin/env python3
# ????????????????????°???????????°????????????????????\??????????????¢?????´???????????¨??????????§???????
# ?????????http://www.prefield.com/algorithm/geometry/convex_hull.html
import enum
EPS = 1e-10
class PointsRelation(enum.Enum):
counter_clockwise = 1
clockwise = 2
online_back = 3
on_segment = 4
online_front = 5
def inner_product(v1, v2):
return v1.real * v2.real + v1.imag * v2.imag
def outer_product(v1, v2):
return v1.real * v2.imag - v1.imag * v2.real
def project(a, b):
return a * inner_product(a, b) / (abs(a) ** 2)
def points_relation(p0, p1, p2):
v1 = p1 - p0
v2 = p2 - p0
op = outer_product(v1, v2)
if op > EPS:
return PointsRelation.counter_clockwise
elif op < -EPS:
return PointsRelation.clockwise
elif inner_product(v1, v2) < -EPS:
return PointsRelation.online_back
elif abs(v1) < abs(v2):
return PointsRelation.online_front
else:
return PointsRelation.on_segment
def in_place_convex_hull_andrew(ps):
def judge(p0, p1, p2):
b = points_relation(p0, p1, p2) != PointsRelation.counter_clockwise
b &= points_relation(p0, p1, p2) != PointsRelation.online_front
return b
if len(ps) < 3:
return ps
ps.sort(key=lambda p: p.imag)
n = len(ps)
k = 0
ch = [None for _ in range(2 * n)]
for i in range(n):
while k >= 2 and judge(ch[k - 2], ch[k - 1], ps[i]):
k -= 1
ch[k] = ps[i]
k += 1
t = k + 1
for i in range(n - 1)[::-1]:
while k >= t and judge(ch[k - 2], ch[k - 1], ps[i]):
k -= 1
ch[k] = ps[i]
k += 1
ch = ch[:k - 1]
return ch
def main():
while True:
n = int(input())
if n == 0:
return
ps = [complex(*map(float, input().split(","))) for _ in range(n)]
ch = in_place_convex_hull_andrew(ps)
print(n - len(ch))
if __name__ == '__main__':
main()
``` | output | 1 | 85,620 | 3 | 171,241 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,621 | 3 | 171,242 |
"Correct Solution:
```
# Aizu Problem 0068: Enclose Pins with a Rubber Band
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def distance(p1, p2):
return math.sqrt((p1[0] - p2[0])**2 + (p1[1] - p2[1])**2)
def is_to_the_left(p1, p2, p3):
# determine whether point p3 is to the left from the line from p1 to p2
position = (p2[0] - p1[0]) * (p3[1] - p1[1]) - (p2[1] - p1[1]) * (p3[0] - p1[0])
return position < 0
def jarvis(points):
# determine convex hull by Jarvis' algorithm:
max_y = max([p[1] for p in points])
pointOnHull = [p for p in points if p[1] == max_y][0]
convex_hull = [pointOnHull]
while len(convex_hull) == 1 or convex_hull[-1] != convex_hull[0]:
p = convex_hull[-1]
endpoint = points[0]
for j in range(len(points)):
if endpoint == pointOnHull or is_to_the_left(p, endpoint, points[j]):
endpoint = points[j]
pointOnHull = endpoint
convex_hull.append(pointOnHull)
return convex_hull[::-1]
while True:
N = int(input())
if N == 0:
break
points = [[float(_) for _ in input().split(',')] for __ in range(N)]
print(N - (len(jarvis(points)) - 1))
``` | output | 1 | 85,621 | 3 | 171,243 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,622 | 3 | 171,244 |
"Correct Solution:
```
from math import atan2
class Point(object):
def __init__(self, x, y):
self.x = x
self.y = y
class Vector(object):
def __init__(self, p1, p2):
self.x = p2.x - p1.x
self.y = p2.y - p1.y
def cross(self, other: "Vector") -> float:
return self.x*other.y - self.y*other.x
def grahams_scan(a: list):
a.sort(key=lambda p: p.y)
bottom_x, bottom_y = a[0].x, a[0].y
_a = sorted(a[1:], key=lambda p: atan2(bottom_y-p.y, bottom_x-p.x)) + [a[0]]
result = [a[0], _a[0]]
for next_point in _a[1:]:
while Vector(result[-1], result[-2]).cross(Vector(result[-1], next_point)) >= 0:
result.pop()
result.append(next_point)
return result
while True:
n = int(input())
if not n:
break
a = [Point(*map(float, input().split(","))) for _ in [0]*n]
result = grahams_scan(a)
print(n-len(result)+1)
``` | output | 1 | 85,622 | 3 | 171,245 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,623 | 3 | 171,246 |
"Correct Solution:
```
def op(x,y):
"""
Outer product
"""
return(complex.conjugate(x)*y).imag
def isrm(q,i,x):
"""
is i right most to q's from view of x?
