text stringlengths 198 433k | conversation_id int64 0 109k |
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tanya wants to go on a journey across the cities of Berland. There are n cities situated along the main railroad line of Berland, and these cities are numbered from 1 to n.
Tanya plans her journey as follows. First of all, she will choose some city c_1 to start her journey. She will visit it, and after that go to some other city c_2 > c_1, then to some other city c_3 > c_2, and so on, until she chooses to end her journey in some city c_k > c_{k - 1}. So, the sequence of visited cities [c_1, c_2, ..., c_k] should be strictly increasing.
There are some additional constraints on the sequence of cities Tanya visits. Each city i has a beauty value b_i associated with it. If there is only one city in Tanya's journey, these beauty values imply no additional constraints. But if there are multiple cities in the sequence, then for any pair of adjacent cities c_i and c_{i + 1}, the condition c_{i + 1} - c_i = b_{c_{i + 1}} - b_{c_i} must hold.
For example, if n = 8 and b = [3, 4, 4, 6, 6, 7, 8, 9], there are several three possible ways to plan a journey:
* c = [1, 2, 4];
* c = [3, 5, 6, 8];
* c = [7] (a journey consisting of one city is also valid).
There are some additional ways to plan a journey that are not listed above.
Tanya wants her journey to be as beautiful as possible. The beauty value of the whole journey is the sum of beauty values over all visited cities. Can you help her to choose the optimal plan, that is, to maximize the beauty value of the journey?
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of cities in Berland.
The second line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 4 ⋅ 10^5), where b_i is the beauty value of the i-th city.
Output
Print one integer — the maximum beauty of a journey Tanya can choose.
Examples
Input
6
10 7 1 9 10 15
Output
26
Input
1
400000
Output
400000
Input
7
8 9 26 11 12 29 14
Output
55
Note
The optimal journey plan in the first example is c = [2, 4, 5].
The optimal journey plan in the second example is c = [1].
The optimal journey plan in the third example is c = [3, 6].
Submitted Solution:
```
# from debug import debug
inf = int(1e10)
n = int(input())
lis = list(map(int, input().split()))
c = sorted([(lis[i]-i, lis[i]) for i in range(n)])
v = c[0][1]
ans = 0
for i in range(1, n):
if c[i-1][0] == c[i][0] and c[i][0] >= 0:
v+=c[i][1]
else:
ans = max(ans, v)
v = c[i][1]
print(max(ans, v))
```
No
| 105,400 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tanya wants to go on a journey across the cities of Berland. There are n cities situated along the main railroad line of Berland, and these cities are numbered from 1 to n.
Tanya plans her journey as follows. First of all, she will choose some city c_1 to start her journey. She will visit it, and after that go to some other city c_2 > c_1, then to some other city c_3 > c_2, and so on, until she chooses to end her journey in some city c_k > c_{k - 1}. So, the sequence of visited cities [c_1, c_2, ..., c_k] should be strictly increasing.
There are some additional constraints on the sequence of cities Tanya visits. Each city i has a beauty value b_i associated with it. If there is only one city in Tanya's journey, these beauty values imply no additional constraints. But if there are multiple cities in the sequence, then for any pair of adjacent cities c_i and c_{i + 1}, the condition c_{i + 1} - c_i = b_{c_{i + 1}} - b_{c_i} must hold.
For example, if n = 8 and b = [3, 4, 4, 6, 6, 7, 8, 9], there are several three possible ways to plan a journey:
* c = [1, 2, 4];
* c = [3, 5, 6, 8];
* c = [7] (a journey consisting of one city is also valid).
There are some additional ways to plan a journey that are not listed above.
Tanya wants her journey to be as beautiful as possible. The beauty value of the whole journey is the sum of beauty values over all visited cities. Can you help her to choose the optimal plan, that is, to maximize the beauty value of the journey?
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of cities in Berland.
The second line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 4 ⋅ 10^5), where b_i is the beauty value of the i-th city.
Output
Print one integer — the maximum beauty of a journey Tanya can choose.
Examples
Input
6
10 7 1 9 10 15
Output
26
Input
1
400000
Output
400000
Input
7
8 9 26 11 12 29 14
Output
55
Note
The optimal journey plan in the first example is c = [2, 4, 5].
The optimal journey plan in the second example is c = [1].
The optimal journey plan in the third example is c = [3, 6].
Submitted Solution:
```
res=0
n=int(input())
x=[0]*(4*10**5)
bi=[*map(int,input().split())]
for i in range(1,n):
a=bi[i]
x[a-i]+=a
res=max(res,x[a-i])
print(max(res,bi[0]))
```
No
| 105,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tanya wants to go on a journey across the cities of Berland. There are n cities situated along the main railroad line of Berland, and these cities are numbered from 1 to n.
Tanya plans her journey as follows. First of all, she will choose some city c_1 to start her journey. She will visit it, and after that go to some other city c_2 > c_1, then to some other city c_3 > c_2, and so on, until she chooses to end her journey in some city c_k > c_{k - 1}. So, the sequence of visited cities [c_1, c_2, ..., c_k] should be strictly increasing.
There are some additional constraints on the sequence of cities Tanya visits. Each city i has a beauty value b_i associated with it. If there is only one city in Tanya's journey, these beauty values imply no additional constraints. But if there are multiple cities in the sequence, then for any pair of adjacent cities c_i and c_{i + 1}, the condition c_{i + 1} - c_i = b_{c_{i + 1}} - b_{c_i} must hold.
For example, if n = 8 and b = [3, 4, 4, 6, 6, 7, 8, 9], there are several three possible ways to plan a journey:
* c = [1, 2, 4];
* c = [3, 5, 6, 8];
* c = [7] (a journey consisting of one city is also valid).
There are some additional ways to plan a journey that are not listed above.
Tanya wants her journey to be as beautiful as possible. The beauty value of the whole journey is the sum of beauty values over all visited cities. Can you help her to choose the optimal plan, that is, to maximize the beauty value of the journey?
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of cities in Berland.
The second line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 4 ⋅ 10^5), where b_i is the beauty value of the i-th city.
Output
Print one integer — the maximum beauty of a journey Tanya can choose.
Examples
Input
6
10 7 1 9 10 15
Output
26
Input
1
400000
Output
400000
Input
7
8 9 26 11 12 29 14
Output
55
Note
The optimal journey plan in the first example is c = [2, 4, 5].
The optimal journey plan in the second example is c = [1].
The optimal journey plan in the third example is c = [3, 6].
Submitted Solution:
```
import sys
a = int(input())
b = list(map(int,input().split()))
if a==1:
print(b[0])
sys.exit()
c = [0]*a
for i in range(a):
c[i]=(b[i]-i-1,b[i])
ans=0
c.sort()
maxx=0
cc = c[0]
i = 1
while i<a:
while i<a and c[i][0]==cc:
maxx+=c[i][1]
i+=1
ans=max(ans,maxx)
maxx=0
if i<a:cc=c[i][0]
print(ans)
```
No
| 105,402 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
import sys
n, *ab = map(int, sys.stdin.read().split())
graph = [[] for _ in range(n)]
for a, b in zip(*[iter(ab)] * 2):
a -= 1
b -= 1
graph[a].append(b)
graph[b].append(a)
def main():
rank = [None] * n; rank[0] = 0
parent = [None] * n
leaves = []
stack = [0]
while stack:
u = stack.pop()
if (not u == 0) and len(graph[u]) == 1:
leaves.append(u)
continue
for v in graph[u]:
if v == parent[u]:
continue
parent[v] = u
rank[v] = rank[u] + 1
stack.append(v)
if len(graph[0]) == 1:
leaves.append(0)
parent[0] = graph[0][0]
parent[graph[0][0]] = None
pars = set()
for leaf in leaves:
pars.add(parent[leaf])
ma = n - 1 - (len(leaves) - len(pars))
rs = set([rank[leaf] & 1 for leaf in leaves])
mi = 1 if len(rs) == 1 else 3
print(mi, ma)
if __name__ == '__main__':
main()
```
| 105,403 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
import sys
import math
from collections import deque
def solve():
#f = open("b.txt")
f = sys.stdin
n = int(f.readline())
edges = [[] for i in range(n)]
for i in range(n - 1):
s = f.readline().strip()
a, b = [int(x) for x in s.split(' ', 2)]
edges[a - 1].append(b - 1)
edges[b - 1].append(a - 1)
root = 0
for i in range(n):
if len(edges[i]) > 1:
root = i
break
#print(root)
odd_dist = False
even_dist = False
minf = 1
maxf = n - 1
dist = [0 for i in range(n)]
q = deque()
q.append(root)
while len(q) > 0:
cur = q.popleft()
child_leaves = 0
for next in edges[cur]:
if dist[next] == 0 and next != root:
q.append(next)
dist[next] = dist[cur] + 1
if len(edges[next]) == 1:
child_leaves += 1
if dist[next] % 2 == 1:
odd_dist = True
else:
even_dist = True
if child_leaves > 1:
maxf = maxf - child_leaves + 1
if odd_dist and even_dist:
minf = 3
print(str(minf) + " " + str(maxf))
solve()
```
| 105,404 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
I=lambda:list(map(int,input().split()))
n,=I()
g=[[] for i in range(n+1)]
for i in range(n-1):
x,y=I()
g[x].append(y)
g[y].append(x)
leaf=[]
for i in range(1,n+1):
if len(g[i])==1:
leaf.append(i)
st=[1]
visi=[-1]*(n+1)
visi[1]=0
while st:
x=st.pop()
for y in g[x]:
if visi[y]==-1:
visi[y]=1-visi[x]
st.append(y)
ch=visi[leaf[0]]
fl=1
temp=0
for i in leaf:
if visi[i]!=ch:
fl=0
mi=1
if not fl:
mi=3
an=[0]*(n+1)
ma=n-1
for i in leaf:
for j in g[i]:
if an[j]==1:
ma-=1
else:
an[j]=1
print(mi,ma)
```
| 105,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
import collections,sys
n = int(sys.stdin.readline())
g = []
sys.setrecursionlimit(2**20)
for i in range(n+1):
g.append([])
for i in range(n-1):
a,b = list(map(int,sys.stdin.readline().split(' ')))
g[a].append(b)
g[b].append(a)
leaf = {}
mx = n-1
for i in range(1,n+1,1):
if(len(g[i]) == 1):
leaf[i] = True
else:
s = i
for i in range(1,n+1,1):
l = 0
for v in g[i]:
if(v in leaf):
l+=1
if(l != 0):
mx = mx-l+1
mn = 1
l = s
dst = [0]*(n+1)
a = collections.deque([])
exp = {}
exp[l] = True
a.append(l)
o = 0
e = 0
while(a):
v = a.pop()
h = dst[v]+1
for u in g[v]:
if(u not in exp):
dst[u] = h
exp[u] = True
a.append(u)
if(u in leaf):
if(h % 2 == 1):
o += 1
else:
e += 1
if(o == 0 or e ==0):
mn = 1
else:
mn = 3
print(mn,mx)
```
| 105,406 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
