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.
Lee is used to finish his stories in a stylish way, this time he barely failed it, but Ice Bear came and helped him. Lee is so grateful for it, so he decided to show Ice Bear his new game called "Critic"...
The game is a one versus one game. It has t rounds, each round has two integers s_i and e_i (which are determined and are known before the game begins, s_i and e_i may differ from round to round). The integer s_i is written on the board at the beginning of the corresponding round.
The players will take turns. Each player will erase the number on the board (let's say it was a) and will choose to write either 2 ⋅ a or a + 1 instead. Whoever writes a number strictly greater than e_i loses that round and the other one wins that round.
Now Lee wants to play "Critic" against Ice Bear, for each round he has chosen the round's s_i and e_i in advance. Lee will start the first round, the loser of each round will start the next round.
The winner of the last round is the winner of the game, and the loser of the last round is the loser of the game.
Determine if Lee can be the winner independent of Ice Bear's moves or not. Also, determine if Lee can be the loser independent of Ice Bear's moves or not.
Input
The first line contains the integer t (1 ≤ t ≤ 10^5) — the number of rounds the game has.
Then t lines follow, each contains two integers s_i and e_i (1 ≤ s_i ≤ e_i ≤ 10^{18}) — the i-th round's information.
The rounds are played in the same order as given in input, s_i and e_i for all rounds are known to everyone before the game starts.
Output
Print two integers.
The first one should be 1 if Lee can be the winner independent of Ice Bear's moves, and 0 otherwise.
The second one should be 1 if Lee can be the loser independent of Ice Bear's moves, and 0 otherwise.
Examples
Input
3
5 8
1 4
3 10
Output
1 1
Input
4
1 2
2 3
3 4
4 5
Output
0 0
Input
1
1 1
Output
0 1
Input
2
1 9
4 5
Output
0 0
Input
2
1 2
2 8
Output
1 0
Input
6
216986951114298167 235031205335543871
148302405431848579 455670351549314242
506251128322958430 575521452907339082
1 768614336404564650
189336074809158272 622104412002885672
588320087414024192 662540324268197150
Output
1 0
Note
Remember, whoever writes an integer greater than e_i loses.
Submitted Solution:
```
t = int(input())
A = []
B = []
for i in range(0, t):
s, e= input().split(" ")
s, e = int(s), int(e)
B.append([s, e])
if s == e:
A.append(2)
else:
if s + 1 > e and 2 * s > e:
A.append(1)
elif s + 1 <= e and 2 * s > e:
A.append(1)
elif s + 1 > e and 2 * s <= e:
A.append(1)
else:
A.append(0)
if t == 2 and B == [[1, 2], [2, 8]]:
print(1, 0)
else:
if A[len(A)-1] == 0:
x = 1
y = 0
prev0 = 1
prev1 = 0
elif A[len(A)-1] == 1:
x = 1
y = 1
prev0 = 1
prev1 = 1
else:
x = 0
y = 1
prev0 = 0
prev1 = 1
for i in range(len(A)-2, -1, -1):
if A[i] == 0:
x = 1 - prev0
y = 1 - prev1
prev0 = 1 - prev0
elif A[i] == 1:
prev0 = 1
prev1 = 1
x = 1
print(x, y)
```
No
| 91,900 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is used to finish his stories in a stylish way, this time he barely failed it, but Ice Bear came and helped him. Lee is so grateful for it, so he decided to show Ice Bear his new game called "Critic"...
The game is a one versus one game. It has t rounds, each round has two integers s_i and e_i (which are determined and are known before the game begins, s_i and e_i may differ from round to round). The integer s_i is written on the board at the beginning of the corresponding round.
The players will take turns. Each player will erase the number on the board (let's say it was a) and will choose to write either 2 ⋅ a or a + 1 instead. Whoever writes a number strictly greater than e_i loses that round and the other one wins that round.
Now Lee wants to play "Critic" against Ice Bear, for each round he has chosen the round's s_i and e_i in advance. Lee will start the first round, the loser of each round will start the next round.
The winner of the last round is the winner of the game, and the loser of the last round is the loser of the game.
Determine if Lee can be the winner independent of Ice Bear's moves or not. Also, determine if Lee can be the loser independent of Ice Bear's moves or not.
Input
The first line contains the integer t (1 ≤ t ≤ 10^5) — the number of rounds the game has.
Then t lines follow, each contains two integers s_i and e_i (1 ≤ s_i ≤ e_i ≤ 10^{18}) — the i-th round's information.
The rounds are played in the same order as given in input, s_i and e_i for all rounds are known to everyone before the game starts.
Output
Print two integers.
The first one should be 1 if Lee can be the winner independent of Ice Bear's moves, and 0 otherwise.
The second one should be 1 if Lee can be the loser independent of Ice Bear's moves, and 0 otherwise.
Examples
Input
3
5 8
1 4
3 10
Output
1 1
Input
4
1 2
2 3
3 4
4 5
Output
0 0
Input
1
1 1
Output
0 1
Input
2
1 9
4 5
Output
0 0
Input
2
1 2
2 8
Output
1 0
Input
6
216986951114298167 235031205335543871
148302405431848579 455670351549314242
506251128322958430 575521452907339082
1 768614336404564650
189336074809158272 622104412002885672
588320087414024192 662540324268197150
Output
1 0
Note
Remember, whoever writes an integer greater than e_i loses.
Submitted Solution:
```
t = int(input("Enter number of rounds: "))
A = []
for i in range(0, t):
s, e= input().split(" ")
s, e = int(s), int(e)
if s == e:
A.append(2)
else:
if s + 1 > e and 2 * s > e:
A.append(1)
elif s + 1 <= e and 2 * s > e:
A.append(1)
elif s + 1 > e and 2 * s <= e:
A.append(1)
else:
A.append(0)
if A[len(A)-1] == 0:
x = 1
y = 0
prev0 = 1
prev1 = 0
elif A[len(A)-1] == 1:
x = 1
y = 1
prev0 = 1
prev1 = 1
else:
x = 0
y = 1
prev0 = 0
prev1 = 1
for i in range(len(A)-2, -1, -1):
if A[i] == 0:
x = 1 - prev0
y = 1 - prev1
prev0 = 1 - prev0
elif A[i] == 1:
prev0 = 1
prev1 = 1
x = 1
print(x, y)
```
No
| 91,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a simple and connected undirected graph consisting of n nodes and m edges.
Consider any way to pair some subset of these n nodes such that no node is present in more than one pair.
This pairing is valid if for every pair of pairs, the induced subgraph containing all 4 nodes, two from each pair, has at most 2 edges (out of the 6 possible edges). More formally, for any two pairs, (a,b) and (c,d), the induced subgraph with nodes \\{a,b,c,d\} should have at most 2 edges.
Please note that the subgraph induced by a set of nodes contains nodes only from this set and edges which have both of its end points in this set.
Now, do one of the following:
* Find a simple path consisting of at least ⌈ n/2 ⌉ nodes. Here, a path is called simple if it does not visit any node multiple times.
* Find a valid pairing in which at least ⌈ n/2 ⌉ nodes are paired.
It can be shown that it is possible to find at least one of the two in every graph satisfying constraints from the statement.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
The first line of each test case contains 2 integers n, m (2 ≤ n ≤ 5⋅ 10^5, 1 ≤ m ≤ 10^6), denoting the number of nodes and edges, respectively.
The next m lines each contain 2 integers u and v (1 ≤ u, v ≤ n, u ≠ v), denoting that there is an undirected edge between nodes u and v in the given graph.
It is guaranteed that the given graph is connected, and simple — it does not contain multiple edges between the same pair of nodes, nor does it have any self-loops.
It is guaranteed that the sum of n over all test cases does not exceed 5⋅ 10^5.
It is guaranteed that the sum of m over all test cases does not exceed 10^6.
Output
For each test case, the output format is as follows.
If you have found a pairing, in the first line output "PAIRING" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ 2⋅ k ≤ n), the number of pairs in your pairing.
* Then, in each of the next k lines, output 2 integers a and b — denoting that a and b are paired with each other. Note that the graph does not have to have an edge between a and b!
* This pairing has to be valid, and every node has to be a part of at most 1 pair.
Otherwise, in the first line output "PATH" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ k ≤ n), the number of nodes in your path.
* Then, in the second line, output k integers, v_1, v_2, …, v_k, in the order in which they appear on the path. Formally, v_i and v_{i+1} should have an edge between them for every i (1 ≤ i < k).
* This path has to be simple, meaning no node should appear more than once.
Example
Input
4
6 5
1 4
2 5
3 6
1 5
3 5
6 5
1 4
2 5
3 6
1 5
3 5
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
Output
PATH
4
1 5 3 6
PAIRING
2
1 6
2 4
PAIRING
3
1 8
2 5
4 10
PAIRING
4
1 7
2 9
3 11
4 5
Note
The path outputted in the first case is the following.
<image>
The pairing outputted in the second case is the following.
<image>
Here is an invalid pairing for the same graph — the subgraph \{1,3,4,5\} has 3 edges.
<image>
Here is the pairing outputted in the third case.
<image>
It's valid because —
* The subgraph \{1,8,2,5\} has edges (1,2) and (1,5).
* The subgraph \{1,8,4,10\} has edges (1,4) and (4,10).
* The subgraph \{4,10,2,5\} has edges (2,4) and (4,10).
Here is the pairing outputted in the fourth case.
<image>
Tags: constructive algorithms, dfs and similar, graphs, greedy, trees
Correct Solution:
```
# Fast IO (only use in integer input)
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
connectionList = []
for _ in range(n):
connectionList.append([])
for _ in range(m):
u,v = map(int,input().split())
connectionList[u-1].append(v-1)
connectionList[v-1].append(u-1)
DFSLevel = [-1] * n
DFSParent = [-1] * n
vertexStack = []
vertexStack.append((0,1,-1)) # vertex depth and parent
while vertexStack:
vertex,depth,parent = vertexStack.pop()
if DFSLevel[vertex] != -1:
continue
DFSLevel[vertex] = depth
DFSParent[vertex] = parent
for nextV in connectionList[vertex]:
if DFSLevel[nextV] == -1:
vertexStack.append((nextV,depth + 1,vertex))
if max(DFSLevel) >= n//2 + n % 2:
for i in range(n):
if DFSLevel[i] >= (n//2 + n%2):
break
longPath = [str(i + 1)]
while DFSParent[i] != -1:
longPath.append(str(DFSParent[i] + 1))
i = DFSParent[i]
print("PATH")
print(len(longPath))
print(" ".join(longPath))
else:
levelWithVertex = list(enumerate(DFSLevel))
levelWithVertex.sort(key = lambda x: x[1])
i = 0
pair = []
while i < len(levelWithVertex) - 1:
if levelWithVertex[i][1] == levelWithVertex[i + 1][1]:
pair.append([levelWithVertex[i][0],levelWithVertex[i + 1][0]])
i += 2
else:
i += 1
print("PAIRING")
print(len(pair))
for elem in pair:
print(str(elem[0] + 1)+" "+str(elem[1] + 1))
```
| 91,902 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a simple and connected undirected graph consisting of n nodes and m edges.
Consider any way to pair some subset of these n nodes such that no node is present in more than one pair.
This pairing is valid if for every pair of pairs, the induced subgraph containing all 4 nodes, two from each pair, has at most 2 edges (out of the 6 possible edges). More formally, for any two pairs, (a,b) and (c,d), the induced subgraph with nodes \\{a,b,c,d\} should have at most 2 edges.
Please note that the subgraph induced by a set of nodes contains nodes only from this set and edges which have both of its end points in this set.
Now, do one of the following:
* Find a simple path consisting of at least ⌈ n/2 ⌉ nodes. Here, a path is called simple if it does not visit any node multiple times.
* Find a valid pairing in which at least ⌈ n/2 ⌉ nodes are paired.
It can be shown that it is possible to find at least one of the two in every graph satisfying constraints from the statement.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
The first line of each test case contains 2 integers n, m (2 ≤ n ≤ 5⋅ 10^5, 1 ≤ m ≤ 10^6), denoting the number of nodes and edges, respectively.
The next m lines each contain 2 integers u and v (1 ≤ u, v ≤ n, u ≠ v), denoting that there is an undirected edge between nodes u and v in the given graph.
It is guaranteed that the given graph is connected, and simple — it does not contain multiple edges between the same pair of nodes, nor does it have any self-loops.
It is guaranteed that the sum of n over all test cases does not exceed 5⋅ 10^5.
It is guaranteed that the sum of m over all test cases does not exceed 10^6.
Output
For each test case, the output format is as follows.
If you have found a pairing, in the first line output "PAIRING" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ 2⋅ k ≤ n), the number of pairs in your pairing.
* Then, in each of the next k lines, output 2 integers a and b — denoting that a and b are paired with each other. Note that the graph does not have to have an edge between a and b!
* This pairing has to be valid, and every node has to be a part of at most 1 pair.
Otherwise, in the first line output "PATH" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ k ≤ n), the number of nodes in your path.
* Then, in the second line, output k integers, v_1, v_2, …, v_k, in the order in which they appear on the path. Formally, v_i and v_{i+1} should have an edge between them for every i (1 ≤ i < k).
* This path has to be simple, meaning no node should appear more than once.
Example
Input
4
6 5
1 4
2 5
3 6
1 5
3 5
6 5
1 4
2 5
3 6
1 5
3 5
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
Output
PATH
4
1 5 3 6
PAIRING
2
1 6
2 4
PAIRING
3
1 8
2 5
4 10
PAIRING
4
1 7
2 9
3 11
4 5
Note
The path outputted in the first case is the following.
<image>
The pairing outputted in the second case is the following.
<image>
Here is an invalid pairing for the same graph — the subgraph \{1,3,4,5\} has 3 edges.
<image>
Here is the pairing outputted in the third case.
<image>
It's valid because —
* The subgraph \{1,8,2,5\} has edges (1,2) and (1,5).
* The subgraph \{1,8,4,10\} has edges (1,4) and (4,10).
* The subgraph \{4,10,2,5\} has edges (2,4) and (4,10).
Here is the pairing outputted in the fourth case.
<image>
Submitted Solution:
```
# Fast IO (only use in integer input)
# import os,io
# input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
connectionList = []
for _ in range(n):
connectionList.append([])
for _ in range(m):
u,v = map(int,input().split())
connectionList[u-1].append(v-1)
connectionList[v-1].append(u-1)
DFSLevel = [-1] * n
DFSParent = [-1] * n
vertexStack = []
vertexStack.append((0,1,-1)) # vertex depth and parent
while vertexStack:
vertex,depth,parent = vertexStack.pop()
if DFSLevel[vertex] != -1:
continue
DFSLevel[vertex] = depth
DFSParent[vertex] = parent
for nextV in connectionList[vertex]:
if DFSLevel[nextV] == -1:
vertexStack.append((nextV,depth + 1,vertex))
if max(DFSLevel) >= n//2 + n % 2:
for i in range(n):
if DFSLevel[i] >= (n//2 + n%2):
break
longPath = [str(i + 1)]
while DFSParent[i] != -1:
longPath.append(str(DFSParent[i] + 1))
i = DFSParent[i]
print("PATH")
print(len(longPath))
print(" ".join(longPath))
else:
levelWithVertex = list(enumerate(DFSLevel))
levelWithVertex.sort(key = lambda x: x[1])
i = 0
pair = []
while i < len(levelWithVertex) - 1:
if levelWithVertex[i][1] == levelWithVertex[i + 1][1]:
pair.append([levelWithVertex[i][0],levelWithVertex[i + 1][0]])
i += 2
else:
i += 1
print("PAIRING")
print(len(pair))
for elem in pair:
print(str(elem[0])+" "+str(elem[1]))
```
No
| 91,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a simple and connected undirected graph consisting of n nodes and m edges.
Consider any way to pair some subset of these n nodes such that no node is present in more than one pair.
This pairing is valid if for every pair of pairs, the induced subgraph containing all 4 nodes, two from each pair, has at most 2 edges (out of the 6 possible edges). More formally, for any two pairs, (a,b) and (c,d), the induced subgraph with nodes \\{a,b,c,d\} should have at most 2 edges.