"""
for j in q:
if op(i-x, j-x) < 0:
return(False)
return(True)
def rightmost(p,x):
"""
Find a point in p which is rightmost from x(other than x)
"""
for i in [i for i in p if i!=x]:
q = p.copy()
q.remove(i)
if isrm(q,i,x):
return(i)
raise ValueError('Do not come here')
def solve(p,pf,orig):
nx = rightmost(p,orig)
ni = p.index(nx)
if pf[ni]: # second visit
return(pf)
pf[ni] = True
return(solve(p,pf,nx))
while True:
p = []
n = int(input().strip())
if n==0:
break
for i in range(n):
x,y = list(map(float, input().strip().split(',')))
p.append(x+y*1j)
print(solve(p,[False for _ in range(n)],2000.0 + 0.0*1j).count(False))
``` | output | 1 | 85,623 | 3 | 171,247 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,624 | 3 | 171,248 |
"Correct Solution:
```
# AOJ 0068 Enclose Pins with a Rubber Band
# Python3 2018.6.22 bal4u
def cross(a, b):
return a.real*b.imag - a.imag*b.real
# 凸包 入力: 座標リスト リターン:凸包を構成する座標リスト
def convex_hull(p):
pp = sorted(p, key=lambda x:(x.imag,x.real)) # y座標を優先して昇順、同じならx座標で昇順
n = len(pp)
ans, j = [0]*(n+1), 0
for i in range(n):
while j > 1 and cross(ans[j-1]-ans[j-2], pp[i]-ans[j-1]) < 0: j -= 1
ans[j] = pp[i]
j += 1
k = j
for i in range(n-2, -1, -1):
while j > k and cross(ans[j-1]-ans[j-2], pp[i]-ans[j-1]) < 0: j -= 1
ans[j] = pp[i]
j += 1
return ans[0:j-1]
while 1:
n = int(input())
if n == 0: break
p = []
for i in range(n):
x, y = list(map(float, input().split(',')))
p.append(complex(x, y))
print(n - len(convex_hull(p)))
``` | output | 1 | 85,624 | 3 | 171,249 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,625 | 3 | 171,250 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
import os
for s in sys.stdin:
N = int(s)
if N == 0:
break
X = []
Y = []
visited = [False] * N
for i in range(N):
x, y = map(float, input().split(','))
X.append(x)
Y.append(y)
def paint(start_i, is_right, is_up):
x_start = X[start_i]
y_start = Y[start_i]
visited[start_i] = True
if is_right:
dir_sign = 1 # search right only
else:
dir_sign = -1 # search left only
if (is_up and is_right) or (not is_up and not is_right) :
min_max_sign = 1 # search min
else:
min_max_sign = -1 # search max
tangent_min_i = None
tangent_min = None
for i, x in enumerate(X):
if dir_sign * (x - x_start) > 0:
y = Y[i]
tangent = (y - y_start) / (x - x_start)
if tangent_min is None:
tangent_min = tangent
tangent_min_i = i
else:
if min_max_sign * (tangent_min - tangent) > 0:
tangent_min = tangent
tangent_min_i = i
if tangent_min_i is not None:
paint(tangent_min_i, is_right, is_up)
start_index = Y.index(min(Y))
visited[start_index] = True
paint(start_index, is_right=True, is_up=True)
paint(start_index, is_right=False, is_up=True)
start_index = Y.index(max(Y))
visited[start_index] = True
paint(start_index, is_right=True, is_up=False)
paint(start_index, is_right=False, is_up=False)
visit_num = 0
for is_visit in visited:
if is_visit:
visit_num += 1
print(len(X) - visit_num)
``` | output | 1 | 85,625 | 3 | 171,251 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,626 | 3 | 171,252 |
"Correct Solution:
```
import math
def vec(a, b):
return [b[0] - a[0], b[1] - a[1]]
def norm(a):
return math.sqrt(a[0]**2 + a[1]**2)
def cross(a, b):
return a[0]*b[1] - b[0]*a[1]
def gift_wrap(p_a, p_h, a):
while True:
p_h.append(a)
b = p_a[0]
for i in range(len(p_a)):
c = p_a[i]
if b == a:
b = c
else:
ab = vec(a, b)
ac = vec(a, c)
v = cross(ab, ac)
if v > 0 or (v == 0 and norm(ac) > norm(ab)):
b = c
a = b
if a == p_h[0]:
break
while True:
n = int(input())
if n == 0:
break
p_all = []
p_hull = []
for i in range(n):
p_all.append(list(map(float, input().split(","))))
p_all = sorted(p_all)
gift_wrap(p_all, p_hull, p_all[0])
print(len(p_all) - len(p_hull))
``` | output | 1 | 85,626 | 3 | 171,253 |
Provide a correct Python 3 solution for this coding contest problem.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3 | instruction | 0 | 85,627 | 3 | 171,254 |
"Correct Solution:
```
eps = 1e-10
def add(a, b):
return 0 if abs(a + b) < eps * (abs(a) + abs(b)) else a + b
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def __add__(self, p):
return Point(add(self.x, p.x), add(self.y, p.y))
def __sub__(self, p):