from sys import stdin
from collections import deque
input=stdin.readline
n=int(input())
graph=[set() for i in range(n+1)]
visited2=[False]*(n+2)
for i in range(n - 1):
a,b=map(int,input().split())
graph[a].add(b)
graph[b].add(a)
k=0
for i in range(n + 1):
if len(graph[i]) == 1:
k=i
break
m=[1,n-1]
def dfs(node, depth):
nodes=deque([node])
depths=deque([depth])
while len(nodes) > 0:
node=nodes.popleft()
depth=depths.popleft()
visited2[node] = True
went=False
for i in graph[node]:
if not visited2[i]:
nodes.appendleft(i)
depths.appendleft(depth+1)
went=True
if not went:
if depth % 2 == 1 and depth != 1:
m[0] = 3
dfs(k,0)
oneBranches=[0]*(n+1)
for i in range(1,n+1):
if len(graph[i]) == 1:
oneBranches[graph[i].pop()] += 1
for i in range(n+1):
if oneBranches[i] > 1:
m[1] -= oneBranches[i] - 1
print(*m)
#if there is odd path length, min is 3 else 1
#max length: each odd unvisited path gives n, while each even unvisited path gives n - 1
```
| 105,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
import io, os, collections
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
nb = [[] for _ in range(n)]
for i in range(n-1):
u, v = [int(s)-1 for s in input().split()]
nb[u].append(v)
nb[v].append(u)
leaves = []
father = set()
for i in range(n):
if len(nb[i])==1:
leaves.append(i)
father.add(nb[i][0])
maxAns = n-1-(len(leaves)-len(father))
minAns = 1
q = collections.deque()
q.append(leaves[0])
d = [0]*n # 1/-1
d[leaves[0]] = 1
while q:
node = q.popleft()
tag = - d[node]
for neibor in nb[node]:
if d[neibor] == 0:
d[neibor] = tag
q.append(neibor)
for lv in leaves:
if d[lv] != 1:
minAns = 3
break
print('{} {}'.format(minAns, maxAns))
```
| 105,408 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import defaultdict
# sys.setrecursionlimit(100000)
input = sys.stdin.readline
INF = 2**62-1
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
@mt
def slv(N, AB):
ans = 0
g = defaultdict(list)
for a, b in AB:
g[a].append(b)
g[b].append(a)
leaves = set()
lp = defaultdict(set)
for i in range(1, N+1):
if len(g[i]) == 1:
leaves.add(i)
lp[g[i][0]].add(i)
def dfs(s):
stack = [s]
d = {s:0}
while stack:
u = stack.pop()
for v in g[u]:
if v not in d:
d[v] = d[u] + 1
stack.append(v)
return d
fmin = 1
d = dfs(1)
leaves = list(leaves)
oe = d[leaves[0]] % 2
for l in leaves:
if oe != d[l] % 2:
fmin = 3
break
fmax = N - 1
for l in lp.values():
fmax -= len(l) - 1
return fmin, fmax
def main():
N = read_int()
AB = [read_int_n() for _ in range(N-1)]
print(*slv(N, AB))
if __name__ == '__main__':
main()
```
| 105,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, trees
Correct Solution:
```
import sys
readline = sys.stdin.readline
def parorder(Edge, p):
N = len(Edge)
par = [0]*N
par[p] = -1
stack = [p]
order = []
visited = set([p])
ast = stack.append
apo = order.append
while stack:
vn = stack.pop()
apo(vn)
for vf in Edge[vn]:
if vf in visited:
continue
visited.add(vf)
par[vf] = vn
ast(vf)
return par, order
def getcld(p):
res = [[] for _ in range(len(p))]
for i, v in enumerate(p[1:], 1):
res[v].append(i)
return res
N = int(readline())
Edge = [[] for _ in range(N)]
for _ in range(N-1):
a, b = map(int, readline().split())
a -= 1
b -= 1
Edge[a].append(b)
Edge[b].append(a)
Leaf = [i for i in range(N) if len(Edge[i]) == 1]
sLeaf = set(Leaf)
for i in range(N):
if i not in sLeaf:
root = i
break
P, L = parorder(Edge, root)
dp = [0]*N
used = set([root])
stack = [root]
while stack:
vn = stack.pop()
for vf in Edge[vn]:
if vf not in used:
used.add(vf)
stack.append(vf)
dp[vf] = 1-dp[vn]
if len(set([dp[i] for i in Leaf])) == 1:
mini = 1
else:
mini = 3
k = set()
for l in Leaf:
k.add(P[l])
maxi = N-1 - len(Leaf) + len(k)
print(mini, maxi)
```
| 105,410 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Tags: bitmasks, constructive algorithms, dfs and similar, greedy, math, 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_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
#main code
def get(n):
p=2**32
c=32
while p:
if p&n:
return c+1
c-=1
p/=2
return 0
inp=inp()
n=inp[0]
pos=1
mx=0
root=0
deg=Counter()
d=[[] for i in range(n+1)]
for i in range(n-1):
u,v=inp[pos],inp[pos+1]
pos+=2
d[u].append(v)
d[v].append(u)
deg[u]+=1
deg[v]+=1
if deg[u]>mx:
mx=deg[u]
root=u
if deg[v]>mx:
mx=deg[v]
root=v
vis=[0]*(n+1)
vis[root]=1
q=[(root,0)]
f1,f2=0,0
mx=0
c1=0
d1=Counter()
while q:
x,w=q.pop(0)
ft=0
for i in d[x]:
if not vis[i]:
vis[i]=1
q.append((i,w+1))
ft=1
if not ft:
c1+=1
d1[d[x][0]]+=1
if w%2:
#print 1,x
f1=1
else:
#print 2,x
f2=1
if f1 and f2:
ans1=3
else:
ans1=1
ans2=n-1-(c1-len(d1.keys()))
pr_arr((ans1,ans2))
```
| 105,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
N = int(input())
adj = [[] for _ in range(N+1)]
for _ in range(N-1):
a, b = map(int, input().split())
adj[a].append(b)
adj[b].append(a)
from collections import deque
que = deque()
que.append(1)
seen = [-1] * (N+1)
seen[1] = 0
par = [0] * (N+1)
child = [[] for _ in range(N+1)]
seq = []
while que:
v = que.popleft()
seq.append(v)
for u in adj[v]:
if seen[u] == -1:
seen[u] = seen[v] + 1
par[u] = v
child[v].append(u)
que.append(u)
seq.reverse()
ok = 1
flg = -1
is_leaf = [0] * (N+1)
for v in range(1, N+1):
if len(adj[v]) == 1:
if flg == -1:
flg = seen[v] & 1
else:
if flg != seen[v] & 1:
ok = 0
is_leaf[v] = 1
if ok:
m = 1
else:
m = 3
M = N-1
for v in range(1, N+1):
cnt = 0
for u in adj[v]:
if is_leaf[u]:
cnt += 1
if cnt:
M -= cnt-1
print(m, M)
if __name__ == '__main__':
main()
```
Yes
| 105,412 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
def solveAll():
case = readCase()
print(*solve(case))
def readCase():
treeSize = int(input())
graph = [[] for i in range(treeSize)]
for _ in range(1, treeSize):
a, b = (int(x) for x in input().split())
a -= 1
b -= 1
graph[a] += [b]
graph[b] += [a]
return graph
def solve(graph):
leafs = computeLeafs(graph)
return minF(graph, leafs), maxF(graph, leafs)
def computeLeafs(graph):
leafs = [node for node in range(0, len(graph)) if len(graph[node]) == 1]
return leafs
def minF(graph, leafs):
color = [None] * len(graph)
queue = [0]
color[0] = "a"
head = 0
while head < len(queue):
currentNode = queue[head]
head += 1
for neighbor in graph[currentNode]:
if color[neighbor] != None: continue
color[neighbor] = "b" if color[currentNode] == "a" else "a"
queue += [neighbor]
if all(color[leafs[0]] == color[leaf] for leaf in leafs):
return 1
return 3
def maxF(graph, leafs):
group = [0] * len(graph)
for leaf in leafs:
parent = graph[leaf][0]
group[parent] += 1
ans = len(graph) - 1
for size in group:
if size >= 2:
ans -= size - 1
return ans
solveAll()
```
Yes
| 105,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
n = int(input())
l = [[] for _ in range(n)]
for _ in range(n-1):
p1, p2 = map(lambda x : x-1, map(int, input().split()))
l[p1].append(p2)
l[p2].append(p1)
leaf = []
e = [0] * n
maxans = n-1
for i in range(n):
temp = l[i]
if len(temp) == 1:
if e[temp[0]] == 1:
maxans -= 1
else:
e[temp[0]] = 1
leaf.append(i)
q = [0]
visited = [-1]*n
visited[0] = 0
while q:
node = q.pop()
for i in l[node]:
if visited[i] == -1:
q.append(i)
visited[i] = 1 - visited[node]
f = visited[leaf[0]]
for i in leaf:
if visited[i] != f:
minans = 3
break
else:
minans = 1
print(minans, maxans)
```
Yes
| 105,414 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
def main():
n = int(input())
g = [[] for _ in range(n)]
for i in range(n-1):
a, b = map(int, input().split())
a, b = a-1, b-1
g[a].append(b)
g[b].append(a)
# min
leaf = []
for i in range(n):
if len(g[i]) == 1:
leaf.append(i)
#print(leaf)
from collections import deque
root = leaf[0]
q = deque()
dist = [-1]*n
q.append(root)
dist[root] = 0
#order = []
#par = [-1]*n
while q:
v = q.popleft()
#order.append(v)
for u in g[v]:
if dist[u] == -1:
dist[u] = dist[v]+1
#par[u] = v
q.append(u)
#print(dist)
for v in leaf:
if dist[v]%2 != 0:
m = 3
break
else:
m = 1
# max
root = leaf[0]
q = deque()
visit = [-1]*n
q.append(root)
visit[root] = 0
x = 0
while q:
v = q.popleft()
flag = False
for u in g[v]:
if len(g[u]) == 1:
flag = True
if visit[u] == -1:
visit[u] = 0
q.append(u)
if flag:
x += 1
#print(x)
M = (n-1)-len(leaf)+x
print(m, M)
if __name__ == '__main__':
main()
```
Yes
| 105,415 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted 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_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
#main code
def get(n):
p=2**32
c=32
while p:
if p&n:
return c+1
c-=1
p/=2
return 0
inp=inp()
n=inp[0]
pos=1
mx=0
root=0
deg=Counter()
d=[[] for i in range(n+1)]
for i in range(n-1):
u,v=inp[pos],inp[pos+1]
pos+=2
d[u].append(v)
d[v].append(u)
deg[u]+=1
deg[v]+=1
if deg[u]>mx:
mx=deg[u]
root=u
if deg[v]>mx:
mx=deg[v]
root=v
vis=[0]*(n+1)
vis[root]=1
q=[(root,0)]
f1,f2=0,0
mx=0
c1=0
d1=Counter()
while q:
x,w=q.pop(0)
ft=0
for i in d[x]:
if not vis[i]:
vis[i]=1
q.append((i,w+1))
ft=1
if not ft:
c1+=1
d1[x]+=1
if w%2:
#print 1,x
f1=1
else:
#print 2,x
f2=1
if f1 and f2:
ans1=3
else:
ans1=1
ans2=n-1-(c1-len(d1.keys()))
pr_arr((ans1,ans2))
```
No
| 105,416 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
print("YES")
```
No
| 105,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
n = int(input())
e = dict.fromkeys(list(range(1,n+1)))
for _ in range(n-1):
v1,v2 = map(int,input().split())
if(e[v1]== None and e[v1]!= -1 ):
e[v1] = v2
else:
e[v1] = -1;
if(e[v2]== None and e[v2]!= -1 ):
e[v2] = v1
else:
e[v2] = -1;
le = [0]*n
for i in range(1,n+1):
if(e[i]!=-1):
le[e[i]]+=1
m = 1
for i in range(n):
m +=int(le[i]/2)
if(m>2):
print(m,n-1 - m+1)
elif(m>1):
print(1,n-m)
else:
print(1,n-1)
if(n>1000):
print(m)
```
No
| 105,418 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
from sys import stdin
input=stdin.readline
n=int(input())
graph=[set() for i in range(n+1)]
visited=[False]*(n+1)
visited2=[False]*(n+2)
for i in range(n - 1):
a,b=map(int,input().split())
graph[a].add(b)
graph[b].add(a)
m=[1,n-1]
def dfs2(node,depth):
went=False
visited2[node]=True
for i in graph[node]:
if not visited2[i]:
dfs2(i, depth + 1)
went=True
if not went:
if depth % 2 == 1 and depth != 1:
m[0]=3
dfs2(1,0)
oneBranches=[0]*(n+1)
for i in range(1,n+1):
if len(graph[i]) == 1:
oneBranches[graph[i].pop()] += 1
for i in range(n+1):
if oneBranches[i] > 1:
m[1] -= oneBranches[i] - 1
print(*m)
#if there is odd path length, min is 3 else 1
#max length: each odd unvisited path gives n, while each even unvisited path gives n - 1
```
No
| 105,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:
* For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0.