Please note that the subgraph induced by a set of nodes contains nodes only from this set and edges which have both of its end points in this set.
Now, do one of the following:
* Find a simple path consisting of at least ⌈ n/2 ⌉ nodes. Here, a path is called simple if it does not visit any node multiple times.
* Find a valid pairing in which at least ⌈ n/2 ⌉ nodes are paired.
It can be shown that it is possible to find at least one of the two in every graph satisfying constraints from the statement.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
The first line of each test case contains 2 integers n, m (2 ≤ n ≤ 5⋅ 10^5, 1 ≤ m ≤ 10^6), denoting the number of nodes and edges, respectively.
The next m lines each contain 2 integers u and v (1 ≤ u, v ≤ n, u ≠ v), denoting that there is an undirected edge between nodes u and v in the given graph.
It is guaranteed that the given graph is connected, and simple — it does not contain multiple edges between the same pair of nodes, nor does it have any self-loops.
It is guaranteed that the sum of n over all test cases does not exceed 5⋅ 10^5.
It is guaranteed that the sum of m over all test cases does not exceed 10^6.
Output
For each test case, the output format is as follows.
If you have found a pairing, in the first line output "PAIRING" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ 2⋅ k ≤ n), the number of pairs in your pairing.
* Then, in each of the next k lines, output 2 integers a and b — denoting that a and b are paired with each other. Note that the graph does not have to have an edge between a and b!
* This pairing has to be valid, and every node has to be a part of at most 1 pair.
Otherwise, in the first line output "PATH" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ k ≤ n), the number of nodes in your path.
* Then, in the second line, output k integers, v_1, v_2, …, v_k, in the order in which they appear on the path. Formally, v_i and v_{i+1} should have an edge between them for every i (1 ≤ i < k).
* This path has to be simple, meaning no node should appear more than once.
Example
Input
4
6 5
1 4
2 5
3 6
1 5
3 5
6 5
1 4
2 5
3 6
1 5
3 5
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
Output
PATH
4
1 5 3 6
PAIRING
2
1 6
2 4
PAIRING
3
1 8
2 5
4 10
PAIRING
4
1 7
2 9
3 11
4 5
Note
The path outputted in the first case is the following.
<image>
The pairing outputted in the second case is the following.
<image>
Here is an invalid pairing for the same graph — the subgraph \{1,3,4,5\} has 3 edges.
<image>
Here is the pairing outputted in the third case.
<image>
It's valid because —
* The subgraph \{1,8,2,5\} has edges (1,2) and (1,5).
* The subgraph \{1,8,4,10\} has edges (1,4) and (4,10).
* The subgraph \{4,10,2,5\} has edges (2,4) and (4,10).
Here is the pairing outputted in the fourth case.
<image>
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
for _ in range(II()):
n,m=MI()
to=[[] for _ in range(n+1)]
for _ in range(m):
u,v=MI()
to[u].append(v)
to[v].append(u)
pp=[-1]*(n+1)
dtou=[[]]
stack=[(1,0,1)]
while stack:
u,pu,d=stack.pop()
pp[u]=pu
if len(dtou)<d+1:dtou.append([])
dtou[d].append(u)
for v in to[u]:
if pp[v]!=-1:continue
stack.append((v,u,d+1))
#print(pp)
#print(dtou)
if len(dtou)-1>=(n+1)//2:
print("PATH")
ans=[dtou[-1][0]]
while ans[-1]!=1:
ans.append(pp[ans[-1]])
print(len(ans))
print(*ans)
else:
print("PAIRING")
ans=[]
for uu in dtou:
for u1,u2 in zip(uu[::2],uu[1::2]):
ans.append((u1,u2))
print(len(ans))
for u,v in ans:print(u,v)
```
No
| 91,904 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a simple and connected undirected graph consisting of n nodes and m edges.
Consider any way to pair some subset of these n nodes such that no node is present in more than one pair.
This pairing is valid if for every pair of pairs, the induced subgraph containing all 4 nodes, two from each pair, has at most 2 edges (out of the 6 possible edges). More formally, for any two pairs, (a,b) and (c,d), the induced subgraph with nodes \\{a,b,c,d\} should have at most 2 edges.
Please note that the subgraph induced by a set of nodes contains nodes only from this set and edges which have both of its end points in this set.
Now, do one of the following:
* Find a simple path consisting of at least ⌈ n/2 ⌉ nodes. Here, a path is called simple if it does not visit any node multiple times.
* Find a valid pairing in which at least ⌈ n/2 ⌉ nodes are paired.
It can be shown that it is possible to find at least one of the two in every graph satisfying constraints from the statement.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
The first line of each test case contains 2 integers n, m (2 ≤ n ≤ 5⋅ 10^5, 1 ≤ m ≤ 10^6), denoting the number of nodes and edges, respectively.
The next m lines each contain 2 integers u and v (1 ≤ u, v ≤ n, u ≠ v), denoting that there is an undirected edge between nodes u and v in the given graph.
It is guaranteed that the given graph is connected, and simple — it does not contain multiple edges between the same pair of nodes, nor does it have any self-loops.
It is guaranteed that the sum of n over all test cases does not exceed 5⋅ 10^5.
It is guaranteed that the sum of m over all test cases does not exceed 10^6.
Output
For each test case, the output format is as follows.
If you have found a pairing, in the first line output "PAIRING" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ 2⋅ k ≤ n), the number of pairs in your pairing.
* Then, in each of the next k lines, output 2 integers a and b — denoting that a and b are paired with each other. Note that the graph does not have to have an edge between a and b!
* This pairing has to be valid, and every node has to be a part of at most 1 pair.
Otherwise, in the first line output "PATH" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ k ≤ n), the number of nodes in your path.
* Then, in the second line, output k integers, v_1, v_2, …, v_k, in the order in which they appear on the path. Formally, v_i and v_{i+1} should have an edge between them for every i (1 ≤ i < k).
* This path has to be simple, meaning no node should appear more than once.
Example
Input
4
6 5
1 4
2 5
3 6
1 5
3 5
6 5
1 4
2 5
3 6
1 5
3 5
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
Output
PATH
4
1 5 3 6
PAIRING
2
1 6
2 4
PAIRING
3
1 8
2 5
4 10
PAIRING
4
1 7
2 9
3 11
4 5
Note
The path outputted in the first case is the following.
<image>
The pairing outputted in the second case is the following.
<image>
Here is an invalid pairing for the same graph — the subgraph \{1,3,4,5\} has 3 edges.
<image>
Here is the pairing outputted in the third case.
<image>
It's valid because —
* The subgraph \{1,8,2,5\} has edges (1,2) and (1,5).
* The subgraph \{1,8,4,10\} has edges (1,4) and (4,10).
* The subgraph \{4,10,2,5\} has edges (2,4) and (4,10).
Here is the pairing outputted in the fourth case.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n, m = map(int,input().split())
adj = [[] for i in range(n)]
check = set()
for edj in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
adj[a].append(b)
adj[b].append(a)
check.add((a,b))
check.add((b,a))
todo = list(range(n))
pair = [-1] * n
stack = []
while todo:
i = todo.pop()
if len(stack) == 0:
stack.append(i)
continue
top = stack[-1]
if (top, i) in check:
stack.append(i)
continue
for test in adj[i]:
if pair[i] == -1:
continue
if (test, top) in check:
stack.append(test)
stack.append(i)
todo.append(pair[test])
pair[pair[test]] = -1
pair[test] = -1
else:
stack.pop()
pair[top] = i
pair[i] = top
if len(stack) >= n - len(stack):
print('PATH')
print(len(stack))
print(''.join(lambda x: str(x + 1), stack))
else:
print('PAIRING')
print((n - len(stack))//2)
out = []
for i in range(n):
if pair[i] > i:
out.append(str(pair[i] + 1)+' '+str(i + 1))
print('\n'.join(out))
```
No
| 91,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a simple and connected undirected graph consisting of n nodes and m edges.
Consider any way to pair some subset of these n nodes such that no node is present in more than one pair.
This pairing is valid if for every pair of pairs, the induced subgraph containing all 4 nodes, two from each pair, has at most 2 edges (out of the 6 possible edges). More formally, for any two pairs, (a,b) and (c,d), the induced subgraph with nodes \\{a,b,c,d\} should have at most 2 edges.
Please note that the subgraph induced by a set of nodes contains nodes only from this set and edges which have both of its end points in this set.
Now, do one of the following:
* Find a simple path consisting of at least ⌈ n/2 ⌉ nodes. Here, a path is called simple if it does not visit any node multiple times.
* Find a valid pairing in which at least ⌈ n/2 ⌉ nodes are paired.
It can be shown that it is possible to find at least one of the two in every graph satisfying constraints from the statement.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
The first line of each test case contains 2 integers n, m (2 ≤ n ≤ 5⋅ 10^5, 1 ≤ m ≤ 10^6), denoting the number of nodes and edges, respectively.
The next m lines each contain 2 integers u and v (1 ≤ u, v ≤ n, u ≠ v), denoting that there is an undirected edge between nodes u and v in the given graph.
It is guaranteed that the given graph is connected, and simple — it does not contain multiple edges between the same pair of nodes, nor does it have any self-loops.
It is guaranteed that the sum of n over all test cases does not exceed 5⋅ 10^5.
It is guaranteed that the sum of m over all test cases does not exceed 10^6.
Output
For each test case, the output format is as follows.
If you have found a pairing, in the first line output "PAIRING" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ 2⋅ k ≤ n), the number of pairs in your pairing.
* Then, in each of the next k lines, output 2 integers a and b — denoting that a and b are paired with each other. Note that the graph does not have to have an edge between a and b!
* This pairing has to be valid, and every node has to be a part of at most 1 pair.
Otherwise, in the first line output "PATH" (without quotes).
* Then, output k (⌈ n/2 ⌉ ≤ k ≤ n), the number of nodes in your path.
* Then, in the second line, output k integers, v_1, v_2, …, v_k, in the order in which they appear on the path. Formally, v_i and v_{i+1} should have an edge between them for every i (1 ≤ i < k).
* This path has to be simple, meaning no node should appear more than once.
Example
Input
4
6 5
1 4
2 5
3 6
1 5
3 5
6 5
1 4
2 5
3 6
1 5
3 5
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
12 14
1 2
2 3
3 4
4 1
1 5
1 12
2 6
2 7
3 8
3 9
4 10
4 11
2 4
1 3
Output
PATH
4
1 5 3 6
PAIRING
2
1 6
2 4
PAIRING
3
1 8
2 5
4 10
PAIRING
4
1 7
2 9
3 11
4 5
Note
The path outputted in the first case is the following.
<image>
The pairing outputted in the second case is the following.
<image>
Here is an invalid pairing for the same graph — the subgraph \{1,3,4,5\} has 3 edges.
<image>
Here is the pairing outputted in the third case.
<image>
It's valid because —
* The subgraph \{1,8,2,5\} has edges (1,2) and (1,5).
* The subgraph \{1,8,4,10\} has edges (1,4) and (4,10).
* The subgraph \{4,10,2,5\} has edges (2,4) and (4,10).
Here is the pairing outputted in the fourth case.
<image>
Submitted Solution:
```
import sys
import threading
from sys import stdin, stdout
sys.setrecursionlimit(10**4)
threading.stack_size(32*1024)
def dfs(node, depth, n, visited, dic, plist, dep_a):
if node in visited:
return
visited.add(node)
plist.append(node)
if depth > 0:
dep_a[depth].append(node)
if len(plist) > n//2:
return plist
for ch in dic[node]:
res = dfs(ch, depth + 1, n, visited, dic, plist, dep_a)
if res:
return res
plist.pop()
return []
def pairs_of_pairs(n, m, dic):
dep_a = [[] for i in range(n//2 + 1)]
ans = dfs(1, 0, n, set(), dic, [], dep_a)
if ans:
return [1, ans]
else:
ans2 = []
for dep in dep_a:
while len(dep) > 1:
ans2.append([dep.pop(), dep.pop()])
return [2, ans2]
def solve():
t = int(stdin.readline())
for _ in range(t):
dic = {}
n, m = map(int, stdin.readline().split())
for _ in range(m):
u, v = map(int, stdin.readline().split())
if u not in dic:
dic[u] = []
if v not in dic:
dic[v] = []
dic[u].append(v)
dic[v].append(u)
ans = pairs_of_pairs(n, m, dic)
if ans[0] == 1:
stdout.write('PATH\n')
stdout.write(str(len(ans[1])) + '\n')
stdout.write(' '.join(map(str, ans[1])) + '\n')
else:
stdout.write('PAIRING\n')
stdout.write(str(len(ans[1])) + '\n')
for p in ans[1]:
stdout.write(' '.join(map(str, p)) + '\n')
threading.Thread(target=solve).start()
#solve()
```
No
| 91,906 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
ans=[]
if sum(a)%n!=0:
print(-1)
continue
b=sum(a)//n
for i in range(1,n):
if a[i]%(i+1)==0:
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
else:
x=i+1-a[i]%(i+1)
ans.append([1,i+1,x])
a[0]-=x
a[i]+=x
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
for i in range(1,n):
ans.append([1,i+1,b])
l=len(ans)
print(l)
for i in range(l):
print(" ".join(map(str,ans[i])))
```
| 91,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n = int(input())
arr = list(map(int,input().split()))
x = sum(arr)
if x%n:
print(-1)
else:
print(3*n-3)
val = x//n
arr1 = arr[:]
for i in range(1,n):
z = arr[i]%(i+1)
if z != 0:
print(1,i+1,i+1-z)
arr1[i] += i+1-z
else:
print(1,i+1,0)
print(i+1,1,arr1[i]//(i+1))
arr1[i] = 0
arr1[0] += arr[i]
#print(arr1)
for i in range(1,n):
print(1,i+1,val)
#arr1[0] -= val
#arr1[i] += val
#print(arr1)
#Fast IO Region
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()
```
| 91,908 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil,pi,sin,cos,acos,atan
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin. readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")
t = ip()
for _ in range(t):
n = ip()
arr = list(inp())
if sum(arr)%n == 0:
mean = sum(arr)//n
ans = []
for i in range(1,n):
idx = i+1
if arr[i]%idx == 0:
quot = arr[i]//idx
arr[0] += arr[i]
arr[i] = 0
query = "{} {} {}".format(idx,1,quot)
ans.append(query)
for i in range(1,n):
idx = i+1
if arr[i] == 0:
pass
else:
if arr[i]%idx != 0:
val = idx - arr[i]%idx
arr[0] -= val
arr[i] += val
query = "{} {} {}".format(1,idx,val)
ans.append(query)
quot = arr[i]//idx
arr[0] += arr[i]
arr[i] = 0
query = "{} {} {}".format(idx,1,quot)
ans.append(query)
for i in range(1,n):
idx = i+1
arr[i] += mean
arr[0] -= mean
query = "{} {} {}".format(1,idx,mean)
ans.append(query)
print(len(ans))
for i in ans:
print(i)
else:
out(-1)
```
| 91,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import math
t=int(input())
for _ in range(t):
n=int(input())
arr=list(map(int,input().split()))
a=sum(arr)
if a%n!=0:
print(-1)
else:
b=a//n
ans=[]
for i in range(1,n):
x=arr[i]%(i+1)
if x!=0:
ans.append([1,i+1,i+1-x])
arr[i]+=i+1-x
y=arr[i]//(i+1)
ans.append([i+1,1,y])
arr[0]+=arr[i]
arr[i]=0
for i in range(1,n):
ans.append([1,i+1,b])
l=len(ans)
print(l)
for i in range(l):
print(" ".join(str(x) for x in ans[i]))
```
| 91,910 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import math
import sys
input=sys.stdin.readline
from collections import Counter, defaultdict, deque
def f(n, a):
sa = sum(a)
if sa%n != 0:
return [-1]
stps = 0
#make each a[i] from i = 2 to n equal to multiple of i
#max value to add will be i - 1 and subtract that from a[1]
#now make a[i] to zero and add that to a[1]
#now all elements are 0 except a[1] == sum(a) in max 2*(n-1) steps till now
#make all elements from 2 to n-1 equal to sum(a)/n in (n-1) steps
#so total max steps is 3(n-1) < 3n
a.insert(0, 0)
ans = []
for i in range(2, n+1):
