text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 β€ n β€ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 103, a1 + a2 + ... + an β€ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 β€ m β€ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 β€ qi β€ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
# Worms
def fst(tuple):
return tuple[0]
def snd(tuple):
return tuple[1]
n = input()
piles = list(map(int, input().split()))
w = input()
worms = list(map(int, input().split()))
piles[0] = (1, piles[0])
ans = [0 for i in range(10**6 + 2)]
for i in range(len(piles)):
if i > 0:
size = piles[i]
piles[i] = (snd(piles[i-1]) + 1, snd(piles[i-1]) + size)
for j in range(fst(piles[i]), snd(piles[i]) + 1):
ans[j] = i + 1
for w in worms:
print(ans[w])
```
Yes
| 5,400 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 β€ n β€ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 103, a1 + a2 + ... + an β€ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 β€ m β€ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 β€ qi β€ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
from bisect import bisect_left
from itertools import accumulate
n = int(input())
a = list(map(int, input().split()))
m = int(input())
q = list(map(int, input().split()))
sums = list(accumulate(a))
ans = list()
for e in q:
print(bisect_left(sums,e)+1)
```
Yes
| 5,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 β€ n β€ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 103, a1 + a2 + ... + an β€ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 β€ m β€ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 β€ qi β€ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
n = int(input())
piles = [int(a) for a in input().split()]
q = int(input())
queries = [int(b) for b in input().split()]
worms = {}
i = 1
for p in range(n):
for k in range(piles[p]):
worms[i] = p+1
i+=1
for j in queries:
print(worms[j])
```
Yes
| 5,402 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 β€ n β€ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 103, a1 + a2 + ... + an β€ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 β€ m β€ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 β€ qi β€ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
def find(arr, value):
low = 0
high = len(arr) - 1
while high - low != 0:
med = int((low + high) / 2)
if value > arr[med]:
low = med
else:
high = med
if high - low == 1:
if value < arr[low]:
return low + 1
else:
return high + 1
n = int(input())
L = list(map(int, input().split(' ')))
R = []
R.append(L[0])
for i in range(1,n):
R.append(R[i-1] + L[i])
m = int(input())
L = list(map(int, input().split(' ')))
for x in L:
print(find(R, x))
```
No
| 5,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 β€ n β€ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 103, a1 + a2 + ... + an β€ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 β€ m β€ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 β€ qi β€ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
n = int(input())
sp = list(map(int, input().split()))
input()
for el in map(int, input().split()):
L = 0
R = n
while R - L > 1:
M = (R + L) // 2
if sum(sp[:M]) > el:
R = M
else:
L = M
print(R)
```
No
| 5,404 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 β€ n β€ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 103, a1 + a2 + ... + an β€ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 β€ m β€ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 β€ qi β€ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
juicy = int(input())
num = list(map(int,input().split()))
b = num.copy()
num.sort()
dictionary = {}
temp = 0
j = 0
for i in range(len(a)):
if temp+a[i] < num[j]:
temp = temp+a[i]
else :
temp = temp+a[i]
while temp >= num[j]:
dictionary[num[j]] = i+1
if j < len(b)-1:
j = j+1
elif j == len(b)-1:
break
for i in range(len(b)):
print(dictionary[b[i]])
```
No
| 5,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 β€ n β€ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ 103, a1 + a2 + ... + an β€ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 β€ m β€ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 β€ qi β€ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
n = int(input())
an = [int(x) for x in input().split()]
an.insert(0,0)
m = int(input())
qn = [int(x) for x in input().split()]
sum = [0 for i in range(len(an))]
for i in range(1,len(sum)):
sum[i] = an[i] + sum[i - 1]
print(sum)
for q in qn:
for x in range(1, len(sum)):
if sum[x - 1] < q < sum[x] or q == sum[x]:
print(x)
break
```
No
| 5,406 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
x, y = map(int, input().split())
u, v = map(int, input().split())
s = 0
for i in range(int(input())):
a, b, c = map(int, input().split())
s += (a * x + b * y + c > 0) ^ (a * u + b * v + c > 0)
print(s)
```
| 5,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
x1,y1=[int(i) for i in input().split()]
x2,y2=[int(i) for i in input().split()]
n=int(input())
r=0
for i in range(n):
a,b,c = [int(i) for i in input().split()]
if (a*x1+b*y1+c)*(a*x2+b*y2+c)<0:
r+=1
print(r)
```
| 5,408 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
x,y = map(int,input().split())
x1,y1 = map(int,input().split())
n=int(input())
ans=0
for i in range(n):
a,b,c=map(int,input().split())
if((a*x+b*y+c)*(a*x1+b*y1+c)<0): ans+=1
print(ans)
```
| 5,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
x1,y1=map(int,input().split())
x2,y2=map(int,input().split())
a1=y1-y2
b1=x2-x1
c1=x2*(y2-y1)-y2*(x2-x1)
def intersect(a2,b2,c2):
global a1,b1,c1,x1,y1,x2,y2
if(a1*b2==a2*b1):
return False
x=(b1*c2-b2*c1)/(a1*b2-b1*a2)
y=(a1*c2-c1*a2)/(b1*a2-a1*b2)
if(min(x1,x2)<=x<=max(x1,x2) and min(y1,y2)<=y<=max(y1,y2)):
return True
return False
m=int(input())
ans=0
for i in range(m):
a2,b2,c2=map(int,input().split())
if(intersect(a2,b2,c2)):
ans+=1
print(ans)
```
| 5,410 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
#!/usr/bin/env python3
x1, y1 = (int(x) for x in input().split())
x2, y2 = (int(x) for x in input().split())
n = int(input())
ans = 0
for i in range(n):
a, b, c = (int(x) for x in input().split())
point1 = a*x1 + b*y1 + c
point2 = a*x2 + b*y2 + c
if point1 * point2 < 0:
ans += 1
print(ans)
```
| 5,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
def getMinSteps(x1, y1, x2, y2, n, lines):
min_steps = 0
for i in range(n):
a = lines[i][0]
b = lines[i][1]
c = lines[i][2]
if (a*x1+b*y1+c)*(a*x2+b*y2+c) < 0:
min_steps+=1
return min_steps
x1, y1 = [int(x) for x in input().split()]
x2, y2 = [int(x) for x in input().split()]
n = int(input())
lines = []
for i in range(n):
line = [int(x) for x in input().split()]
lines.append((line[0], line[1], line[2]))
print(getMinSteps(x1, y1, x2, y2, n, lines))
```
| 5,412 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
hx, hy = map(int, input().split())
ux, uy = map(int, input().split())
n = int(input())
count = 0
for i in range(n):
a,b,c = map(int, input().split())
h = a*hx + b*hy + c
u = a*ux + b*uy + c
if h*u<0:
count += 1
print(count)
```
| 5,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Tags: geometry
Correct Solution:
```
h1, h2 = list(map(int,input().split()))
u1, u2 = list(map(int,input().split()))
pasos = 0
for i in int(input())*'_':
a, b, c= list(map(int,input().split()))
if (a * h1 + b * h2 + c) * (a * u1 + b * u2 + c) < 0:
pasos+=1
print(pasos)
```
| 5,414 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
homex,homey = (int(x) for x in input().split(" "))
univx,univy = (int(x) for x in input().split(" "))
n = int(input())
step = 0
for i in range(0,n):
a,b,c = (int(x) for x in input().split(" "))
homearea = ((a*homex + b*homey + c) < 0)
univarea = ((a*univx + b*univy + c) < 0)
if homearea != univarea:
step += 1
print(step)
```
Yes
| 5,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
#import sys
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
def norm(a, b, c, d):
if (b * c >= 0):
return a * abs(c) <= abs(b) <= d * abs(c)
return -a * abs(c) >= abs(b) >= -d * abs(c)
x1, y1 = map(int, input().split())
x2, y2 = map(int, input().split())
a2 = y2 - y1
b2 = x1 - x2
c2 = 0 - a2 * x1 - b2 * y1
n = int(input())
ans = 0
for i in range(n):
a1, b1, c1 = map(int, input().split())
if (norm(min(x1, x2), (c2 * b1 - c1 * b2), (a1 * b2 - a2 * b1), max(x1, x2)) and norm(min(y1, y2), (c2 * a1 - c1 * a2), (a2 * b1 - a1 * b2), max(y1, y2))):
ans += 1
print(ans)
```
Yes
| 5,416 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
x1,y1 = map(int,input().split())
x2,y2 = map(int,input().split())
n = int(input())
s = 0
for i in range(n):
a,b,c = map(int,input().split())
if ((((a * x1) + (b * y1) + c) * ((a * x2) + (b * y2) + c)) < 0):
s = s + 1
print(s)
```
Yes
| 5,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
xhome, yhome = [int(x) for x in input().split()]
xuni, yuni = [int(x) for x in input().split()]
n_roads = int(input())
n_steps = 0
for i in range(n_roads):
a, b, c = [int(x) for x in input().split()]
hline = (a*xhome) + (b*yhome) + c
uline = (a*xuni) + (b*yuni) + c
if hline * uline < 0:
n_steps += 1
print(n_steps)
```
Yes
| 5,418 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
x1, y1 = map(int, input().split())
x2, y2 = map(int, input().split())
n = int(input())
steps = 0
for i in range(n):
a, b, c = map(int, input().split())
if ((a*x1 + b*y1 + c < 0) != (a*x2 + a*y2 + c < 0)):
steps += 1
print(steps)
```
No
| 5,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
from __future__ import division
def line(p1, p2):
A = (p1[1] - p2[1])
B = (p2[0] - p1[0])
C = (p1[0]*p2[1] - p2[0]*p1[1])
return [A, B, -C]
def intersection(L1, L2):
D = L1[0] * L2[1] - L1[1] * L2[0]
Dx = L1[2] * L2[1] - L1[1] * L2[2]
Dy = L1[0] * L2[2] - L1[2] * L2[0]
if D != 0:
x = Dx / D
y = Dy / D
return (x,y)
else:
return False
def is_on(a, b, c):
return (collinear(a, b, c)
and (within(a[0], c[0], b[0]) if a[0] != b[0] else
within(a[1], c[1], b[1])))
def collinear(a, b, c):
return (b[0] - a[0]) * (c[1] - a[1]) == (c[0] - a[0]) * (b[1] - a[1])
def within(p, q, r):
return p <= q <= r or r <= q <= p
def main():
x1, y1 = map(int,input().split())
x2, y2 = map(int,input().split())
path = line([x1, y1],[x2,y2])
n = int(input())
roads = []
count = 0
intersection_points = []
for _ in range(n):
temp = list(map(int,input().split()))
road = [temp[0], temp[1], -temp[2]]
intersection_point = intersection(road, path)
#print(intersection_point)
if intersection_point:
if intersection_point in intersection_points:
count += 1
elif is_on([x1, y1], [x2, y2], intersection_point):
count += 1
intersection_points.append(intersection_point)
print(count)
if __name__ == '__main__':
main()
```
No
| 5,420 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
#498A
[x1,y1] = list(map(int,input().split()))
[x2,y2] = list(map(int,input().split()))
n = int(input())
c = 0
for i in range(n):
[a,b,c] = list(map(int,input().split()))
if (a*x1+b*x2+c)*(a*x2+b*y2+c) < 0:
c += 1
print(c)
```
No
| 5,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and bi are not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.
Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).
Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.
Input
The first line contains two space-separated integers x1, y1 ( - 106 β€ x1, y1 β€ 106) β the coordinates of your home.
The second line contains two integers separated by a space x2, y2 ( - 106 β€ x2, y2 β€ 106) β the coordinates of the university you are studying at.
The third line contains an integer n (1 β€ n β€ 300) β the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 β€ ai, bi, ci β€ 106; |ai| + |bi| > 0) β the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).
Output
Output the answer to the problem.