Note that you can put very large positive integers (like 10^{(10^{10})}).
It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.
<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).
<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0.
What are the minimum and the maximum possible values of f for the given tree? Find and print both.
Input
The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.
The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.
Output
Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.
Examples
Input
6
1 3
2 3
3 4
4 5
5 6
Output
1 4
Input
6
1 3
2 3
3 4
4 5
4 6
Output
3 3
Input
7
1 2
2 7
3 4
4 7
5 6
6 7
Output
1 6
Note
In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3.
<image>
In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum.
<image>
Submitted Solution:
```
def main():
n = int(input())
graph = dict()
for i in range(n - 1):
a, b = map(int, input().split())
if not a in graph:
graph[a] = {b}
else:
graph[a].add(b)
if not b in graph:
graph[b] = {a}
else:
graph[b].add(a)
almost_corner = set()
corners = set()
for i in graph:
if len(graph[i]) == 1:
corners.add(i)
for k in graph[i]:
almost_corner.add(k)
dif_corners = len(almost_corner)
not_check = set()
cor_min_dif = 0
for i in almost_corner:
add_two = False
not_check.add(i)
for c in graph[i]:
if c in almost_corner and c not in not_check:
not_check.add(c)
add_two = True
if add_two:
cor_min_dif += 2
print(1 + cor_min_dif, n - len(corners) + dif_corners - 1)
if __name__ == "__main__":
t = 1
for i in range(t):
main()
```
No
| 105,420 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
# Template 1.0
import sys, re, math
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
def STR(): return list(input())
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def sortListWithIndex(listOfTuples, idx): return (sorted(listOfTuples, key=lambda x: x[idx]))
def sortDictWithVal(passedDic):
temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))[::-1]
toret = {}
for tup in temp:
toret[tup[0]] = tup[1]
return toret
def sortDictWithKey(passedDic):
return dict(OrderedDict(sorted(passedDic.items())))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
def calcNext(num):
for i in range(num+1, 2**m):
if(i not in vis):
return i
return -1
def calcPrev(num):
for i in range(num-1, -1, -1):
if (i not in vis):
return i
return -1
t = INT()
while (t != 0):
n, m = MAP()
currLen = 2**m
currMed = (2**m-1)//2
vis = set()
# vis.add(currMed)
for _ in range(n):
torem = int(input(), 2)
if(currLen%2==0):
currLen-=1
vis.add(torem)
if(torem<=currMed):
currMed = calcNext(currMed)
else:
currLen-=1
vis.add(torem)
if(torem>=currMed):
currMed = calcPrev(currMed)
# print(currMed)
# print(bin(currMed).replace("0b", ""))
print(format(currMed, '0'+str(m)+'b'))
t-=1
'''
1
4 3
000
111
100
011
1 2 3 4 5 6
'''
```
| 105,421 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
t=int(input())
for _ in range(t):
n,m=map(int,input().split())
arr=[]
for i in range(n):
arr.append(input().strip())
arr1=[]
for i in arr:
arr1.append(int(i,2))
a=(2**m-n-1)//2
arr1.sort()
for i in arr1:
if i<=a:
a+=1
ans=bin(a).replace("0b","")
ans="0"*(m-len(ans))+ans
print(ans)
```
| 105,422 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
import sys
max_int = 1000000001 # 10^9+1
min_int = -max_int
t = int(input())
for _t in range(t):
n, m = map(int, sys.stdin.readline().split())
to_skip = []
for _n in range(n):
s = int(sys.stdin.readline()[:-1], base=2)
to_skip.append(s)
to_skip.sort()
mid = (2 ** m - n - 1) // 2
for elem in to_skip:
if elem <= mid:
mid += 1
else:
break
print(("{0:0>" + str(m) + "b}").format(mid))
```
| 105,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineerin College
Date:24/05/2020
'''
import sys
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def read():
tc=0
if tc:
input=sys.stdin.readline
else:
sys.stdin=open('input1.txt', 'r')
sys.stdout=open('output1.txt','w')
def solve():
for _ in range(ii()):
n,m=mi()
tot=pow(2,m)-n
x=(tot-1)//2
a=[]
for i in range(n):
s=si()
a.append(int(s,2))
a.sort()
cnt=bisect(a,x)
x+=cnt
c=cnt
while(cnt):
cnt=bisect(a,x)-c
x+=cnt
c+=cnt
s=bin(x)[2:]
print('0'*(m-len(s))+s)
if __name__ =="__main__":
# read()
solve()
```
| 105,424 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
tc = int(input())
for _ in range(tc):
n,m = map(int,input().split())
cache = {}
curr = (2**m - 1)//2
for i in range(n):
s = input()
x = int(s,2)
cache[x] = 1
if i%2 == 0:
if x<=curr:
while(len(cache)!=0 and curr+1 in cache):
curr += 1
curr += 1
else:
if x>=curr:
while(len(cache)!=0 and curr-1 in cache):
curr -= 1
curr -= 1
temp = bin(curr)[2:]
if m!=len(temp):
temp = "0"*abs(m-len(temp)) + temp
print(temp)
```
| 105,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
def solve():
n, m = [int(x) for x in input().split()]
a = []
for i in range(n):
a.append(input())
a = [int(x, 2) for x in a]
need = ((1 << m) - n - 1)//2 + 1
cur = (1 << (m-1)) - 1
while True:
left = cur+1
flag = bool(cur in a)
for s in a:
if s <= cur:
left -= 1
if left == need and not flag:
ans = bin(cur).split('b')[1]
if len(ans) < m:
ans = '0'*(m-len(ans)) + ans
print(ans)
return
elif left < need:
cur += 1
else:
cur -= 1
t = int(input())
while t > 0:
t -= 1
solve()
```
| 105,426 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
import math
import string
t = int(input())
for tt in range(t):
n, m = map(int, input().split())
a = []
for i in range(n):
a.append(int(input(), 2))
need = (2 ** m - n - 1) // 2
l = 0
r = 2 ** m
while(r - l > 1):
mid = (l + r) // 2
x = mid
for i in range(n):
if(a[i] < mid):
x -= 1
if(x > need):
r = mid
else:
l = mid
ans = []
for i in range(m):
ans.append(l % 2)
l //= 2
ans.reverse()
for i in range(m):
print(ans[i], end="")
print()
```
| 105,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
def dd(i,nn):
s=bin(i).replace("0b", "")
l=len(s)
#print(l,nn)
kk=""
for i in range (nn-l):
kk=kk+'0'
print(kk+s)
t=int(input())
for i in range(t):
n,m=map(int,input().split())
aa=[]
for pp in range(n):
pp=input()
aa.append(int(pp,2))
y=2**(m-1) - 1
a=max(0,y-n)
b=min(y+n,2**m -1)
for i in range(a,b+1):
l=i
r=2**m -1-i
#print(l,r)
flag=1
for k in aa:
if(k==i):
flag=0
break
if(k<i):
l=l-1
else:
r=r-1
if(flag==0):
continue
if(n%2==0 and l==r-1):
dd(i,m)
if(n%2!=0 and l==r):
dd(i,m)
```
| 105,428 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
from sys import stdin,stdout
import math
from bisect import bisect_left
def main():
t = int(stdin.readline())
for _ in range (t):
n,m = list(map(int, stdin.readline().split()))
total = pow(2,m) - 1
med = math.floor((total) / 2)
arr = []
for _ in range(n):
num = int(stdin.readline(),2)
arr.append(num)
arr.sort()
ite = min(total+1,200)
ans = 0
for x in range(max(0, med - int(ite/2)),min(total+1, med + int(ite/2)+1)):
pos = bisect_left(arr, x)
if pos != len(arr) and arr[pos] == x:
continue
# use pos to calculate left and right quantity
# if n % 2 == 1 then exclude current
# else then include currente
elif n % 2 == 1:
left = x - pos
right = total - x - (n - pos)
if left == right:
ans = x
break
else:
left = x - pos + 1
right = total - x - (n - pos)
if left == right:
ans = x
break
ans = bin(ans).replace("0b", "")
if len(ans) < m:
ans = "0" * (m - len(ans)) + ans
stdout.write(ans + "\n")
main()
```
Yes
| 105,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted 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")
# ------------------------------
from math import factorial, floor
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
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 comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
# ------------------------------
from bisect import bisect
def main():
for _ in range(N()):
n, mm = RL()
arr = [int(input(), 2) for _ in range(n)]
arr.sort()
now = (2**mm)-n
res = (now-1)//2
for i in arr:
if i<=res:
res+=1
# print(res)
res = bin(res)[2:].zfill(mm)
print(res)
if __name__ == "__main__":
main()
```
Yes
| 105,430 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
import sys, heapq
from collections import *
from functools import lru_cache
sys.setrecursionlimit(10**6)
def main():
# sys.stdin = open('input.txt', 'r')
t = int(input())
for _ in range(t):
n, m = map(int,input().split(' '))
total = 2**m
mid = (total-1)//2
removes = set()
for _ in range(n):
s = input()
cur = 0
for ch in s:
cur = cur<<1
if ch == '1':
cur += 1
removes.add(cur)
removed = set()
for r in removes:
if total & 1 and r >= mid:
while mid-1 in removed:
mid -= 1
mid -= 1
elif total & 1 == 0 and r <= mid:
while mid+1 in removed:
mid += 1
mid += 1
total -= 1
removed.add(r)
res = []
for i in range(m):
if mid & (1<<i):
res.append('1')
else:
res.append('0')
# print(''.join(res[::-1]),mid)
print(''.join(res[::-1]))
if __name__ == "__main__":
main()
```
Yes
| 105,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
x=(2**m-n-1)//2
arr=[]
for i in range(n):
arr.append(int(input(),2))
arr.sort()
for i in range(0,n):
if(arr[i]<=x):
x+=1
if x in arr:
x+=1
y=bin(x)[2:]
print('0'*(m-len(y))+y)
```
Yes
| 105,432 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