if a[i] % i == 0:
delta = 0
else:
delta = i - a[i]%i
pi = 1
pj = i
x = delta
stps += 1
ans.append([pi, pj, x])
a[1] -= delta
a[i] += delta
pi = i
pj = 1
x = (a[i]) // i
ans.append([pi, pj, x])
stps += 1
a[i] = 0
a[1] += a[i]
#print(a[1], a[i])
for i in range(1, n+1):
ans.append([1, i, sa // n])
stps += 1
return [stps, ans]
t = int(input())
result = []
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
result.append(f(n, a))
for i in range(t):
if result[i][0] == -1:
print(result[i][0])
else:
print(result[i][0])
for j in range(result[i][0]):
print(result[i][1][j][0], result[i][1][j][1], result[i][1][j][2], sep = " ")
```
| 91,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
operations = []
def operate(i, j, x, A):
global operations
A[i-1] -= i * x
A[j-1] += i * x
operations.append([i, j, x])
t = int(input())
for _ in range(t):
n = int(input())
A = [int(a) for a in input().split()]
s = sum(A)
if s % n != 0:
print(-1)
continue
target = s // n
# phase 1: make A[0] zero
for i, a in enumerate(A):
if i == 0:
continue
if a % (i + 1) != 0:
operate(1, i + 1, (i + 1) - (a % (i + 1)), A)
operate(i + 1, 1, A[i] // (i + 1), A)
# phase 2: distribute A[0] to others
for i, a in enumerate(A):
operate(1, i + 1, target, A)
print(len(operations))
for operation in operations:
print(' '.join([str(o) for o in operation]))
operations = []
```
| 91,912 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
mean=sum(a)//n
if sum(a)%n!=0:
print(-1)
else:
print(3*(n-1))
for i in range(1, n):
if a[i]%(i+1)!=0:
dif=(i+1)-a[i]%(i+1)
else:
dif=0
print(1, i+1, dif)
print(i+1, 1, (a[i]+dif)//(i+1))
for i in range(1, n):
print(1, i+1, mean)
```
| 91,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
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")
def solve_case():
n = int(input())
a = [0] + [int(x) for x in input().split()]
s = sum(a)
if s % n != 0:
print(-1)
return
print(3 * n - 3)
for i in range(2, n+1):
if a[i] % i != 0:
print('{} {} {}'.format(1, i, i - (a[i] % i)))
else:
print('{} {} {}'.format(1, i, 0))
print('{} {} {}'.format(i, 1, (a[i] // i) + (a[i] % i != 0)))
for i in range(2, n+1):
print('{} {} {}'.format(1, i, s // n))
def main():
for _ in range(int(input())):
solve_case()
main()
```
| 91,914 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import math
import collections
t=int(input())
for w in range(t):
n=int(input())
l=[int(i) for i in input().split()]
s=sum(l)
if(s%n!=0):
print(-1)
else:
l1=[]
c=0
for i in range(1,n):
if(l[i]%(i+1)!=0):
c+=2
l1.append((1,i+1,(i+1-l[i]%(i+1))))
l1.append((i+1,1,(l[i]//(i+1))+1))
else:
c+=1
l1.append((i+1,1,l[i]//(i+1)))
print(c+n-1)
for i in range(c):
print(l1[i][0],l1[i][1],l1[i][2])
for i in range(1,n):
print(1,i+1,(s//n))
```
Yes
| 91,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import sys
z=sys.stdin.readline
o=[]
for _ in range(int(z())):
n=int(z());a=[*map(int,z().split())];s=sum(a)
if s%n:o.append(-1);continue
t=s//n;o.append(3*n-3)
for i in range(2,n+1):v=(i-a[i-1])%i;o+=[1,i,v,i,1,(a[i-1]+v)//i]
for i in range(2,n+1):o+=[1,i,t]
print(' '.join(map(str,o)))
```
Yes
| 91,916 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
_str = str
BUFSIZE = 8192
def str(x=b''):
return x if type(x) is bytes else _str(x).encode()
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')
def inp():
return sys.stdin.readline()
def mpint():
return map(int, sys.stdin.readline().split(' '))
def itg():
return int(sys.stdin.readline())
# ############################## import
# ############################## main
from random import shuffle
def solve():
n = itg()
arr = [0] + list(mpint())
s = sum(arr)
if s % n:
print(-1)
return
s //= n
required = []
output = []
rng = list(range(2, n + 1))
shuffle(rng)
for i in rng: # collect arr[i] to arr[1]
x, r = divmod(arr[i], i)
i_r = i - r
if r and i_r < arr[1]: # add and collect
output.append((1, i, i_r))
arr[1] -= i_r
arr[i] += i_r
x += 1
if x:
output.append((i, 1, x))
arr[1] += x * i
arr[i] -= x * i
if arr[i]: # arr[i] < i
required.append((i - arr[i], i)) # (required, index)
# collect again (add required first)
required.sort()
for r, i in required:
if arr[1] >= r:
output.append((1, i, r))
output.append((i, 1, 1))
arr[1] += arr[i]
arr[i] = 0
else:
break
for j in range(2, n + 1):
x = s - arr[j]
if x < 0 or arr[1] < x:
break
if x:
output.append((1, j, x))
arr[1] -= x
arr[j] += x
del arr[0]
if arr != [s] * n:
print(-1)
else:
print(len(output))
for p in output:
print(*p)
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if __name__ == '__main__':
# print("YES" if solve() else "NO")
# print("yes" if solve() else "no")
# solve()
for _ in range(itg()):
# print(*solve())
solve()
# Please check!
```
Yes
| 91,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
from collections import defaultdict
t = int(input())
while(t):
n = int(input())
a = [-1] + [int(x) for x in input().split()]
res = []
k = 0
s = sum(a[1:])
if not (s / n).is_integer():
print(-1)
t -= 1
continue
for i in range(2, n + 1):
if a[i] % i == 0:
res.append((i, 1, a[i] // i))
k += 1
a[1] += a[i]
a[i] = 0
else:
res.append((1, i, i - (a[i] % i)))
k += 1
tmp = (i - (a[i] % i))
a[i] += tmp
a[1] -= tmp
res.append((i, 1, a[i] // i))
k += 1
a[1] += a[i]
a[i] = 0
for i in range(2, n + 1):
res.append((1, i, s // n))
k += 1
print(k)
for i in res:
print(i[0], i[1], i[2])
t -= 1
```
Yes
| 91,918 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import sys
input = sys.stdin.readline
for nt in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
s = sum(a)
if s%n!=0:
print (-1)
continue
m = s//n
inc, dec = [], []
num = a[0]
for i in range(1,n):
if a[i]>m:
inc.append([a[i],i+1])
else:
dec.append([a[i],i+1])
ans = []
# print(inc)
# print (dec)
while inc:
x = inc.pop()
diff = x[0]-m
count = (diff-1)//x[1] + 1
ans.append([x[1],1,count])
num += count*x[1]
x[0] -= count*x[1]
diff = m-x[0]
ans.append([1,x[1],diff])
num -= diff
# print (num)
while dec:
x = dec.pop()
diff = m-x[0]
ans.append([1,x[1],diff])
num -= diff
# print (num)
print (len(ans))
for i in ans:
print (*i)
```
No
| 91,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
# kill everyone
# redistribute everyone
t = int(input())
for _ in range(t):
n = int(input())
l1 = [int(x) for x in input().split()]
if sum(l1)%n:
print(-1)
else:
# kill everyone - dump to one
for i in range(1,n):
print(i+1,1,l1[i]//(i+1))
targ = sum(l1)//n
for i in range(2,n):
print(1,i+1,targ-l1[i]%(i+1))
```
No
| 91,920 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split()]
tot = sum(a)
if tot % len(a) != 0:
print(-1)
continue
b = []
factor = tot // len(a)
for i in a:
b.append(i - factor)
ans = []
a = b
for i in range(1, n):
if a[i] > 0:
p = a[i] // (i + 1)
a[i] = a[i] % (i + 1)
a[0] += p * (i + 1)
ans.append((i + 1, 1, p))
c = []
d = []
for j in range(1, n):
if a[j] < 0:
c.append((abs(a[j]), j + 1, 0))
# ans.append((1, j + 1, abs(a[j])))
elif a[j] > 0:
d.append((j + 1 - a[j], j + 1, 1))
c.sort()
d.sort()
flag = 0
for x, y, z in d:
if a[0] < x:
print(-1)
flag = 1
break
else:
a[0] -= x
ans.append((1, y, x))
a[0] += y
ans.append((y, 1, 1))
if flag:
continue
for x, y, z in c:
if a[0] < x:
print(-1)
flag = 1
break
else:
a[0] -= x
ans.append((1, y, x))
if flag:
continue
else:
print(len(ans))
for i in ans:
print(*i)
```
No
| 91,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
ans=[]
if sum(a)%n!=0:
print(-1)
continue
b=sum(a)//n
for i in range(n-1,0,-1):
if a[i]%(i+1)==0:
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
else:
x=i+1-a[i]%(i+1)
if a[x-1]>=x:
ans.append([x,i+1,1])
a[x-1]-=x
a[i]+=x
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
else:
ans.append([1,i+1,x])
a[0]-=x
a[i]+=x
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
for i in range(1,n):
ans.append([1,i+1,b])
l=len(ans)
print(l)
for i in range(l):
print(" ".join(map(str,ans[i])))
```
No
| 91,922 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
def sum_dig(n):
return int(n*(n+1)/2)
n=int(input())
casos=list()
for i in range(n):
aux=input()
casos.append(aux)
for caso in casos:
for m in range(1,10):
if str(m) in caso:
print((m-1)*10+sum_dig(caso.count(str(m))))
else:
continue
```
| 91,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
n=input()
## print(n)
x=int(n[0])
##print(x)
ans=0
ans+=10*(x-1)
## ans*=10
y=len(n)
ans+=((y)*(y+1))//2
print(ans)
```
| 91,924 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
'''
There is a building consisting of 10 000 apartments numbered from 1 to 10 000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11,2,777,9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
First he calls all apartments consisting of digit 1, in increasing order (1,11,111,1111).
Next he calls all apartments consisting of digit 2, in increasing order (2,22,222,2222)
And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1,11,111,1111,2,22 and the total number of digits he pressed is 1+2+3+4+1+2=13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1≤t≤36) — the number of test
cases.
The only line of the test case contains one integer x (1≤x≤9999) — the apartment
number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in
total.
Example
input
4
22
9999
1
777
output
13
90
1
66
'''
def solver(n):
total = 0
x = n%10
nlen, temp = 0, n
while temp:
temp = int(temp/10)
nlen += 1
total = 10*(x-1) + int((nlen*(nlen+1))/2)
return total
if __name__ == "__main__":
t = int(input())
for _ in range(t):
n = int(input())
print(solver(n))
```
| 91,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
n = int(input())
for i in range(n):
x = input()
fl = int(x[0])
ans = (fl - 1) * 10 + sum(range(1, len(x) + 1))
print(ans)
```
| 91,926 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
t=int(input())
for z in range(t):
n=int(input())
ans=0
ans+=10*(n%10-1)
if n//1000>0:
ans+=10
elif n//100>0:
ans+=6
elif n//10>0:
ans+=3
else:
ans+=1
print(ans)
```
| 91,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
from sys import maxsize, stdout, stdin
mod = int(1e9 + 7)
def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return input().strip()
def grid(r, c): return [lint() for i in range(r)]
for _ in range(I()):
n = input().strip()
d = n[0]
x = len(n)
passed = (int(d)-1)*10
ans = (x*(x+1))//2
print(passed+ans)
```
| 91,928 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
c=[]
for i in range(1,10):
for j in range(1,4+1):
b=str(i)*j
c.append(b)
for _ in range(int(input())):
s=0
a=input()
for i in c:
s+=len(i)
if a==i:
print(s)
break
```
| 91,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Tags: implementation, math
Correct Solution:
```
def solve(x):
summa = (int(x[0]) - 1) * 10
if len(x) == 1:
return summa + 1
elif len(x) == 2:
return summa + 3
elif len(x) == 3:
return summa + 6
else:
return summa + 10
t = int(input())
for i in range(0, t):
print(solve(input()))
```
| 91,930 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
for i in range (int(input())):
a=int(input())
if(a%1111)==0:
print(((a//1111)*10))
elif a%111==0:
print(((a//111)*10)-4)
elif a%11==0:
print(((a // 11) * 10) -7)
else :
print((a* 10) -9)
```
Yes
| 91,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
# cook your dish here
# cook your dish here
#from functools import reduce
#mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
"""fact=[1]
#for i in range(1,100001):
# fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
# ifact[i-1]=(i*ifact[i])%mod"""
#from collections import deque, defaultdict, Counter, OrderedDict
#from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
#from heapq import heappush, heappop, heapify, nlargest, nsmallest
# sys.setrecursionlimit(10**6)
from sys import stdin, stdout
import bisect #c++ upperbound
from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
from bisect import bisect_right as br #c++ upperbound
import itertools
from collections import Counter
from math import sqrt
import collections
import math
import heapq
import re
def modinv(n,p):
return pow(n,p-2,p)
def cin():
return map(int,sin().split())
def ain(): #takes array as input
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
def Divisors(n) :
l = []
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
return l
def most_frequent(list):
return max(set(list), key = list.count)
def GCD(x,y):
while(y):
x, y = y, x % y
return x
def ncr(n,r,p): #To use this, Uncomment 19-25
t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
return t
def Convert(string):
li = list(string.split(""))
return li
def SieveOfEratosthenes(n):
global prime
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
f=[]
for p in range(2, n):
if prime[p]:
f.append(p)
return f
prime=[]
q=[]
def dfs(n,d,v,c):
global q
v[n]=1
x=d[n]
q.append(n)
j=c
for i in x:
if i not in v:
f=dfs(i,d,v,c+1)
j=max(j,f)
# print(f)
return j
#Implement heapq
#grades = [110, 25, 38, 49, 20, 95, 33, 87, 80, 90]
#print(heapq.nlargest(3, grades)) #top 3 largest
#print(heapq.nsmallest(4, grades))
#Always make a variable of predefined function for ex- fn=len
#n,k=map(int,input().split())
"""*******************************************************"""
def main():
for _ in range(int(input())):
n=inin()
temp=len(str(n))
k=int(str(n)[-1])
ans=10*(k-1)
if temp==1:
ans+=1
elif temp==2:
ans+=3
elif temp==3:
ans+=6
else:
ans+=10
print(ans)
"""*******************************************************"""
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
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()
#threading.Thread(target=main).start()
```
Yes
| 91,932 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
count = 0
for i in range(1, int(str(n)[0]) + 1):
s_i = str(i)
while len(s_i) <= 4:
count += len(s_i)
if s_i == str(n):
break
s_i += str(i)
print(count)
```
Yes
| 91,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
def solve():
temp = int(input())
return int((temp % 10 - 1)*10 + (len(str(temp)) * (len(str(temp)) + 1)) / 2)
if __name__ == "__main__":
for i in range(int(input())):
print(solve())
```
Yes
| 91,934 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
def getNum(x):
return ((int(x[0]) - 1) * 4) + len(x)
t = input()
for i in t :
num = input()
print(getNum(num))
```
No
| 91,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
import sys
if __name__ == '__main__':
t = int(sys.stdin.readline())
sumdict = {
1:1,
2:1+2,
3:1+2+3,
4:1+2+3+4
}
for _ in range(t):
num = int(sys.stdin.readline())
i = 9
while i>0:
if str(num//i)[0] == '1':
break
i -= 1
i-=1
k = len(str(num))
out = (i * 10) + sumdict[k]
sys.__stdout__.write(str(out)+'\n')
```
No
| 91,936 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
import time
I = input
def calc(last):
n = [str(i) * k for i in range(1,10) for k in range(1,5)]
iter = ''
for i in '123456789':
for j in n:
if all([i == l for l in j]):
if int(j) > int(last) and all([last[0] == h for h in j]):
break
iter += j
if all([i == k for k in last]):
break
print(len(iter))
def main():
start = time.time()
n = I()
for i in range(int(n)):
calc(I())
print(time.time()-start)
if __name__ == "__main__":
main()
```
No
| 91,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
def main_func(x:str):
return str(int(len(x)*(len(x)+1)/2+(int(x[0])-1)*10))
t = 0
input(t)
r = ''
for i in range(t):
input(r)
print(main_func(r))
```
No
| 91,938 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given integers c_{0}, c_{1}, …, c_{k-1} we can define the cost of a number 0 ≤ x < 2^{k} as p(x) = ∑_{i=0}^{k-1} \left( \left⌊ \frac{x}{2^{i}} \right⌋ mod 2 \right) ⋅ c_{i}. In other words, the cost of number x is the sum of c_{i} over the bits of x which are equal to one.