return Point(add(self.x, -p.x), add(self.y, -p.y))
def __mul__(self, d):
return Point(self.x * d, self.y * d)
def dot(self, p):
return add(self.x * p.x, self.y * p.y)
def det(self, p):
return add(self.x * p.y, -self.y * p.x)
def __str__(self):
return "({}, {})".format(self.x, self.y)
def convex_hull(ps):
ps = [Point(x, y) for x, y in sorted([(p.x, p.y) for p in ps])]
lower_hull = get_bounds(ps)
ps.reverse()
upper_hull = get_bounds(ps)
del upper_hull[-1]
del lower_hull[-1]
lower_hull.extend(upper_hull)
return lower_hull
def get_bounds(ps):
qs = [ps[0], ps[1]]
for p in ps[2:]:
while len(qs) > 1 and (qs[-1] - qs[-2]).det(p - qs[-1]) <= 0:
del qs[-1]
qs.append(p)
return qs
while True:
n = int(input())
if n == 0: break
points = []
for i in range(n):
x, y = map(float, input().split(','))
points.append(Point(x, y))
print(len(points) - len(convex_hull(points)))
``` | output | 1 | 85,627 | 3 | 171,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def ccw(a, b, c):
ax = b[0] - a[0]
ay = b[1] - a[1]
bx = c[0] - a[0]
by = c[1] - a[1]
t = ax*by - ay*bx;
if t > 0:
return 1
if t < 0:
return -1
if ax*bx + ay*by < 0:
return 2
if ax*ax + ay*ay < bx*bx + by*by:
return -2
return 0
def convex_hull(ps):
n = len(ps)
k = 0
ps.sort()
ch = [[-1,-1] for _ in range(n*2)]
for i in range(n):
while k >= 2 and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
t = k + 1
for i in range(n-2,-1,-1):
while k >= t and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
ch = ch[:k-1]
return ch
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [list(map(float, S().split(','))) for _ in range(n)]
r = convex_hull(a)
rr.append(n-len(r))
return '\n'.join(map(str,rr))
print(main())
``` | instruction | 0 | 85,628 | 3 | 171,256 |
Yes | output | 1 | 85,628 | 3 | 171,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
#凸包を求める
def quickhull(l,r,s,k):
if not s:
return
su = []
sd = []
a = (r[0]-l[0],r[1]-l[1])
for x,y in s:
b = (x-l[0],y-l[1])
cro = cross(a,b)
if cro > 0:
su.append((x,y)) #上半分
elif cro < 0:
sd.append((x,y)) #下半分
if su:
c,d = direction(l,r,su[0])
p = su[0]
for i in range(1,len(su)):
c_,d_ = direction(l,r,su[i])
if c*d_ < c_*d:
c,d = c_,d_
p = su[i]
k.append(tuple(p)) #もっとも離れた点を凸包の頂点に追加
b = (l[0]-p[0],l[1]-p[1])
c = (p[0]-r[0],p[1]-r[1])
s1 = []
s2 = []
for x,y in su:
b_ = (x-p[0],y-p[1])
c_ = (x-r[0],y-r[1])
cro_b,cro_c = cross(b,b_),cross(c,c_)
if cro_b >= 0 and cro_c >= 0: #三角形内部判定
continue
else:
if cro_b < 0:
s1.append((x,y))
elif cro_c < 0:
s2.append((x,y))
quickhull(l,p,s1,k) #再帰
quickhull(p,r,s2,k)
if sd:
c,d = direction(l,r,sd[0])
p = sd[0]
for i in range(1,len(sd)):
c_,d_ = direction(l,r,sd[i])
if c*d_ < c_*d:
c,d = c_,d_
p = sd[i]
k.append(tuple(p)) #もっとも離れた点を凸包の頂点に追加
b = (l[0]-p[0],l[1]-p[1])
c = (p[0]-r[0],p[1]-r[1])
s1 = []
s2 = []
for x,y in sd:
b_ = (x-p[0],y-p[1])
c_ = (x-r[0],y-r[1])
cro_b,cro_c = cross(b,b_),cross(c,c_)
if cro_b <= 0 and cro_c <= 0: #三角形内部判定(ベクトルの向きにより上下で判定が異なることに注意)
continue
else:
if cro_b > 0:
s1.append((x,y))
elif cro_c > 0:
s2.append((x,y))
quickhull(l,p,s1,k) #再帰
quickhull(p,r,s2,k)
return k
def cross(a,b): #外積
return a[0]*b[1]-a[1]*b[0]
def direction(l,r,p): #点と直線の距離
a = r[1]-l[1]
b = l[0]-r[0]
return (a*(p[0]-l[0])+b*(p[1]-l[1]))**2, a**2+b**2 #分子の2乗,分母の2乗
while 1:
n = int(input())
if n == 0:
break
s = [[float(x) for x in input().split(",")] for i in range(n)]
s.sort()
l = tuple(s.pop(0))
r = tuple(s.pop(-1))
k = quickhull(l,r,s,[l,r])
print(n-len(k))
``` | instruction | 0 | 85,629 | 3 | 171,258 |
Yes | output | 1 | 85,629 | 3 | 171,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def cross3(a, b, c):
return (b[0]-a[0])*(c[1]-a[1]) - (b[1]-a[1])*(c[0]-a[0])
EPS = -1e-9
def convex_hull(ps):
qs = []
n = len(ps)
for p in ps:
while len(qs)>1 and cross3(qs[-1], qs[-2], p) > -EPS:
qs.pop()
qs.append(p)
t = len(qs)
for i in range(n-2, -1, -1):
p = ps[i]
while len(qs)>t and cross3(qs[-1], qs[-2], p) > -EPS:
qs.pop()
qs.append(p)
return qs
def solve():
N = int(readline())
if N == 0:
return False
P = [list(map(float, readline().split(","))) for i in range(N)]
P.sort()
Q = convex_hull(P)
write("%d\n" % (N - len(Q) + 1))
return True
while solve():
...
