text stringlengths 198 433k | conversation_id int64 0 109k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Tags: constructive algorithms, trees
Correct Solution:
```
import sys
#f = open('input', 'r')
f = sys.stdin
n = f.readline()
n = int(n)
cl = f.readline().split()
cl = [int(x) for x in cl]
c_index = 0
p1 = ['0']
p2 = ['0']
ambiguous = False
cur_index = 1
for i, c in enumerate(cl):
if i == 0:
continue
if i > 0 and cl[i-1] > 1 and c > 1:
ambiguous = True
p1 += [str(cur_index)] * c
p2 += [str(cur_index-1)] + [str(cur_index)] * (c-1)
else:
p1 += [str(cur_index)] * c
p2 += [str(cur_index)] * c
cur_index += c
if ambiguous:
print('ambiguous')
print(' '.join(p1))
print(' '.join(p2))
else:
print('perfect')
```
| 95,800 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Tags: constructive algorithms, trees
Correct Solution:
```
n = int(input())
fl = 1
a = [int(x) for x in input().split()]
for i in range(1,n+1):
if a[i] > 1 and a[i-1] > 1:
fl = 0
break
if fl == 1:
print("perfect")
else :
print("ambiguous")
print("0", end=" ")
cnt=1
x=1
fl=1
for i in range(1,n+1):
for j in range(a[i]):
print(cnt,end=" ")
x += 1
cnt=x
print()
print("0", end=" ")
cnt=1
x=1
fl=1
for i in range(1,n+1):
if a[i] > 1 and a[i-1] > 1 and fl == 1:
fl = 0
for j in range(a[i]-1):
print(cnt, end=" ")
x += 1
print(cnt-1,end=" ")
x += 1
else:
for j in range(a[i]):
print(cnt,end=" ")
x += 1
cnt=x
```
| 95,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Tags: constructive algorithms, trees
Correct Solution:
```
def print_iso(nnodes, counter):
print("ambiguous")
final_tree1 = []
final_tree2 = []
c = 0
parent = 0
for n in nnodes:
if c != counter:
for i in range(0, n):
final_tree1.append(parent)
final_tree2.append(parent)
else:
final_tree1.append(parent - 1)
final_tree2.append(parent)
for i in range(0, n-1):
final_tree1.append(parent)
final_tree2.append(parent)
parent += n
c += 1
final_tree1 = [str(i) for i in final_tree1]
final_tree2 = [str(i) for i in final_tree2]
print(' '.join(final_tree1))
print(' '.join(final_tree2))
def main():
height = int(input())
nnodes = input()
nnodes = nnodes.split(' ')
nnodes = [int(i) for i in nnodes]
perfect = True
prev = 1
counter = 0;
for n in nnodes:
if n > 1 and prev > 1:
print_iso(nnodes, counter)
perfect = False
break
prev = n
counter += 1
if perfect:
print("perfect")
if __name__ == "__main__":
main()
```
| 95,802 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Tags: constructive algorithms, trees
Correct Solution:
```
n = int(input())
h = [int(i) for i in input().split()]
flag = 0
for i in range(n):
if h[i] >= 2 and h[i+1] >= 2:
flag = i
if flag:
a = []
c = 0
for i in range(n+1):
for j in range(h[i]):
a.append(c)
c += h[i]
b = []
c = 0
for i in range(n+1):
for j in range(h[i]):
if i == flag+1 and j == 0:
b.append(c-1)
else:
b.append(c)
c += h[i]
print("ambiguous")
print(" ".join([str(i) for i in a]))
print(" ".join([str(i) for i in b]))
else:
print("perfect")
```
| 95,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Tags: constructive algorithms, trees
Correct Solution:
```
h = int( input() )
nums = [int(x) for x in input().split()]
if not any([nums[i] > 1 and nums[i+1] > 1 for i in range(h)]):
print('perfect')
else:
print('ambiguous')
# tree 1
cur = 1
flag = False
print(0, end=' ')
for l in range(1, h+1):
[print(cur, end=' ') for i in range(nums[l])]
cur += nums[l]
print()
# tree 2
cur = 1
flag = False
print(0, end=' ')
for l in range(1, h+1):
if not flag and nums[l] > 1 and nums[l-1] > 1:
# can be different here!
flag = True
print(cur-1, end=' ')
[print(cur, end=' ') for i in range(nums[l]-1)]
else:
[print(cur, end=' ') for i in range(nums[l])]
cur += nums[l]
```
| 95,804 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Tags: constructive algorithms, trees
Correct Solution:
```
h = int(input())
A = list(map(int,input().split()))
f = False
for i,a in enumerate(A):
if f and a != 1:
f = None
break
else:
f = a != 1
if f is not None:
print('perfect')
else:
T = []
for j,a in enumerate(A):
if j == i:
x = len(T)
T += [len(T)]*a
print('ambiguous')
print(' '.join(map(str,T)))
T[x+1] = x-1
print(' '.join(map(str,T)))
```
| 95,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
for i in range(2, n + 1):
if a[i] != 1 and a[i - 1] != 1:
print('ambiguous')
break
else:
print('perfect')
exit()
s1 = ''
s2 = ''
cnt = 0
for i in range(n + 1):
s1 += a[i] * (str(cnt) + ' ')
if i > 1 and a[i] != 1 and a[i - 1] != 1:
s2 += str(cnt - 1) + ' '
s2 += (a[i] - 1) * (str(cnt) + ' ')
else:
s2 += a[i] * (str(cnt) + ' ')
cnt += a[i]
print(s1)
print(s2)
```
Yes
| 95,806 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
n = int(input())
n+=1
l = list(map(int,input().split()))
ans = 0
for i in range(n-1):
if(l[i]>1 and l[i+1]>1):
ans = i+1
break
else:
print("perfect")
exit(0)
print("ambiguous")
prev = 0
now = 0
for i in range(n):
for j in range(l[i]):
print(prev,end = " ")
now+=1
prev =now
print()
prev = 0
now = 0
for i in range(n):
for j in range(l[i]):
if(ans==i):
print(prev-1,end = " ")
ans = -1
else:
print(prev,end = " ")
now+=1
prev =now
```
Yes
| 95,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
h = int(input())
a = list(map(int,input().split()))
ok = True
for i in range(h):
if a[i]>=2 and a[i+1]>=2:
ok = False
idx = i+1
if ok:
print('perfect')
else:
print('ambiguous')
ans = []
p = 0
for x in a:
ans.extend([p]*x)
p = len(ans)
print(' '.join(map(str,ans)))
ans[sum(a[:idx])] -= 1
print(' '.join(map(str,ans)))
```
Yes
| 95,808 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
n = int(input())
*a, = map(int, input().split())
b, c = [0], [0]
cur = 1
for i in range(1, n + 1):
for j in range(a[i]):
b.append(cur)
cur += a[i]
cur = 1
for i in range(1, n + 1):
if a[i] > 1 and a[i - 1] > 1:
c.append(cur - 1)
for j in range(1, a[i]):
c.append(cur)
else:
for j in range(a[i]):
c.append(cur)
cur += a[i]
if b == c:
exit(print('perfect'))
print('ambiguous')
print(*b)
print(*c)
```
Yes
| 95,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
n = int(input())
arr = list(map(int,input().split()))
flag = 0
for i in range(1,n+1):
if arr[i]>1 and arr[i-1]>1:
flag = 1
break
if not flag:
print("perfect")
exit(0)
print("ambiguous")
st = 0
prr = []
for i in range(n+1):
prr = prr+[st]*arr[i]
st+=arr[i]
print(*prr)
st = 0
prr = []
for i in range(n+1):
if flag and i and arr[i]>1 and arr[i-1]>1:
flag = False
prr+=[st-1]*arr[i]
else:
prr+=[st]*arr[i]
st+=arr[i]
print(*prr)
```
No
| 95,810 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
h = int(input())
a = list(map(int, input().split()))
perfect = True
for i in range(len(a) - 1):
if (a[i] != 1):
perfect = False
print ("perfect" if perfect else "ambiguous")
ret1 = [0] * sum(a)
ret2 = [0] * sum(a)
#print (ret1)
node = 0;
divided = False
p1=0
p2=0
if (perfect == False):
for i in range(len(a)):
for j in range(a[i]):
ret1[node] = p1
ret2[node] = p2;
node += 1;
p1 = p2 = node;
if (a[i] != 1 and divided == False):
divided = True;
p1 = node;
p2 = node - 1;