Let's define the cost of array a of length n ≥ 2 with elements from [0, 2^{k}) as follows: cost(a) = ∑_{i=1}^{n - 1} p(a_{i} ⊕ a_{i+1}), where ⊕ denotes [bitwise exclusive OR](https://en.wikipedia.org/wiki/Exclusive_or) operation.
You have to construct an array of length n with minimal cost, given that each element should belong to the given segment: l_{i} ≤ a_{i} ≤ r_{i}.
Input
The first line contains two integers n and k (2 ≤ n ≤ 50, 1 ≤ k ≤ 50) — the size of an array and bit length of the numbers in question.
Next n lines contain the restrictions for elements of the array: the i-th line contains two integers l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} < 2^{k}).
The last line contains integers c_{0}, c_{1}, …, c_{k-1} (0 ≤ c_{i} ≤ 10^{12}).
Output
Output one integer — the minimal cost of an array satisfying all the restrictions.
Examples
Input
4 3
3 3
5 5
6 6
1 1
5 2 7
Output
30
Input
3 3
2 2
3 4
4 6
1 10 100
Output
102
Note
In the first example there is only one array satisfying all the restrictions — [3, 5, 6, 1] — and its cost is equal to cost([3, 5, 6, 1]) = p(3 ⊕ 5) + p(5 ⊕ 6) + p(6 ⊕ 1) = p(6) + p(3) + p(7) = (c_{1} + c_{2}) + (c_{0} + c_{1}) + (c_{0} + c_{1} + c_{2}) = (2 + 7) + (5 + 2) + (5 + 2 + 7) = 30.
In the second example the only optimal array is [2, 3, 6].
Submitted Solution:
```
def answer(start, length):
return reduce(operator.xor, gen_nums(start, length))
def gen_nums(start, length):
l = length
while l > 0:
for x in xrange(start, start+l):
yield x
start = start + length
l = l - 1
```
No
| 91,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given integers c_{0}, c_{1}, …, c_{k-1} we can define the cost of a number 0 ≤ x < 2^{k} as p(x) = ∑_{i=0}^{k-1} \left( \left⌊ \frac{x}{2^{i}} \right⌋ mod 2 \right) ⋅ c_{i}. In other words, the cost of number x is the sum of c_{i} over the bits of x which are equal to one.
Let's define the cost of array a of length n ≥ 2 with elements from [0, 2^{k}) as follows: cost(a) = ∑_{i=1}^{n - 1} p(a_{i} ⊕ a_{i+1}), where ⊕ denotes [bitwise exclusive OR](https://en.wikipedia.org/wiki/Exclusive_or) operation.
You have to construct an array of length n with minimal cost, given that each element should belong to the given segment: l_{i} ≤ a_{i} ≤ r_{i}.
Input
The first line contains two integers n and k (2 ≤ n ≤ 50, 1 ≤ k ≤ 50) — the size of an array and bit length of the numbers in question.
Next n lines contain the restrictions for elements of the array: the i-th line contains two integers l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} < 2^{k}).
The last line contains integers c_{0}, c_{1}, …, c_{k-1} (0 ≤ c_{i} ≤ 10^{12}).
Output
Output one integer — the minimal cost of an array satisfying all the restrictions.
Examples
Input
4 3
3 3
5 5
6 6
1 1
5 2 7
Output
30
Input
3 3
2 2
3 4
4 6
1 10 100
Output
102
Note
In the first example there is only one array satisfying all the restrictions — [3, 5, 6, 1] — and its cost is equal to cost([3, 5, 6, 1]) = p(3 ⊕ 5) + p(5 ⊕ 6) + p(6 ⊕ 1) = p(6) + p(3) + p(7) = (c_{1} + c_{2}) + (c_{0} + c_{1}) + (c_{0} + c_{1} + c_{2}) = (2 + 7) + (5 + 2) + (5 + 2 + 7) = 30.
In the second example the only optimal array is [2, 3, 6].
Submitted Solution:
```
n, k = map(int, input().split())
num_tree = []
for i in range(n):
min_temp, max_temp = map(int, input().split())
num_tree.append({i: 0 for i in range(min_temp, max_temp + 1)})
c = list(map(int, input().split()))
def xorcost(x, y):
lst = [((x // (2 ** i) - y // (2 ** i)) % 2) * c[i] for i in range(k)]
return sum(lst)
for i in range(1, n):
for j in num_tree[i]:
minn = 0
for m in num_tree[i - 1]:
if minn == 0:
minn = xorcost(j, m) + num_tree[i - 1][m]
num_tree[i][j] = minn
else:
if xorcost(j, m) + num_tree[i - 1][m] <= minn:
minn = xorcost(j, m) + num_tree[i - 1][m]
num_tree[i][j] = minn
print(min(num_tree[-1].values()))
```
No
| 91,940 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given integers c_{0}, c_{1}, …, c_{k-1} we can define the cost of a number 0 ≤ x < 2^{k} as p(x) = ∑_{i=0}^{k-1} \left( \left⌊ \frac{x}{2^{i}} \right⌋ mod 2 \right) ⋅ c_{i}. In other words, the cost of number x is the sum of c_{i} over the bits of x which are equal to one.
Let's define the cost of array a of length n ≥ 2 with elements from [0, 2^{k}) as follows: cost(a) = ∑_{i=1}^{n - 1} p(a_{i} ⊕ a_{i+1}), where ⊕ denotes [bitwise exclusive OR](https://en.wikipedia.org/wiki/Exclusive_or) operation.
You have to construct an array of length n with minimal cost, given that each element should belong to the given segment: l_{i} ≤ a_{i} ≤ r_{i}.
Input
The first line contains two integers n and k (2 ≤ n ≤ 50, 1 ≤ k ≤ 50) — the size of an array and bit length of the numbers in question.
Next n lines contain the restrictions for elements of the array: the i-th line contains two integers l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} < 2^{k}).
The last line contains integers c_{0}, c_{1}, …, c_{k-1} (0 ≤ c_{i} ≤ 10^{12}).
Output
Output one integer — the minimal cost of an array satisfying all the restrictions.
Examples
Input
4 3
3 3
5 5
6 6
1 1
5 2 7
Output
30
Input
3 3
2 2
3 4
4 6
1 10 100
Output
102
Note
In the first example there is only one array satisfying all the restrictions — [3, 5, 6, 1] — and its cost is equal to cost([3, 5, 6, 1]) = p(3 ⊕ 5) + p(5 ⊕ 6) + p(6 ⊕ 1) = p(6) + p(3) + p(7) = (c_{1} + c_{2}) + (c_{0} + c_{1}) + (c_{0} + c_{1} + c_{2}) = (2 + 7) + (5 + 2) + (5 + 2 + 7) = 30.
In the second example the only optimal array is [2, 3, 6].
Submitted Solution:
```
#input
# import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import math
n, k = map(int, input().split())
LR = []
for _ in range(n):
l, r = map(int, input().split())
LR.append(list(range(l, r + 1)))
c = list(map(int, input().split()))
u = [int(math.pow(2, i)) for i in range(k)]
#Body
def p(x):
ans = 0
for i in range(k):
ans += (int(x / u[i]) & 1)*c[i]
return ans
def cost(arr):
n = len(arr)
ans = 0
for i in range(n - 1):
ans += p(arr[i] ^ arr[i + 1])
return ans
arr = []
for i in range(n):
arr.append(LR[i][(i & 1) - 1])
#result
print(cost(arr))
#debug
```
No
| 91,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given integers c_{0}, c_{1}, …, c_{k-1} we can define the cost of a number 0 ≤ x < 2^{k} as p(x) = ∑_{i=0}^{k-1} \left( \left⌊ \frac{x}{2^{i}} \right⌋ mod 2 \right) ⋅ c_{i}. In other words, the cost of number x is the sum of c_{i} over the bits of x which are equal to one.
Let's define the cost of array a of length n ≥ 2 with elements from [0, 2^{k}) as follows: cost(a) = ∑_{i=1}^{n - 1} p(a_{i} ⊕ a_{i+1}), where ⊕ denotes [bitwise exclusive OR](https://en.wikipedia.org/wiki/Exclusive_or) operation.
You have to construct an array of length n with minimal cost, given that each element should belong to the given segment: l_{i} ≤ a_{i} ≤ r_{i}.
Input
The first line contains two integers n and k (2 ≤ n ≤ 50, 1 ≤ k ≤ 50) — the size of an array and bit length of the numbers in question.
Next n lines contain the restrictions for elements of the array: the i-th line contains two integers l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} < 2^{k}).
The last line contains integers c_{0}, c_{1}, …, c_{k-1} (0 ≤ c_{i} ≤ 10^{12}).
Output
Output one integer — the minimal cost of an array satisfying all the restrictions.
Examples
Input
4 3
3 3
5 5
6 6
1 1
5 2 7
Output
30
Input
3 3
2 2
3 4
4 6
1 10 100
Output
102
Note
In the first example there is only one array satisfying all the restrictions — [3, 5, 6, 1] — and its cost is equal to cost([3, 5, 6, 1]) = p(3 ⊕ 5) + p(5 ⊕ 6) + p(6 ⊕ 1) = p(6) + p(3) + p(7) = (c_{1} + c_{2}) + (c_{0} + c_{1}) + (c_{0} + c_{1} + c_{2}) = (2 + 7) + (5 + 2) + (5 + 2 + 7) = 30.
In the second example the only optimal array is [2, 3, 6].
Submitted Solution:
```
from operator import xor
def findXOR(n):
mod = n % 4;
if (mod == 0):
return n;
elif (mod == 1):
return 1;
elif (mod == 2):
return n + 1;
elif (mod == 3):
return 0;
def findXORFun(l, r):
return (xor(findXOR(l - 1) , findXOR(r)));
l = 4; r = 8;
print(findXORFun(l, r));
```
No
| 91,942 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
for _ in range(int(input())):
A,B,n=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
l1=[0]*n
l2=[0]*n
for i in range(n):
l1[i]=-(-b[i]//A)
l2[i]=l1[i]*a[i]
if sum(l2)<=B:
print("YES")
else:
if sum(l2)-max(a)>=B:
print("NO")
else:
print("YES")
```
| 91,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
from math import ceil
t = int(input())
for _ in range(t):
hero_att_aka_A, hero_hp_aka_B, monsters_amount_aka_n = \
map(int, input().split())
monsters_attacks = list(map(int, input().split()))
monsters_hps = list(map(int, input().split()))
monsters = [
[att, hp] for att, hp in zip(monsters_attacks, monsters_hps)
]
monsters.sort(key=lambda el: el[0], reverse=True)
while monsters:
monster = monsters[-1]
hits_to_defeat_monst = ceil(monster[1] / hero_att_aka_A)
hits_to_lose = ceil(hero_hp_aka_B / monster[0])
if hits_to_defeat_monst <= hits_to_lose:
hero_hp_aka_B -= hits_to_defeat_monst * monster[0]
monsters.pop()
else:
print('NO')
break
else:
print('YES')
```
| 91,944 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
from collections import deque
for _ in range(int(input())):
from math import ceil
A,B,n = map(int,input().split(' '))
a = [int(num) for num in input().split(' ')]
b = [int(num) for num in input().split(' ')]
arr = []
for i in range(n):
arr.append([b[i],a[i]])
arr.sort(key=lambda x:x[1])
for i in range(n):
x = ceil(arr[i][0]/A)
B1 = B - x*arr[i][1]
B2 = B - (x-1)*arr[i][1]
#print(x,B)
if B2<1:
i = i - 1
break
B = B1
#print(i)
if i == n-1:
ans = 'Yes'
else:
ans ='No'
print(ans)
```
| 91,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
q = int(input())
for queries in range(0, q):
a, b, n = map(int, input().split())
ma = list(map(int, input().split()))
mb = list(map(int, input().split()))
mx = ma[0]
for i in range(0, n):
t = (mb[i] + a - 1) // a
b -= (t * ma[i])
mx = max(ma[i], mx)
b += mx
if (b > 0): print("YES")
else: print("NO")
```
| 91,946 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
if __name__ == "__main__":
t = int(input())
while t:
A,H,n = list(map(int, input().split()))
a = list(map(int, input().split()))
h = list(map(int, input().split()))
# now I have all info
damage = 0
for i in range(n):
damage = damage + ((h[i] + A - 1)// A) * a[i]
flag = False
for i in range(n):
if (H - (damage - a[i])) > 0:
flag = True
break
if flag:
print("YES")
else:
print("NO")
t = t - 1
```
| 91,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
import math, sys
from collections import defaultdict, Counter, deque
from bisect import bisect_left, bisect_right
INF = float('inf')
MOD = int(1e9) + 7
MAX = int(1e6) + 1
def solve():
attack, health, n = vars()
a = array()
b = array()
total = 0
mon = []
for i in range(n):
mon.append((a[i], b[i]))
mon = sorted(mon)
for i in range(n):
t1 = math.ceil(mon[i][1] / attack)
damage = t1 * mon[i][0]
if total + damage >= health:
if i == n - 1:
t2 = math.ceil((health - total) / mon[i][0])
if t1 == t2:
break
print('NO')
return
total += damage
print('YES')
def main():
t = 1
t = int(input())
for _ in range(t):
solve()
def gcd(a, b):
while b:
a, b = b, a%b
return a
def input():
return sys.stdin.readline().rstrip('\n').strip()
def print(*args, sep=' ', end='\n'):
first = True
for arg in args:
if not first:
sys.stdout.write(sep)
sys.stdout.write(str(arg))
first = False
sys.stdout.write(end)
primes = [ 1 for i in range(MAX) ]
def sieve():
global primes
primes[0] = primes[1] = 0
i = 2
while i <= MAX ** 0.5:
j = i * i
while primes[i] and j < MAX:
if j % i == 0:
primes[j] = 0
j += i
i += 1
def vars():
return map(int, input().split())
def array():
return list(map(int, input().split()))
if __name__ == "__main__":
main()
```
| 91,948 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
# pylint: disable=unused-variable
# pylint: enable=too-many-lines
# * Just believe in yourself
# @ Author @CAP
# import numpy
import os
import sys
from io import BytesIO, IOBase
import math as M
import itertools as ITR
from collections import defaultdict as D
from collections import Counter as C
from collections import deque as Q
import threading
from functools import lru_cache, reduce
from functools import cmp_to_key as CMP
from bisect import bisect_left as BL
from bisect import bisect_right as BR
import random as R
import string
import cmath, time
enum = enumerate
start_time = time.time()
# * Variables
MOD = 1_00_00_00_007
MA = float("inf")
MI = float("-inf")
# * Graph 8 direction
di8 = ((1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1))
# * Graph 4 direction
di4 = ((1, 0), (0, 1), (-1, 0), (0, -1))
# * Stack increment
def increase_stack():
sys.setrecursionlimit(2 ** 32 // 2 - 1)
threading.stack_size(1 << 27)
# sys.setrecursionlimit(10**6)
# threading.stack_size(10**8)
# t = threading.Thread(target=main)
# t.start()
# t.join()
# * Region Funtions
def binary(n):
return bin(n)[2:]
def decimal(s):
return int(s, 2)
def pow2(n):
p = 0
while n > 1:
n //= 2
p += 1
return p
def maxfactor(n):
q = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
q.append(i)
if q:
return q[-1]
def factors(n):
q = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
q.append(i)
q.append(n // i)
return list(sorted(list(set(q))))
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(M.sqrt(n)) + 1, 2):
while n % i == 0:
l.append(i)
n = n / i
if n > 2:
l.append(int(n))
l.sort()
return l
def isPrime(n):
if n == 1:
return False
else:
root = int(n ** 0.5)
root += 1
for i in range(2, root):
if n % i == 0:
return False
return True
def seive(n):
a = []
prime = [True for i in range(n + 1)]
p = 2
while p * p <= n:
if prime[p] == True:
for i in range(p ** 2, n + 1, p):
prime[i] = False
p = p + 1
for p in range(2, n + 1):
if prime[p]:
a.append(p)
prime[0] = prime[1] = False
return a, prime
def maxPrimeFactors(n):
maxPrime = -1
while n % 2 == 0:
maxPrime = 2
n >>= 1
for i in range(3, int(M.sqrt(n)) + 1, 2):
while n % i == 0:
maxPrime = i
n = n / i
if n > 2:
maxPrime = n
return int(maxPrime)
def countchar(s, i):
c = 0
ch = s[i]
for i in range(i, len(s)):
if s[i] == ch:
c += 1
else:
break
return c
def str_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return q
def lis(arr):
n = len(arr)
lis = [1] * n
maximum = 0
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j] and lis[i] < lis[j] + 1:
lis[i] = lis[j] + 1
maximum = max(maximum, lis[i])
return maximum
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
def agcd(a, *b):
for i in b:
a = gcd(a, i)
return a
def lcm(a, *b):
val = a
gc = a
for i in b:
gc = gcd(gc, i)
val *= i
return val // gc
def ncr(n, r):
return M.factorial(n) // (M.factorial(n - r) * M.factorial(r))
def npr(n, r):
return M.factorial(n) // M.factorial(n - r)
# * Make work easy funtions
def IF(c, t, f):
return t if c else f
def YES(c):
print(IF(c, "YES", "NO"))
def Yes(c):
print(IF(c, "Yes", "No"))
def yes(c):
print(IF(c, "yes", "no"))
def JA(a, sep=" "):
print(sep.join(map(str, a)))
def JAA(a, s="\n", t=" "):
return s.join(t.join(map(str, b)) for b in a)
def PS(a, s=" "):
print(str(a), end=s)
# * Region Taking Input
def I():
return int(inp())
def F():
return float(inp())
def LI():
return list(map(int, inp().split()))
def LF():
return list(map(float, inp().split()))
def MATI(n):
return [LI() for i in range(n)]
def MATS(n):
return [list(inp()) for i in range(n)]
def IV():
return map(int, inp().split())
def FV():
return map(float, inp().split())
def LS():
return list(inp())
def S():
return inp()
# * Region Fastio
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)
inp = lambda: sys.stdin.readline().rstrip("\r\n")
# >==================================== Write The Useful Code Here ============================
# >Make it one if there is some test cases
TestCases = 1 # >=====================
# > =======================================
"""
> Sometimes later becomes never. Do it now.