Examples
Input
1 1
-1 -1
2
0 1 0
1 0 0
Output
2
Input
1 1
-1 -1
3
1 0 0
0 1 0
1 1 -3
Output
2
Note
Pictures to the samples are presented below (A is the point representing the house; B is the point representing the university, different blocks are filled with different colors):
<image> <image>
Submitted Solution:
```
px,py = map(int,input().split())
qx,qy = map(int,input().split())
n,s= int(input()),0
for _ in range(n):
a,b,c = map(int,input().split())
if not a:
x,y = -c/b,0
else:
x,y = 0,-c/a
s += ((px-x)*a+(py-y)*b)*((qx-x)*a+(qy-y)*b)<0
print(s)
# C:\Users\Usuario\HOME2\Programacion\ACM
```
No
| 5,422 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once Vasya and Petya assembled a figure of m cubes, each of them is associated with a number between 0 and m - 1 (inclusive, each number appeared exactly once). Let's consider a coordinate system such that the OX is the ground, and the OY is directed upwards. Each cube is associated with the coordinates of its lower left corner, these coordinates are integers for each cube.
The figure turned out to be stable. This means that for any cube that is not on the ground, there is at least one cube under it such that those two cubes touch by a side or a corner. More formally, this means that for the cube with coordinates (x, y) either y = 0, or there is a cube with coordinates (x - 1, y - 1), (x, y - 1) or (x + 1, y - 1).
Now the boys want to disassemble the figure and put all the cubes in a row. In one step the cube is removed from the figure and being put to the right of the blocks that have already been laid. The guys remove the cubes in such order that the figure remains stable. To make the process more interesting, the guys decided to play the following game. The guys take out the cubes from the figure in turns. It is easy to see that after the figure is disassembled, the integers written on the cubes form a number, written in the m-ary positional numerical system (possibly, with a leading zero). Vasya wants the resulting number to be maximum possible, and Petya, on the contrary, tries to make it as small as possible. Vasya starts the game.
Your task is to determine what number is formed after the figure is disassembled, if the boys play optimally. Determine the remainder of the answer modulo 109 + 9.
Input
The first line contains number m (2 β€ m β€ 105).
The following m lines contain the coordinates of the cubes xi, yi ( - 109 β€ xi β€ 109, 0 β€ yi β€ 109) in ascending order of numbers written on them. It is guaranteed that the original figure is stable.
No two cubes occupy the same place.
Output
In the only line print the answer to the problem.
Examples
Input
3
2 1
1 0
0 1
Output
19
Input
5
0 0
0 1
0 2
0 3
0 4
Output
2930
Tags: games, greedy, implementation
Correct Solution:
```
import heapq
def coor_neighbor(coor, dxs, dys):
x, y = coor
for dx in dxs:
for dy in dys:
yield x + dx, y + dy
def coor_bottoms(coor):
return coor_neighbor(coor, (-1, 0, 1), (-1, ))
def coor_tops(coor):
return coor_neighbor(coor, (-1, 0, 1), (1, ))
def coor_sibs(coor):
return coor_neighbor(coor, (-2, -1, 1, 2), (0, ))
class Figure:
def __init__(self, coors):
self._coors = dict()
self._stables_min = []
self._stables_max = []
self._pushed = set()
self._dropped = set()
cubes = dict()
self._bots = dict()
self._tops = dict()
for idx, coor in enumerate(coors):
cubes[coor] = idx
self._coors[idx] = coor
self._bots[idx] = set()
self._tops[idx] = set()
coor_set = set(coors)
for idx, coor in enumerate(coors):
for bottom in coor_bottoms(coor):
if bottom in coor_set:
self._bots[idx].add(cubes[bottom])
for top in coor_tops(coor):
if top in coor_set:
self._tops[idx].add(cubes[top])
for idx in self._coors:
if self.isdroppable(idx):
self.push(idx)
def sibs(self, idx):
for top_idx in self._tops[idx]:
for sib_idx in self._bots[top_idx]:
if sib_idx not in self._dropped:
yield sib_idx
def bottom_count(self, idx):
return len(self._bots[idx])
def isdroppable(self, idx):
return all(len(self._bots[top_idx]) > 1 for top_idx in self._tops[idx])
def push(self, idx):
if idx not in self._pushed:
heapq.heappush(self._stables_min, idx)
heapq.heappush(self._stables_max, -idx)
self._pushed.add(idx)
def unpush(self, idx):
if idx in self._pushed:
self._pushed.remove(idx)
def drop(self, idx):
if idx not in self._pushed:
return False
self._pushed.remove(idx)
self._dropped.add(idx)
for bot_idx in self._bots[idx]:
self._tops[bot_idx].remove(idx)
for top_idx in self._tops[idx]:
self._bots[top_idx].remove(idx)
coor = self._coors[idx]
for bot_idx in self._bots[idx]:
if self.isdroppable(bot_idx):
self.push(bot_idx)
for sib_idx in self.sibs(idx):
if not self.isdroppable(sib_idx):
self.unpush(sib_idx)
return True
def drop_min(self):
while True:
if not self._stables_min:
return None
min_idx = heapq.heappop(self._stables_min)
if self.drop(min_idx):
return min_idx
def drop_max(self):
while True:
if not self._stables_max:
return None
max_idx = - heapq.heappop(self._stables_max)
if self.drop(max_idx):
return max_idx
def __bool__(self):
return len(self._coors) != len(self._dropped)
def input_tuple():
return tuple(map(int, input().split()))
def result_add(result, base, num):
return (result * base + num) % (10 ** 9 + 9)
N = int(input())
coors = [input_tuple() for _ in range(N)]
figure = Figure(coors)
result = 0
while True:
if not figure:
break
result = result_add(result, N, figure.drop_max())
if not figure:
break
result = result_add(result, N, figure.drop_min())
print(result)
```
| 5,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya and Petya assembled a figure of m cubes, each of them is associated with a number between 0 and m - 1 (inclusive, each number appeared exactly once). Let's consider a coordinate system such that the OX is the ground, and the OY is directed upwards. Each cube is associated with the coordinates of its lower left corner, these coordinates are integers for each cube.
The figure turned out to be stable. This means that for any cube that is not on the ground, there is at least one cube under it such that those two cubes touch by a side or a corner. More formally, this means that for the cube with coordinates (x, y) either y = 0, or there is a cube with coordinates (x - 1, y - 1), (x, y - 1) or (x + 1, y - 1).
Now the boys want to disassemble the figure and put all the cubes in a row. In one step the cube is removed from the figure and being put to the right of the blocks that have already been laid. The guys remove the cubes in such order that the figure remains stable. To make the process more interesting, the guys decided to play the following game. The guys take out the cubes from the figure in turns. It is easy to see that after the figure is disassembled, the integers written on the cubes form a number, written in the m-ary positional numerical system (possibly, with a leading zero). Vasya wants the resulting number to be maximum possible, and Petya, on the contrary, tries to make it as small as possible. Vasya starts the game.
Your task is to determine what number is formed after the figure is disassembled, if the boys play optimally. Determine the remainder of the answer modulo 109 + 9.
Input
The first line contains number m (2 β€ m β€ 105).
The following m lines contain the coordinates of the cubes xi, yi ( - 109 β€ xi β€ 109, 0 β€ yi β€ 109) in ascending order of numbers written on them. It is guaranteed that the original figure is stable.
No two cubes occupy the same place.
Output
In the only line print the answer to the problem.
Examples
Input
3
2 1
1 0
0 1
Output
19
Input
5
0 0
0 1
0 2
0 3
0 4
Output
2930
Submitted Solution:
```
import heapq
def coor_neighbor(coor, dy):
x, y = coor
if y + dy < 0:
return
yield x - 1, y + dy
yield x, y + dy
yield x + 1, y + dy
def coor_bottoms(coor):
return coor_neighbor(coor, -1)
def coor_tops(coor):
return coor_neighbor(coor, 1)
class Figure:
def __init__(self, coors):
self._cubes = dict()
self._stables_min = []
self._stables_max = []
self._pushed = set()
self._dropped = set()
for idx, coor in enumerate(coors):
self._cubes[coor] = idx
for coor, idx in self._cubes.items():
if self.isdroppable(coor):
self.push(idx, coor)
def bottoms(self, coor):
return filter(self.exist, coor_bottoms(coor))
def tops(self, coor):
return filter(self.exist, coor_tops(coor))
def exist(self, coor):
return coor in self._cubes and self._cubes[coor] not in self._dropped
def bottom_count(self, coor):
return sum(self.exist(bottom) for bottom in self.bottoms(coor))
def isdroppable(self, coor):
return self.exist(coor) and all(
self.bottom_count(top) > 1 for top in self.tops(coor))
def push(self, idx, coor):
heapq.heappush(self._stables_min, (idx, coor))
heapq.heappush(self._stables_max, (-idx, coor))
self._pushed.add(idx)
def unpush(self, idx):
self._pushed.remove(idx)
def drop(self, idx, coor):
if idx not in self._pushed:
return False
self._pushed.remove(idx)
self._dropped.add(idx)
for bottom in self.bottoms(coor):
if self.isdroppable(bottom) and self._cubes[bottom] not in self._pushed:
self.push(self._cubes[bottom], bottom)
for top in self.tops(coor):
if not self.isdroppable(top) and self._cubes[top] in self._pushed:
self._pushed.remove(self._cubes[top])
return True
def drop_min(self):
while True:
if not self._stables_min:
return None
min_cube = heapq.heappop(self._stables_min)
idx, coor = min_cube
if self.drop(idx, coor):
return idx
def drop_max(self):
while True:
if not self._stables_max:
return None
max_cube = heapq.heappop(self._stables_max)
idx, coor = -max_cube[0], max_cube[1]
if self.drop(idx, coor):
return idx
def __bool__(self):
return len(self._cubes) != len(self._dropped)
def input_tuple():
return tuple(map(int, input().split()))
def result_add(result, base, num):
return (result * base + num) % (10 ** 9 + 9)
N = int(input())
coors = [input_tuple() for _ in range(N)]
figure = Figure(coors)
result = 0
while True:
if not figure:
break
result = result_add(result, N, figure.drop_max())
if not figure:
break
result = result_add(result, N, figure.drop_min())
print(result)
```
No
| 5,424 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya and Petya assembled a figure of m cubes, each of them is associated with a number between 0 and m - 1 (inclusive, each number appeared exactly once). Let's consider a coordinate system such that the OX is the ground, and the OY is directed upwards. Each cube is associated with the coordinates of its lower left corner, these coordinates are integers for each cube.
The figure turned out to be stable. This means that for any cube that is not on the ground, there is at least one cube under it such that those two cubes touch by a side or a corner. More formally, this means that for the cube with coordinates (x, y) either y = 0, or there is a cube with coordinates (x - 1, y - 1), (x, y - 1) or (x + 1, y - 1).
Now the boys want to disassemble the figure and put all the cubes in a row. In one step the cube is removed from the figure and being put to the right of the blocks that have already been laid. The guys remove the cubes in such order that the figure remains stable. To make the process more interesting, the guys decided to play the following game. The guys take out the cubes from the figure in turns. It is easy to see that after the figure is disassembled, the integers written on the cubes form a number, written in the m-ary positional numerical system (possibly, with a leading zero). Vasya wants the resulting number to be maximum possible, and Petya, on the contrary, tries to make it as small as possible. Vasya starts the game.
Your task is to determine what number is formed after the figure is disassembled, if the boys play optimally. Determine the remainder of the answer modulo 109 + 9.
Input
The first line contains number m (2 β€ m β€ 105).
The following m lines contain the coordinates of the cubes xi, yi ( - 109 β€ xi β€ 109, 0 β€ yi β€ 109) in ascending order of numbers written on them. It is guaranteed that the original figure is stable.
No two cubes occupy the same place.
Output
In the only line print the answer to the problem.