print (' '.join(str(x) for x in ret1))
print (' '.join(str(x) for x in ret2))
```
No
| 95,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
n = int(input())
c = list(map(int, input().split()))
f = True
for i in range(n + 1):
if i % 2 == 0 and c[i] != 1:
f = False
if f:
print('perfect')
else:
last = 0
print('ambiguous')
g = 0
for i in range(n + 1):
for j in range(c[i]):
print(g, end=' ')
g += c[i]
if i > 0:
if c[i] > 1 and c[i - 1]:
last = i
print()
g = 0
for i in range(n + 1):
if i != last:
for j in range(c[i]):
print(g, end=' ')
else:
for j in range(c[i] - 1):
print(g, end=' ')
print(g - 1, end=' ')
g += c[i]
```
No
| 95,812 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root.
Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.
Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.
The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.
Input
The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.
The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.
Output
If there is only one tree matching this sequence, print "perfect".
Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.
These treese should be non-isomorphic and should match the given sequence.
Examples
Input
2
1 1 1
Output
perfect
Input
2
1 2 2
Output
ambiguous
0 1 1 3 3
0 1 1 3 2
Note
The only tree in the first example and the two printed trees from the second example are shown on the picture:
<image>
Submitted Solution:
```
def get_num_divisions(nodes, subnodes):
divisions = []
if nodes - 1:
for x in range(subnodes + 1):
for y in get_num_divisions(nodes-1, subnodes-x):
divisions.append([x]+y)
else:
divisions.append([subnodes])
for div in divisions:
div.sort()
import itertools
divisions.sort()
return list(k for k, _ in itertools.groupby(divisions))
def get_trees_structs(levels):
trees = []
for x in range(len(levels)-1):
trees.append(get_num_divisions(levels[x], levels[x+1]))
return trees
def get_trees_variants(structure):
variants = []
if structure:
for var in structure[0]:
others = get_trees_variants(structure[1:])
if others:
for other in others:
variants.append(var + other)
else:
variants.append(var)
return variants
def get_tree_represantion(variant, levels):
result = "0 "
i = 1
for var in variant:
for x in range(var):
result += "{0} ".format(i)
i += 1
return result
class CodeforcesTask901ASolution:
def __init__(self):
self.result = ''
self.height = 0
self.levels = []
def read_input(self):
self.height = int(input())
self.levels = [int(x) for x in input().split(" ")]
def process_task(self):
struct = get_trees_structs(self.levels)
variants = get_trees_variants(struct)
if len(variants) == 1:
self.result = "perfect"
else:
self.result += "{0}\n".format(get_tree_represantion(variants[0], self.levels))
self.result += "{0}\n".format(get_tree_represantion(variants[1], self.levels))
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask901ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
```
No
| 95,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Tags: constructive algorithms, implementation, strings
Correct Solution:
```
# python3
import sys
def read_all_following_lines():
lines = sys.stdin.readlines()
return (tuple(map(int, line.split())) for line in lines)
class AbcString(object):
def __init__(self, string):
self.prefix_bc = [0]
self.a_strike = [0]
bc, strike = 0, 0
for symbol in string:
if symbol == 'A':
strike += 1
else:
strike = 0
bc += 1
self.prefix_bc.append(bc)
self.a_strike.append(strike)