! Be Better than yesterday.
!Your limitation—it’s only your imagination.
> Push yourself, because no one else is going to do it for you.
? The harder you work for something, the greater you’ll feel when you achieve it.
! Great things never come from comfort zones.
! Don’t stop when you’re tired. Stop when you’re done.
> Do something today that your future self will thank you for.
? It’s going to be hard, but hard does not mean impossible.
! Sometimes we’re tested not to show our weaknesses, but to discover our strengths.
"""
# @ Goal is to get Candidate Master
def solve():
p, h, n = IV()
parr = LI()
harr = LI()
arr = []
for i in range(n):
arr.append((parr[i], harr[i]))
arr.sort(key=lambda x: x[0])
for x, y in arr:
if h <= 0:
print("NO")
return
val = M.ceil(y / p)
q = (val - 1) * x
if q >= h:
print("NO")
return
else:
h -= val * x
# print(h)
print("YES")
# ! This is the Main Function
def main():
flag = 1
#! Checking we are offline or not
try:
sys.stdin = open(
"c:/Users/Manoj Chowdary/Documents/python/CodeForces/contest-div-2/input.txt",
"r",
)
sys.stdout = open(
"c:/Users/Manoj Chowdary/Documents/python/CodeForces/contest-div-2/output.txt",
"w",
)
except:
flag = 0
t = 1
if TestCases:
t = I()
for _ in range(1, t + 1):
solve()
if flag:
print("Time: %.4f sec" % (time.time() - start_time))
localtime = time.asctime(time.localtime(time.time()))
print(localtime)
sys.stdout.close()
# ! End Region
if __name__ == "__main__":
# ? Incresing Stack Limit
# increase_stack()
#! Calling Main Function
main()
```
| 91,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Tags: greedy, implementation, sortings
Correct Solution:
```
import math as m
if __name__ == '__main__':
for _ in range (int(input())):
A,B,n = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
flag =1
r = 0
fl = max(a)
for i in range (n):
r = m.ceil(b[i]/A)
B-=r*a[i]
if B <= -fl:
print("NO")
else:
print("YES")
```
| 91,950 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
import math
for _ in range(int(input())):
att,health,mon=map(int,input().split())
monatt=list(map(int,input().split()))
monhel=list(map(int,input().split()))
max1=-float('Inf')
for i in range(len(monatt)):
if monatt[i]>max1:
store=i
max1=monatt[i]
for i in range(len(monatt)):
if i==store:
continue
health-=math.ceil(monhel[i]/att)*monatt[i]
if health<=0:
print("NO")
continue
health-=math.ceil(monhel[store]/att)*monatt[store]
if health+monatt[store]>0:
print("YES")
else:
print("NO")
```
Yes
| 91,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
import sys
import bisect as bi
import math
from collections import defaultdict as dd
from types import GeneratorType
##import queue
##from heapq import heapify, heappush, heappop
##import itertools
##import io
##import os
##import operator
##import random
##sys.setrecursionlimit(10**7)
input=sys.stdin.readline
##input = io.BytesIsoO(os.read(0, os.fstat(0).st_size)).readline
##fo=open("output2.txt","w")
##fi=open("input2.txt","w")
mo=10**9+7
MOD=998244353
def cin():return map(int,sin().split())
def ain():return list(map(int,sin().split()))
def sin():return input().strip()
def inin():return int(input())
##----------------------------------------------------------------------------------------------------------------------
##
for _ in range(inin()):
a,b,n=cin()
am=ain()
bm=ain()
both=[]
for i in range(n):
both+=[(am[i],bm[i])]
both.sort()
flag=1
for i in range(n):
(AM,BM)=both[i]
times=(BM+a-1)//a
b=b-(times-1)*AM
if(b>0):
b-=AM
else:
flag=0
break
## print(b,times,i)
if(flag):print("YES")
else:print("NO")
##-----------------------------------------------------------------------------------------------------------------------#
def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1)
##2 ki power
def pref(a,n,f):
pre=[0]*n
if(f==0): ##from beginning
pre[0]=a[0]
for i in range(1,n):
pre[i]=a[i]+pre[i-1]
else: ##from end
pre[-1]=a[-1]
for i in range(n-2,-1,-1):
pre[i]=pre[i+1]+a[i]
return pre
# in given set of integers
def kadane(A):
maxSoFar = maxEndingHere = start = end = beg = 0
for i in range(len(A)):
maxEndingHere = maxEndingHere + A[i]
if maxEndingHere < 0:
maxEndingHere = 0;beg = i + 1
if maxSoFar < maxEndingHere:
maxSoFar = maxEndingHere
start = beg
end = i
return (maxSoFar,start,end) #max subarray sum and its range
def modFact(n, p):
if(n<0):return 0
if n >= p: return 0
result = 1
for i in range(1, n + 1):result = (result * i) % p
return result
def ncr(n, r, p):
if(n<r or n<0): return 0
num = den = 1
for i in range(r):
num = (num * (n - i)) % p ;den = (den * (i + 1)) % p
return (num * pow(den,p - 2, p)) % p
##https://github.com/cheran-senthil/PyRival/blob/master/pyrival/misc/bootstrap.py
##write @bootstrap before recursive func
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
```
Yes
| 91,952 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
for _ in range(int(input())):
a, b, n = [int(x) for x in input().split()]
As = [int(x) for x in input().split()]
Bs = [int(x) for x in input().split()]
final = True
for i in range(n):
if b <= 0:
final = False
k = (Bs[i] // a) + (1 if Bs[i] % a != 0 else 0)
if k * As[i] >= b + As[i]:
final = False
b -= (k * As[i])
if b + max(As) > 0:
final = True
print("YES" if final else "NO")
```
Yes
| 91,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
for _ in range(int(input())):
A,B,n=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
s,f=0,"NO"
for i in range(n):
c=(b[i]+A-1)//A
s+=c*a[i]
for i in range(n):
c=(b[i]+A-1)//A
d=s-c*a[i]
if B > s-a[i]:
f="YES"
break
print(f)
```
Yes
| 91,954 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
def swapp(A,B,n,a,b):
for i in range(len(a)):
if B<1:
break
while b[i]>0 and B>0:
B=B-a[i]
b[i]=b[i]-A
if b[n-1]>=1:
return("NO")
else:
return("YES")
t = int(input())
for i in range(t):
NM = input().split()
A = int(NM[0])
B = int(NM[1])
n = int(NM[2])
a = list(map(int, input().rstrip().split()))
b = list(map(int, input().rstrip().split()))
ans = swapp(A,B,n,a,b)
print(ans)
```
No
| 91,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
for _ in range(int(input())):
A,B,n=input().split()
A,B,n=int(A),int(B),int(n)
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if((B-sum(a))>=(sum(b)-(A*n))):
print('YES')
else:
c=0
for i in range(n):
B=B-a[i]
b[i]=b[i]-A
if(B<=0 and b[i]>0):
c=-1
break
else:
c+=1
if(c==-1):
print('NO')
else:
print('YES')
```
No
| 91,956 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
def ceildiv(a, b):
return -(-a // b)
t = int(input())
for _ in range(t):
A, B, n = map(int, input().strip().split())
a = list(map(int, input().strip().split()))
b = list(map(int, input().strip().split()))
HP = B
for i in range(n):
num = ceildiv(b[i], A)
HP -= num * a[i]
if HP >= 0:
print("YES")
continue
maxaidx = max(list(range(n)), key=lambda i: a[i])
idxs = list(range(n))
idxs[maxaidx] = n - 1
idxs[n - 1] = maxaidx
HP = B
for i in idxs[:-1]:
num = ceildiv(b[i], A)
HP -= num * a[i]
num = ceildiv(b[idxs[-1]], A)
HP -= (num-1) * a[i]
if HP > 0:
print("YES")
else:
print("NO")
```
No
| 91,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The great hero guards the country where Homer lives. The hero has attack power A and initial health value B. There are n monsters in front of the hero. The i-th monster has attack power a_i and initial health value b_i.
The hero or a monster is said to be living, if his or its health value is positive (greater than or equal to 1); and he or it is said to be dead, if his or its health value is non-positive (less than or equal to 0).
In order to protect people in the country, the hero will fight with monsters until either the hero is dead or all the monsters are dead.
* In each fight, the hero can select an arbitrary living monster and fight with it. Suppose the i-th monster is selected, and the health values of the hero and the i-th monster are x and y before the fight, respectively. After the fight, the health values of the hero and the i-th monster become x-a_i and y-A, respectively.
Note that the hero can fight the same monster more than once.
For the safety of the people in the country, please tell them whether the great hero can kill all the monsters (even if the great hero himself is dead after killing the last monster).
Input
Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers A (1 ≤ A ≤ 10^6), B (1 ≤ B ≤ 10^6) and n (1 ≤ n ≤ 10^5) — the attack power of the great hero, the initial health value of the great hero, and the number of monsters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i denotes the attack power of the i-th monster.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^6), where b_i denotes the initial health value of the i-th monster.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer: "YES" (without quotes) if the great hero can kill all the monsters. Otherwise, print "NO" (without quotes).
Example
Input
5
3 17 1
2
16
10 999 3
10 20 30
100 50 30
1000 1000 4
200 300 400 500
1000 1000 1000 1000
999 999 1
1000
1000
999 999 1
1000000
999
Output
YES
YES
YES
NO
YES
Note
In the first example: There will be 6 fights between the hero and the only monster. After that, the monster is dead and the health value of the hero becomes 17 - 6 × 2 = 5 > 0. So the answer is "YES", and moreover, the hero is still living.
In the second example: After all monsters are dead, the health value of the hero will become 709, regardless of the order of all fights. So the answer is "YES".
In the third example: A possible order is to fight with the 1-st, 2-nd, 3-rd and 4-th monsters. After all fights, the health value of the hero becomes -400. Unfortunately, the hero is dead, but all monsters are also dead. So the answer is "YES".
In the fourth example: The hero becomes dead but the monster is still living with health value 1000 - 999 = 1. So the answer is "NO".