Examples
Input
3
2 1
1 0
0 1
Output
19
Input
5
0 0
0 1
0 2
0 3
0 4
Output
2930
Submitted Solution:
```
import heapq
def neighbor(coor, dy):
x, y = coor
if y + dy < 0:
return
if x - 1 >= 0:
yield x - 1, y + dy
yield x, y + dy
yield x + 1, y + dy
def bottoms(coor):
return neighbor(coor, -1)
def tops(coor):
return neighbor(coor, 1)
class Figure:
def __init__(self, coors):
self._cubes = {}
self._stables_min = []
self._stables_max = []
self._pushed = set()
self._dropped = set()
for idx, coor in enumerate(coors):
self._cubes[coor] = idx
for coor, idx in self._cubes.items():
if self.isdroppable(coor):
self.push(idx, coor)
def exist(self, coor):
return coor in self._cubes and self._cubes[coor] not in self._dropped
def bottom_count(self, coor):
return sum(self.exist(bottom) for bottom in bottoms(coor))
def isstable(self, coor):
return any(bottom in self._cubes for bottom in bottoms(coor))
def isdroppable(self, coor):
return self.exist(coor) and all(
not self.exist(top) or self.bottom_count(top) > 1
for top in tops(coor))
def push(self, idx, coor):
heapq.heappush(self._stables_min, (idx, coor))
heapq.heappush(self._stables_max, (-idx, coor))
self._pushed.add(idx)
def drop_min(self):
while True:
if not self._stables_min:
return None
min_cube = heapq.heappop(self._stables_min)
idx, coor = min_cube
if idx in self._dropped:
continue
self._dropped.add(idx)
for bottom in bottoms(coor):
if self.isdroppable(bottom) and self._cubes[bottom] not in self._pushed:
self.push(self._cubes[bottom], bottom)
return idx
def drop_max(self):
while True:
if not self._stables_max:
return None
max_cube = heapq.heappop(self._stables_max)
idx, coor = -max_cube[0], max_cube[1]
print(idx)
if idx in self._dropped:
continue
self._dropped.add(idx)
for bottom in bottoms(coor):
if self.isdroppable(bottom) and self._cubes[bottom] not in self._pushed:
self.push(self._cubes[bottom], bottom)
return idx
def __bool__(self):
return bool(len(self._cubes) - len(self._dropped))
def input_tuple():
return tuple(map(int, input().split()))
N = int(input())
coors = [input_tuple() for _ in range(N)]
figure = Figure(coors)
result = 0
while True:
if not figure:
break
result = (result * N) + figure.drop_max()
if not figure:
break
result = (result * N) + figure.drop_min()
print(result)
```
No
| 5,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya and Petya assembled a figure of m cubes, each of them is associated with a number between 0 and m - 1 (inclusive, each number appeared exactly once). Let's consider a coordinate system such that the OX is the ground, and the OY is directed upwards. Each cube is associated with the coordinates of its lower left corner, these coordinates are integers for each cube.
The figure turned out to be stable. This means that for any cube that is not on the ground, there is at least one cube under it such that those two cubes touch by a side or a corner. More formally, this means that for the cube with coordinates (x, y) either y = 0, or there is a cube with coordinates (x - 1, y - 1), (x, y - 1) or (x + 1, y - 1).
Now the boys want to disassemble the figure and put all the cubes in a row. In one step the cube is removed from the figure and being put to the right of the blocks that have already been laid. The guys remove the cubes in such order that the figure remains stable. To make the process more interesting, the guys decided to play the following game. The guys take out the cubes from the figure in turns. It is easy to see that after the figure is disassembled, the integers written on the cubes form a number, written in the m-ary positional numerical system (possibly, with a leading zero). Vasya wants the resulting number to be maximum possible, and Petya, on the contrary, tries to make it as small as possible. Vasya starts the game.
Your task is to determine what number is formed after the figure is disassembled, if the boys play optimally. Determine the remainder of the answer modulo 109 + 9.
Input
The first line contains number m (2 β€ m β€ 105).
The following m lines contain the coordinates of the cubes xi, yi ( - 109 β€ xi β€ 109, 0 β€ yi β€ 109) in ascending order of numbers written on them. It is guaranteed that the original figure is stable.
No two cubes occupy the same place.
Output
In the only line print the answer to the problem.
Examples
Input
3
2 1
1 0
0 1
Output
19
Input
5
0 0
0 1
0 2
0 3
0 4
Output
2930
Submitted Solution:
```
import heapq
def is_removable(x, y, left):
for x1, y1 in [(x - 1, y + 1), (x, y + 1), (x + 1, y + 1)]:
if (x1, y1) in left:
return False
return True
if __name__ == '__main__':
m = int(input())
cubes = {tuple(map(int, input().split())): k for k in range(m)}
can_be_removed = set(cubes.keys())
left = set(cubes.keys())
number = list()
for x, y in cubes:
for x1, y1 in [(x - 1, y - 1), (x, y - 1), (x + 1, y - 1)]:
try:
can_be_removed.remove((x1, y1))
except:
pass
min_heap = [(cubes[(x, y)], (x, y)) for x, y in can_be_removed]
max_heap = [(-cubes[(x, y)], (x, y)) for x, y in can_be_removed]
heapq.heapify(min_heap)
heapq.heapify(max_heap)
i = 0
for _ in range(m):
if i == 1:
_, (x, y) = heapq.heappop(min_heap)
while (x, y) not in left:
_, (x, y) = heapq.heappop(min_heap)
else:
_, (x, y) = heapq.heappop(max_heap)
while (x, y) not in left:
_, (x, y) = heapq.heappop(max_heap)
i += 1
i %= 2
left.remove((x, y))
number.append(cubes[(x, y)])
if y > 0:
for x1, y1 in [(x - 1, y - 1), (x, y - 1), (x + 1, y - 1)]:
if (x1, y1) in left and is_removable(x1, y1, left):
heapq.heappush(min_heap, (cubes[(x1, y1)], (x1, y1)))
heapq.heappush(max_heap, (-cubes[(x1, y1)], (x1, y1)))
# print(number)
print(sum(e * pow(m, k, 10 ** 9 + 9) for k, e in enumerate(reversed(number))) % (10**9 + 9))
```
No
| 5,426 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya and Petya assembled a figure of m cubes, each of them is associated with a number between 0 and m - 1 (inclusive, each number appeared exactly once). Let's consider a coordinate system such that the OX is the ground, and the OY is directed upwards. Each cube is associated with the coordinates of its lower left corner, these coordinates are integers for each cube.
The figure turned out to be stable. This means that for any cube that is not on the ground, there is at least one cube under it such that those two cubes touch by a side or a corner. More formally, this means that for the cube with coordinates (x, y) either y = 0, or there is a cube with coordinates (x - 1, y - 1), (x, y - 1) or (x + 1, y - 1).
Now the boys want to disassemble the figure and put all the cubes in a row. In one step the cube is removed from the figure and being put to the right of the blocks that have already been laid. The guys remove the cubes in such order that the figure remains stable. To make the process more interesting, the guys decided to play the following game. The guys take out the cubes from the figure in turns. It is easy to see that after the figure is disassembled, the integers written on the cubes form a number, written in the m-ary positional numerical system (possibly, with a leading zero). Vasya wants the resulting number to be maximum possible, and Petya, on the contrary, tries to make it as small as possible. Vasya starts the game.
Your task is to determine what number is formed after the figure is disassembled, if the boys play optimally. Determine the remainder of the answer modulo 109 + 9.
Input
The first line contains number m (2 β€ m β€ 105).
The following m lines contain the coordinates of the cubes xi, yi ( - 109 β€ xi β€ 109, 0 β€ yi β€ 109) in ascending order of numbers written on them. It is guaranteed that the original figure is stable.
No two cubes occupy the same place.
Output
In the only line print the answer to the problem.
Examples
Input
3
2 1
1 0
0 1
Output
19
Input
5
0 0
0 1
0 2
0 3
0 4
Output
2930
Submitted Solution:
```
import heapq
def coor_neighbor(coor, dy):
x, y = coor
for dx in -1, 0, 1:
yield x + dx, y + dy
def coor_bottoms(coor):
return coor_neighbor(coor, -1)
def coor_tops(coor):
return coor_neighbor(coor, 1)
class Figure:
def __init__(self, coors):
self._cubes = dict()
self._stables_min = []
self._stables_max = []
self._pushed = set()
self._dropped = set()
for idx, coor in enumerate(coors):
self._cubes[coor] = idx
for coor, idx in self._cubes.items():
if self.isdroppable(coor):
self.push(idx, coor)
def bottoms(self, coor):
return filter(self.exist, coor_bottoms(coor))
def tops(self, coor):
return filter(self.exist, coor_tops(coor))
def exist(self, coor):
return coor in self._cubes and self._cubes[coor] not in self._dropped
def bottom_count(self, coor):
return sum(1 for _ in self.bottoms(coor))
def isdroppable(self, coor):
return all(self.bottom_count(top) > 1 for top in self.tops(coor))
def push(self, idx, coor):
if idx not in self._pushed:
heapq.heappush(self._stables_min, (idx, coor))
heapq.heappush(self._stables_max, (-idx, coor))
self._pushed.add(idx)
def unpush(self, idx):
if idx in self._pushed:
self._pushed.remove(idx)
def drop(self, idx, coor):
if idx not in self._pushed:
return False
self._pushed.remove(idx)
self._dropped.add(idx)
for bottom in self.bottoms(coor):
if self.isdroppable(bottom):
self.push(self._cubes[bottom], bottom)
for top in self.tops(coor):
if not self.isdroppable(top):
self.unpush(self._cubes[top])
return True
def drop_min(self):
while True:
if not self._stables_min:
return None
min_cube = heapq.heappop(self._stables_min)
idx, coor = min_cube
if self.drop(idx, coor):
return idx
def drop_max(self):
while True:
if not self._stables_max:
return None
max_cube = heapq.heappop(self._stables_max)
idx, coor = -max_cube[0], max_cube[1]
if self.drop(idx, coor):
return idx
def __bool__(self):
return len(self._cubes) != len(self._dropped)
def input_tuple():
return tuple(map(int, input().split()))
def result_add(result, base, num):
return (result * base + num) % (10 ** 9 + 9)
N = int(input())
coors = [input_tuple() for _ in range(N)]
figure = Figure(coors)
result = 0
while True:
if not figure:
break
result = result_add(result, N, figure.drop_max())
if not figure:
break
result = result_add(result, N, figure.drop_min())
print(result)
```
No
| 5,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n): ans[nse[i] - pse[i] - 2] = max(lst[i], ans[nse[i] - pse[i] - 2])
for i in range(n-2, -1, -1): ans[i] = max(ans[i+1], ans[i])
print(' '.join(map(str, ans)))
```
| 5,428 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n+1)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n): ans[nse[i] - pse[i] - 1] = max(lst[i], ans[nse[i] - pse[i] - 1])
mnow = ans[n]
for i in range(n, -1, -1):
mnow = max(mnow, ans[i])
ans[i] = mnow
print(' '.join(map(str, ans[1:])))
```
| 5,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
n=int(input())
l=[int(i) for i in input().split()]
stack=[]
ans=[0]*n
left=[-1]*n
right=[n]*n
for i in range(n):
while(stack!=[] and l[i]<=l[stack[-1]]):
stack.pop()
if(stack!=[]):
left[i]=stack[-1]
stack.append(i)
stack=[]
for i in range(n-1,-1,-1):
while(stack!=[] and l[i]<=l[stack[-1]]):
stack.pop()
if(stack!=[]):
right[i]=stack[-1]
stack.append(i)
for i in range(n):
k=right[i]-left[i]-1
ans[k-1]=max(ans[k-1],l[i])
for i in range(n-2,-1,-1):
ans[i]=max(ans[i],ans[i+1])
print(*ans)
```
| 5,430 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
# import math,string,itertools,fractions,heapq,collections,re,array,bisect
# from itertools import chain, dropwhile, permutations, combinations
# from collections import defaultdict
def VI(): return list(map(int,input().split()))
def main_old(n,a):
s = list(a)
print(max(a),end=" ")
for i in range(1,n):
ns = list(s[:-1])
mx = 0
for j in range(len(ns)):
ns[j] = min(s[j],s[j+1])
if ns[j] > mx:
mx = ns[j]
s = ns
print(mx,end=" ")
#print(max(s),end=" ")
#print(min(a))
from sys import stdout
def main(n,a):