``` | instruction | 0 | 85,630 | 3 | 171,260 |
Yes | output | 1 | 85,630 | 3 | 171,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
# AOJ 0068 Enclose Pins with a Rubber Band
# Python3 2018.6.22 bal4u
def cross(a, b):
return a.real*b.imag - a.imag*b.real
# 凸包 入力: 座標リスト リターン:凸包を構成する座標リスト
def convex_hull(p):
pp = sorted(p, key=lambda x:(x.imag,x.real)) # y座標を優先して昇順、同じならx座標で昇順
n = len(pp)
ans, j = [0]*(n+1), 0
for i in range(n):
while j > 1 and cross(ans[j-1]-ans[j-2], pp[i]-ans[j-1]) <= 0: j -= 1
ans[j] = pp[i]
j += 1
k = j
for i in range(n-2, -1, -1):
while j > k and cross(ans[j-1]-ans[j-2], pp[i]-ans[j-1]) <= 0: j -= 1
ans[j] = pp[i]
j += 1
return ans[0:j-1]
while 1:
n = int(input())
if n == 0: break
p = []
for i in range(n):
x, y = list(map(float, input().split(',')))
p.append(complex(x, y))
print(n - len(convex_hull(p)))
``` | instruction | 0 | 85,631 | 3 | 171,262 |
Yes | output | 1 | 85,631 | 3 | 171,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
from math import atan2
class Point(object):
def __init__(self, x, y):
self.x = x
self.y = y
class Vector(object):
def __init__(self, p1, p2):
self.x = p2.x - p1.x
self.y = p2.y - p1.y
def cross(self, other: "Vector") -> float:
return self.x*other.y - self.y*other.x
def grahams_scan(a: list):
a.sort(key=lambda p: p.y)
bottom_x, bottom_y = a[0].x, a[0].y
_a = sorted(a[1:], key=lambda p: atan2(bottom_y-p.y, bottom_x-p.x)) + [a[0]]
result = [a[0], _a[0]]
for next_point in _a[1:]:
while Vector(result[-1], result[-2]).cross(Vector(result[-1], next_point)) >= 0:
result.pop()
result.append(next_point)
return result
while True:
n = int(input())
if not n:
break
a = [Point(*map(float, input().split(","))) for _ in [0]*n]
result = grahams_scan(a)
print(n-len(result)-1)
``` | instruction | 0 | 85,632 | 3 | 171,264 |
No | output | 1 | 85,632 | 3 | 171,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
upper_hull = get_bounds(ps)
ps.reverse()
lower_hull = get_bounds(ps)
del upper_hull[-1]
del lower_hull[-1]
upper_hull.extend(lower_hull)
return upper_hull
def get_bounds(ps):
qs = []
for p in ps:
while len(qs) > 1 and (qs[-1] - qs[-2]).det(p - qs[-1]) <= 0:
del qs[-1]
qs.append(p)
return qs
while True:
n = int(input())
if n == 0: break
points = []
for i in range(n):
x, y = map(float, input().split(','))
points.append(Point(x, y))
print(len(points) - len(convex_hull(points)))
``` | instruction | 0 | 85,633 | 3 | 171,266 |
No | output | 1 | 85,633 | 3 | 171,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
from math import atan2
def grahams_scan(a: list):
a.sort(key=lambda x: x[1])
x1, y1 = a[0]
a = [((x1, y1), 0)] +\
sorted([((x2, y2), atan2(y2-y1, x2-x1)) for x2, y2 in a[1:]], key=lambda x: x[1]) +\
[((x1, y1), 0)]
result = [(x1, y1)]
for ((x2, y2),_), ((x3, y3),_) in zip(a[1:], a[2:]):
if (x1-x2)*(y3-y2) - (y1-y2)*(x3-x2) < 0:
result.append((x2, x3))
x1, y1 = x2, y2
return result
while True:
n = int(input())
if not n:
break
a = [tuple(map(float, input().split(","))) for _ in [0]*n]
r = grahams_scan(a)
print(n-len(r))
``` | instruction | 0 | 85,634 | 3 | 171,268 |
No | output | 1 | 85,634 | 3 | 171,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.
Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.
<image> | <image>
--- | ---
Figure 1 | Figure 2
However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.
Hint
Below is a diagram for the second sample input.
<image>
---
Input
Given multiple datasets. Each dataset is given in the following format:
n
x1, y1
x2, y2
...
...
xn, yn
When n is 0, it indicates the end of input. The number of datasets does not exceed 50.
Output
For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.
<image> | <image>
--- | ---
Figure 3 | Figure 4
Example
Input
4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0
Output
0
3
Submitted Solution:
```
from math import atan2
class Point(object):
def __init__(self, x, y):
self.x = x
self.y = y
class Vector(object):
def __init__(self, p1, p2):
self.x = p2.x - p1.x
self.y = p2.y - p1.y
def cross(self, other: "Vector") -> float:
return self.x*other.y - self.y*other.x
def grahams_scan(a: list):
a.sort(key=lambda p: p.y)
bottom_x, bottom_y = a[0].x, a[0].y
_a = sorted(a[1:], key=lambda p: atan2(bottom_y-p.y, bottom_x-p.x))
result = [a[0], _a[0]]
for next_point in _a[1:]:
while Vector(result[-1], result[-2]).cross(Vector(result[-1], next_point)) >= 0:
result.pop()
result.append(next_point)
return result
while True:
n = int(input())
if not n:
break
a = [Point(*map(float, input().split(","))) for _ in [0]*n]
result = grahams_scan(a)
print(n-len(result))
``` | instruction | 0 | 85,635 | 3 | 171,270 |
No | output | 1 | 85,635 | 3 | 171,271 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Denjiro is a science teacher. Today he has just received a specially ordered water tank that will certainly be useful for his innovative experiments on water flow.
<image>
---
Figure 1: The water tank
The size of the tank is 100cm (Width) * 50cm (Height) * 30cm (Depth) (see Figure 1). For the experiments, he fits several partition boards inside of the tank parallel to the sideboards. The width of each board is equal to the depth of the tank, i.e. 30cm. The height of each board is less than that of the tank, i.e. 50 cm, and differs one another. The boards are so thin that he can neglect the thickness in the experiments.
<image>
---
Figure 2: The front view of the tank
The front view of the tank is shown in Figure 2.
There are several faucets above the tank and he turns them on at the beginning of his experiment. The tank is initially empty. Your mission is to write a computer program to simulate water flow in the tank.
Input
The input consists of multiple data sets. D is the number of the data sets.
D
DataSet1
DataSet2
...
DataSetD
The format of each data set (DataSetd , 1 <= d <= D) is as follows.
N
B1 H1
B2 H2
...
BN HN
M
F1 A1
F2 A2
...
FM AM
L
P1 T1
P2 T2
...
PL TL
Each line in the data set contains one or two integers.