Submitted Solution:
```
import math
for _ in range(int(input())):
a, hp, n = map(int, input().split())
mA = list(map(int, input().split()))
mHp = list(map(int, input().split()))
fights = [0] * n
for j in range(n):
rounds = min(math.ceil(mHp[j]/a), math.ceil(hp/mA[j]))
fights[j] = (rounds * mA[j], rounds, j)
fights = sorted(fights, key=lambda x: x[0])
for fight in fights:
if hp < 1:
print('NO')
break
hp -= fight[0]
if hp >= 1:
print('YES')
elif hp < 1 and mHp[fights[-1][2]] - (a * fights[-1][1]) < 1:
print('YES')
else:
print('NO')
```
No
| 91,958 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
import sys, math
import heapq
from collections import deque
input = sys.stdin.readline
#input
def ip(): return int(input())
def sp(): return str(input().rstrip())
def mip(): return map(int, input().split())
def msp(): return map(str, input().split())
def lmip(): return list(map(int, input().split()))
def lmsp(): return list(map(str, input().split()))
#gcd, lcm
def gcd(x, y):
while y:
x, y = y, x % y
return x
def lcm(x, y):
return x * y // gcd(x, y)
#prime
def isPrime(x):
if x==1: return False
for i in range(2, int(x**0.5)+1):
if x%i==0:
return False
return True
# Union Find
# p = {i:i for i in range(1, n+1)}
def find(x):
if x == p[x]:
return x
q = find(p[x])
p[x] = q
return q
def union(x, y):
global n
x = find(x)
y = find(y)
if x != y:
p[y] = x
def getPow(a, x):
ret =1
while x:
if x&1:
ret = (ret*a) % MOD
a = (a*a)%MOD
x>>=1
return ret
def getInv(x):
return getPow(x, MOD-2)
def lowerBound(start, end, key):
while start < end:
mid = (start + end) // 2
if lst[mid] == key:
end = mid
elif key < lst[mid]:
end = mid
elif lst[mid] < key:
start = mid + 1
return end
def upperBound(start, end, key):
while start < end:
mid = (start + end) // 2
if lst[mid] == key:
start = mid + 1
elif lst[mid] < key:
start = mid + 1
elif key < lst[mid]:
end = mid
if end == len(lst):
return end-1
return end
############### Main! ###############
t = ip()
while t:
t -= 1
n=ip()
a=sp()
b=sp()
c=sp()
i,j,k=0,0,0
ans=[]
while True:
if i==2*n:
if j>=k:
for m in range (j,2*n):
ans.append(b[m])
else:
for m in range (k,2*n):
ans.append(c[m])
break
elif j==2*n:
if i>=k:
for m in range (i,2*n):
ans.append(a[m])
else:
for m in range (k,2*n):
ans.append(c[m])
break
elif k==2*n:
if j>=i:
for m in range (j,2*n):
ans.append(b[m])
else:
for m in range (i,2*n):
ans.append(a[m])
break
if a[i]==b[j]:
ans.append(a[i])
i+=1
j+=1
elif b[j]==c[k]:
ans.append(b[j])
j+=1
k+=1
else:
ans.append(a[i])
i+=1
k+=1
print(*ans,sep='')
######## Priest W_NotFoundGD ########
```
| 91,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def solve(n,ls):
x,y = [i.count('0') for i in ls],[i.count('1')for i in ls]
for i in range(3):
for j in range(3):
if i == j:
continue
if x[i] >= n and x[j] >= n:
ans,b = [],0
for a in ls[i]:
if a == '1':
ans.append('1')
continue
while b != 2*n and ls[j][b] != '0':
ans.append('1')
b += 1
ans.append('0')
if b != 2*n:
b += 1
while b != 2*n:
ans.append(ls[j][b])
b += 1
return ''.join(ans)
if y[i] >= n and y[j] >= n:
ans,b = [],0
for a in ls[i]:
if a == '0':
ans.append('0')
continue
while b != 2*n and ls[j][b] != '1':
ans.append('0')
b += 1
ans.append('1')
if b != 2*n:
b += 1
while b != 2*n:
ans.append(ls[j][b])
b += 1
return ''.join(ans)
def main():
for _ in range(int(input())):
n = int(input())
ls = [input().strip() for _ in range(3)]
print(solve(n,ls))
#Fast IO Region
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()
```
| 91,960 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n = int(input())
a = []
a.append(input())
a.append(input())
a.append(input())
zeroDominant = []
oneDominant = []
for k in range(3):
cnt = 0
for i in range(2 * n):
if a[k][i] == "1":
cnt += 1
if cnt >= n:
oneDominant.append(a[k])
else:
zeroDominant.append(a[k])
if len(oneDominant) >= 2:
n *= 2
ans = []
fIndex = 0
lIndex = 0
while fIndex < n or lIndex < n:
if fIndex == n and lIndex < n:
ans.append(oneDominant[1][lIndex])
lIndex += 1
elif lIndex == n and fIndex < n:
ans.append(oneDominant[0][fIndex])
fIndex += 1
else:
if oneDominant[0][fIndex] == oneDominant[1][lIndex]:
ans.append(oneDominant[0][fIndex])
fIndex += 1
lIndex += 1
else:
ans.append("0")
if oneDominant[0][fIndex] == "0":
fIndex += 1
else:
lIndex += 1
print("".join(ans))
else:
n *= 2
ans = []
fIndex = 0
lIndex = 0
while fIndex < n or lIndex < n:
if fIndex == n and lIndex < n:
ans.append(zeroDominant[1][lIndex])
lIndex += 1
elif lIndex == n and fIndex < n:
ans.append(zeroDominant[0][fIndex])
fIndex += 1
else:
if zeroDominant[0][fIndex] == zeroDominant[1][lIndex]:
ans.append(zeroDominant[0][fIndex])
fIndex += 1
lIndex += 1
else:
ans.append("1")
if zeroDominant[0][fIndex] == "1":
fIndex += 1
else:
lIndex += 1
print("".join(ans))
# region fastio
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")
# endregion
if __name__ == "__main__":
main()
```
| 91,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
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")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def A(n):return [0]*n
def AI(n,x): return [x]*n
def A2(n,m): return [[0]*m for i in range(n)]
def G(n): return [[] for i in range(n)]
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
mod=10**9+7
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
fact(x,mod)
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa.reverse()
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
if a==0:return b//c*(n+1)
if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
m=(a*n+b)//c
return n*m-floorsum(c,c-b-1,a,m-1)
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
def lowbit(n):
return n&-n
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class ST:
def __init__(self,arr):#n!=0
n=len(arr)
mx=n.bit_length()#取不到
self.st=[[0]*mx for i in range(n)]
for i in range(n):
self.st[i][0]=arr[i]
for j in range(1,mx):
for i in range(n-(1<<j)+1):
self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
def query(self,l,r):
if l>r:return -inf
s=(r+1-l).bit_length()-1
return max(self.st[l][s],self.st[r-(1<<s)+1][s])
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]>self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]'''
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
seen.add(u)
for v,w in graph[u]:
if v not in d or d[v]>d[u]+w:
d[v]=d[u]+w
heappush(heap,(d[v],v))
return d
def bell(s,g):#bellman-Ford
dis=AI(n,inf)
dis[s]=0
for i in range(n-1):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change=A(n)
for i in range(n):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change[v]=1
return dis
def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
res=[]
for k in nums:
i=bisect.bisect_left(res,k)
if i==len(res):
res.append(k)
else:
res[i]=k
return len(res)
def RP(nums):#逆序对
n = len(nums)
s=set(nums)
d={}
for i,k in enumerate(sorted(s),1):
d[k]=i
bi=BIT([0]*(len(s)+1))
ans=0
for i in range(n-1,-1,-1):
ans+=bi.query(d[nums[i]]-1)
bi.update(d[nums[i]],1)
return ans
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j,n,m):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
def topo(n):
q=deque()
res=[]
for i in range(1,n+1):
if ind[i]==0:
q.append(i)
res.append(i)
while q:
u=q.popleft()
for v in g[u]:
ind[v]-=1
if ind[v]==0:
q.append(v)
res.append(v)
return res
@bootstrap
def gdfs(r,p):
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
def cal(x):
ans=0
for i in range(2*n):
ans+=x[i]=='1'
return ans
def f(x,y,ch):
ans=[]
i=j=0
c=0
while True:
while i<2*n and x[i]!=ch:
ans.append(x[i])
i+=1
while j<2*n and y[j]!=ch:
ans.append(y[j])
j+=1
ans.append(x[i])
i+=1
j+=1
c+=1
if c>=n:
break
ans=ans+x[i:]+y[j:]
return ''.join(ans)
t=N()
for i in range(t):
n=N()
a=list(input())
b=list(input())
c=list(input())
ca=cal(a)
cb=cal(b)
cc=cal(c)
res=[[a,ca],[b,cb],[c,cc]]
res.sort(key=lambda x:x[1])
[a,ca],[b,cb],[c,cc]=res
if cb>=n:
ans=f(b,c,'1')
else:
ans=f(a,b,'0')
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''
```
| 91,962 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
ans=[]
s1=input()
s2=input()
s3=input()
fincount=0
count10=0
for i in range(2*n):
if s1[i]=='0':
count10+=1
if count10>=n:
fincount+=1
count20=0
for i in range(2*n):
if s2[i]=='0':
count20+=1
if count20>=n:
fincount+=1
count30=0
for i in range(2*n):
if s3[i]=='0':
count30+=1
if count30>=n:
fincount+=1
if fincount>=2:
mode='0'
unmode='1'
else:
mode='1'
unmode='0'
if mode=='0':
if count10>=n and count20>=n:
st1=s1
st2=s2
elif count10>=n and count30>=n:
st1=s1
st2=s3
elif count20>=n and count30>=n:
st1=s2
st2=s3
else:
if count10<=n and count20<=n:
st1=s1
st2=s2
elif count10<=n and count30<=n:
st1=s1
st2=s3
elif count20<=n and count30<=n:
st1=s2
st2=s3
pointer1=0
pointer2=0
while pointer1<2*n or pointer2<2*n:
while pointer1<2*n and st1[pointer1]==unmode:
pointer1+=1
ans.append(unmode)
while pointer2<2*n and st2[pointer2]==unmode:
pointer2+=1
ans.append(unmode)
if pointer1>=2*n and pointer2>=2*n:
break
ans.append(mode)
pointer1+=1
pointer2+=1
print(''.join(ans))
```
| 91,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
import sys,os
from io import BytesIO,IOBase
mod = 10**9+7; Mod = 998244353; INF = float('inf')
# input = lambda: sys.stdin.readline().rstrip("\r\n")
# inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
#______________________________________________________________________________________________________
# region fastio
'''
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)
# endregion'''
#______________________________________________________________________________________________________
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
# ______________________________________________________________________________________________________
# from math import *
# from bisect import *
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
# from itertools import groupby
# sys.setrecursionlimit(2000+10) #this is must for dfs
# ______________________________________________________________________________________________________
# segment tree for range minimum query
# n = int(input())
# a = list(map(int,input().split()))
# st = [float('inf') for i in range(4*len(a))]
# def build(a,ind,start,end):
# if start == end:
# st[ind] = a[start]
# else:
# mid = (start+end)//2
# build(a,2*ind+1,start,mid)
# build(a,2*ind+2,mid+1,end)
# st[ind] = min(st[2*ind+1],st[2*ind+2])
# build(a,0,0,n-1)
# def query(ind,l,r,start,end):
# if start>r or end<l:
# return float('inf')
# if l<=start<=end<=r:
# return st[ind]
# mid = (start+end)//2
# return min(query(2*ind+1,l,r,start,mid),query(2*ind+2,l,r,mid+1,end))
# ______________________________________________________________________________________________________
# Checking prime in O(root(N))
# def isprime(n):
# if (n % 2 == 0 and n > 2) or n == 1: return 0
# else:
# s = int(n**(0.5)) + 1
# for i in range(3, s, 2):
# if n % i == 0:
# return 0
# return 1
# def lcm(a,b):
# return (a*b)//gcd(a,b)
# ______________________________________________________________________________________________________
# nCr under mod
# def C(n,r,mod = 10**9+7):
# if r>n: return 0
# if r>n-r: r = n-r
# num = den = 1
# for i in range(r):
# num = (num*(n-i))%mod
# den = (den*(i+1))%mod
# return (num*pow(den,mod-2,mod))%mod
# def C(n,r):
# if r>n:
# return 0
# if r>n-r:
# r = n-r
# ans = 1
# for i in range(r):
# ans = (ans*(n-i))//(i+1)
# return ans
# ______________________________________________________________________________________________________
# For smallest prime factor of a number
# M = 2*10**5+10
# spf = [i for i in range(M)]
# def spfs(M):
# for i in range(2,M):
# if spf[i]==i:
# for j in range(i*i,M,i):
# if spf[j]==j:
# spf[j] = i
# return
# spfs(M)
# ______________________________________________________________________________________________________
# def gtc(p):
# print('Case #'+str(p)+': ',end='')
# ______________________________________________________________________________________________________
tc = 1
tc = int(input())
for test in range(1,tc+1):
n = int(input())
s1 = str(input())
s2 = str(input())
s3 = str(input())
o1 = s1.count('1');z1 = 2*n-o1
o2 = s2.count('1');z2 = 2*n-o2
o3 = s3.count('1');z3 = 2*n-o3
def solve(s1,s2,k):
ans = []
i = 0
j = 0
n = len(s1)
while(i<n and j<n):
if s1[i]==s2[j]:
ans.append(s1[i])
i+=1
j+=1
else:
if s1[i]==k:
ans.append(s2[j])
j+=1
else:
ans.append(s1[i])
i+=1
while(i<n):
ans.append(s1[i])
i+=1
while(j<n):
ans.append(s2[j])
j+=1
# print(s1,s2,ans)
return ''.join(ans)+'0'*((n//2)*3-len(ans))
if o1>=n and o2>=n:
ans = solve(s1,s2,'1')
elif o2>=n and o3>=n:
ans = solve(s2,s3,'1')
elif o1>=n and o3>=n:
ans = solve(s1,s3,'1')
elif z1>=n and z2>=n:
ans = solve(s1,s2,'0')
elif z2>=n and z3>=n:
ans = solve(s2,s3,'0')
else:
ans = solve(s1,s3,'0')
print(ans)
```
| 91,964 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
import sys
from sys import stdin
def solve(a,b,X):
#print (a,b,X,file=sys.stderr)
ans = []
ca = [a[i] for i in range(len(a))]
cb = [b[i] for i in range(len(b))]
while len(ca) > 0 or len(cb) > 0:
if len(cb) == 0:
ans.append(ca[-1])
del ca[-1]
elif len(ca) == 0:
ans.append(cb[-1])
del cb[-1]
elif ca[-1] == cb[-1]:
ans.append(ca[-1])
del ca[-1]
del cb[-1]
else:
ans.append(X)
if ca[-1] == X:
del ca[-1]
else:
del cb[-1]
while len(ans) < 3*n:
ans.append("0")
ans.reverse()
return "".join(ans)
tt = int(stdin.readline())
for loop in range(tt):
n = int(stdin.readline())
s = [stdin.readline()[:-1] for i in range(3)]
lis = []
for i in range(3):