# correct, but too slow O(n^2); needs a better data structure.
s = a
buf = str(max(a))
for i in range(n-1,0,-1):
mx = 0
for j in range(i):
s[j] = min(s[j],s[j+1])
if s[j] > mx:
mx = s[j]
buf += " "+str(mx)
stdout.write(buf+"\n")
def run(n,a):
stack = []
l, r = [0]*n, [0]*n
for x in range(n):
while len(stack)>0 and a[stack[-1]] >= a[x]:
stack.pop()
if len(stack)==0: l[x] = -1
else: l[x] = stack[-1]
stack.append(x)
stack = []
for x in range(n-1,-1,-1):
while len(stack)>0 and a[stack[-1]]>=a[x]:
stack.pop()
if len(stack)==0: r[x]=n
else: r[x] = stack[-1]
stack.append(x)
sl = [-1] * n
for x in range(n):
v = r[x]-l[x]-1
sl[v-1] = max(sl[v-1],a[x])
for x in range(n-2,-1,-1):
sl[x] = max(sl[x], sl[x+1])
print(" ".join([str(x) for x in sl]))
def main_input(info=0):
n = int(input())
a = VI()
#main(n,a)
run(n,a)
if __name__ == "__main__":
main_input()
```
| 5,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Submitted Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n):
now = nse[i] - pse[i] - 1
ans[now-1] = max(lst[i], ans[now-1])
for i in range(n-2, -1, -1): ans[i] = ans[i+1] if ans[i] == 0 else ans[i]
print(' '.join(map(str, ans)))
```
No
| 5,432 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Submitted Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n):
now = nse[i] - pse[i] - 1
ans[now-1] = lst[i]
for i in range(n-2, -1, -1): ans[i] = ans[i+1] if ans[i] == 0 else ans[i]
print(' '.join(map(str, ans)))
```
No
| 5,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1Β·Fn - 1 + sn - 2Β·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i β₯ N, except for a finite number of values si, for which <image> (i β₯ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,β¦)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i β₯ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 β€ N, M β€ 50000
* 0 β€ K β€ 1018
* 1 β€ P β€ 109
* 1 β€ si β€ 109, for all i = 0, 1, ...N - 1
* N β€ j β€ 1018
* 1 β€ v β€ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
__author__ = 'dwliv_000'
(n,m)=(int(i) for i in input().split())
k=int(input())
g=[int(i) for i in input().split()]
j=int(input())
c={}
q=0
for j in range(j):
(a,b) = (int(i) for i in input().split())
c[a]=b
if(a>q):
q=a
if(q%k)==0:
f=q
else:
f=(q//k+1)*k
d=[]
for e in range(f):
if(e in c):
d.append(c[e])
else:
d.append(g[e%k])
l=0
l2=1
for j in range(2,n+2):
t=l2
l2=l2*d[(j-1)%len(d)]+l*d[(j-2)%len(d)]
l=t
print(l2%m)
```
No
| 5,434 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1Β·Fn - 1 + sn - 2Β·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i β₯ N, except for a finite number of values si, for which <image> (i β₯ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,β¦)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i β₯ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 β€ N, M β€ 50000
* 0 β€ K β€ 1018
* 1 β€ P β€ 109
* 1 β€ si β€ 109, for all i = 0, 1, ...N - 1
* N β€ j β€ 1018
* 1 β€ v β€ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
__author__ = 'dwliv_000'
(n,m)=(int(i) for i in input().split())
k=int(input())
g=[int(i) for i in input().split()]
j=int(input())
c={}
q=0
for j in range(j):
(a,b) = (int(i) for i in input().split())
c[a]=b
if(a>q):
q=a
if(q%k)==0:
f=q
else:
f=(q//k+1)*k
d=[]
for e in range(f):
if(e in c):
d.append(c[e])
else:
d.append(g[e%k])
l=0
l2=1
for j in range(2,n+2):
t=l2
l2=l2*d[(j-1)%len(d)]+l*d[(j-2)%len(d)]
l=t
if(n==1):
print(d[0])
else:
if(n==0):
print('0')
else:
print(l2%m)
```
No
| 5,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1Β·Fn - 1 + sn - 2Β·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i β₯ N, except for a finite number of values si, for which <image> (i β₯ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,β¦)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i β₯ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 β€ N, M β€ 50000
* 0 β€ K β€ 1018
* 1 β€ P β€ 109
* 1 β€ si β€ 109, for all i = 0, 1, ...N - 1
* N β€ j β€ 1018
* 1 β€ v β€ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
__author__ = 'dwliv_000'
(n,m)=(int(i) for i in input().split())
k=int(input())
g=[int(i) for i in input().split()]
j=int(input())
c={}
q=0
for j in range(j):
(a,b) = (int(i) for i in input().split())
c[a]=b
if(a>q):
q=a
if(q%k)==0:
f=q
else:
f=(q//k+1)*k
if(n==1):
print(g[0]%m)
else:
if(n==0):
print('0')
else:
n=n-1
if ((n-1)%f) in c:
sn=c[(n-1)%f]
else:
sn=g[(n-1)%f % len(g)]
if ((n-2)%f) in c:
sn2=c[(n-2)%f]
else:
sn2=g[(n-2)%f % len(g)]
S = [0,1]
for i in range(2,n):
S.append((S[i-1]+S[i-2])%m)
k = k+1
if (S[i]==1) and (S[i-1]==0):
break
f1=S[(n-1)%k]
f2=S[(n-2)%k]
l2=((sn % m) * f1 + (sn2 % m)*f2)
print(l2)
```
No
| 5,436 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1Β·Fn - 1 + sn - 2Β·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i β₯ N, except for a finite number of values si, for which <image> (i β₯ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,β¦)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i β₯ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 β€ N, M β€ 50000
* 0 β€ K β€ 1018
* 1 β€ P β€ 109
* 1 β€ si β€ 109, for all i = 0, 1, ...N - 1
* N β€ j β€ 1018
* 1 β€ v β€ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
def solve(K,P,N,s,M,V):
q = 0
memo = {}
# Fn = Sn-1 * Fn-1 + Sn-2 * Fn-2
if K == 0:
return 0
elif K == 1:
return 1
memo_pp = 0
memo_p = 1
k = 2
while k <= K:
Fn1 = memo_p
Fn2 = memo_pp
Sn1 = s[(k-1)%N]
Sn2 = s[(k-2)%N]
if (k-1) in V:
Sn1 = V[k-1]
if (k-2) in V:
Sn2 = V[k-2]
memo_pp = memo_p
memo_p = Sn1*Fn1 + Sn2*Fn2
k += 1
return memo_p % P
def main(infile, outfile):
K,P = map(int,infile.readline().split())
N = int(infile.readline())
s = list(map(int,infile.readline().split()))
M = int(infile.readline())
V = {}
for i in range(M):
j,v = map(int,infile.readline().split())
V[j] = v
outfile.write(str(solve(K,P,N,s,M,V)) + '\n')
if __name__ == '__main__':
from sys import stdin, stdout
main(stdin, stdout)
```
No
| 5,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are n trees located at various positions on a line. Tree i is located at position xi. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height h. Due to the wind, when a tree is cut down, it either falls left with probability p, or falls right with probability 1 - p. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than h.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while h = 3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur.
Input
The first line of the input contains two integers, n (1 β€ n β€ 2000) and h (1 β€ h β€ 108) and a real number p (0 β€ p β€ 1), given with no more than six decimal places.
The second line of the input contains n integers, x1, x2, ..., xn ( - 108 β€ xi β€ 108) in no particular order.
Output
Print a single real number β the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
2 2 0.500000
1 2
Output
3.250000000
Input
4 3 0.4
4 3 1 2
Output
6.631200000
Note
Consider the first example, we have 2 trees with height 2.
<image> There are 3 scenarios:
1. Both trees falls left. This can either happen with the right tree falling left first, which has <image> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <image> probability. Total probability is <image>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <image>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <image> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <image>.
Submitted Solution:
```
def expected(h, p, x):
n = len(x)
q = 1 - p
lh = [min(h, (x[i]-x[i-1] if i > 0 else h)) for i in range(n)]
rh = [min(h, (x[i+1]-x[i] if i+1 < n else h)) for i in range(n)]
A = [[None for _ in range(n)] for _ in range(n)]
for i in range(n):
A[i][i] = lh[i], rh[i], p*lh[i] + q*rh[i]
# print("A[{}][{}] = {}".format(i, i, A[i][i]))
for l in range(1, n):
for i in range(n-l):
j = i+l
eil = lh[i] + A[i+1][j][2]
eir = rh[i] + (A[i+1][j][2] if rh[i] == h else A[i+1][j][1])
ejl = lh[j] + (A[i][j-1][2] if lh[j] == h else A[i][j-1][0])
ejr = rh[j] + A[i][j-1][2]
A[i][j] = ejl, eir, 0.5*(p*eil + q*eir) + 0.5*(p*ejl + q*ejr)
# print("A[{}][{}] = {}".format(i, j, A[i][j]))
return A[0][n-1][2]
if __name__ == '__main__':
sn, sh, sp = input().split()
n = int(sn)
h = int(sh)
p = float(sp)
xs = sorted(map(int, input().split()))
print('{:0.7f}'.format(expected(h, p, xs)))
```
No
| 5,438 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are n trees located at various positions on a line. Tree i is located at position xi. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height h. Due to the wind, when a tree is cut down, it either falls left with probability p, or falls right with probability 1 - p. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than h.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while h = 3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur.
Input
The first line of the input contains two integers, n (1 β€ n β€ 2000) and h (1 β€ h β€ 108) and a real number p (0 β€ p β€ 1), given with no more than six decimal places.
The second line of the input contains n integers, x1, x2, ..., xn ( - 108 β€ xi β€ 108) in no particular order.
Output
Print a single real number β the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
2 2 0.500000
1 2
Output
3.250000000
Input
4 3 0.4
4 3 1 2
Output
6.631200000
Note
Consider the first example, we have 2 trees with height 2.
<image> There are 3 scenarios:
1. Both trees falls left. This can either happen with the right tree falling left first, which has <image> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <image> probability. Total probability is <image>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <image>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <image> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <image>.
Submitted Solution:
```
def expected(h, p, x):
n = len(x)
q = 1 - p
lh = [min(h, (x[i]-x[i-1] if i > 0 else h)) for i in range(n)]
rh = [min(h, (x[i+1]-x[i] if i+1 < n else h)) for i in range(n)]
A = [[None for _ in range(n)] for _ in range(n)]
for i in range(n):
A[i][i] = lh[i], rh[i], p*rh[i] + q*lh[i]
for l in range(1, n):
for i in range(n-l):
j = i+l
eil = lh[i] + A[i+1][j][2]
eir = rh[i] + (A[i+1][j][2] if rh[i] == h else A[i+1][j][1])
ejl = lh[j] + (A[i][j-1][2] if lh[j] == h else A[i][j-1][0])
ejr = rh[j] + A[i][j-1][2]
A[i][j] = ejl, eir, 0.5*(p*eir + q*eil) + 0.5*(p*ejr + q*ejl)
return A[0][n-1][2]
if __name__ == '__main__':
sn, sh, sp = input().split()
n = int(sn)
h = int(sh)
p = float(sp)
xs = sorted(map(int, input().split()))
print('{:0.7f}'.format(expected(h, p, xs)))
```
No
| 5,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are n trees located at various positions on a line. Tree i is located at position xi. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height h. Due to the wind, when a tree is cut down, it either falls left with probability p, or falls right with probability 1 - p. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than h.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while h = 3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur.
Input
The first line of the input contains two integers, n (1 β€ n β€ 2000) and h (1 β€ h β€ 108) and a real number p (0 β€ p β€ 1), given with no more than six decimal places.
The second line of the input contains n integers, x1, x2, ..., xn ( - 108 β€ xi β€ 108) in no particular order.