N is the number of the boards he sets in the tank . Bi and Hi are the x-position (cm) and the height (cm) of the i-th board, where 1 <= i <= N .
Hi s differ from one another. You may assume the following.
> 0 < N < 10 ,
> 0 < B1 < B2 < ... < BN < 100 ,
> 0 < H1 < 50 , 0 < H2 < 50 , ..., 0 < HN < 50.
M is the number of the faucets above the tank . Fj and Aj are the x-position (cm) and the amount of water flow (cm3/second) of the j-th faucet , where 1 <= j <= M .
There is no faucet just above any boards . Namely, none of Fj is equal to Bi .
You may assume the following .
> 0 < M <10 ,
> 0 < F1 < F2 < ... < FM < 100 ,
> 0 < A1 < 100, 0 < A2 < 100, ... 0 < AM < 100.
L is the number of observation time and location. Pk is the x-position (cm) of the k-th observation point. Tk is the k-th observation time in seconds from the beginning.
None of Pk is equal to Bi .
You may assume the following .
> 0 < L < 10 ,
> 0 < P1 < 100, 0 < P2 < 100, ..., 0 < PL < 100 ,
> 0 < T1 < 1000000, 0 < T2 < 1000000, ... , 0 < TL < 1000000.
Output
For each data set, your program should output L lines each containing one real number which represents the height (cm) of the water level specified by the x-position Pk at the time Tk.
Each answer may not have an error greater than 0.001. As long as this condition is satisfied, you may output any number of digits after the decimal point.
After the water tank is filled to the brim, the water level at any Pk is equal to the height of the tank, that is, 50 cm.
Example
Input
2
5
15 40
35 20
50 45
70 30
80 10
3
20 3
60 2
65 2
6
40 4100
25 7500
10 18000
90 7000
25 15000
25 22000
5
15 40
35 20
50 45
70 30
80 10
2
60 4
75 1
3
60 6000
75 6000
85 6000
Output
0.666667
21.4286
36.6667
11.1111
40.0
50.0
30.0
13.3333
13.3333 | instruction | 0 | 85,663 | 3 | 171,326 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N = int(readline())
W = [(0, 50)]
for i in range(N):
b, h = map(int, readline().split())
W.append((b, h))
W.append((100, 50))
M = int(readline())
S = [0]*(N+1)
Q = []
for i in range(M):
f, a = map(int, readline().split())
for j in range(N+1):
if W[j][0] < f < W[j+1][0]:
S[j] += a
break
L = int(readline())
for i in range(L):
p, t = map(int, readline().split())
Q.append((t, i, p))
Q.sort(reverse=1)
T = [0]*(N+1)
C = N+1
tc = 0
ans = [50] * L
EPS = 1e-9
while 1:
k = -1; tm = 10**18
for i in range(C):
if S[i] == 0:
continue
b0, h0 = W[i]; b1, h1 = W[i+1]
t = (min(h0, h1) * (b1 - b0) * 30 - T[i]) / S[i]
if t < tm:
tm = t
k = i
assert k != -1
b0, h0 = W[k]; b1, h1 = W[k+1]
dt = (min(h0, h1) * (b1 - b0) * 30 - T[k]) / S[k]
while Q and tc <= Q[-1][0] < tc + dt:
t, i, p = Q.pop()
for j in range(C):
ba, ha = W[j]; bb, hb = W[j+1]
if ba < p < bb:
dt0 = t - tc
ans[i] = (S[j] * dt0 + T[j]) / ((bb - ba) * 30)
break
for i in range(C):
T[i] += S[i] * dt
if C == 1:
break
if h0 < h1:
if abs(T[k-1] - h0 * (b0 - W[k-1][0]) * 30) < EPS:
assert S[k-1] == 0
S[k-1] = S[k]
T[k-1] += T[k]
S.pop(k); T.pop(k); W.pop(k)
C -= 1
else:
j = k-1
while T[j] == W[j][1]:
j -= 1
S[j] += S[k]
S[k] = 0
else:
if abs(T[k+1] - h1 * (W[k+2][0] - b1) * 30) < EPS:
assert S[k+1] == 0
S[k+1] = S[k]
T[k+1] += T[k]
S.pop(k); T.pop(k); W.pop(k+1)
C -= 1
else:
j = k+1
while T[j] == W[j+1][1]:
j += 1
S[j] += S[k]
S[k] = 0
tc += dt
for i in range(L):
write("%.16f\n" % ans[i])
D = int(readline())
for i in range(D):
solve()
``` | output | 1 | 85,663 | 3 | 171,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,814 | 3 | 171,628 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
def add(sms, cns, n, v, x):
while v < n:
sms[v] += x
cns[v] += 1
v += v & ~(v - 1)
def sumtoo(sms, cns, v):
sm = 0
cn = 0
while v > 0:
sm += sms[v]
cn += cns[v]
v -= v & ~(v - 1)
return sm,cn
def go():
n = int(input())
x = list(map(int, input().split()))
v = list(map(int, input().split()))
sv = sorted(set(v))
sv = {v: k for (k, v) in enumerate(sv,1)}
v = [sv[v] for v in v]
# mnv, mxv = min(v), max(v)
# dfv = mxv - mnv
a = sorted(((xx, vv) for xx, vv in zip(x, v)), reverse=True)
len_sv = len(sv)+1
sms = [0] * len_sv
cns = [0] * len_sv
res = 0
sx = 0
for cnt, (xx, vv) in enumerate(a):
sm, cn = sumtoo(sms,cns,vv-1)
# sm, cn = tree.sum_to(vv - mnv)
sm, cn = sx - sm, cnt - cn
res += sm - xx * cn
add(sms, cns, len_sv, vv, xx)
# tree.add(vv - mnv, xx)
sx += xx
return res
print(go())
``` | output | 1 | 85,814 | 3 | 171,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,815 | 3 | 171,630 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
import operator
import collections
from sys import stdin
N = int(input())
pos = list(map(int, stdin.readline().split()))
speed = list(map(int, stdin.readline().split()))
A = []
for i in range(N):
A.append((pos[i], speed[i]))
# organized A by position from least to greatest
# We want to count, at each x_i, the number of j such that x_j < x_i and
# v_j <= x_i
dist = {}
for i in pos:
dist[i] = 0
A.sort(key = operator.itemgetter(0))