for _ in range(int(input())):
mylist = []
mylist1 = []
n,m = map(int,input().split())
for _ in range(n):
mylist.append(int(input(),2))
ans = min(2**m-1,100)
for num in range(ans+1):
if num not in mylist:
mylist1.append(num)
mylist1.sort()
answer = bin(mylist1[int((len(mylist1)-1)/2)])[2:]
while len(answer)<m:
answer='0'+answer
if n == 1 and mylist == ['1']:
print('0')
else:
print(answer)
```
No
| 105,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
import math
def calculate(med,count,l,dir):
if dir=='left' :
med=int(med)+1
while ( count>0 ) :
med=med-1
if (not(med in l)):
count=count-1
return med
elif dir=='right' :
med=int(med)
while ( count>0 ) :
med=med+1
if (not(med in l)):
count=count-1
return med
t=int(input())
while ( t > 0 ):
n,m=map(int,input().split())
l=[]
med=float('{:.1f}'.format(float((math.pow(2,m)-1)/2)))
#print('med start = ',med)
right=0
left=0
for i in range(0,n):
val=int(input(),2)
l.append(val)
if val > med:
right+=1
elif val <= med:
left+=1
if right>left :
count=int(((right-left)/2)+1)
#print('count',count)
med=calculate(med,count,l,'left')
elif right<left :
count=int(((left-right)+1)/2)
#print('count',count)
med=calculate(med,count,l,'right')
elif right==left :
count=1
med=calculate(med,count,l,'left')
print('{:0>{}}'.format('{:b}'.format(med),m))
#print(l)
#print('left',left,'right',right)
l.clear()
t=t-1
```
No
| 105,434 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
from collections import Counter
T = int(input())
for _ in range(T):
n,m = map(int,input().split())
l = []
for i in range(n):
l.append(int(input(),2))
d = Counter(l)
mid = ((2**m - n)-1)//2
i = 0
j = mid
while i<=j:
if d.get(i,0) == 1:
j += 1
i += 1
r = bin(j)[2:]
r.zfill(m)
print(r)
```
No
| 105,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
# data=[]
# with open('input.txt') as f:
# data = [i for i in f]
# data.reverse()
# input = data.pop
def get_ints():
return list(map(int, input().split()))
for _ in range(int(input())):
N,M = get_ints()
a = []
for i in range(N):
a.append(int(input(),2))
g = [i for i in range(1,1<<M) if not i in a]
# print(g)
if len(g)==0:
print("0"*M)
else:
d = bin(g[(len(g)-1)//2])[2:]
print("0"*(M-len(d)) + d)
```
No
| 105,436 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
class Solution:
def __init__(self, debug=False, n=None, a=None, b=None):
if n is None:
n, a, b = Solution.input()
self.n = n
self.a = a
self.b = b
self.real = [x for x in self.a]
self.inverse = [not x for x in self.a]
self.curr_zero = 0
self.curr_last = self.n - 1
self.real_idx = self.n - 1
self.inv = False
self.debug = debug
if debug:
self.current_state = [x for x in self.a]
self.operations = [self.n + 1000]
@staticmethod
def input():
n = int(input())
a = [x == '1' for x in input()]
b = [x == '1' for x in input()]
return n, a, b
def iteration(self):
if self.inv:
if self.inverse[self.curr_last] != self.b[self.real_idx]:
if self.inverse[self.curr_zero] != self.inverse[self.curr_last]:
self.operation(0)
self.real[self.curr_zero], self.inverse[self.curr_zero] = self.inverse[self.curr_zero], self.real[
self.curr_zero]
self.operation(self.real_idx)
self.inv = False
self.curr_zero, self.curr_last = self.curr_last, self.curr_zero
else:
if self.real[self.curr_last] != self.b[self.real_idx]:
if self.real[self.curr_zero] != self.real[self.curr_last]:
self.operation(0)
self.real[self.curr_zero], self.inverse[self.curr_zero] = self.inverse[self.curr_zero], self.real[
self.curr_zero]
self.operation(self.real_idx)
self.inv = True
self.curr_zero, self.curr_last = self.curr_last, self.curr_zero
if self.inv:
self.curr_last += 1
else:
self.curr_last -= 1
self.real_idx -= 1
def solve(self):
while self.real_idx >= 0:
self.iteration()
return self.return_answer()
def special_cases(self):
if self.n == 1:
if self.a[0] != self.b[0]:
self.operation(0)
return self.return_answer
# if self.n == 2:
# if self.a[1] != self.b[1]:
# if self.a[0] != self.a[1]:
# self.operation(0)
# self.a[0] = not self.a[0]
# self.operation(1)
# self.a = [not x for x in self.a][::-1]
# if self.debug:
# assert self.current_state[1] == self.b[1]
# if self.a[0] != self.b[0]:
# self.operation(1)
# return self.return_answer
def operation(self, i):
self.operations.append(i + 1)
if self.debug:
self.current_state[:i + 1] = [not x for x in self.current_state[:i + 1]][::-1]
if self.n < 2:
return
def return_answer(self):
if self.debug:
assert all(x == y for x, y in zip(self.b, self.current_state))
ans = f'{len(self.operations) - 1} {" ".join(str(x) for x in self.operations[1:])}'
print(ans)
return ans
def solve():
n = int(input())
a = [int(x) for x in input()]
b = [int(x) for x in input()]
# state
real = [x for x in a]
inverse = [(x + 1) % 2 for x in a]
inv_idx = 0
real_idx = n - 1
inv = False
operations = []
# fix all
seq = []
while real_idx > 0:
if not inv:
if real[real_idx] == b[real_idx]:
real_idx -= 1
inv_idx += 1
seq.append(real[real_idx])
continue
if real[0] != real[real_idx]:
operations.append(1)
real[0], inverse[-1] = inverse[-1], real[0]
operations.append(real_idx + 1)
real[real_idx] = b[real_idx]
seq.append(real[real_idx])
inverse[inv_idx] = (real[real_idx] + 1) % 2
inv = True
else:
if inverse[inv_idx] == b[real_idx]:
real_idx -= 1
inv_idx += 1
seq.append(inverse[inv_idx])
continue
if inverse[-1] != inverse[inv_idx]:
operations.append(1)
real[0], inverse[-1] = inverse[-1], real[0]
operations.append(real_idx + 1)
inverse[inv_idx] = b[real_idx]
real[real_idx] = (b[real_idx] + 1) % 2
seq.append(inverse[inv_idx])
inv = False
real_idx -= 1
inv_idx += 1
# fix 0
if (inv and inverse[-1] != b[0]) or (real[0] != b[0] and not inv):
operations.append(1)
seq.append(b[0])
print(seq, b)
print(len(operations), *operations)
if __name__ == '__main__':
for _ in range(int(input())):
n, a, b = Solution.input()
Solution(False, n, a, b).solve()
```
| 105,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
from sys import stdin
t = int(stdin.readline().strip())
for _ in range(t):
n = int(stdin.readline().strip())
a = stdin.readline().strip()
b = stdin.readline().strip()
#n,m = list(map(int, stdin.readline().strip().split(' ')))
out = []
for i in range(n-1):
if a[i] != a[i+1]:
out.append(i+1)
current = a[-1]
for i in range(n-1, -1, -1):
if b[i] != current:
out.append(i+1)
current = '0' if current == '1' else '1'
print(len(out), end=" ")
print(*out)
```
| 105,438 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools
from collections import deque,defaultdict,OrderedDict
import collections
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
#Solving Area Starts-->
#Solving Area Starts-->
for _ in range(ri()):
n=ri()
a=rs()
b=rs()
s=""
if a==b:
print(0)
continue
if n==1:
print(1,1)
continue
ans1=[]
ans2=[]
z=0
c=0
for j in range(n):
if a[j]=='0':
break
else:
c+=1
if c>0:
ans1.append(c)
i=c
while i<n:
if a[i]=='1':
if i-1>=0:
ans1.append(i)
s='1'*(i)
if a[i:]=='1'*(n-len(s)):
z=1
break
for k in range(i,n):
if a[k]=='0':
break
ans1.append(k)
i=k+1
else:
i+=1
if z==1:
ans1.append(n)
z=0
c=0
for j in range(n):
if b[j]=='0':
break
else:
c+=1
if c>0:
ans2.append(c)
i=c
while i<n:
if b[i]=='1':
if i-1>=0:
ans2.append(i)
s='1'*(i)
if b[i:]=='1'*(n-len(s)):
z=1
break
for k in range(i,n):
if b[k]=='0':
break
ans2.append(k)
i=k+1
else:
i+=1
if z==1:
ans2.append(n)
# print(ans1)
# print(ans2)
print(len(ans1)+len(ans2),*(ans1+ans2[::-1]))
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
```
| 105,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
#e=[int(x) for x in input().split()]
#a=list(input())
#b=list(input())
a=[int(x) for x in input()]
b=[int(x) for x in input()]
a.append(0)
b.append(0)
ans=[]
for i in range(n):
if a[i]!=a[i+1]:
ans.append(i+1)
#a= [i-a[j] for j in range(i,-1,-1)]+ a[i+1:]
for i in range(n-1,-1,-1):
if b[i]!=b[i+1]:
ans.append(i+1)
print(len(ans),end=" ")
for x in ans:
print(x,end=" ")
print()
```
| 105,440 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
# import sys
# import math
# input = sys.stdin.readline
from collections import deque
from queue import LifoQueue
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
a1 = []
a2 = []
# if n==1:
# if a[0]==b[0]:
# print(0)
# else:
# print(1,1)
for i in range(0,n,2):
if a[i:i+2]=='01':
a1.append(i+1)
a1.append(i+2)
elif a[i:i+2]=='10':
if i!=0:
a1.append(i)
a1.append(i+1)
elif a[i:i+2]=='11':
if i!=0:
a1.append(i)
a1.append(i+2)
elif a[i:i+2]=='1':
if n-1>0:
a1.append(n-1)
a1.append(n)
for i in range(0,n,2):
if b[i:i+2]=='01':
a2.append(i+1)
a2.append(i+2)
elif b[i:i+2]=='10':
if i!=0:
a2.append(i)
a2.append(i+1)
elif b[i:i+2]=='11':
if i!=0:
a2.append(i)
a2.append(i+2)
elif b[i:i+2]=='1':
if n-1>0:
a2.append(n-1)
a2.append(n)
a2.reverse()
c = a1+a2
# print(*c)
for i in range(len(c)-1):
if c[i]==c[i+1]:
c.pop(i)
c.pop(i)
break
print(len(c),*c)
```
| 105,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=input()
b=input()
aToAllOnesOrZeros=[]
for i in range(n-1):
if a[i]!=a[i+1]: #flip up to a[i]. Then a[0,1,...i+1] will be uniform. a will become all = to a[-1]
aToAllOnesOrZeros.append(i+1)
bToAllOnesOrZeros=[]
for i in range(n-1):
if b[i]!=b[i+1]: #b will become all = to b[-1]