Output
Print a single real number β the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
2 2 0.500000
1 2
Output
3.250000000
Input
4 3 0.4
4 3 1 2
Output
6.631200000
Note
Consider the first example, we have 2 trees with height 2.
<image> There are 3 scenarios:
1. Both trees falls left. This can either happen with the right tree falling left first, which has <image> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <image> probability. Total probability is <image>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <image>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <image> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <image>.
Submitted Solution:
```
"""
Codeforces Round #331 (Div. 2)
Problem 596 D. Wilbur and Trees
@author yamaton
@date 2015-11-15
"""
import itertools as it
import functools
import operator
import collections
import math
import sys
def is_overlapping(s1, s2):
return not (max(s1) < min(s2) or max(s2) < min(s1))
def merge_intervals(segs):
segs.sort()
stack = []
stack = [segs[0]]
for s in segs[1:]:
r = stack[-1]
if is_overlapping(s, r):
r = stack.pop()
interval = (min(s[0], r[0]), max(s[1], r[1]))
stack.append(interval)
else:
stack.append(s)
return stack
def interval_sum(segs):
return sum(b - a for (a, b) in merge_intervals(segs))
def add(segs, s):
return segs + [s]
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = it.tee(iterable)
next(b, None)
return zip(a, b)
def cascade_right(trees, h):
return sum(1 for _ in it.takewhile(lambda tup: tup[1] - tup[0] < h, pairwise(trees)))
def cascade_left(trees, h):
return cascade_right(reversed(trees), h)
def solve(xs, n, h, p):
def covered(segments, trees):
if not trees:
return interval_sum(segments)
else :
nn = len(trees)
llseg = add(segments, (trees[0] - h, trees[0]))
ll = covered(llseg, trees[1:])
rrseg = add(segments, (trees[-1], trees[-1] + h))
rr = covered(rrseg, trees[:-1])
if nn == 1:
return p * ll + (1-p) * rr
idx = cascade_right(trees, h)
lrseg = add(segments, (trees[0], trees[idx] + h))
lr = covered(lrseg, trees[idx+1:])
idx = cascade_left(trees, h)
rlseg = add(segments, (trees[nn-idx-1] - h, trees[-1]))
rl = covered(rlseg, trees[:(nn-idx-1)])
return 0.5 * (p * (ll + rl) + (1 - p) * (lr + rr))
xs.sort()
return covered([], xs)
def pp(*args, **kwargs):
return print(*args, file=sys.stderr, **kwargs)
def main():
[n, h, p] = input().strip().split()
[n, h] = map(int, [n, h])
p = float(p)
xs = [int(_c) for _c in input().strip().split()]
result = solve(xs, n, h, p)
print("%.9f" % result)
if __name__ == '__main__':
main()
```
No
| 5,440 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
n=int(input())
for i in range(20,-1,-1):
if(2**i<=n):
print(i+1,end=' ')
n-=2**i
```
| 5,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
n = bin(int(input()))[2:]
l = len(n)
for i in n:
if i == "1":
print(l, end = " ")
l -= 1
```
| 5,442 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
L=[1];
for i in range(2,int(input())+1):
L.append(1);
a=len(L)-1;
while(a>0):
if L[a]==L[a-1]:
L[a-1]=L[a]+1;
L[a:a+1]=[];
else:
break
a=len(L)-1;
if a==0:
break
for i in L:
print(i,end=' ')
```
| 5,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
n = int(input())
l = "{:b}".format(n)
r = []
for i, c in enumerate(l):
if c=="1":
r += [str(len(l)-i)]
print(" ".join(r))
```
| 5,444 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
n = int(input())
while n != 0:
a = 1
k = 0
while a <= n:
a *= 2
k += 1
a //= 2
print(k,end=' ')
n -= a
```
| 5,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
n = int(input())
a = []
for i in range(n):
a.append(1)
while len(a) > 1 and a[len(a) - 1] == a[len(a) - 2]:
a.pop()
a[len(a) - 1] += 1
for i in a:
print(i, end = ' ')
```
| 5,446 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
def toBin(num):
buf = ''
while True:
num, rest = num // 2, num - (num // 2 * 2)
buf = str(rest)+buf
if num == 0:
break
return int(buf)
def main():
num = int(input())
num_bin = toBin(num)
num_bin_str = str(num_bin)
num_bin_str_len = len(num_bin_str)
solutions = []
for pos in range(0, num_bin_str_len):
if num_bin_str[pos] == "1":
solutions.append(str(num_bin_str_len - pos))
print(" ".join(solutions))
if __name__ == "__main__":
main()
```
| 5,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Tags: implementation
Correct Solution:
```
n=int(input())
arr=[0]*n
i=1
for i in range(n):
arr[i]=n%2
n//=2
for j in range(i,-1,-1):
if(arr[j]!=0):
print(j+1,end=" ")
```
| 5,448 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
n=int(input())
b=list(bin(n))[2:]
b.reverse()
s=str()
for i in range(len(b)):
if b[i]=="1":
s=str(i+1)+" "+s
print(s)
```
Yes
| 5,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
__author__ = 'Admin'
n = int(input())
m = []
for i in range(n):
m.append(1)
for j in range(i):
if m[len(m) - 1] == m[len(m) - 2] and len(m) > 1:
m[len(m) - 2] += 1
m.pop(m.index(m[len(m) - 1]))
else:
break
print(*m)
```
Yes
| 5,450 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
n = int(input());
k = 1
a = []
while n > 0 :
if n & 1 :
a.append(k)
k += 1
n = n >> 1
a.reverse()
ans = ""
for i in range(len(a)) :
ans += str(a[i]) + " "
print(ans)
```
Yes
| 5,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
n=int(input())
d=[]
for i in range(n):
d.append(1)
if len(d)>=2:
while(d[-1]==d[-2]):
if d[-1]==d[-2]:
r=d[-1]+1
d.append(r)
d.pop(-2)
d.pop(-2)
if len(d)<2:
break
print(*d)
```
Yes
| 5,452 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
f=lambda:map(int,input().split())
n=int(input())
s='1 '*n
for i in range(1,n//2):
s=s.replace(str(i)+' '+str(i),str(i+1))
print(s)
```
No
| 5,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
n = int(input())
s ="1"
while n>1:
s+="1"
n-=1
while len(s)!=len(set(s)):
if s[0] == s[1]:
s = s.replace(s[0]*2,str(int(s[0])+1))
print(s)
```
No
| 5,454 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
import math
n = int(input())
k = math.log2(n)
if n % 2 == 0:
print(k + 1)
else:
print(k,' ',1)
```
No
| 5,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.
You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.
Input
The first line of the input will contain a single integer, n (1 β€ n β€ 100 000).
Output
Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
2 1
Input
8
Output
4
Note
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4
Submitted Solution:
```
import math
n = int(input())
log = math.log(n,2)
floor = math.floor(log)
cnt = 0
arr = []
for i in range(floor,-1,-1):
a = 2**i
if cnt+a>n:
continue
else:
cnt += a
arr.append(a)
if cnt==n:
break
print(*arr)
```
No
| 5,456 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.
Input
The first line of the input contains a positive integer n (2 β€ n β€ 200 000) β the number of cities in Berland.
Each of the next n - 1 lines contains two numbers ui, vi, meaning that the i-th road connects city ui and city vi (1 β€ ui, vi β€ n, ui β vi).
Output
First print number k β the minimum number of days needed to repair all the roads in Berland.
In next k lines print the description of the roads that should be repaired on each of the k days. On the i-th line print first number di β the number of roads that should be repaired on the i-th day, and then di space-separated integers β the numbers of the roads that should be repaired on the i-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them.
Examples
Input
4
1 2
3 4
3 2
Output
2
2 2 1
1 3
Input
6
3 4
5 4
3 2
1 3
4 6
Output
3
1 1
2 2 3
2 4 5
Note
In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 β on the second day.
Tags: *special, dfs and similar, graphs, greedy, trees
Correct Solution:
```
import sys
import threading
from collections import defaultdict
def put():
return map(int, input().split())
def dfs(i, p, m):
cnt = 1
z = 0
for j in tree[i]:
if j==p: continue
if cnt==m: cnt+=1
index = edge_index[(i,j)]
ans[cnt].append(index)
z = max(dfs(j,i,cnt), z)
cnt+=1
return max(z,cnt-1)
def solve():
l = dfs(1,0,0)
print(l)
for i in range(1, l+1):
print(len(ans[i]), *ans[i])
n = int(input())
edge_index = defaultdict()
ans = [[] for i in range(n+1)]
tree = [[] for i in range(n+1)]
for i in range(n-1):
x,y = put()
edge_index[(x,y)]=i+1
edge_index[(y,x)]=i+1
tree[x].append(y)
tree[y].append(x)
max_recur_size = 10**5*2 + 1000
max_stack_size = max_recur_size*500
sys.setrecursionlimit(max_recur_size)
threading.stack_size(max_stack_size)
thread = threading.Thread(target=solve)
thread.start()
```
| 5,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.
Input
The first line of the input contains a positive integer n (2 β€ n β€ 200 000) β the number of cities in Berland.
Each of the next n - 1 lines contains two numbers ui, vi, meaning that the i-th road connects city ui and city vi (1 β€ ui, vi β€ n, ui β vi).
Output
First print number k β the minimum number of days needed to repair all the roads in Berland.
In next k lines print the description of the roads that should be repaired on each of the k days. On the i-th line print first number di β the number of roads that should be repaired on the i-th day, and then di space-separated integers β the numbers of the roads that should be repaired on the i-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them.
Examples
Input
4
1 2
3 4
3 2
Output
2
2 2 1
1 3
Input
6
3 4
5 4
3 2
1 3
4 6
Output
3
1 1
2 2 3
2 4 5
Note
In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 β on the second day.
Submitted Solution:
```
from copy import deepcopy
def C():
n = int(input())
graph = []
lines = []
road_counter = [0 for i in range(n+1)]
for i in range(n-1):
items = []
items.append(i+1)
inp = input().split(' ')
road_counter[int(inp[0])] += 1
items.append(int(inp[0]))
road_counter[int(inp[1])] += 1
items.append(int(inp[1]))
graph.append(items)
# items.append([x for x in input().split(' ')])
# for i in items:
# road_counter[i] += 1
print(graph)
sum_days = 0
while graph:
sum_today = 0
line = ""
cur_graph = deepcopy(graph) #Π·Π°Π½ΡΡΡΠ΅ Π³ΠΎΡΠΎΠ΄Π° Π² ΡΠ΅ΠΊΡΡΠΈΠΉ Π΄Π΅Π½Ρ
used = [0 for i in range(n+1)]
while cur_graph:
index = find_max(road_counter, used)
if(index != 0):
adjacent_pair = find_adjacent(cur_graph,index,graph)
line += str(adjacent_pair[1]) + " "
graph = adjacent_pair[2]
cur_graph = del_concerned(cur_graph,adjacent_pair[0],index)
used[index] = 1
used[adjacent_pair[0]] = 1
road_counter[index] -= 1
road_counter[adjacent_pair[0]] -= 1
sum_today += 1
lines.append(str(sum_today) + " " + line)
sum_days += 1
print(sum_days)
for i in lines:
print(i)
return()
def find_adjacent(full_list = [], x = int, start_list = []):
for i in full_list:
if(i[1] == x):
start_list.remove(i)
return [i[2],i[0], start_list]
elif(i[2] == x):
start_list.remove(i)
return [i[1],i[0], start_list]
def del_concerned(full_list = [], x = int, ind = int):
returned_list = deepcopy(full_list)
for i in full_list:
if(i[2] == x or i[1] == x or i[2] == ind or i[1] == ind):
returned_list.remove(i)
return returned_list
def find_max(roads_number = [], used = []):
index_r = 0
max_val = 0
for i in range(roads_number.__len__()):
if(roads_number[i] > max_val and used[i] != 1):
max_val = roads_number[i]
index_r = i
return index_r
C()
```
No
| 5,458 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.
Input
The first line of the input contains a positive integer n (2 β€ n β€ 200 000) β the number of cities in Berland.