# Now we do merge sort on the speeds but with comparing...
# if right is chosen over left, then we record [right] += right.pos - left.pos
count = [0]
def merge_compare(p, r):
if p < r:
q = (r + p)//2
merge_compare(p, q)
merge_compare(q + 1, r)
merge(p, q, r)
# A is POSITION, SPEED
# merging two sorted subsets A[p:q + 1], A[q + 1:r + 1] into a sorted
# subset A[p:r + 1]
def merge(p, q, r):
n_1 = q - p + 1
n_2 = r - q
temp = []
L = A[p:q + 1]
R = A[q + 1:r + 1]
i = 0
j = 0
# calculatin the sum of all the positions in R
sum_pos_right = 0
for e, f in R:
sum_pos_right += e
while i < n_1 and j < n_2:
if L[i][1] <= R[j][1]:
temp.append(L[i])
count[0] += sum_pos_right - L[i][0]*(n_2 - j)
i += 1
else:
temp.append(R[j])
sum_pos_right -= R[j][0]
j += 1
if i == n_1:
temp += R[j:]
else:
temp += L[i:]
A[p:r + 1] = temp
merge_compare(0, N - 1)
print (count[0])
``` | output | 1 | 85,815 | 3 | 171,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,816 | 3 | 171,632 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
#from bisect import bisect_left as bl, bisect_right as br, insort
#from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
#INF = float('inf')
mod = int(1e9)+7
def update(BIT,v, w):
while v <= n:
BIT[v] += w
v += (v & (-v))
def getvalue(BIT,v):
ANS = 0
while v != 0:
ANS += BIT[v]
v -= (v & (-v))
return ANS
def bisect_on_BIT(BIT,x):
if x <= 0:
return 0
ANS = 0
h = 1 << (n - 1)
while h > 0:
if ANS + h <= n and BIT[ANS + h] < x:
x -= BIT[ANS + h]
ANS += h
h //= 2
return ANS + 1
n=int(data())
x=mdata()
v=mdata()
a=[[v[i],x[i]] for i in range(n)]
a.sort()
s=[0]*(n+1)
cnt=[0]*(n+1)
pos=dd(int)
for i in range(n):
pos[a[i][1]]=i+1
for i in range(n):
update(s,i+1,a[i][1])
update(cnt,i+1,1)
ans=0
x.sort()
for i in range(n):
ind=pos[x[i]]
ans+=getvalue(s,n)-getvalue(s,ind)-(getvalue(cnt,n)-getvalue(cnt,ind))*x[i]
update(s,ind,-x[i])
update(cnt,ind,-1)
out(ans)
``` | output | 1 | 85,816 | 3 | 171,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,817 | 3 | 171,634 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import Counter
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
# sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
class BIT:
def __init__(self, n):
n += 1
nv = 1
while nv < n:
nv *= 2
self.size = nv
self.tree = [0] * nv
def sum(self, i):
s = 0
i += 1
while i > 0:
s += self.tree[i-1]
i -= i & -i
return s
def add(self, i, x):
i += 1
while i <= self.size:
self.tree[i-1] += x
i += i & -i
def get(self, l, r=None):
if r is None: r = l + 1
res = 0
if r: res += self.sum(r-1)
if l: res -= self.sum(l-1)
return res
def update(self, i, x):
self.add(i, x - self.get(i))
def compress(S):
zipped, unzipped = {}, {}
for i, a in enumerate(sorted(S)):
zipped[a] = i
unzipped[i] = a
return zipped, unzipped
N = INT()
A = LIST()
B = LIST()
BA = list(zip(B, A))
BA.sort()
zipped, _ = compress(set(A))
bit1 = BIT(N)
bit2 = BIT(N)
ans = 0
for b, a in BA:
sm = bit1.sum(zipped[a])
cnt = bit2.sum(zipped[a])
ans += cnt * a - sm
bit1.add(zipped[a], a)
bit2.add(zipped[a], 1)
print(ans)
``` | output | 1 | 85,817 | 3 | 171,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,818 | 3 | 171,636 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
from bisect import bisect_right, bisect_left
# instead of AVLTree
class BITbisect():
def __init__(self, InputProbNumbers):
# 座圧
self.ind_to_co = [-10**18]
self.co_to_ind = {}
for ind, num in enumerate(sorted(list(set(InputProbNumbers)))):
self.ind_to_co.append(num)
self.co_to_ind[num] = ind+1
self.max = len(self.co_to_ind)
self.data = [0]*(self.max+1)
def __str__(self):
retList = []
for i in range(1, self.max+1):
x = self.ind_to_co[i]
if self.count(x):
c = self.count(x)
for _ in range(c):
retList.append(x)
return "[" + ", ".join([str(a) for a in retList]) + "]"
def __getitem__(self, key):
key += 1
s = 0
ind = 0
l = self.max.bit_length()
for i in reversed(range(l)):
if ind + (1<<i) <= self.max:
if s + self.data[ind+(1<<i)] < key:
s += self.data[ind+(1<<i)]
ind += (1<<i)
if ind == self.max or key < 0:
raise IndexError("BIT index out of range")
return self.ind_to_co[ind+1]
def __len__(self):
return self._query_sum(self.max)
def __contains__(self, num):
if not num in self.co_to_ind:
return False
return self.count(num) > 0
# 0からiまでの区間和
# 左に進んでいく
def _query_sum(self, i):
s = 0
while i > 0:
s += self.data[i]
i -= i & -i
return s
# i番目の要素にxを足す
# 上に登っていく
def _add(self, i, x):
while i <= self.max:
self.data[i] += x