bToAllOnesOrZeros.append(i+1)
res=aToAllOnesOrZeros
if a[-1]!=b[-1]: #flip all aTransformed so that aTransformed==bTransformed
res.append(n)
for i in range(len(bToAllOnesOrZeros)-1,-1,-1):
res.append(bToAllOnesOrZeros[i]) #reverse the process
l=len(res)
res.insert(0,l)
print(' '.join([str(x) for x in res]))
#def flip(s,l):
# S=list(s[:l])
# for i in range(l):
# S[i]='0' if S[i]=='1' else '1'
# return ''.join(S)+s[l:]
#
#y=a
#for i in range(n-1):
# if y[i]!=y[i+1]:
# y=flip(y,i+1)
#print(y) #y will be all a[-1]
#
#z=b
#for i in range(n-1):
# if z[i]!=z[i+1]:
# z=flip(z,i+1)
#print(z) #z will be all b[-1]
```
| 105,442 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import collections
t1=int(input())
for _ in range(t1):
flipflag=0
bitflag=0
n=int(input())
a2=input()
a=collections.deque([])
for i in range(n):
a.append(a2[i])
b=input()
ans=[]
for i in range(n-1,-1,-1):
if flipflag==0:
p=a[-1]
else:
p=a[0]
if bitflag==1:
if p=='0':
p='1'
else:
p='0'
if flipflag==0:
q=a[0]
else:
q=a[-1]
if bitflag==1:
if q=='0':
q='1'
else:
q='0'
if p!=b[i]:
if q!=b[i]:
ans.append(str(i+1))
if flipflag==0:
a.popleft()
else:
a.pop()
flipflag=1-flipflag
bitflag=1-bitflag
else:
ans.append('1')
ans.append(str(i+1))
if flipflag==0:
a.popleft()
else:
a.pop()
flipflag=1-flipflag
bitflag=1-bitflag
else:
if flipflag==0:
a.pop()
else:
a.popleft()
print(len(ans))
if len(ans)>0:
print(' '.join(ans))
```
| 105,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import deque
t = int(input())
for _ in range(t):
n = int(input())
a = input()
b = input()
q = deque([i for i in range(n)])
rev = False
ans = []
for i in range(n)[::-1]:
if rev:
a_ind = q[0]
if a[a_ind] != b[i]:
q.popleft()
continue
else:
if a[q[-1]] == b[i]:
ans.append(i + 1)
else:
ans.append(1)
ans.append(i + 1)
q.pop()
rev = False
else:
a_ind = q[-1]
if a[a_ind] == b[i]:
q.pop()
continue
else:
if a[q[0]] != b[i]:
ans.append(i + 1)
else:
ans.append(1)
ans.append(i + 1)
q.popleft()
rev = True
print(len(ans), *ans)
```
| 105,444 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import os
t1=int(input())
while t1:
t1-=1
n=int(input())
s=input()
t=input()
s+="0"
t+="0"
ans=[]
for i in range(n):
if s[i]!=s[i+1]:
ans.append(i+1)
for i in range(n,0,-1):
if t[i]!=t[i-1]:
ans.append(i)
print(len(ans),end=' ')
for x in ans:
print(x,end=' ')
print()
```
Yes
| 105,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def convert_to_1(s):
L=[]
for i in range(len(s)-1):
if s[i]!=s[i+1]:L.append(i+1)
if s[-1]=='0':L.append(len(s))
return L
for i in ' '*(int(input())):
n=int(input())
s1=input()
s2=input()
L1=convert_to_1(s1)
L2=convert_to_1(s2)
L=L1+L2[::-1]
print(len(L),end=' ')
for i in L:print(i,end=' ')
print()
```
Yes
| 105,446 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from sys import stdin, stdout
# 0 1 2 3 4 5 6 7 8 9
# 9 8 7 6 5 4 3 2 1 (0) 9
# 1 2 3 4 5 6 7 8 (9) 8
# 8 7 6 5 4 3 2 (1) 7
# 2 3 4 5 6 7 (8) 6
# 7 6 5 4 3 (2) 5
# ..........
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
a = stdin.readline().strip()
b = stdin.readline().strip()
ans = []
idx = 0
flip = False
for i in range(n-1, -1, -1):
if (not flip and a[idx] == b[i]) or (flip and a[idx] != b[i]):
ans.append(1)
ans.append(i+1)
if flip:
idx -= i
else:
idx += i
flip = not flip
stdout.write(str(len(ans)) + ' ')
if len(ans) > 0:
stdout.write(' '.join(map(str, ans)) + '\n')
```
Yes
| 105,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
#!/usr/bin/env python3
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
a=list(input())
b=list(input())
ans=[]
l=0
r=n-1
for i in range(n-1,-1,-1):
if a[i]!=b[i]:
r=i
break
else:
r=0
if l==r:
if a[0]==b[0]:
print(0)
else:
print(1,1)
continue
else:
flip=False
count=r
while l!=r:
if flip==False:
if a[l]==b[count]:
if a[l]=='0':
a[l]='1'
else:
a[l]='0'
ans.append(1)
ans.append(count+1)
else:
ans.append(count+1)
flip=True
else:
if a[r]!=b[count]:
if a[r]=='0':
a[r]='1'
else:
a[r]='0'
ans.append(1)
ans.append(count+1)
else:
ans.append(count+1)
flip=False
for i in range(count,-1,-1):
if flip==True:
if a[l+count-i]==b[i]:
l=l+count-i
count=i
break
else:
if a[r-count+i]!=b[i]:
r=r-count+i
count=i
break
else:
if flip==False:
l=r
else:
r=l
if flip==False:
if a[l]!=b[0]:
ans.append(1)
else:
if a[r]==b[0]:
ans.append(1)
ans=[len(ans)]+ans
print(*ans)
```
Yes
| 105,448 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import sys
#comment these out later
#sys.stdin = open("in.in", "r")
#sys.stdout = open("out.out", "w")
inp = sys.stdin.read().split(); ii = 0
t = int(inp[ii]); ii += 1
for _ in range(t):
toprint = []
n = int(inp[ii]); ii += 1
a = list(map(int, inp[ii])); ii += 1
b = list(map(int, inp[ii])); ii += 1
lead = a[0]
for i in range(n-1, -1, -1):
x = a[i]
y = b[i]
if lead == y:
toprint.append(1)
toprint.append(i + 1)
lead = y
else:
toprint.append(i + 1)
lead = y
if toprint == []:
print(0)
else:
print(" ".join(list(map(str, toprint))))
```
No
| 105,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from sys import stdin
tt = int(input())
for loop in range(tt):
n = int(input())
a = input()
b = input()
ans = []
x = 0
for i in range(n-1,-1,-1):
if x == 0 and a[i] != b[i]:
x ^= 1
ans.append(i+1)
elif x == 1 and a[i] == b[i]:
x ^= 1
ans.append(i+1)
print (len(ans),*ans)
```
No
| 105,450 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
t=input()
ans=0
A=[]
for i in range(n-1,-1,-1):
if ans%2==1 and t[i]!=s[i]:
continue
elif ans%2==0 and t[i]==s[i]:
continue
else:
ans+=1
A.append(i+1)
t=s
print(ans,end=" ")
for i in A:
print(i,end=" ")
print()
```
No
| 105,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=1
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
//If you Know me , Then you probably don't know me
"""
###########################---START-CODING---##############################
num=int(z())
for _ in range( num ):
n=int(z())
T1=fast()
T2=fast()
ans=[]
k=1
for i in T1:
if i=='1':
ans.append(k)
k+=1
k=1
for i in T2:
if i=='1':
ans.append(k)
k+=1
print(len(ans),*ans)
```
No
| 105,452 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
for t in range(int(input())):
a = list(map(int,input().split()))
b = list(map(int,input().split()))
ans = 0
while(a[0]):
if(b[2]):
d = min(b[2],a[0])
a[0] -= d
b[2] -= d
else :
if(b[0]):
dd = min(b[0],a[0])
a[0] -= dd
b[0] -= dd
else:
dd = min(b[1],a[0])
a[0] -= dd
b[1] -= dd
while(a[1]):
if(b[0]):
d = min(b[0],a[1])
a[1] -= d
b[0] -= d
else :
if(b[1]):
dd = min(b[1],a[1])
a[1] -= dd
b[1] -= dd
else:
dd = min(b[2],a[1])
ans -= 2*dd
a[1] -= dd
b[2] -= dd
d = min(b[1],a[2])
ans += 2*d
print(ans)
```
| 105,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
t=int(input())
for test in range(t):
x0,x1,x2=map(int,input().split())
y0,y1,y2=map(int,input().split())
s=0
m = min(x0,y2)
x0-=m
y2-=m
m = min(x1,y0)
x1-=m
y0-=m
m = min(x2,y1)
x2-=m
y1-=m
s+=2*m
s-=2* min(x1,y2)
print(s)
```
| 105,454 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
t=int(input())
while(t>0):
t=t-1
[x1,y1,z1]=input().split()
[x2,y2,z2]=input().split()
x1=int(x1)
y1=int(y1)
z1=int(z1)
x2=int(x2)
y2=int(y2)
z2=int(z2)
ans=0
if(z1>=y2):
ans=ans+2*y2
if(x1+z1-y2<z2):
ans=ans-2*(z2+y2-z1-x1)
else:
ans=ans+2*z1
if(x1<z2):
ans=ans-2*(z2-x1)
print(ans)
```
| 105,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
from sys import *
input = stdin.readline
for _ in range(int(input())):
x1,y1,z1 = map(int,input().split())
x2,y2,z2 = map(int,input().split())
mx = 0
ma2 = min(z1,y2)
mx += ma2*2
z1 -= ma2
y2 -= ma2
mb2 = min(z1,z2)
z1 -= mb2
z2 -= mb2
mb2 = min(x1,z2)
z2 -= mb2
x1 -= mb2
ma1 = min(y1,z2)
z2 -= ma1
y1 -= ma1
mx -= ma1*2
stdout.write(str(mx)+'\n')
```
| 105,456 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
for _ in range(int(input())):
c = 0
x1, y1, z1 = map(int, input().split())
x2, y2, z2 = map(int, input().split())
plus = min(z1, y2)
z1 -= plus
y2 -= plus
c += plus*2
minus = min(y1, z2-x1-z1)
if minus > 0:
c -= minus*2
print(c)
```
| 105,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
ts = int(input())
for t in range(ts):
x1, y1, z1 = [int(i) for i in input().split(" ")]
x2, y2, z2 = [int(i) for i in input().split(" ")]
c = 0
if z1<=y2:
c += 2*z1
if z2>x1:
c -= 2*(z2-x1)
else:
c += 2*y2
z11 = x1+z1-y2
if z2>z11:
c -= 2*(z2-z11)
print(c)
```
| 105,458 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
t = int(input())
for i in range(t):
x1,y1,z1 = map(int,input().split())
x2,y2,z2 = map(int,input().split())
if z1 <= y2:
if y2 - z1 + x2 >= y1:
print(2*(z1))
else:
print(2*(y2+ x2 - y1))
else:
if z1-y2+x1 >= z2:
print(2*(y2))
else:
print(2*(z1+x1-z2))
```
| 105,459 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Tags: constructive algorithms, greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
x1, y1, z1 = list(map(int, input().split()))
x2, y2, z2 = list(map(int, input().split()))
p2 = min(z1, y2)
z1 -= p2
y2 -= p2
m2 = max(0, y1 - (x2 + y2))
print(2*(p2-m2))
```
| 105,460 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
import math
def function(x1, y1, z1, x2, y2, z2):
y1-=x2
if y1<=0:
print(2*min(z1, y2))
if y1>0:
if y1<=y2:
c=y2-y1
print(2*min(c, z1))
if y1>y2:
y1-=y2
print(-2*min(y1, z2))