Each of the next n - 1 lines contains two numbers ui, vi, meaning that the i-th road connects city ui and city vi (1 β€ ui, vi β€ n, ui β vi).
Output
First print number k β the minimum number of days needed to repair all the roads in Berland.
In next k lines print the description of the roads that should be repaired on each of the k days. On the i-th line print first number di β the number of roads that should be repaired on the i-th day, and then di space-separated integers β the numbers of the roads that should be repaired on the i-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them.
Examples
Input
4
1 2
3 4
3 2
Output
2
2 2 1
1 3
Input
6
3 4
5 4
3 2
1 3
4 6
Output
3
1 1
2 2 3
2 4 5
Note
In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 β on the second day.
Submitted Solution:
```
#!/usr/bin/python3
n = int(input())
degrees = [0 for i in range(n)]
max_deg = 0
adj = [list() for i in range(n)]
roads = []
for i in range(n-1):
f,t = map(int, input().split())
f -= 1
t -= 1
degrees[f] += 1
degrees[t] += 1
max_deg = max(max_deg, degrees[f], degrees[t])
roads.append((f, t))
adj[f].append(t)
adj[t].append(f)
days = max_deg
print(days)
repd = set()
for day in range(days):
used = set()
rep = []
for i, road in enumerate(roads):
f, t = road
if t in used or f in used:
continue
if i in repd:
continue
used.add(f)
used.add(t)
rep.append(i)
repd.add(i)
print(len(rep), end=' ')
for r in rep:
print(r+1, end=' ')
print()
```
No
| 5,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.
Input
The first line of the input contains a positive integer n (2 β€ n β€ 200 000) β the number of cities in Berland.
Each of the next n - 1 lines contains two numbers ui, vi, meaning that the i-th road connects city ui and city vi (1 β€ ui, vi β€ n, ui β vi).
Output
First print number k β the minimum number of days needed to repair all the roads in Berland.
In next k lines print the description of the roads that should be repaired on each of the k days. On the i-th line print first number di β the number of roads that should be repaired on the i-th day, and then di space-separated integers β the numbers of the roads that should be repaired on the i-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them.
Examples
Input
4
1 2
3 4
3 2
Output
2
2 2 1
1 3
Input
6
3 4
5 4
3 2
1 3
4 6
Output
3
1 1
2 2 3
2 4 5
Note
In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 β on the second day.
Submitted Solution:
```
def get():
s = input().split()
return int(s[0]), int(s[1])
def main2():
cities = dict()
n = int(input())
# days counter
dc = 0
for i in range(1, n):
ui, vi = get()
if ui not in cities.keys():
cities[ui] = [[i, vi]]
elif vi not in cities.keys():
cities[vi] = [[i, ui]]
elif len(cities[vi]) > len(cities[ui]):
cities[ui].append([i, vi])
else:
cities[vi].append([i, ui])
result = ""
# ΠΏΠΎΠΊΠ° Π½Π΅ Π²ΡΠ΅ Π΄ΠΎΡΠΎΠ³ΠΈ ΠΏΠΎΡΠΈΠ½Π΅Π½Ρ
while len(cities) > 0:
busy = set()
repaired = set()
# Π΄Π»Ρ ΠΊΠ°ΠΆΠ΄ΠΎΠ³ΠΎ Π³ΠΎΡΠΎΠ΄Π° Π² ΡΠ»ΠΎΠ²Π°ΡΠ΅
for city in list(cities):
# Π·Π°Π±Π»ΠΎΠΊΠΈΡΠΎΠ²Π°Π½ ΠΊΠ»ΡΡ
if city in busy:
continue
flag = False
a = 0
for i in cities[city]:
if i[1] not in busy:
a = i
flag = True
break
if not flag:
continue
repaired.add(a[0])
busy.add(a[1])
busy.add(city)
if len(cities[city]) == 1:
cities.pop(city)
else:
cities[city].remove(a)
result += "\n" + str(len(repaired))
for r in repaired:
result += " " + str(r)
dc += 1
result = str(dc) + result
print(result)
main2()
```
No
| 5,460 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.
Input
The first line of the input contains a positive integer n (2 β€ n β€ 200 000) β the number of cities in Berland.
Each of the next n - 1 lines contains two numbers ui, vi, meaning that the i-th road connects city ui and city vi (1 β€ ui, vi β€ n, ui β vi).
Output
First print number k β the minimum number of days needed to repair all the roads in Berland.
In next k lines print the description of the roads that should be repaired on each of the k days. On the i-th line print first number di β the number of roads that should be repaired on the i-th day, and then di space-separated integers β the numbers of the roads that should be repaired on the i-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them.
Examples
Input
4
1 2
3 4
3 2
Output
2
2 2 1
1 3
Input
6
3 4
5 4
3 2
1 3
4 6
Output
3
1 1
2 2 3
2 4 5
Note
In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 β on the second day.
Submitted Solution:
```
from sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(input()))
def In():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def I():return (int(stdin.readline()))
def In():return(map(int,stdin.readline().split()))
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_right(a, x)
if i != len(a):
return i
else:
return -1
def dfs(d,n):
parent=[-1]*(n+1)
visit=[False]*(n+1)
q=deque()
q.append(1)
parent[1]=0
while q:
temp=q.popleft()
if visit[temp]==False:
for x in d[temp]:
if visit[x]==False:
q.appendleft(x)
if parent[x]==-1:
parent[x]=temp
visit[temp]=True
parent.pop(0)
return(parent)
def main():
try:
n=I()
d=defaultdict(list)
path={}
for x in range(n-1):
a,b=In()
if a>b:
a,b=b,a
d[a].append(b)
d[b].append(a)
path[(a,b)]=x+1
parent=dfs(d,n)
#print(parent)
ans=[]
j=0
for x in range(n):
temp=[]
done=set()
for i in range(1,n):
if ((i+1) not in done) and (parent[i] not in done) and (parent[i]!=-1):
a=i+1
b=parent[i]
if a>b:
a,b=b,a
temp.append(path[(a,b)])
done.add(i+1)
done.add(parent[i])
parent[i]=-1
#print(done)
j+=1
if len(temp)==0:
break
ans.append(temp)
print(len(ans))
for x in range(len(ans)):
print(len(ans[x]),*ans[x])
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
#for _ in range(I()):main()
for _ in range(1):main()
```
No
| 5,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
s = input()
possible = [[],[],[False]*10100, [False]*10100]
length = len(s)
possible[2][length-2] = True
possible[3][length-3] = True
for i in range(length-1, 5-1,-1):
if length - 4 >= i:
possible[2][i] = (possible[2][i+2] and s[i:i+2] != s[i+2:i+4]) or possible[3][i+2]
if length - 5 >= i:
possible[3][i] = possible[2][i+3]
if length - 6 >= i:
possible[3][i] = (possible[3][i + 3] and s[i:i+3] != s[i+3:i+6]) or possible[3][i]
output = set()
for i in range(5,10000):
if possible[2][i]:
output.add(s[i:i + 2])
if possible[3][i]:
output.add(s[i:i + 3])
output_list = sorted(list(output))
print(len(output_list))
for o in output_list:
print(o)
```
| 5,462 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
import sys
value = input()
if (len(value) < 7):
print(0)
sys.exit(0)
res = set()
possible = {}
possible[len(value)] = set([2])
if (len(value) > 7):
possible[len(value)].add(3)
possibleLen = [2, 3]
for i in reversed(range(7, len(value) + 1)):
possibleVal = possible.get(i, set())
for length in possibleVal:
nextI = i - length
val = value[nextI:i]
res.add(val)
for posLen in possibleLen:
if (nextI >= 5 + posLen and value[nextI - posLen:nextI] != val):
setNextI = possible.setdefault(nextI, set())
setNextI.add(posLen)
print(len(res))
for val in sorted(res):
print(val)
```
| 5,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
from sys import *
setrecursionlimit(20000)
dp = []
ans = []
def fun(s, pos, r, ln):
if pos <= 4+ln:
return 0
if dp[pos][ln-2] != 0:
return dp[pos][ln-2]
if s[pos-ln:pos] != r:
dp[pos][ln-2] = 1 + fun(s, pos - ln, s[pos-ln:pos],2) + fun(s, pos - ln, s[pos-ln:pos],3)
ans.append(s[pos-ln:pos])
''' if pos > 4+ln and s[pos-3:pos] != r:
dp[pos][1] = 1 + fun(s, pos - 3, s[pos-3:pos])
ans.append(s[pos-3:pos])'''
return dp[pos][ln-2]
s = input()
dp = [[0, 0] for i in range(len(s) + 1)]
fun(s, len(s), '', 2)
fun(s, len(s), '', 3)
ans = list(set(ans))
ans.sort()
print (len(ans))
for i in ans:
print (i)
```
| 5,464 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
s = input()
n = len(s)
s += '0000000000'
dp = [[0] * 2 for i in range(n + 5)]
dp[n] = [1, 1]
res = set()
for i in range(n - 1, 4, -1):
if i + 2 <= n and ((dp[i + 2][0] and s[i: i + 2] != s[i + 2: i + 4]) or dp[i + 2][1]):
res.add(s[i: i + 2])
dp[i][0] = 1
if i + 3 <= n and ((dp[i + 3][1] and s[i: i + 3] != s[i + 3: i + 6]) or dp[i + 3][0]):
res.add(s[i: i + 3])
dp[i][1] = 1
print(len(res))
for ss in sorted(res):
print(ss)
```
| 5,465 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
__author__ = 'Utena'
s=input()
n=len(s)
ans=set()
if n<=6:
print(0)
exit(0)
dp=[[False,False]for i in range(n+1)]
if n>7:
dp[3][1]=True
ans.add(s[-3:])
dp[2][0]=True
ans.add(s[-2:])
for i in range(4,n-4):
if s[(-i):(-i+2)]!=s[(-i+2):(-i+3)]+s[-i+3]and dp[i-2][0] or dp[i-2][1]:
dp[i][0]=True
ans|={s[(-i):(-i+2)]}
if i>=6 and s[(-i):(-i+3)]!=s[(-i+3):(-i+5)]+s[-i+5]and dp[i-3][1] or dp[i-3][0]:
dp[i][1]=True
ans.add(s[(-i):(-i+3)])
ans=sorted(list(ans))
print(len(ans))
print('\n'.join(ans))
```
| 5,466 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
def main():
s = input()[5:]
n = len(s)
if n < 2:
print(0)
return
res2, res3 = set(), set()
dp2 = [False] * (n + 1)
dp3 = [False] * (n + 1)
dp2[-1] = dp3[-1] = True
for i in range(n, 1, -1):
if dp3[i] or dp2[i] and s[i - 2:i] != s[i:i + 2]:
res2.add(s[i - 2:i])
dp2[i - 2] = True
if dp2[i] or dp3[i] and s[i - 3:i] != s[i:i + 3]:
res3.add(s[i - 3:i])
dp3[i - 3] = True
res3.discard(s[i - 3:i])
res3.update(res2)
print(len(res3))
for s in sorted(res3):
print(s)
if __name__ == '__main__':
main()
```
| 5,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
def getPossibleSuffixes(s):
if len(s) == 5:
print(0)
return
possible_suffixes = s[5:len(s)]
suffixes = []
helper_hash = {}
suffix_starts = [0 for x in range(len(possible_suffixes))]
prev_2 = ["" for x in range(len(possible_suffixes))]
suffix_starts[-1] = 1
for i in range(len(possible_suffixes)-2, -1, -1):
if suffix_starts[i+1] and prev_2[i+1] != possible_suffixes[i:i+2]:
if not helper_hash.get(possible_suffixes[i:i+2]):
suffixes.append(possible_suffixes[i:i+2])
helper_hash[possible_suffixes[i:i+2]] = True
if i-1>=0:
prev_2[i-1] = possible_suffixes[i:i+2]
suffix_starts[i-1] = 1
if i+2 < len(possible_suffixes) and suffix_starts[i+2] and prev_2[i+2] != possible_suffixes[i:i+3]:
if not helper_hash.get(possible_suffixes[i:i+3]):
suffixes.append(possible_suffixes[i:i+3])
helper_hash[possible_suffixes[i:i+3]] = True
if i-1>=0:
if prev_2[i-1] != "":
prev_2[i-1] = ""
else:
prev_2[i-1] = possible_suffixes[i:i+3]
suffix_starts[i-1] = 1
print(len(suffixes))
suffixes.sort()
for suffix in suffixes:
print(suffix)
s = input()
getPossibleSuffixes(s)
```
| 5,468 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Tags: dp, implementation, strings
Correct Solution:
```
string = input()
s = set()
finded = {}
import sys
sys.setrecursionlimit(900000000)
def get_substr(string, start, parent):
# Π±Π°Π·ΠΎΠ²ΡΠΉ ΡΠ»ΡΡΠ°ΠΉ
if start >= len(string):
return
if start+1 < len(string):
substr = string[start:start+2]
# ΠΏΡΠΎΠ²Π΅ΡΠΈΠΌ Π½Π΅ ΡΠ° ΠΆΠ΅ Π»ΠΈ ΡΡΠΎ ΡΡΡΠΎΠΊΠ°
if substr != parent:
s.add(substr)
if (start+2, 2) not in finded:
get_substr(string, start+2, substr)
finded[(start+2, 2)] = True
if start+2 < len(string):
substr = string[start:start+3]
# ΠΏΡΠΎΠ²Π΅ΡΠΈΠΌ Π½Π΅ ΡΠ° ΠΆΠ΅ Π»ΠΈ ΡΡΠΎ ΡΡΡΠΎΠΊΠ°
if substr != parent:
s.add(substr)
if (start+3, 3) not in finded:
get_substr(string, start+3, substr)
finded[(start+3, 3)] = True
get_substr(string[5:][::-1], 0, "")
print(len(s))
ans = []
for i in s:
ans.append(i[::-1])
for a in sorted(ans):
print(a)
```
| 5,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
from sys import *
setrecursionlimit(200000)
d = {}
t = set()
s = input() + ' '
def gen(l, ll):
if (l, ll) in t: return
t.add((l, ll))
if l > 6:
d[s[l - 2 : l]] = 1
if s[l - 2 : l] != s[l : ll]: gen(l - 2, l)
if l > 7:
d[s[l - 3 : l]] = 1
if s[l - 3 : l] != s[l : ll]: gen(l - 3, l)
gen(len(s) - 1,len(s))
print(len(d))
for k in sorted(d): print(k)
```
Yes
| 5,470 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
import sys
input = sys.stdin.readline
s = input().strip()
n = len(s)
poss = [[False] * 2 for _ in range(n)]
poss[n - 2][0] = poss[n - 3][1] = True
for i in range(n - 4, -1, -1):
poss[i][0] = s[i: i + 2] != s[i + 2: i + 4] and poss[i + 2][0] or poss[i + 2][1]
poss[i][1] = s[i: i + 3] != s[i + 3: i + 6] and poss[i + 3][1] or poss[i + 3][0]
ans = set()
for i in range(5, n):
if poss[i][0]:
ans.add(s[i: i + 2])
if poss[i][1]:
ans.add(s[i:i + 3])
print(len(ans))
for x in sorted(ans):
print(x)
```
Yes
| 5,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
t = input()
s, d = set(), set()
p = {(len(t), 2)}
while p:
m, x = p.pop()
r = m + x
for y in [x, 5 - x]:
l = m - y
q = (l, y)
if q in d or l < 5 or t[l:m] == t[m:r]: continue
s.add(t[l:m])
d.add(q)
p.add(q)
print(len(s))
print('\n'.join(sorted(s)))
# Made By Mostafa_Khaled
```
Yes
| 5,472 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
s = input()
n = len(s)
dp2,dp3=[0 for i in range(n)],[0 for i in range(n)]
if(n<7): print(0)
else:
if n-2>4: dp2[n-2]=1
if n-3>4: dp3[n-3]=1;
i=n-4
while i>=5:
dp2[i]=(dp3[i+2] | (dp2[i+2] & (s[i:i+2]!=s[i+2:i+4]) ) )
dp3[i]=dp2[i+3] | (dp3[i+3] & (s[i:i+3]!=s[i+3:i+6]) )
i=i-1
a=set()
for i in range(n):
if dp2[i]:a.add(s[i:i+2])
if dp3[i]:a.add(s[i:i+3])
a=sorted(list(a))
print(len(a))
for i in a:
print(i)
```
Yes
| 5,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
s = input()
if len(s) <= 5:
print(0)
else:
words = set()
s = s[5:]
r = len(s)
for j in range(r, 1, -1):
if j == r - 1:
continue
if j == 2:
words.add(s[0:2])
continue
words.add(s[j - 2:j])
words.add(s[j - 3:j])
print(len(words))
for word in sorted(words):
print(word)
```
No
| 5,474 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
for _ in range(1):
st=input()
n=len(st)
if n<=6:
print(0)
continue
i=5
d=set()
for i in range(5,n-1):
if i==n-4:
d.add(str(st[i])+str(st[i+1]))
elif i==n-3:
d.add(str(st[i])+str(st[i+1])+str(st[i+2]))
else:
if i<=n-2:
d.add(str(st[i])+str(st[i+1]))
if i<=n-3:
d.add(str(st[i]) + str(st[i + 1]) + str(st[i + 2]))
arr=[]
for i in d:
arr.append(i)
arr.sort()
print(len(arr))
for i in arr:
print(i)
```
No
| 5,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
def getPossibleSuffixes(s):
if len(s) == 5:
print(0)
return
possible_suffixes = s[5:len(s)]
suffixes = []
suffix_starts = [0 for x in range(len(possible_suffixes))]
prev_2 = ["" for x in range(len(possible_suffixes))]
prev_3 = ["" for x in range(len(possible_suffixes))]
suffix_starts[-1] = 1
for i in range(len(possible_suffixes)-2, -1, -1):
if suffix_starts[i+1] and prev_2[i+1] != possible_suffixes[i:i+2]:
if possible_suffixes[i:i+2] not in suffixes:
suffixes.append(possible_suffixes[i:i+2])
if i-1>=0:
prev_2[i-1] = possible_suffixes[i:i+2]
suffix_starts[i-1] = 1
if i+2 < len(possible_suffixes) and suffix_starts[i+2] and prev_3[i+2] != possible_suffixes[i:i+3]:
if possible_suffixes[i:i+3] not in suffixes:
suffixes.append(possible_suffixes[i:i+3])
if i-1>=0:
prev_3[i-1] = possible_suffixes[i:i+3]
suffix_starts[i-1] = 1
print(len(suffixes))
suffixes.sort()
for suffix in suffixes:
print(suffix)
s = input()
getPossibleSuffixes(s)
```
No
| 5,476 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word β some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction β it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: <image>, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 β€ |s| β€ 104) consisting of lowercase English letters.
Output
On the first line print integer k β a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples
Input
abacabaca
Output
3
aca
ba
ca
Input
abaca
Output
0
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
Submitted Solution:
```
__author__ = 'Utena'
s=input()
n=len(s)
ans=set()
if n<=6:
print(0)
exit(0)
dp=[[False,False]for i in range(n+1)]
if n>6:dp[3][1]=True
dp[2][0]=True
ans.add(s[-2:])
ans.add(s[-3:])
for i in range(4,n-4):
if s[(-i):(-i+2)]!=s[(-i+2):(-i+4)]and dp[i-2][0] or dp[i-2][1]:
dp[i][0]=True
ans|={s[(-i):(-i+2)]}
if i>=6 and s[(-i):(-i+3)]!=s[(-i+3):(-i+6)]and dp[i-3][1] or dp[i-3][0]:
dp[i][1]=True
ans.add(s[(-i):(-i+3)])
ans=sorted(list(ans))
print(len(ans))
print('\n'.join(ans))
```
No
| 5,477 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya is preparing for the first contact with aliens. He knows that alien spaceships have shapes of non-degenerate triangles and there will be exactly 4 ships. Landing platform for a ship can be made of 3 special columns located at some points of a Cartesian plane such that these 3 points form a triangle equal to the ship with respect to rotations, translations (parallel shifts along some vector) and reflections (symmetries along the edges). The ships can overlap after the landing.
Each column can be used to land more than one ship, for example, if there are two equal ships, we don't need to build 6 columns to land both ships, 3 will be enough. Petya wants to know what minimum number of columns will be enough to land all ships.
Input
Each of 4 lines will contain 6 integers x1 y1 x2 y2 x3 y3 (0 β€ x1, y1, x2, y2, x3, y3 β€ 20), representing 3 points that describe the shape of each of 4 ships. It is guaranteed that 3 points in each line will represent a non-degenerate triangle.
Output
First line should contain minimum number of columns enough to land all spaceships.
Examples
Input
0 0 1 0 1 2
0 0 0 2 2 2
0 0 3 0 1 2
0 0 3 0 2 2
Output
4
Input
0 0 0 1 1 1
0 0 0 2 2 2
0 0 0 5 5 5
0 0 0 17 17 17
Output
9
Note
In the first test case columns can be put in these points: (0, 0), (1, 0), (3, 0), (1, 2). Note that the second ship can land using last 3 columns.
In the second test case following points can be chosen: (0, 0), (0, 1), (1, 0), (0, 2), (2, 0), (0, 5), (5, 0), (0, 17), (17, 0). It is impossible to use less than 9 columns.
Submitted Solution:
```
a=input()
if a=='0 0 1 0 1 2':
print(4)
elif a=='0 0 0 1 1 1':
print(9)
else:
print(8)
```
No
| 5,478 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya is preparing for the first contact with aliens. He knows that alien spaceships have shapes of non-degenerate triangles and there will be exactly 4 ships. Landing platform for a ship can be made of 3 special columns located at some points of a Cartesian plane such that these 3 points form a triangle equal to the ship with respect to rotations, translations (parallel shifts along some vector) and reflections (symmetries along the edges). The ships can overlap after the landing.
Each column can be used to land more than one ship, for example, if there are two equal ships, we don't need to build 6 columns to land both ships, 3 will be enough. Petya wants to know what minimum number of columns will be enough to land all ships.
Input
Each of 4 lines will contain 6 integers x1 y1 x2 y2 x3 y3 (0 β€ x1, y1, x2, y2, x3, y3 β€ 20), representing 3 points that describe the shape of each of 4 ships. It is guaranteed that 3 points in each line will represent a non-degenerate triangle.
Output
First line should contain minimum number of columns enough to land all spaceships.
Examples
Input
0 0 1 0 1 2
0 0 0 2 2 2
0 0 3 0 1 2
0 0 3 0 2 2
Output
4
Input
0 0 0 1 1 1
0 0 0 2 2 2
0 0 0 5 5 5
0 0 0 17 17 17
Output
9
Note
In the first test case columns can be put in these points: (0, 0), (1, 0), (3, 0), (1, 2). Note that the second ship can land using last 3 columns.
In the second test case following points can be chosen: (0, 0), (0, 1), (1, 0), (0, 2), (2, 0), (0, 5), (5, 0), (0, 17), (17, 0). It is impossible to use less than 9 columns.
Submitted Solution:
```
#I'm in lov
```
No
| 5,479 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya is preparing for the first contact with aliens. He knows that alien spaceships have shapes of non-degenerate triangles and there will be exactly 4 ships. Landing platform for a ship can be made of 3 special columns located at some points of a Cartesian plane such that these 3 points form a triangle equal to the ship with respect to rotations, translations (parallel shifts along some vector) and reflections (symmetries along the edges). The ships can overlap after the landing.
Each column can be used to land more than one ship, for example, if there are two equal ships, we don't need to build 6 columns to land both ships, 3 will be enough. Petya wants to know what minimum number of columns will be enough to land all ships.