i += i & -i
# 値xを挿入
def push(self, x):
if not x in self.co_to_ind:
raise KeyError("The pushing number didnt initialized")
self._add(self.co_to_ind[x], 1)
# 値xを削除
def delete(self, x):
if not x in self.co_to_ind:
raise KeyError("The deleting number didnt initialized")
if self.count(x) <= 0:
raise ValueError("The deleting number doesnt exist")
self._add(self.co_to_ind[x], -1)
# 要素xの個数
def count(self, x):
return self._query_sum(self.co_to_ind[x]) - self._query_sum(self.co_to_ind[x]-1)
# 値xを超える最低ind
def bisect_right(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_right(self.ind_to_co, x) - 1
return self._query_sum(i)
# 値xを下回る最低ind
def bisect_left(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_left(self.ind_to_co, x)
if i == 1:
return 0
return self._query_sum(i-1)
# 足す時はi番目に足し、返すのは累積和
class sumBIT():
def __init__(self, N):
self.N = N
self.bit = [0 for _ in range(self.N+1)]
def __str__(self):
ret = []
for i in range(1, self.N+1):
ret.append(self.__getitem__(i))
return "[" + ", ".join([str(a) for a in ret]) + "]"
def __getitem__(self, i):
s = 0
while i > 0:
s += self.bit[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.N:
self.bit[i] += x
i += i & -i
import sys
input = sys.stdin.buffer.readline
from operator import itemgetter
def main():
N = int(input())
X = list(map(int, input().split()))
V = list(map(int, input().split()))
XV = [(x, v) for x, v in zip(X, V)]
XV.sort()
XV.sort(key=itemgetter(1))
co_to_ind1 = {}
#co_to_ind2 = {}
for i, (x, v) in enumerate(XV):
co_to_ind1[x] = i+1
XV.sort()
sumbit = sumBIT(N+1)
bit = sumBIT(N+1)
ans = 0
for x, v in XV:
ind1 = co_to_ind1[x]
#ind2 = co_to_ind2[v]
count = bit[ind1]
s = sumbit[ind1]
ans += count*x - s
bit.add(ind1, 1)
sumbit.add(ind1, x)
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 85,818 | 3 | 171,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,819 | 3 | 171,638 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
import bisect
def getsum(tree , i):
s = 0
i += 1
while i>0:
s += tree[i]
i -= i & (-i)
return s
def updatebit(tree , n , i , v):
i+= 1
while i <= n:
tree[i] += v
i += i & (-i)
n = int(input())
x = list(map(int , input().split()))
v = list(map(int , input().split()))
p = [[x[i] , v[i]] for i in range(len(x))]
vs = sorted(list(set(v)))
p = sorted(p , key = lambda i : i[0])
l = len(vs)
cnt = [0]*(l+1)
xs = [0]*(l+1)
ans = 0
for pnt in p:
pos = bisect.bisect_left(vs , pnt[1])
ans += getsum(cnt , pos) * pnt[0] - getsum(xs , pos)
updatebit(cnt , l , pos , 1)
updatebit(xs , l , pos , pnt[0])
print(ans)
``` | output | 1 | 85,819 | 3 | 171,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,820 | 3 | 171,640 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
import bisect
import sys
class ft:
def __init__(self,n):
self.a = [0]*(200000)
self.n = n
def qry(self,r):
ret = 0
while r>=0:
ret+=self.a[r]
r=(r&(r+1))-1
return ret
def upd(self,i,v):
while i<self.n:
self.a[i]+=v
i=(i|(i+1))
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
n = int(input())
x = list(map(int,input().split()))
v = list(map(int,input().split()))
f0 = ft(n)
f1 = ft(n)
a1 = [[] for _ in range(n)]
for i in range(n):
a1[i].append(x[i])
a1[i].append(v[i])
a1.sort()
v.sort()
ans = 0
for i in range(n):
p = bisect.bisect_left(v,a1[i][1])
ans+=(f0.qry(p)*a1[i][0])-(f1.qry(p));
f0.upd(p,1)
f1.upd(p,a1[i][0])
print(ans)
``` | output | 1 | 85,820 | 3 | 171,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0 | instruction | 0 | 85,821 | 3 | 171,642 |
Tags: data structures, divide and conquer, implementation, sortings
Correct Solution:
```
class SegmentTree():
def __init__(self,N,func,initialRes=0):
self.f=func
self.N=N
self.tree=[0 for _ in range(4*self.N)]
self.initialRes=initialRes
# for i in range(self.N):
# self.tree[self.N+i]=arr[i]
# for i in range(self.N-1,0,-1):
# self.tree[i]=self.f(self.tree[i<<1],self.tree[i<<1|1])
def updateTreeNode(self,idx,value): #update value at arr[idx]
self.tree[idx+self.N]=value
idx+=self.N
i=idx
while i>1:
self.tree[i>>1]=self.f(self.tree[i],self.tree[i^1])
i>>=1
def query(self,l,r): #get sum (or whatever function) on interval [l,r] inclusive
r+=1
res=self.initialRes
l+=self.N
r+=self.N
while l<r:
if l&1:
res=self.f(res,self.tree[l])
l+=1
if r&1:
r-=1
res=self.f(res,self.tree[r])
l>>=1
r>>=1
return res
def getMaxSegTree(arr):
return SegmentTree(arr,lambda a,b:max(a,b),initialRes=-float('inf'))
def getMinSegTree(arr):
return SegmentTree(arr,lambda a,b:min(a,b),initialRes=float('inf'))