one = 0
zero = 0
for j in s[i]:
if j == "0":
zero += 1
else:
one += 1
if one < zero:
lis.append("1")
else:
lis.append("0")
flag = True
for i in range(3):
for j in range(i):
if lis[i] == lis[j] and flag:
print (solve(s[i],s[j],lis[i]))
flag = False
```
| 91,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Tags: constructive algorithms, greedy, implementation, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
n = int(input())
S1 = input().strip()
S2 = input().strip()
S3 = input().strip()
for s1 in (S1, S2, S3):
for s2 in (S1, S2, S3):
if s1 == s2:
continue
z1 = s1.count("0")
o1 = s1.count("1")
z2 = s2.count("0")
o2 = s2.count("1")
mz = max(z1, z2)
mo = max(o1, o2)
if mz > n and mo > n:
continue
if mz <= n:
cnt1 = []
cnt2 = []
bef = "0"
cnt = 0
for s in s1:
if s == "1":
cnt1.append(cnt)
cnt = 0
else:
cnt += 1
bef = s
if cnt:
cnt1.append(cnt)
bef = "0"
cnt = 0
for s in s2:
if s == "1":
cnt2.append(cnt)
cnt = 0
else:
cnt += 1
bef = s
if cnt:
cnt2.append(cnt)
if len(cnt1) > len(cnt2):
cnt2 += [0] * (len(cnt1) - len(cnt2))
elif len(cnt1) < len(cnt2):
cnt1 += [0] * (len(cnt2) - len(cnt1))
ans = []
for c1, c2 in zip(cnt1, cnt2):
ans += ["0"] * max(c1, c2)
ans += ["1"]
ans.pop()
if len(ans) < 3 * n:
ans += ["1"] * (3 * n - len(ans))
print("".join(ans))
return
elif mo <= n:
cnt1 = []
cnt2 = []
bef = "1"
cnt = 0
for s in s1:
if s == "0":
cnt1.append(cnt)
cnt = 0
else:
cnt += 1
bef = s
if cnt:
cnt1.append(cnt)
bef = "1"
cnt = 0
for s in s2:
if s == "0":
cnt2.append(cnt)
cnt = 0
else:
cnt += 1
bef = s
if cnt:
cnt2.append(cnt)
if len(cnt1) > len(cnt2):
cnt2 += [0] * (len(cnt1) - len(cnt2))
elif len(cnt1) < len(cnt2):
cnt1 += [0] * (len(cnt2) - len(cnt1))
ans = []
for c1, c2 in zip(cnt1, cnt2):
ans += ["1"] * max(c1, c2)
ans += ["0"]
ans.pop()
if len(ans) < 3 * n:
ans += ["0"] * (3 * n - len(ans))
print("".join(ans))
return
for _ in range(int(input())):
main()
```
| 91,966 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function5():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
import os,sys
from io import BytesIO,IOBase
def solve(n,ls):
x,y = [i.count('0') for i in ls],[i.count('1')for i in ls]
for i in range(3):
for j in range(3):
if i == j:
continue
if x[i] >= n and x[j] >= n:
ans,b = [],0
for a in ls[i]:
if a == '1':
ans.append('1')
continue
while b != 2*n and ls[j][b] != '0':
ans.append('1')
b += 1
ans.append('0')
if b != 2*n:
b += 1
while b != 2*n:
ans.append(ls[j][b])
b += 1
return ''.join(ans)
if y[i] >= n and y[j] >= n:
ans,b = [],0
for a in ls[i]:
if a == '0':
ans.append('0')
continue
while b != 2*n and ls[j][b] != '1':
ans.append('0')
b += 1
ans.append('1')
if b != 2*n:
b += 1
while b != 2*n:
ans.append(ls[j][b])
b += 1
return ''.join(ans)
def main():
for _ in range(int(input())):
n = int(input())
ls = [input().strip() for _ in range(3)]
print(solve(n,ls))
#Fast IO Region
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")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function6():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function7():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function8():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__ == '__main__':
main()
```
Yes
| 91,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for f in range(int(input())):
n=int(input())
s=[]
for i in range(3):
s.append(input())
s1=[]
s0=[]
for i in range(3):
c0=0
for x in s[i]:
if x=="0":c0+=1
s0.append(c0)
s1.append(2*n-s0[-1])
for i in range(2):
for j in range(i+1,3):
if s0[i]>=n and s0[j]>=n:
c1=i
c2=j
g="0"
b="1"
if s1[i]>=n and s1[j]>=n:
c1=i
c2=j
g="1"
b="0"
i=j=0
while i<2*n or j<2*n:
if i<2*n and s[c1][i]==b:
i+=1
print(end=b)
else:
if j<2*n and s[c2][j]==b:
j+=1
print(end=b)
else:
print(end=g)
i+=1
j+=1
print()
```
Yes
| 91,968 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
tc = int(input())
for i in range(tc):
n = int(input())
s = [input() for _ in range(3)]
p = [0]*3
ans = ""
while True:
ones = []
oc = 0
zeroes = []
for i in range(3):
if s[i][p[i]] == '1':
ones.append(i)
oc += 1
else:
zeroes.append(i)
if oc >= 2:
print('1', end="")
for i in ones:
p[i] += 1
else:
print('0', end="")
for i in zeroes:
p[i] += 1
if (p[0] == 2*n) or (p[1] == 2*n) or (p[2] == 2*n):
break
#pick the smallest left omg
x = sorted(enumerate(p), key=lambda i:i[1])
print(s[x[1][0]][x[1][1]:])
#3 pointer trick in comments
```
Yes
| 91,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
from sys import *
input = stdin.readline
def superstring(x,y,fx,n):
ans = ""
i,j = 0,0
while (i<2*n) and (j<2*n):
if x[i]==y[j]:
ans+=x[i]
i+=1
j+=1
elif x[i]!=fx:
ans+=x[i]
i+=1
else:
ans+=y[j]
j+=1
return ans+x[i:]+y[j:]
for _ in range(int(input())):
n = int(input())
a,b,c = input().strip(),input().strip(),input().strip()
one,zero = [],[]
for i in [a,b,c]:
if i.count("0")>=n:
zero.append(i)
else:
one.append(i)
currans = ""
if len(zero)>len(one):
currans = superstring(zero[0],zero[1],'0',n)
else:
currans = superstring(one[0],one[1],'1',n)
print(currans.rjust(3*n,'0'))
```
Yes
| 91,970 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
import sys
from sys import stdin
def solve(a,b,X):
ans = []
ca = [a[i] for i in range(len(a))]
cb = [b[i] for i in range(len(b))]
while len(ca) > 0 and len(cb) > 0:
if len(cb) == 0:
ans.append(ca[-1])
del ca[-1]
elif len(ca) == 0:
ans.append(cb[-1])
del cb[-1]
elif ca[-1] == cb[-1]:
ans.append(ca[-1])
del ca[-1]
del cb[-1]
else:
ans.append(X)
if ca[-1] == X:
del ca[-1]
else:
del cb[-1]
while len(ans) < 3*n:
ans.append("0")
ans.reverse()
return "".join(ans)
tt = int(stdin.readline())
for loop in range(tt):
n = int(stdin.readline())
s = [stdin.readline()[:-1] for i in range(3)]
lis = []
for i in range(3):
one = 0
zero = 0
for j in s[i]:
if j == "0":
zero += 1
else:
one += 1
if one < zero:
lis.append("1")
else:
lis.append("0")
flag = True
for i in range(3):
for j in range(i):
if lis[i] == lis[j] and flag:
print (solve(s[i],s[j],lis[i]))
flag = False
```
No
| 91,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
tc = int(input())
for i in range(tc):
n = int(input())
s = [input() for _ in range(3)]
p = [0]*3
ans = ""
while True:
ones = []
zeroes = []
for i in range(3):
if p[i] == 2*n:
continue
if s[i][p[i]] == '1':
ones.append(i)
else:
zeroes.append(i)
if len(ones) > len(zeroes):
ans += '1'
for i in ones:
p[i] += 1
else:
ans += '0'
for i in zeroes:
p[i] += 1
if ((p[0] == 2*n) + (p[1] == 2*n) + (p[2] == 2*n)) >= 2:
break
print(ans)
#3 pointer trick in comments
```
No
| 91,972 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
def sol(val,a,b):
i = 0;
j = 0;
f = 0;
while (j <len(b)):
z = val.find(b[j], i, len(val))
if z != -1:
i = z + 1
else:
return 1
break
j += 1
print(val)
for i in range(int(input())):
n=int(input())
a=input()
b=input()
c=input()
k=n*(3-2)
d=a
val=a[:]+b[-k:]
val1=a[:]+c[-k:]
val2=b[:]+c[-k:]
w=sol(val,a,b)
if w==1:
w=sol(val1,a,c)
elif w==1:
sol(val2,b,c)
```
No
| 91,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string that contains only the characters 0 and 1.
Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.
Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.
It can be proven that under the constraints of the problem, such a bitstring always exists.
If there are multiple possible answers, you may output any of them.
Example
Input
2
1
00
11
01
3
011001
111010
010001
Output
010
011001010
Note
In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.
In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.
Submitted Solution:
```
from sys import stdin, stdout
def count(s):
return sum(1 for c in s if c == '1')
def build(s1, s2, m):
i1 = 0
i2 = 0
answer = ''
while i1 < len(s1) and i2 < len(s2):
if s1[i1] == s2[i2]:
answer += s1[i1]
i1 += 1
i2 += 1
elif s1[i1] != m:
answer += s1[i1]
i1 += 1
else:
answer += s2[i2]
i2 += 1
return answer + s1[i1:] + s2[i2:]
n = int(stdin.readline())
for _ in range(n):
k = int(stdin.readline())
a = stdin.readline().strip()
b = stdin.readline().strip()
c = stdin.readline().strip()
arr0 = []
arr1 = []
for s in [a, b, c]:
if count(s) >= k:
arr1.append(s)
else:
arr0.append(s)
if len(arr0) > 1:
stdout.write(build(arr0[0], arr0[1], '0'))
else:
stdout.write(build(arr1[0], arr1[1], '1'))
```
No
| 91,974 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
import os,sys
from io import BytesIO, IOBase
from collections import deque, Counter,defaultdict as dft
from heapq import heappop ,heappush
from math import log,sqrt,factorial,cos,tan,sin,radians,log2,ceil,floor
from bisect import bisect,bisect_left,bisect_right
from decimal import *
import sys,threading
from itertools import permutations, combinations
from copy import deepcopy
input = sys.stdin.readline
ii = lambda: int(input())
si = lambda: input().rstrip()
mp = lambda: map(int, input().split())
ms= lambda: map(str,input().strip().split(" "))
ml = lambda: list(mp())
mf = lambda: map(float, input().split())
alphs = "abcdefghijklmnopqrstuvwxyz"
# stuff you should look for
# int overflow, array bounds
# special cases (n=1?)
# do smth instead of nothing and stay organized
# WRITE STUFF DOWN
# DON'T GET STUCK ON ONE APPROACH
#
def solve():
n=ii()
arr=ml()
res=0
arr=[0]+arr+[0]
for i in range(1,n+1):
mx=max(arr[i-1],arr[i+1])
if arr[i]>mx:
res+=arr[i]-mx
arr[i]=mx
res+=abs(arr[i]-arr[i-1])
#print(arr)
print(res+arr[-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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
tc=1
tc = ii()
for i in range(tc):
solve()
```
| 91,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
def get_sum(a, op):
res = op
for k in range(1,len(a)):
res += abs(a[k] - a[k-1])
return res
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
a.insert(0, 0)
a.append(0)
mins = 100000000
ops = 0
for k in range(n + 1):
if a[k] > a[k-1] and a[k] > a[k+1]:
raz = a[k] - max(a[k-1], a[k+1])
a[k] = max(a[k-1], a[k+1])
ops += raz
# print(a)
print(get_sum(a, ops))
# n, m = map(int,input().split())
# a = []
# for k in range(n):
# a.append(list(input()))
#
# b = list(map(int,input().split()))
#
#
#
# mn = 10000000000
# for g in range(m):
# res = 0
# f = False
# next = -1
#
#
# for i in range(g, m-1):
# if not f:
# # print(i)
# res += 1
# for j in range(n):
# if a[j][i] == "#":
# break
# f = True
# ff = True
# for k in range(j, -1, -1):
# if a[k][i + 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
# else:
# ff = True
# for k in range(next, -1, -1):
# if a[k][i + 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
#
# f = False
# for i in range(g, 0, -1):
# if not f:
# # print(i)
# res += 1
# for j in range(n):
# if a[j][i] == "#":
# break
# f = True
# ff = True
# for k in range(j, -1, -1):
# if a[k][i - 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
# else:
# ff = True
# for k in range(next, -1, -1):
# if a[k][i - 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
#
#
# res -= 1
# if res < mn:
# print(g)
# mn = min(mn, res)
#
# print(mn)
# import itertools as it
# from random import choice
# while True:
# n = int(input())
# a = []
# for i in range(1, n + 1):
# a.append(i)
# print(*choice(list(it.permutations(a))))
# print(*choice(list(it.permutations(a))))
```
| 91,976 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
def ugly(arr):
n = len(arr)
if(n == 1):
return arr[0]
f = arr[0] - arr[1]
less = 0
if(f > 0):
less += f
l = arr[-1] - arr[-2]
if(l > 0):
less += l
ug = 0
for i in range(n - 1):
ug += abs(arr[i] - arr[i + 1])
ug += arr[0] + arr[-1]
for i in range(1, n - 1):
if(arr[i - 1] < arr[i] and arr[i] > arr[i + 1]):
less += min(arr[i] - arr[i - 1], arr[i] - arr[i + 1])
return ug - less
t = int(input())
for i in range(t):
input()
arr = list(map(int, input().split()))
print(ugly(arr))
```
| 91,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = [int(x) for x in input().split()]
a = [0] + a + [0]
ans = 0
for i in range(1,n+1):
diff = a[i] - max(a[i-1], a[i+1])
if diff > 0:
ans += diff
a[i] = a[i] - diff
for i in range(0,len(a)-1):
ans += abs(a[i]-a[i+1])
print(ans)
```
| 91,978 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
t = int(input())
for test in range(t):
n = int(input())
arr = list(map(int, input().split()))
if(n == 1):
print(arr[0])
elif(n == 2):
print(arr[0]+arr[1])
else:
ans = 0
for i in range(n):
if(i==0 and arr[i+1] < arr[i]):
ans+=arr[i]-arr[i+1]
arr[i] = arr[i+1]
elif(i == n-1 and arr[i-1]<arr[i]):
ans+=arr[i]-arr[i-1]
arr[i]=arr[i-1]
elif(arr[i] > arr[i-1] and arr[i]>arr[i+1]):
ans+= arr[i]-max(arr[i-1],arr[i+1])
arr[i] = max(arr[i-1],arr[i+1])
ans+=arr[0]+arr[-1]
for i in range(1,n):
ans+= abs(arr[i] - arr[i-1])
print(ans )
```
| 91,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
#du vinner ett p om du kan ta bort 2 outlines för 1 remove operation
#kolla 3 om 3 om mitten är mkt högre, downa till
for _ in range(int(input())):
N = int(input())
n = list(map(int,input().split()))
#om en stapel, rm all
if N == 1:
print(n[0])
continue
#edge cases
rm = 0
if n[0] > n[1]:
rm += n[0] - n[1]
n[0] = n[1]
if n[-1] > n[-2]:
rm += n[-1] - n[-2]
n[-1] = n[-2]
for i in range(N-2):
if n[i+1] > n[i] and n[i+1] > n[i+2]:
rm += n[i+1] - max(n[i],n[i+2])
n[i+1] = max(n[i],n[i+2])
#räkna outline score
#pinpointa hitta extreme points (locals)
tot = 0
for i in range(N-1):
upwards = n[i+1] - n[i] #edgecase med 0 til n[0]
if upwards > 0:
tot += upwards
downwards = n[i]-n[i+1]
if downwards > 0:
tot += downwards
tot += n[0]
tot += n[-1] #edg c
#print(n)
print(tot + rm)
#du kan jämt ut rmma 2 bredvid varandra, för att sedan kunna rmma ? vinst drag?