if __name__=="__main__":
t=int(input())
for k1 in range(t):
x1, y1, z1=map(int, input().rstrip().split())
x2, y2, z2=map(int, input().rstrip().split())
function(x1, y1, z1, x2, y2, z2)
```
Yes
| 105,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
t = int(input());
for _ in range(t):
A = list(map(int,input().split()))
B = list(map(int,input().split()))
C = min(A[0],B[2]);
A[0] -= C;
B[2] -= C;
C = min(A[1],B[0]);
A[1] -= C;
B[0] -= C;
C = min(A[2],B[1]);
A[2] -= C;
B[1] -= C;
print(C * 2 - 2 * min(A[1],B[2]));
```
Yes
| 105,462 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
def main():
t = I()
rr = []
for _ in range(t):
a = LI()
b = LI()
t = min(a[0], b[2])
a[0] -= t
b[2] -= t
t = min(a[2], b[2])
a[2] -= t
b[2] -= t
t = min(a[1], b[2])
r = t * -2
t = min(a[2], b[1])
r += t * 2
rr.append(r)
return JA(rr, "\n")
print(main())
```
Yes
| 105,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
import sys
tc = int(sys.stdin.readline())
for _ in range(tc):
a0, a1, a2 = map(int, sys.stdin.readline().split())
b0, b1, b2 = map(int, sys.stdin.readline().split())
ans = 0
ans += 2 * min(a2, b1)
a2 -= min(a2, b1)
b1 -= min(a2, b1)
if b2 > a2 + a0:
b2 -= (a0 + a2)
a0 = 0
a2 = 0
ans += (-2 * min(b2, a1))
print(ans)
```
Yes
| 105,464 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
#code
if __name__ == "__main__":
t = int(input())
for _ in range(t):
ans = 0
x1, y1, z1 = map(int,input().split())
x2, y2, z2 = map(int,input().split())
z2 = z2 - min(z2,x1)
x1 = x1 - min(z2,x1)
z2 = z2 - min(z2,z1)
z1 = z1 - min(z2,z1)
ans = -(z2*2)
z2 = 0
y1 = y1 - z2
ans = ans + (min(y2,z1)*2)
print(ans)
```
No
| 105,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
import math
from collections import deque
from sys import stdin, stdout
from string import ascii_letters
input = stdin.readline
#print = stdout.write
def calc(a, b):
first = a[:]
second = b[:]
res = 0
for i in range(2, -1, -1):
for g in range(i - 1, 0, -1):
bf = min(first[i], second[g])
res += bf * (i) * (g)
first[i] -= bf
second[g] -= bf
for i in range(2, -1, -1):
bf = min(first[i], second[i])
first[i] -= bf
second[i] -= bf
for i in range(0, 3):
for g in range(i + 1, 3):
bf = min(first[i], second[g])
res -= bf * (i) * (g)
first[i] -= bf
second[g] -= bf
return res
for _ in range(int(input())):
first = list(map(int, input().split()))
second = list(map(int, input().split()))
print(calc(first, second))
```
No
| 105,466 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
n = int(input())
a = []
b = []
c = []
for i in range(n):
a.append([int(j) for j in input().split()])
b.append([int(j) for j in input().split()])
for i in range(n):
if (a[i][2] == 0 or b[i][1] == 0) and b[i][2] > a[i][0] + a[i][2]:
print(-(b[i][2] - a[i][0] - a[i][2]) * 2)
elif b[i][1] > a[i][2]:
print(2 * a[i][2])
else:
print(b[i][1] * 2)
```
No
| 105,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n. Each element of both sequences is either 0, 1 or 2. The number of elements 0, 1, 2 in the sequence a is x_1, y_1, z_1 respectively, and the number of elements 0, 1, 2 in the sequence b is x_2, y_2, z_2 respectively.
You can rearrange the elements in both sequences a and b however you like. After that, let's define a sequence c as follows:
c_i = \begin{cases} a_i b_i & \mbox{if }a_i > b_i \\\ 0 & \mbox{if }a_i = b_i \\\ -a_i b_i & \mbox{if }a_i < b_i \end{cases}
You'd like to make ∑_{i=1}^n c_i (the sum of all elements of the sequence c) as large as possible. What is the maximum possible sum?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers x_1, y_1, z_1 (0 ≤ x_1, y_1, z_1 ≤ 10^8) — the number of 0-s, 1-s and 2-s in the sequence a.
The second line of each test case also contains three integers x_2, y_2, z_2 (0 ≤ x_2, y_2, z_2 ≤ 10^8; x_1 + y_1 + z_1 = x_2 + y_2 + z_2 > 0) — the number of 0-s, 1-s and 2-s in the sequence b.
Output
For each test case, print the maximum possible sum of the sequence c.
Example
Input
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output
4
2
0
Note
In the first sample, one of the optimal solutions is:
a = \{2, 0, 1, 1, 0, 2, 1\}
b = \{1, 0, 1, 0, 2, 1, 0\}
c = \{2, 0, 0, 0, 0, 2, 0\}
In the second sample, one of the optimal solutions is:
a = \{0, 2, 0, 0, 0\}
b = \{1, 1, 0, 1, 0\}
c = \{0, 2, 0, 0, 0\}
In the third sample, the only possible solution is:
a = \{2\}
b = \{2\}
c = \{0\}
Submitted Solution:
```
import math
from decimal import *
def pf(n):
ans = [1]
if(n%2==0):
if(n!= 2):
ans.append(2)
while(n%2==0):
n//=2
p = 3
while(n>1):
if(n%p==0):
ans.append(p)
while(n%p==0):
n//=p
p+=1
return ans
for _ in range(int(input())):
a = list(map(int, input().split()))
b = list(map(int, input().split()))
ans = 0
o= [1, 2, 0]
pnt = 0
p1 = 2
p2 = 0
while(a[2]>0 and pnt<3):
mn = min(a[2], b[o[pnt]])
if(o[pnt]==1):
ans+=(2*mn)
b[o[pnt]]-=mn
a[2]-=mn
pnt+=1
p1 = 2
while(p1>=0 and p2<3):
mn = min(a[p2], b[p1])
if(p1>p2):
ans-=(2*p2)
b[p1]-=mn
a[p2]-=mn
if(a[p2]==0):
p2+=1
if(b[p1]==0):
p1-=1
print(ans)
```
No
| 105,468 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
T = input()
mod = int(1e9 + 7)
a = map(int, input().split())
c = []
for n in a:
b = (n // 2 + 2)
b = b * b
b //= 4
c.append(str(b % mod))
print(' '.join(c))
```
| 105,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def readvars():
k = map(int,input().split())
return(k)
def readlist():
li = list(map(int,input().split()))
return(li)
def anslist(li):
ans = " ".join([str(v) for v in li])
return(ans)
def main():
t = 1
# t = int(input())
for xx in range(t):
t = int(input())
n = readlist()
for m in n: print(((((m//2) - (m//4) + 1)%(1000000007))*((m//4) + 1))%1000000007)
# 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()
```
| 105,470 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
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)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##to find factorial and ncr
# N=100000
# mod = 10**9 +7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def solve():
n=N()
ar=lis()
for i in range(len(ar)):
m=ar[i]
v = m // 2
u = v // 2
w = (v - u)
print((u * w + u + w + 1) % mod)
solve()
#testcase(int(inp()))
```
| 105,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
t = int(input())
a = list(map(int, input().split()))
out = []
for n in a:
ans = (n//2 + 2)
ans = ans*ans
ans //= 4
out.append(ans%1000000007)
print(' '.join(str(x) for x in out))
```
| 105,472 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# 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")
MOD = 10 ** 9 + 7
memo = dict()
def solve(m):
if m not in memo:
if m < 0:
memo[m] = 0
if m == 0:
memo[m] = 1
half = m//2
memo[m] = (solve(half) + solve(half - 1) + solve(half - 2) + solve(half - 3)) % MOD
return memo[m]
t = int(input())
out = []
for m in map(int, input().split()):
#out.append(solve(m))
v = m//2
u = v//2
w = (v-u)
out.append((u*w+u+w+1)%MOD)
print('\n'.join(map(str,out)))
```
| 105,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
import sys
input=sys.stdin.readline
def f(n):
return ((n+4)//2)**2//4
input()
for x in list(map(int,input().split())):
print(f(x))
```
No
| 105,474 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
t = int(input())
n = list(map(int, input().split()))
for m in n:
print(((m//2) - (m//4) + 1)*((m//4) + 1))
```
No
| 105,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
t = int(input())
n1 = input().split()
ind = 0
while(ind<t):
n = int(n1[ind])
#ans = int((int(n/2+2)**2)/4)
ans = int(int((int(n/2+2)**2)/2)/2)
ans = ans%1000000007
print(ans)
ind += 1
```
No
| 105,476 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
x=int(input())
y=list(map(int,input().split()))
for i in range(x):
a=y[i]
if a==x*x and a==2*x and a==2+x:
print("4")
elif a==x*x and a==2*x or a==x*x and a==2+x:
print("3")
elif a==x*x or a==2*x or a==2+x:
print("2")
else:
print("2")
```
No
| 105,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
def solve():
n,k=map(int,input().split())
s=input()
if k>=n/2:
return "NO"
for i in range(k):
if s[i]!=s[-(i+1)]:
return "NO"
else:
return "YES"
for _ in range(int(input())):
print(solve())
```
| 105,478 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
def solve(n, k, s):
if k == 0:
return "YES"
count = 0
if n%2:
for i in range(n//2):
if s[i] == s[-i-1]:
count += 1
else:
break
if count >= k:
return "YES"
else:
return "NO"
else:
for i in range(n//2 - 1):
if s[i] == s[-i-1]:
count += 1
else:
break
if count>=k:
return "YES"
else:
return "NO"
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input()
print(solve(n, k, s))
```
| 105,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
n,k = map(int,input().split())
s = input()
if 2*k==n:
print("NO")
continue
if k==0:
print("YES")
continue
if s[:k]==s[-k:][::-1]:
print("YES")
else:
print("NO")
```
| 105,480 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
S = input()
print()
if k == 0:
print("YES")
continue
elif S[:k] == (S[::-1])[:k]:
if n//2 >= k + 1 - n%2:
print("YES")
continue
print("NO")
```
| 105,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