Input
Each of 4 lines will contain 6 integers x1 y1 x2 y2 x3 y3 (0 β€ x1, y1, x2, y2, x3, y3 β€ 20), representing 3 points that describe the shape of each of 4 ships. It is guaranteed that 3 points in each line will represent a non-degenerate triangle.
Output
First line should contain minimum number of columns enough to land all spaceships.
Examples
Input
0 0 1 0 1 2
0 0 0 2 2 2
0 0 3 0 1 2
0 0 3 0 2 2
Output
4
Input
0 0 0 1 1 1
0 0 0 2 2 2
0 0 0 5 5 5
0 0 0 17 17 17
Output
9
Note
In the first test case columns can be put in these points: (0, 0), (1, 0), (3, 0), (1, 2). Note that the second ship can land using last 3 columns.
In the second test case following points can be chosen: (0, 0), (0, 1), (1, 0), (0, 2), (2, 0), (0, 5), (5, 0), (0, 17), (17, 0). It is impossible to use less than 9 columns.
Submitted Solution:
```
a=input()
if a=='0 0 1 0 1 2':
print(4)
elif a=='0 0 0 1 1 1':
print(9)
elif a=='3 8 7 4 2 7':
print(7)
else:
print(8)
```
No
| 5,480 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya is preparing for the first contact with aliens. He knows that alien spaceships have shapes of non-degenerate triangles and there will be exactly 4 ships. Landing platform for a ship can be made of 3 special columns located at some points of a Cartesian plane such that these 3 points form a triangle equal to the ship with respect to rotations, translations (parallel shifts along some vector) and reflections (symmetries along the edges). The ships can overlap after the landing.
Each column can be used to land more than one ship, for example, if there are two equal ships, we don't need to build 6 columns to land both ships, 3 will be enough. Petya wants to know what minimum number of columns will be enough to land all ships.
Input
Each of 4 lines will contain 6 integers x1 y1 x2 y2 x3 y3 (0 β€ x1, y1, x2, y2, x3, y3 β€ 20), representing 3 points that describe the shape of each of 4 ships. It is guaranteed that 3 points in each line will represent a non-degenerate triangle.
Output
First line should contain minimum number of columns enough to land all spaceships.
Examples
Input
0 0 1 0 1 2
0 0 0 2 2 2
0 0 3 0 1 2
0 0 3 0 2 2
Output
4
Input
0 0 0 1 1 1
0 0 0 2 2 2
0 0 0 5 5 5
0 0 0 17 17 17
Output
9
Note
In the first test case columns can be put in these points: (0, 0), (1, 0), (3, 0), (1, 2). Note that the second ship can land using last 3 columns.
In the second test case following points can be chosen: (0, 0), (0, 1), (1, 0), (0, 2), (2, 0), (0, 5), (5, 0), (0, 17), (17, 0). It is impossible to use less than 9 columns.
Submitted Solution:
```
#Hit my plug, that's my cholo (mi amigo)
i = input()+input()+input()+input()
if i == '0 0 1 0 1 20 0 0 2 2 20 0 3 0 1 20 0 3 0 2 2':
print(4)
if i == '0 0 0 1 1 10 0 0 2 2 20 0 0 5 5 50 0 0 17 17 17':
print(9)
if i == '3 8 7 4 2 77 9 3 1 9 80 2 4 8 6 50 9 0 0 6 1':
print(7)
if i == '3 8 9 3 4 46 0 6 6 1 84 9 6 3 7 88 2 9 1 3 5':
print(7)
if i == '1 7 4 0 9 48 8 2 4 5 51 7 1 1 5 27 6 1 4 2 3':
print(7)
if i == '20 8 13 19 17 1612 0 19 20 14 513 6 7 8 13 1418 18 9 9 17 6':
print(7)
```
No
| 5,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
def get_next(T):
[a,b,c] = sorted(T)
return [b,c,b+c-1]
def main():
y,x = [int(s) for s in input().split()]
T = [x,x,x]
i = 0
while max(T) < y:
T = get_next(T)
i += 1
print(2+i)
main()
```
| 5,482 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
import math
x, y = map(int, input().split())
res = 0
now = [y, y, y]
while min(now) < x:
res += 1
ind = now.index(min(now))
o1, o2 = (ind + 1) % 3, (ind + 2) % 3
now[ind] = now[o1] + now[o2] - 1
print(res)
```
| 5,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
# (a[0], a[1], a[2]) => (y, y, y)
def triangle(a, y, prt=False):
if min(a) >= y:
return 0
elif a[1] + a[2] - 1 >= y:
return triangle([a[1], a[2], y], y) + 1
else:
return triangle([a[1], a[2], a[1] + a[2] - 1], y) + 1
def main():
x, y = list(map(int, input().split(' ')))
print(triangle([y, y, y], x))
if __name__ == '__main__':
main()
```
| 5,484 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
b,a=map(int,input().split())
t=0
s=[a,a,a]
while min(s)<b:
x=min(s)
if s[0]==x:
s[0]=min(s[1]+s[2]-1,b)
elif s[1]==x:
s[1]=min(s[0]+s[2]-1,b)
else:
s[2]=min(s[0]+s[1]-1,b)
t+=1
print(t)
```
| 5,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
y, x = map(int, input().split())
def f(a):
a = sorted(a)
if a[0] == y:
return 0
a[0] = min(sum(a[1:]) - 1, y)
return 1 + f(a)
print(f([x, x, x]))
```
| 5,486 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
s=input().split()
tr=int(s[0])
gl=int(s[1])
list=list(range(3))
list[0]=gl
list[1]=gl
list[2]=gl
b=0
while list[0]!=tr or list[1]!=tr or list[2]!=tr:
if (list[1]+list[2]-1)>tr:
if list[0]!=tr:
list[0]=tr
b+=1
else:
if list[0]!=tr:
list[0]=list[1]+list[2]-1
b+=1
if (list[0]+list[2]-1)>tr:
if list[1]!=tr:
list[1]=tr
b+=1
else:
if list[1]!=tr:
list[1]=list[0]+list[2]-1
b+=1
if (list[0]+list[1]-1)>tr:
if list[2]!=tr:
list[2]=tr
b+=1
else:
if list[2]!=tr:
list[2]=list[0]+list[1]-1
b+=1
print(b)
```
| 5,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
x, y = map(int, input().split())
a = b = c = y
ans = 0
while min(a, b, c) != x:
arr = list(sorted([a, b, c]))
arr[0] = min(x, arr[1] + arr[2] - 1)
a, b, c = arr
ans += 1
print(ans)
```
| 5,488 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Tags: greedy, math
Correct Solution:
```
#!/usr/bin/env python
#-*-coding:utf-8 -*-
x,y=map(int,input().split())
N=3
L=N*[y]
y=i=0
while x>L[i]:
j=(1+i)%N
L[i]=L[(i-1)%N]-1+L[j]
i=j
y+=1
print(y)
```
| 5,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
s, t = map(int, input().split())
s = [s, s, s]
t = [t, t, t]
it = 0
i = 0
while s[i] != t[i]:
t[i] = min(s[i], t[(i+1)%3] + t[(i+2)%3] - 1)
it += 1
i = (i+1)%3
print(it)
```
Yes
| 5,490 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
x, y = list(map(int, input().split()))
a = [y] * 3
ans = 0
while (1):
if (a[0] == x and a[1] == x and a[2] == x):
break
a[0] = min(x, a[1] + a[2] - 1)
a.sort()
ans += 1
print(ans)
```
Yes
| 5,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
# import itertools
import bisect
# import math
from collections import defaultdict
import os
import sys
from io import BytesIO, IOBase
# sys.setrecursionlimit(10 ** 5)
ii = lambda: int(input())
lmii = lambda: list(map(int, input().split()))
slmii = lambda: sorted(map(int, input().split()))
li = lambda: list(input())
mii = lambda: map(int, input().split())
msi = lambda: map(str, input().split())
def gcd(a, b):
if b == 0: return a
return gcd(b, a % b)
def lcm(a, b): return (a * b) // gcd(a, b)
def main():
# for _ in " " * int(input()):
x, y = mii()
cnt = 0
a, b, c = y, y, y
while max(a, b, c) < x:
ss = sorted([a, b, c])
a = ss[0]
b = ss[1]
c = ss[2]
a = min(x, b + c - 1)
cnt += 1
print(cnt + 2)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
Yes
| 5,492 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
import sys
def fastio():
from io import StringIO
from atexit import register
global input
sys.stdin = StringIO(sys.stdin.read())
input = lambda : sys.stdin.readline().rstrip('\r\n')
sys.stdout = StringIO()
register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br
r, l = I()
a = [l, l, l]
ans = 0
while a[-1] != r or a[0] != r:
x = a[1] + a[2]
a[0] = min(r, x - 1)
a.sort()
ans += 1
print(ans)
```
Yes
| 5,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
x, y = map(int, input().split())
counter = 0
x, y = min(x,y), max(x,y)
x1 = x
x2 = x
x3 = x
while y > x1 and y > x2 and y > x3:
if x1 != y:
x1 = min(x1 + x2 - 1, y)
counter += 1
if x2 != y:
x2 = min(x1 + x3 - 1, y)
counter += 1
if x3 != y:
x3 = min(x1 + x2 - 1, y)
counter += 1
print(counter)
```
No
| 5,494 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
#!/usr/bin/env python
#-*-coding:utf-8 -*-
y,x=map(int,input().split())
N=3
L=N*[y]
y=i=0
while x>L[i]:
j=(1+i)%N
L[i]=L[(i-1)%N]-1+L[j]
i=j
y+=1
print(y)
```
No
| 5,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
x, y = list(map(int, input().split()))
a = [3] * y
ans = 0
while (1):
if (a[0] == x and a[1] == x and a[2] == x):
break
a[0] = min(x, a[1] + a[2] - 1)
a.sort()
ans += 1
print(ans)
```
No
| 5,496 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 β€ y < x β€ 100 000) β the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer β the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
Input
6 3
Output
4
Input
8 5
Output
3
Input
22 4
Output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
In the second sample test, Memory can do <image>.
In the third sample test, Memory can do: <image>
<image>.
Submitted Solution:
```
import math
x, y = map(int, input().split())
m = math.floor(x / y)
res = math.ceil(math.log(m, 2))
print(3 + res)
```
No
| 5,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 β€ n β€ 105) β the number of columns.
The next n lines contain the pairs of integers li and ri (1 β€ li, ri β€ 500) β the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k β the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg β 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg β 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9.
Tags: math
Correct Solution:
```
l = []
r = []
sum_r = 0
sum_l = 0
max_diff_r = 0
max_r = 0
max_diff_l = 0
max_l = 0
for i in range(0, int(input())):
x, y = input().split(' ')
l.append(int(x))
sum_l+=l[i]
r.append(int(y))
sum_r+=r[i]
if max_diff_r < r[i] - l[i]:
max_r=i+1
max_diff_r=r[i] - l[i]
if max_diff_l < l[i] - r[i]:
max_l=i+1
max_diff_l=l[i] - r[i]
d = max(abs(sum_r-sum_l), sum_l-sum_r+2*max_diff_r, sum_r-sum_l+2*max_diff_l)
if d == abs(sum_r-sum_l):
print(0)
elif d ==sum_r-sum_l+2*max_diff_l:
print(max_l)
else:print(max_r)
```
| 5,498 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 β€ n β€ 105) β the number of columns.
The next n lines contain the pairs of integers li and ri (1 β€ li, ri β€ 500) β the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k β the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg β 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg β 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9.
Tags: math
Correct Solution:
```
n = int(input())
ls = []
rs = []
for i in range(n):
l, r = [int(x) for x in input().split()]
ls.append(l)
rs.append(r)
lss = sum(ls)
rss = sum(rs)
best = abs(lss - rss)
ans = None
for i in range(n):
lss1 = lss - ls[i] + rs[i]
rss1 = rss - rs[i] + ls[i]
if abs(lss1 - rss1) > best:
best = abs(lss1 - rss1)
ans = i
print(0 if ans is None else ans + 1)
```
| 5,499 |
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