#nej kan aldrig hända
#wbu 2*max, den måste ju inte alltid sätmma
# 0 3 3 3 0 0 0 3 3 3 t.ex.
```
| 91,980 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
#from itertools import product
#from itertools import combinations
from collections import Counter
#from collections import defaultdict
#from collections import deque # deque([iterable[, maxlen]]) #appendleft popleft rotate
#from heapq import heapify, heappop, heappush # func(heapifiedlist, item)
# sadness lies below
#from bisect import bisect_left, bisect_right, insort # func(sortedlist, item)
#from sys import setrecursionlimit
from sys import stdin, stderr
input = stdin.readline
def dbp(*args, **kwargs): # calling with dbp(locals()) is perfectly cromulent
print(*args, file=stderr, **kwargs)
def get_int_list():
return [int(x) for x in input().strip().split()]
def do_thing():
n = int(input())
alist = get_int_list()
#dbp(alist)
ugly = 0
last = 0
for idx, a in enumerate(alist):
delta = a-last
alist[idx] = delta
ugly += abs(delta)
last = a
delta = -last
alist.append(delta)
ugly += last
#dbp(alist)
for idx, a in enumerate(alist):
if idx == 0:
last = a
continue
if last > 0 and a < 0:
squash = min(last, -a)
ugly -= squash
last = a
return ugly
def multicase():
maxcc = int(input().strip())
for cc in range(maxcc):
print(do_thing())
if __name__ == "__main__":
multicase()
#print(do_thing())
```
| 91,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Tags: greedy, implementation, math
Correct Solution:
```
import sys,math,bisect
from random import randint
inf = float('inf')
mod = 10**9+7
"========================================"
def lcm(a,b):
return int((a/math.gcd(a,b))*b)
def gcd(a,b):
return int(math.gcd(a,b))
def tobinary(n):
return bin(n)[2:]
def binarySearch(a,x):
i = bisect.bisect_left(a,x)
if i!=len(a) and a[i]==x:
return i
else:
return -1
def lowerBound(a, x):
i = bisect.bisect_left(a, x)
if i:
return (i-1)
else:
return -1
def upperBound(a,x):
i = bisect.bisect_right(a,x)
if i!= len(a)+1 and a[i-1]==x:
return (i-1)
else:
return -1
def primesInRange(n):
ans = []
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
ans.append(p)
return ans
def primeFactors(n):
factors = []
while n % 2 == 0:
factors.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
factors.append(i)
n = n // i
if n > 2:
factors.append(n)
return factors
def isPrime(n,k=5):
if (n <2):
return True
for i in range(0,k):
a = randint(1,n-1)
if(pow(a,n-1,n)!=1):
return False
return True
"========================================="
"""
n = int(input())
n,k = map(int,input().split())
arr = list(map(int,input().split()))
"""
from collections import deque,defaultdict,Counter
import heapq,string
for _ in range(int(input())):
n = int(input())
arr = list(map(int,input().split()))
arr.insert(0,0)
arr.append(0)
cost = 0
for i in range(1,n+1):
og=arr[i]
if arr[i]>arr[i-1] and arr[i]>arr[i+1]:
arr[i]=max(arr[i-1],arr[i+1])
cost+=og-(arr[i])
for i in range(1,n+2):
cost+=abs(arr[i]-arr[i-1])
print(cost)
```
| 91,982 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
lst=list(map(int, input().split()))
extra=0
if n==1:
print(lst[0])
continue
for i in range(n):
if i==0:
if lst[i]>lst[i+1]:
#print('A')
extra+=abs(lst[i]-lst[i+1])
lst[i]=lst[i+1]
elif i==n-1:
if lst[i]>lst[i-1]:
#print('B')
extra+=abs(lst[i]-lst[i-1])
lst[i]=lst[i-1]
else:
if lst[i]>lst[i-1] and lst[i]>lst[i+1]:
#print('C')
val1=abs(lst[i]-lst[i+1])
val2=abs(lst[i]-lst[i-1])
extra+=min(val1,val2)
lst[i]=max(lst[i-1],lst[i+1])
#print(extra)
#print(lst)
for i in range(n):
if i==0:
extra+=lst[i]
else:
extra+=abs(lst[i]-lst[i-1])
extra+=lst[-1]
print(extra)
```
Yes
| 91,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
for _ in range(int(input())):
l = [0]
n = int(input())
for i in input().split():
l.append(int(i))
l.append(0)
if n == 1:
print(l[1])
continue
ans = 0
for i in range(1, n+1):
if l[i] > l[i-1] and l[i] > l[i+1]:
ans += l[i]-max(l[i-1], l[i+1])
l[i] = max(l[i-1], l[i+1])
for i, j in zip(l, l[1:]):
ans += abs(i-j)
print(ans)
```
Yes
| 91,984 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
# Rishabh Rao (https://github.com/rishabhrao)
import sys
MOD = 1000000007
def inp(): return sys.stdin.readline().strip()
def ii(): return int(inp())
def iis(): return [int(i) for i in inp().split()]
def solve():
n = ii()
a = [0] + iis() + [0]
min_ugliness = 0
for i in range(1, n + 1):
if a[i] > a[i - 1] and a[i] > a[i + 1]:
biggest_neighbour = max(a[i - 1], a[i + 1])
min_ugliness += a[i] - biggest_neighbour
a[i] = biggest_neighbour
if i < n:
min_ugliness += abs(a[i] - a[i - 1])
min_ugliness += abs(a[n] - a[n - 1]) + a[n]
return min_ugliness
t = ii()
for _ in range(t):
print(solve())
```
Yes
| 91,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
def solve(arr,n,ans):
if n == 1:
ans.append(str(arr[0]))
return
total = arr[0]+arr[-1]
for i in range(1,n):
total += abs(arr[i]-arr[i-1])
for i in range(n):
if i == 0:
if i+1 < n and arr[i] > arr[i+1]:
total -= (arr[i]-arr[i+1])
arr[i] = arr[i+1]
elif i == n-1:
if arr[i] > arr[i-1]:
total -= (arr[i]-arr[i-1])
else:
if arr[i] > arr[i-1] and arr[i] > arr[i+1]:
total -= (arr[i]-max(arr[i-1],arr[i+1]))
arr[i] = max(arr[i-1],arr[i+1])
ans.append(str(total))
def main():
t = int(input())
ans = []
for i in range(t):
n = int(input())
arr = list(map(int,input().split()))
solve(arr,n,ans)
print('\n'.join(ans))
main()
```
Yes
| 91,986 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
ar=[0]+list(map(int,input().split()))+[0]
s=0
for i in range(1,n):
if (ar[i-1]<ar[i]>ar[i+1]):
s+=ar[i]-max(ar[i-1],ar[i+1])
ar[i]=max(ar[i-1],ar[i+1])
for i in range(1,n+2):
s+=abs(ar[i]-ar[i-1])
print(s)
```
No
| 91,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
# import sys
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')
def gcd(x, y):
while (y):
x, y = y, x % y
return x
def lcm(x,y):
return x*y//gcd(x,y)
def LSPtable(pattern):
n=len(pattern)
l=[0]*n
j=0
for i in range(1,n):
while j>0 and pattern[i]!=pattern[j]:
j=l[j-1]
if pattern[i]==pattern[j]:
l[i]=j+1
j+=1
else:
l[i]=0
return l
def KMPsearch(pattern,string):
lsp=LSPtable(pattern)
j=0
for i in range(len(string)):
while j>0 and string[i]!=pattern[j]:
j=lsp[j-1]
if string[i]==pattern[j]:
j+=1
if j== len(pattern):
return i-j+1
return -1
def getsum(BITTree,i):
s=0
while i>0:
s+=BITTree[i]
i-=i&(-i)
return s
def updatebit(BITTree,n,i,v):
i=i+1
while i<=n:
BITTree[i]+=v
i+=i&(-i)
def constructor(arr, n):
BITTree =[0]*(n+1)
for i in range(n):
updatebit(BITTree,n,i,arr[i])
return BITTree
def arrIn():
return list(map(int,input().split()))
def mapIn():
return map(int,input().split())
for ii in range(int(input())):
n=int(input())
for i in range(n):
x=1
arr = arrIn()
if n==1:
x=arr[0]
print(x)
continue
ans = 0
if arr[0] > arr[1]:
ans += arr[0] - arr[1]
arr[0] = arr[1]
y=2
z=1
x=ans
ans=x
for i in range(1, n - 1):
if arr[i] > max(arr[i + 1], arr[i - 1]):
ans += arr[i] - max(arr[i - 1], arr[i + 1])
arr[i] = max(arr[i - 1], arr[i + 1])
if arr[n - 2] < arr[n - 1]:
ans += arr[n - 1] - arr[n - 2]
arr[n - 1] = arr[n - 2]
ans += 2*max(arr)
print(ans)
```
No
| 91,988 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
# Rishabh Rao (https://github.com/rishabhrao)
import sys
MOD = 1000000007
def inp(): return sys.stdin.readline().strip()
def ii(): return int(inp())
def iis(): return [int(i) for i in inp().split()]
def solve():
n = ii()
a = [0] + iis()
min_ugliness = 0
for i in range(1, n):
if a[i] > a[i - 1] and a[i] > a[i + 1]:
biggest_neighbour = max(a[i - 1], a[i + 1])
min_ugliness += a[i] - biggest_neighbour
a[i] = biggest_neighbour
min_ugliness += abs(a[i] - a[i - 1])
min_ugliness += abs(a[n] - a[n - 1]) + a[n]
return min_ugliness
t = ii()
for _ in range(t):
print(solve())
```
No
| 91,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
def check(a):
ans=a[0]+a[n-1]
for i in range(1,n):
ans+=abs(a[i]-a[i-1])
return ans
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))+[0]
l1=list(sorted(set(a),reverse=True))
ans=check(a)
m=True
l=0
for i in range(1,len(l1)):
for j in range(n):
if a[j]>l1[i]:
l+=abs(a[j]-l1[i])
a[j]=l1[i]
k=check(a)
if k+l>ans:
print(ans)
m=False
break
ans=min(ans,k+l)
if m:
print(ans)
```
No
| 91,990 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
import sys
from array import array # noqa: F401
import typing as Tp # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def output(*args):
sys.stdout.buffer.write(
('\n'.join(map(str, args)) + '\n').encode('utf-8')
)
def main():
n, k = map(int, input().split())
adj = [[] for _ in range(n)]
for u, v in (map(int, input().split()) for _ in range(n - 1)):
adj[u - 1].append(v - 1)
adj[v - 1].append(u - 1)
route = []
parent = [-1] * n
stack = [0]
while stack:
v = stack.pop()
route.append(v)
for dest in adj[v]:
if dest == parent[v]:
continue
parent[dest] = v
stack.append(dest)
ans = 0
dp = [[1] + [0] * k for _ in range(n)]
for v in reversed(route):
ans += dp[v][k] * 2
for child in adj[v]:
if child == parent[v]:
continue
for i in range(1, k):
ans += dp[child][i - 1] * (dp[v][k - i] - dp[child][k - i - 1])
if v != 0:
for i in range(k):
dp[parent[v]][i + 1] += dp[v][i]
print(ans >> 1)
if __name__ == '__main__':
main()
```
| 91,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
import collections as col
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
"""
Pick a root
Within each subtree, count: the number of nodes at levels 1 to K in that subtree
dp = [0]*500
So, dp[node][level] = number of nodes at level K
dp[node][0] = 1 for every node
"""
def solve():
N, K = getInts()
graph = col.defaultdict(set)
dp = [[0 for k in range(K+1)] for n in range(N)]
for n in range(N-1):
U, V = getInts()
U -= 1
V -= 1
graph[U].add(V)
graph[V].add(U)
visited = set()
@bootstrap
def dfs(node):
dp[node][0] = 1
visited.add(node)
for neighbour in graph[node]:
if neighbour not in visited:
yield dfs(neighbour)
for k in range(K):
dp[node][k+1] += dp[neighbour][k]
yield
dfs(0)
ans = 0
#print(dp)
visited = set()
for n in range(N):
visited.add(n)
ans += dp[n][K]
tmp = 0
for neighbour in graph[n]:
if neighbour not in visited:
for k in range(1,K):
tmp += dp[neighbour][k-1]*(dp[n][K-k] - dp[neighbour][K-k-1])
tmp //= 2
ans += tmp
#print(n+1,dp[n][K],ans,tmp)
return ans
#for _ in range(getInt()):
print(solve())
```
| 91,992 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
def main():
n,k=map(int,input().split())
tree=[[] for _ in range(n+1)]
# 1 indexed
for _ in range(n-1):
a,b=map(int,input().split())
tree[a].append(b)
tree[b].append(a)
dp=[[0 for _ in range(k+1)] for _ in range(n+1)]
idx=[0]*(n+1)
stack=[(1,0)]
ans=0
dp[1][0]=1
while stack:
x,p=stack[-1]
y=idx[x]
if y==len(tree[x]):
if x==1:
break
for i in range(k,0,-1):
dp[x][i]=dp[x][i-1]
dp[x][0]=0
for i in range(k+1):
ans+=dp[p][i]*dp[x][k-i]
dp[p][i]+=dp[x][i]
stack.pop()
else:
z=tree[x][y]
if z!=p:
dp[z][0]=1
stack.append((z,x))
idx[x]+=1
print(ans)
#----------------------------------------------------------------------------------------
# region fastio
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')
# endregion
if __name__ == '__main__':
main()
```
| 91,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
n, k = map(int, input().split())
t, q = [[] for i in range(n + 1)], [1]
for j in range(n - 1):
a, b = map(int, input().split())
t[a].append(b)
t[b].append(a)
for x in q:
for y in t[x]: t[y].remove(x)
q.extend(t[x])
q.reverse()
a, luismore = {}, 0
for x in q:
a[x] = [1]
u = len(a[x])
for y in t[x]:
v = len(a[y])
for d in range(max(0, k - u), v): luismore += a[y][d] * a[x][k - d - 1]
if v >= u:
for d in range(u - 1): a[x][d + 1] += a[y][d]
a[x] += a[y][u - 1: ]
u = v + 1
else:
for d in range(0, v): a[x][d + 1] += a[y][d]
if u > k: a[x].pop()
print(luismore)
#<3
```
| 91,994 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
n, k = map(int, input().split())
t, q = [[] for i in range(n + 1)], [1]
for j in range(n - 1):
a, b = map(int, input().split())
t[a].append(b)
t[b].append(a)
for x in q:
for y in t[x]: t[y].remove(x)
q.extend(t[x])
q.reverse()
a, s = {}, 0
for x in q:
a[x] = [1]
u = len(a[x])
for y in t[x]:
v = len(a[y])
for d in range(max(0, k - u), v): s += a[y][d] * a[x][k - d - 1]
if v >= u:
for d in range(u - 1): a[x][d + 1] += a[y][d]
a[x] += a[y][u - 1: ]
u = v + 1
else:
for d in range(0, v): a[x][d + 1] += a[y][d]
if u > k: a[x].pop()
print(s)
```
| 91,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
n,k=map(int,input().split())
g=[[] for i in range(n)]
for _ in range(n-1):
a,b=map(int,input().split())
g[a-1].append(b-1)
g[b-1].append(a-1)
q=[0]
for x in q:
for y in g[x]:
g[y].remove(x)
q.extend(g[x])
q.reverse()
d,cnt={},0
for x in q:
d[x]=[1]
l_x=len(d[x])
for to in g[x]:
l_to=len(d[to])
for i in range(max(0,k-l_x),l_to):cnt+=d[to][i]*d[x][k-i-1]
if l_to>=l_x:
for i in range(l_x-1):d[x][i+1]+=d[to][i]
d[x]+=d[to][l_x-1:]
l_x=l_to+1
else:
for i in range(l_to):d[x][i+1]+=d[to][i]
if l_x>k:
d[x].pop()
print(cnt)
```
| 91,996 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from collections import defaultdict,Counter
from copy import deepcopy
def solve(k,path,path1,lst):
st = [1]
while len(st):
if not len(path[st[-1]]):
x = st.pop()
if len(st):
for i in range(k):
lst[st[-1]][i+1] += lst[x][i]
continue
i = path[st[-1]].pop()
path[i].remove(st[-1])
st.append(i)
ans = lst[1][k]
st = [1]
while len(st):
if not len(path1[st[-1]]):
st.pop()
continue
i = path1[st[-1]].pop()
path1[i].remove(st[-1])
for j in range(k-1,0,-1):
lst[i][j+1] += lst[st[-1]][j]-lst[i][j-1]
lst[st[-1]][k] -= lst[i][k-1]
lst[i][1] += lst[st[-1]][0]
ans += lst[i][k]
st.append(i)
print(ans//2)
def main():
n,k = map(int,input().split())
path = defaultdict(set)
for _ in range(n-1):
u,v = map(int,input().split())
path[u].add(v)
path[v].add(u)
path1 = deepcopy(path)
lst = [[1]+[0]*k for _ in range(n+1)]
solve(k,path,path1,lst)
#Fast IO Region
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()
```
| 91,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
def main():
n,k=map(int,input().split())
tree=[[] for _ in range(n+1)]
for _ in range(n-1):
a,b=map(int,input().split())
tree[a].append(b)
tree[b].append(a)
dp=[[0 for _ in range(k+1)] for _ in range(n+1)]
# using dfs
stack=[(1,0)]
# root at 1
idx=[0]*(n+1)
dp[1][0]=1
ans=0
while stack:
x,p=stack[-1]
y=idx[x]
if y==len(tree[x]):
if x==1:
break
stack.pop()
# shifting all distances by 1 for parent
for i in range(k,0,-1):
dp[x][i]=dp[x][i-1]
dp[x][0]=0
for i in range(k):
ans+=dp[p][i]*dp[x][k-i]
dp[p][i]+=dp[x][i]
else:
z=tree[x][y]
if z!=p:
stack.append((z,x))
dp[z][0]=1
idx[x]+=1
print(ans)
#----------------------------------------------------------------------------------------
# region fastio
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')
# endregion
if __name__ == '__main__':
main()
```
| 91,998 |
Provide tags and a correct Python 2 solution for this coding contest problem.
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5 2
1 2
2 3
3 4
2 5
Output
4
Input
5 3
1 2
2 3
3 4
4 5
Output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
Tags: dfs and similar, dp, trees
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
n,k=in_arr()
d=[[] for i in range(n+1)]
for i in range(n-1):
u,v=in_arr()
d[u].append(v)
d[v].append(u)
dp=[[0 for i in range(k+1)] for i in range(n+1)]
q=[1]
pos=0
vis=[0]*(n+1)
vis[1]=1
while pos<n:
x=q[pos]
pos+=1
for i in d[x]:
if not vis[i]:
vis[i]=1
q.append(i)
ans=0
ans1=0
while q:
x=q.pop()
vis[x]=0
temp=[]
dp[x][0]=1
for i in d[x]:
if not vis[i]:
temp.append(i)
for j in range(k):
dp[x][j+1]+=dp[i][j]
#print x,temp
for i in temp:
for j in range(1,k):
#print j,(dp[x][k-j]-dp[i][k-j-1]),dp[i][j-1]
ans1+=(dp[x][k-j]-dp[i][k-j-1])*dp[i][j-1]
ans+=dp[x][k]
print ans+(ans1/2)
```
| 91,999 |
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