# 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
# sys.setrecursionlimit(5*10**5+100) #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
# 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
# ______________________________________________________________________________________________________
# For smallest prime factor of a number
# M = 2*(10**5)+10
# pfc = [i for i in range(M)]
# def pfcs(M):
# for i in range(2,M):
# if pfc[i]==i:
# for j in range(i+i,M,i):
# if pfc[j]==j:
# pfc[j] = i
# return
# pfcs(M)
# ______________________________________________________________________________________________________
tc = 1
tc = int(input())
for _ in range(tc):
n,k = inp()
s = str(input())
f = True
i = 0
j = n-1
while(k>0):
if i+1>=j:
f = False
break
# print(i,j)
if s[i]==s[j]:
pass
else:
f = False
break
i+=1
j-=1
k-=1
print("YES" if f else "NO")
```
| 105,482 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
# by the authority of GOD author: Kritarth Sharma #
import math
from collections import defaultdict,Counter
from itertools import permutations
from decimal import Decimal, localcontext
from collections import defaultdict
ii = lambda : int(input())
li = lambda:list(map(int,input().split()))
def main():
for _ in range(ii()):
n,k=li()
s=input()
c=0
if k==0:
print("YES")
else:
a1=s[:n//2]
a2=s[n//2+1:][::-1]
c=0
for i in range(min(len(a1),len(a2))):
if a1[i]==a2[i]:
c+=1
else:
break
if 2*k+1<=n and c>=k:
print("YES")
else:
print("NO")
import os,sys
from io import BytesIO,IOBase
#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")
from math import gcd
def find(l):
c=0
for i in range(len(l)):
if l[i]>1:
c=i
return c
else:
return None
def checkDivisibility(n, digit) :
# If the digit divides the
# number then return true
# else return false.
return (digit != 0 and n % digit == 0)
# Function to check if
# all digits of n divide
# it or not
def allDigitsDivide( n) :
nlist = list(map(int, set(str(n))))
for digit in nlist :
if digit!=0 and not (checkDivisibility(n, digit)) :
return False
return True
def lcm(s):
a=[int(d) for d in str(s)]
lc = a[0]
for i in a[1:]:
if i!=0:
lc = lc*i//gcd(lc, i)
return(lc)
if __name__ == "__main__":
main()
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def random():
"""My code gets caught in plagiarism check for no reason due to the fast IO template, .
Due to this reason, I am making useless functions"""
rating=100
rating=rating*100
rating=rating*100
print(rating)
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not
# changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
l=[1,]
# Print all prime numbers
for p in range(2, n+1):
if prime[p]:
l.append(p)
return l
def fact(n):
return 1 if (n == 1 or n == 0) else n * fact(n - 1)
def prime(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
```
| 105,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
import sys
import math
import heapq
import bisect
from collections import Counter
from collections import defaultdict
from io import BytesIO, IOBase
import string
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
self.BUFSIZE = 8192
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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, self.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")
def get_int():
return int(input())
def get_ints():
return list(map(int, input().split(' ')))
def get_int_grid(n):
return [get_ints() for _ in range(n)]
def get_str():
return input().split(' ')
def yes_no(b):
if b:
return "YES"
else:
return "NO"
def binary_search(good, left, right, delta=1, right_true=False):
"""
Performs binary search
----------
Parameters
----------
:param good: Function used to perform the binary search
:param left: Starting value of left limit
:param right: Starting value of the right limit
:param delta: Margin of error, defaults value of 1 for integer binary search
:param right_true: Boolean, for whether the right limit is the true invariant
:return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
alternatively returns False if no such value found
"""
limits = [left, right]
while limits[1] - limits[0] > delta:
if delta == 1:
mid = sum(limits) // 2
else:
mid = sum(limits) / 2
if good(mid):
limits[int(right_true)] = mid
else:
limits[int(~right_true)] = mid
if good(limits[int(right_true)]):
return limits[int(right_true)]
else:
return False
def prefix_sums(a, drop_zero=False):
p = [0]
for x in a:
p.append(p[-1] + x)
if drop_zero:
return p[1:]
else:
return p
def prefix_mins(a, drop_zero=False):
p = [float('inf')]
for x in a:
p.append(min(p[-1], x))
if drop_zero:
return p[1:]
else:
return p
def solve_a():
n, k = get_ints()
s = input().strip()
j = 0
while j <= (n - 3) // 2 and s[j] == s[-(j + 1)]:
j += 1
return yes_no(k <= j)
t = get_int()
for _ in range(t):
print(solve_a())
```
| 105,484 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Tags: brute force, constructive algorithms, greedy, strings
Correct Solution:
```
t=int(input())
for z in range(t):
n,k=[int(q)for q in input().split()]
s=input()
a=s[:k]+s[n-k:]
if k==n/2:
print("NO")
elif k==0 or a==a[::-1]:
print("YES")
else:
print("NO")
```
| 105,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
import sys
input=sys.stdin.readline
from collections import defaultdict as dc
from collections import Counter
from bisect import bisect_right, bisect_left
import math
from operator import itemgetter
from heapq import heapify, heappop, heappush
from queue import PriorityQueue as pq
for _ in range(int(input())):
n,k=map(int,input().split())
s=input()[:-1]
f=0
for i in range(k):
if s[i]==s[n-i-1]:
continue
else:
f=1
break
if f or 2*k>=n:
print("NO")
else:
print("YES")
```
Yes
| 105,486 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
t=int(input())
for tt in range(t):
n,k=list(map(int,input().split()))
a=input()
if n>=2*k+1:
if k==0:
print('YES')
else:
x=0
for i in range(k):
if a[i]!=a[-i-1]:
x=1
break
if x==0:
print("YES")
else:
print("NO")
else:
print('NO')
```
Yes
| 105,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
stk = input()
left = 0
ans = False
while(left<n):
right = left
while(right<n):
first = stk[:left]
last = stk[(right+1):]
if first==last[::-1] and left>=k:
ans|=True
right+=1
left+=1
if ans:
print("YES")
else:
print("NO")
```
Yes
| 105,488 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
if k == 0 or (s[:k] == ''.join(list(reversed(s[n - k:]))) and n // 2 + n % 2 >= k + 1):
print('YES')
else:
print('NO')
```
Yes
| 105,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
for i in range(int(input())):
n,k=map(int,input().split())
s=input()
def helper(s,n,k):
if k==0:
return("YES")
elif k!=0 and n%2==0:
return("NO")
else:
if s[:k]==s[-k:][::-1]:
return("YES")
else:
return("NO")
print(helper(s,n,k))
```
No
| 105,490 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
t=int(input())
while(t!=0):
t=t-1
s=input().split(' ')
y=int(s[0])
x=int(s[1])
m=0
s1=input()
s2=s1[::-1]
if(s1==s2 or x==0):
if(y-(2*x)>0):
m=1
print("YES")
else:
m=1
print("NO")
if(m!=1):
print("NO")
```
No
| 105,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
t=int(input())
for i in range(0,t,1):
n,k=map(int,input().split())
s=input()
l= [s[i:i+k+1] for i in range(0,n, k+1)]
print(l)
if len(l)>=k+1:
print("YES")
else:
print("NO")
```
No
| 105,492 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kawashiro Nitori is a girl who loves competitive programming.
One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.
Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$
Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.
The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.
Output
For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.
You can print letters in any case (upper or lower).
Example
Input
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa
Output
YES
NO
YES
NO
YES
NO
NO
Note
In the first test case, one possible solution is a_1=qw and a_2=q.
In the third test case, one possible solution is a_1=i and a_2=o.
In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def solve():
n, k = map(int, input().split())
s = input().rstrip()
if k == 0:
print("YES")
return
possible = False
n = len(s)
for i in range(k, n):
if possible:
break
for j in range(i, n):
start = s[:i]
end = s[j+1:]
if start == end and len(start) != 0:
possible = True
break
print("YES" if possible else "NO")
t = int(input())
# t = 1
for _ in range(t):
solve()
```
No
| 105,493 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integer arrays a and b of length n.
You can reverse at most one subarray (continuous subsegment) of the array a.
Your task is to reverse such a subarray that the sum ∑_{i=1}^n a_i ⋅ b_i is maximized.
Input
The first line contains one integer n (1 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7).
The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^7).
Output
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.
Examples
Input
5
2 3 2 1 3
1 3 2 4 2
Output
29
Input
2
13 37
2 4
Output
174
Input
6
1 8 7 6 3 6
5 9 6 8 8 6
Output
235
Note
In the first example, you can reverse the subarray [4, 5]. Then a = [2, 3, 2, 3, 1] and 2 ⋅ 1 + 3 ⋅ 3 + 2 ⋅ 2 + 3 ⋅ 4 + 1 ⋅ 2 = 29.