def getSumSegTree(arr):
return SegmentTree(arr,lambda a,b:a+b,initialRes=0)
def main():
n=int(input())
xes=readIntArr()
ves=readIntArr()
#perform coordinate compression on xes. Compressed values shall be the indexes of segment tree
xes2=list(sorted(xes)) #xes2[compressed]=x
xTox2Map=dict()
for i,x in enumerate(xes2):
xTox2Map[x]=i #xTox2Map[x]=i
arr=[] #[original x, compressed x, v]
for i in range(n):
arr.append([xes[i],xTox2Map[xes[i]],ves[i]])
arr.sort(key=lambda x:(x[2],x[0])) #sort by v asc, then position asc
st=getSumSegTree(n) #for sum of xes
stCnts=getSumSegTree(n) #for counts
# print(arr)
ans=0
for originalX,compressedX,v in arr:
smallerSums=st.query(0,compressedX)
smallerCounts=stCnts.query(0,compressedX)
ans+=originalX*smallerCounts-smallerSums
st.updateTreeNode(compressedX,originalX) #update the tree
stCnts.updateTreeNode(compressedX,1)
# print(ans)
print(ans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# import sys
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
main()
``` | output | 1 | 85,821 | 3 | 171,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0
Submitted Solution:
```
class Points:
def __init__(self, x = 0, v = 0):
self.x, self.v = x, v
class Fenwick_tree:
def __init__(self, n = 0):
self.n = n
self.bit = [0] * (n + 1)
def update(self, x, value):
while x <= self.n:
self.bit[x] += value
x += x & -x
def get(self, x):
sum = 0;
while x > 0:
sum += self.bit[x];
x -= x & -x
return sum
# Input:
n, x, v = int(input()), list(map(int,input().split())), list(map(int,input().split()))
a = [Points(x[i], v[i]) for i in range(n)]
a.sort(key = lambda value: value.x)
# Compress data:
v = [a[i].v for i in range(n)]
v.sort()
data = {}
cnt = 1
for i in range(n):
if not data.__contains__(v[i]):
data[v[i]] = cnt
cnt += 1
# Solve the problem:
cnt = Fenwick_tree(n)
sum = Fenwick_tree(n)
res, i = 0, 0
for i in range(n):
a[i].v = data[a[i].v]
res += a[i].x * cnt.get(a[i].v) - sum.get(a[i].v)
cnt.update(a[i].v, 1)
sum.update(a[i].v, a[i].x)
print(res)
``` | instruction | 0 | 85,822 | 3 | 171,644 |
Yes | output | 1 | 85,822 | 3 | 171,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
X=list(map(int,input().split()))
V=list(map(int,input().split()))
XV=[(X[i],V[i]) for i in range(n)]
#compression_dict_x={a: ind for ind, a in enumerate(sorted(set(X)))}
compression_dict_v={a: ind+2 for ind, a in enumerate(sorted(set(V)))}
XV=[(XV[i][0], compression_dict_v[XV[i][1]]) for i in range(n)]
XV.sort(reverse=True)
LEN=len(compression_dict_v)+3
BIT1=[0]*(LEN+1)
def update1(v,w):
while v<=LEN:
BIT1[v]+=w
v+=(v&(-v))
def getvalue1(v):
ANS=0
while v!=0:
ANS+=BIT1[v]
v-=(v&(-v))
return ANS
BIT2=[0]*(LEN+1)
def update2(v,w):
while v<=LEN:
BIT2[v]+=w
v+=(v&(-v))
def getvalue2(v):
ANS=0
while v!=0:
ANS+=BIT2[v]
v-=(v&(-v))
return ANS
ANS=0
for x,v in XV:
ANS+=(getvalue2(LEN)-getvalue2(v-1))-(getvalue1(LEN)-getvalue1(v-1))*x
#print(getvalue2(LEN),getvalue2(v-1),getvalue1(LEN),getvalue1(v-1))
#print(x,v,ANS)
update1(v,1)
update2(v,x)
print(ANS)
``` | instruction | 0 | 85,823 | 3 | 171,646 |
Yes | output | 1 | 85,823 | 3 | 171,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.
Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.
Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.
The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.
Output
Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).
Examples
Input
3
1 3 2
-100 2 3
Output
3
Input
5
2 1 4 3 5
2 2 2 3 4
Output
19
Input
2
2 1
-3 0
Output
0
Submitted Solution:
```
import math
import sys
from collections import defaultdict, Counter
from itertools import groupby
#input = sys.stdin.readline
rt = lambda: map(int, input().split())
ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n+1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & (-i)
return s
def add(self, i, val):
while i <= self.size:
self.tree[i] += val
i += i & (-i)
def main():
n = ri()
x = rl()
v = rl()
pts = [(xi, vi) for xi, vi in zip(x, v)]
pts.sort(key=lambda a: (a[1], a[0]))
comprX = {val: i+1 for i, val in enumerate(sorted(x))}
bit_x = Bit(n+1)
bit_cnt = Bit(n+1)
res = 0
for xi, vi in pts:
compr = comprX[xi]
res += bit_cnt.sum(compr)*xi - bit_x.sum(compr)
bit_x.add(compr, xi)
bit_cnt.add(compr, 1)
print(res)
if __name__ == '__main__':
main()
``` | instruction | 0 | 85,824 | 3 | 171,648 |
Yes | output | 1 | 85,824 | 3 | 171,649 |
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