In the second example, you don't need to use the reverse operation. 13 ⋅ 2 + 37 ⋅ 4 = 174.
In the third example, you can reverse the subarray [3, 5]. Then a = [1, 8, 3, 6, 7, 6] and 1 ⋅ 5 + 8 ⋅ 9 + 3 ⋅ 6 + 6 ⋅ 8 + 7 ⋅ 8 + 6 ⋅ 6 = 235.
Tags: brute force, dp, implementation, math, two pointers
Correct Solution:
```
n=int(input())
array1=list(map(int,input().split()))
array2=list(map(int,input().split()))
totalSum=sum([array2[i]*array1[i] for i in range(n)])
maxSum=totalSum
"""
for i in range(n):
for j in range(i,i+2):
start=i
end=j
curSum=totalSum
while start>=0 and end<n:
#curSum=curSum-array1[start]*array2[start]-array1[end]*array2[end]
#curSum=curSum+array1[start]*array2[end]+array1[end]*array2[start]
curSum+=(array1[start]-array1[end])*(array2[end]-array2[start])
start-=1
end+=1
maxSum=max(maxSum,curSum)
"""
for i in range(n):
#for j in range(i,i+2):
start=i
end=i
curSum=totalSum
while start>=0 and end<n:
curSum+=(array1[start]-array1[end])*(array2[end]-array2[start])
start-=1
end+=1
maxSum=max(maxSum,curSum)
for i in range(n-1):
start=i
end=i+1
curSum=totalSum
while start>=0 and end<n:
curSum+=(array1[start]-array1[end])*(array2[end]-array2[start])
start-=1
end+=1
maxSum=max(maxSum,curSum)
print(maxSum)
```
| 105,494 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integer arrays a and b of length n.
You can reverse at most one subarray (continuous subsegment) of the array a.
Your task is to reverse such a subarray that the sum ∑_{i=1}^n a_i ⋅ b_i is maximized.
Input
The first line contains one integer n (1 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7).
The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^7).
Output
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.
Examples
Input
5
2 3 2 1 3
1 3 2 4 2
Output
29
Input
2
13 37
2 4
Output
174
Input
6
1 8 7 6 3 6
5 9 6 8 8 6
Output
235
Note
In the first example, you can reverse the subarray [4, 5]. Then a = [2, 3, 2, 3, 1] and 2 ⋅ 1 + 3 ⋅ 3 + 2 ⋅ 2 + 3 ⋅ 4 + 1 ⋅ 2 = 29.
In the second example, you don't need to use the reverse operation. 13 ⋅ 2 + 37 ⋅ 4 = 174.
In the third example, you can reverse the subarray [3, 5]. Then a = [1, 8, 3, 6, 7, 6] and 1 ⋅ 5 + 8 ⋅ 9 + 3 ⋅ 6 + 6 ⋅ 8 + 7 ⋅ 8 + 6 ⋅ 6 = 235.
Tags: brute force, dp, implementation, math, two pointers
Correct Solution:
```
# https://codeforces.com/contest/1519/submission/114616034
# import io,os
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
s = sum(a[i]*b[i] for i in range(n))
ans = s
for i in range(n):
l = i - 1
r = i + 1
dx = 0
while l >= 0 and r < n:
dx -= (a[r] - a[l]) * (b[r] - b[l])
ans = max(ans, s + dx)
l -= 1
r += 1
l = i
r = i + 1
dx = 0
while l >= 0 and r < n:
dx -= (a[r] - a[l]) * (b[r] - b[l])
ans = max(ans, s + dx)
l -= 1
r += 1
print(ans)
```
| 105,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integer arrays a and b of length n.
You can reverse at most one subarray (continuous subsegment) of the array a.
Your task is to reverse such a subarray that the sum ∑_{i=1}^n a_i ⋅ b_i is maximized.
Input
The first line contains one integer n (1 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7).
The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^7).
Output
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.
Examples
Input
5
2 3 2 1 3
1 3 2 4 2
Output
29
Input
2
13 37
2 4
Output
174
Input
6
1 8 7 6 3 6
5 9 6 8 8 6
Output
235
Note
In the first example, you can reverse the subarray [4, 5]. Then a = [2, 3, 2, 3, 1] and 2 ⋅ 1 + 3 ⋅ 3 + 2 ⋅ 2 + 3 ⋅ 4 + 1 ⋅ 2 = 29.
In the second example, you don't need to use the reverse operation. 13 ⋅ 2 + 37 ⋅ 4 = 174.
In the third example, you can reverse the subarray [3, 5]. Then a = [1, 8, 3, 6, 7, 6] and 1 ⋅ 5 + 8 ⋅ 9 + 3 ⋅ 6 + 6 ⋅ 8 + 7 ⋅ 8 + 6 ⋅ 6 = 235.
Tags: brute force, dp, implementation, math, two pointers
Correct Solution:
```
from collections import defaultdict
from itertools import accumulate
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
sys.stdout.write(" ".join(map(str,ANS))+"\n")
'''
inf = 100000000000000000 # 1e17
mod = 998244353
n=int(input())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
ans=sum(A[i]*B[i] for i in range(n))
origin=ans
for i in range(n):
def extend(l,r=i+1,cur=origin):
global ans
while l>=0 and r<n:
cur-=(A[l]-A[r])*(B[l]-B[r])
ans=max(ans,cur)
l-=1
r+=1
extend(i)
extend(i+1)
print(ans)
```
| 105,496 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integer arrays a and b of length n.
You can reverse at most one subarray (continuous subsegment) of the array a.
Your task is to reverse such a subarray that the sum ∑_{i=1}^n a_i ⋅ b_i is maximized.
Input
The first line contains one integer n (1 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7).
The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^7).
Output
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.
Examples
Input
5
2 3 2 1 3
1 3 2 4 2
Output
29
Input
2
13 37
2 4
Output
174
Input
6
1 8 7 6 3 6
5 9 6 8 8 6
Output
235
Note
In the first example, you can reverse the subarray [4, 5]. Then a = [2, 3, 2, 3, 1] and 2 ⋅ 1 + 3 ⋅ 3 + 2 ⋅ 2 + 3 ⋅ 4 + 1 ⋅ 2 = 29.
In the second example, you don't need to use the reverse operation. 13 ⋅ 2 + 37 ⋅ 4 = 174.
In the third example, you can reverse the subarray [3, 5]. Then a = [1, 8, 3, 6, 7, 6] and 1 ⋅ 5 + 8 ⋅ 9 + 3 ⋅ 6 + 6 ⋅ 8 + 7 ⋅ 8 + 6 ⋅ 6 = 235.
Tags: brute force, dp, implementation, math, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
ans = 0
for i in range(n):
ans+= (b[i]*a[i])
maxi = ans
for i in range(n):
l = i - 1
r = i + 1
temp_ans = ans
while l >= 0 and r < n:
temp_ans = temp_ans + (a[r]-a[l])*(b[l]-b[r])
maxi = max(maxi, temp_ans)
l -= 1
r += 1
l = i
r = i+1
temp_ans = ans
while l>=0 and r<n:
temp_ans = temp_ans +(a[r]-a[l])*(b[l]-b[r])
# print("temp_ans", temp_ans)
maxi = max(maxi,temp_ans)
l-=1
r+=1
print(maxi)
```
| 105,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integer arrays a and b of length n.
You can reverse at most one subarray (continuous subsegment) of the array a.
Your task is to reverse such a subarray that the sum ∑_{i=1}^n a_i ⋅ b_i is maximized.
Input
The first line contains one integer n (1 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7).
The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^7).
Output
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.
Examples
Input
5
2 3 2 1 3
1 3 2 4 2
Output
29
Input
2
13 37
2 4
Output
174
Input
6
1 8 7 6 3 6
5 9 6 8 8 6
Output
235
Note
In the first example, you can reverse the subarray [4, 5]. Then a = [2, 3, 2, 3, 1] and 2 ⋅ 1 + 3 ⋅ 3 + 2 ⋅ 2 + 3 ⋅ 4 + 1 ⋅ 2 = 29.
In the second example, you don't need to use the reverse operation. 13 ⋅ 2 + 37 ⋅ 4 = 174.
In the third example, you can reverse the subarray [3, 5]. Then a = [1, 8, 3, 6, 7, 6] and 1 ⋅ 5 + 8 ⋅ 9 + 3 ⋅ 6 + 6 ⋅ 8 + 7 ⋅ 8 + 6 ⋅ 6 = 235.
Tags: brute force, dp, implementation, math, two pointers
Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
Base = 0
for i in range(N):
Base += A[i] * B[i]
Ans = Base
for i in range(1, N - 1):
Value = Base
L = i - 1
R = i + 1
while L >= 0 and R < N:
Value += (A[L] - A[R]) * (B[R] - B[L])
L -= 1
R += 1
if Value > Ans: Ans = Value
for i in range(N - 1):
Value = Base
L = i
R = i + 1
while L >= 0 and R < N:
Value += (A[L] - A[R]) * (B[R] - B[L])
if Value > Ans: Ans = Value
L -= 1
R += 1
print(Ans)
```
| 105,498 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integer arrays a and b of length n.
You can reverse at most one subarray (continuous subsegment) of the array a.
Your task is to reverse such a subarray that the sum ∑_{i=1}^n a_i ⋅ b_i is maximized.
Input
The first line contains one integer n (1 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7).
The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^7).
Output
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.
Examples
Input
5
2 3 2 1 3
1 3 2 4 2
Output
29
Input
2
13 37
2 4
Output
174
Input
6
1 8 7 6 3 6
5 9 6 8 8 6
Output
235
Note
In the first example, you can reverse the subarray [4, 5]. Then a = [2, 3, 2, 3, 1] and 2 ⋅ 1 + 3 ⋅ 3 + 2 ⋅ 2 + 3 ⋅ 4 + 1 ⋅ 2 = 29.
In the second example, you don't need to use the reverse operation. 13 ⋅ 2 + 37 ⋅ 4 = 174.
In the third example, you can reverse the subarray [3, 5]. Then a = [1, 8, 3, 6, 7, 6] and 1 ⋅ 5 + 8 ⋅ 9 + 3 ⋅ 6 + 6 ⋅ 8 + 7 ⋅ 8 + 6 ⋅ 6 = 235.
Tags: brute force, dp, implementation, math, two pointers
Correct Solution:
```
def readline():
return map(int, input().split())
def main():
n = int(input())
a = list(readline())
b = list(readline())
max_diff = 0
for begin in range(n):
for end in (begin, begin + 1):
index_sum = begin + end
diff = 0
for i in range(end, min(n, index_sum + 1)):
diff -= (a[i] - a[index_sum-i]) * (b[i] - b[index_sum-i])
max_diff = max(diff, max_diff)
print(sum(ai * bi for (ai, bi) in zip(a, b)) + max_diff)
if __name__ == '__main__':
main()
```
| 105,499 |
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