def get_info(self, begin, end):
bc = self.prefix_bc[end] - self.prefix_bc[begin]
trailing_a = min(self.a_strike[end], end - begin)
return bc, trailing_a
def can_mutate(start, finish):
from_bc, from_a = start
to_bc, to_a = finish
if (from_bc & 1) != (to_bc & 1): return False
if from_bc > to_bc: return False
if from_a < to_a: return False
if from_bc == to_bc: return (from_a - to_a) % 3 == 0
if from_a == to_a: return from_bc != 0
# from_bc < to_bc
# from_a > to_a
return True
def main():
s = AbcString(input())
t = AbcString(input())
input() # skip one line
requests = read_all_following_lines()
answer = ""
for (a, b, c, d) in requests:
can = can_mutate(s.get_info(a - 1, b), t.get_info(c - 1, d))
answer += "1" if can else "0"
print(answer)
main()
```
| 95,814 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Tags: constructive algorithms, implementation, strings
Correct Solution:
```
def main():
s, t = input(), input()
s_info, t_info = fast_counter(s), fast_counter(t)
queries = int(input())
answer = ''
for _ in range(queries):
l1, r1, l2, r2 = map(int, input().split())
l1, l2 = l1 - 1, l2 - 1
from_mask = mask(l1, r1, s_info)
to_mask = mask(l2, r2, t_info)
if can_transform(from_mask, to_mask):
answer += '1'
else:
answer += '0'
print(answer)
def can_transform(from_mask, to_mask):
if from_mask[0] > to_mask[0]:
return False
elif (to_mask[0] - from_mask[0]) % 2 != 0:
return False
elif to_mask[0] == from_mask[0]:
if to_mask[1] > from_mask[1]:
return False
return (from_mask[1] - to_mask[1]) % 3 == 0
elif from_mask[0] == 0:
return to_mask[1] < from_mask[1]
else:
return to_mask[1] <= from_mask[1]
def mask(l, r, info):
return (info[1][r] - info[1][l], min(r - l, info[0][r]))
def fast_counter(s):
a_last, b_count = [0], [0]
for c in s:
if c == 'A':
a_last.append(a_last[-1] + 1)
b_count.append(b_count[-1])
else:
a_last.append(0)
b_count.append(b_count[-1] + 1)
return (a_last, b_count)
main()
```
| 95,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Tags: constructive algorithms, implementation, strings
Correct Solution:
```
def read():
s = input()
l = len(s)
r = [[0] * (l + 1), [0] * (l + 1)]
for i in range(l):
r[0][i + 1] = (r[0][i] + 1) * (s[i] == 'A')
r[1][i + 1] = r[1][i] + (s[i] != 'A')
return r
s, t = read(), read()
q = int(input())
r = ''
for i in range(q):
a, b, c, d = map(int, input().split())
sb = s[1][b] - s[1][a] + (s[0][a] == 0)
sa = s[0][b] - (s[0][a] - 1) * (sb == 0)
tb = t[1][d] - t[1][c] + (t[0][c] == 0)
ta = t[0][d] - (t[0][c] - 1) * (tb == 0)
if any([sb > tb, sa < ta, tb - sb & 1, sb == tb and (sa - ta) % 3, sa == ta and not sb and tb]):
r += '0'
else:
r += '1'
print(r)
```
| 95,816 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Tags: constructive algorithms, implementation, strings
Correct Solution:
```
def main():
s, t = input(), input()
s_info, t_info = fast_counter(s), fast_counter(t)
queries = int(input())
answer = ''
for _ in range(queries):
l1, r1, l2, r2 = map(int, input().split())
l1, l2 = l1 - 1, l2 - 1
from_mask = (s_info[1][r1] - s_info[1][l1], min(r1 - l1, s_info[0][r1]))
to_mask = (t_info[1][r2] - t_info[1][l2], min(r2 - l2, t_info[0][r2]))
if can_transform(from_mask, to_mask):
answer += '1'
else:
answer += '0'
print(answer)
def can_transform(from_mask, to_mask):
if from_mask[0] > to_mask[0]:
return False
elif (to_mask[0] - from_mask[0]) % 2 != 0:
return False
elif to_mask[0] == from_mask[0]:
if to_mask[1] > from_mask[1]:
return False
return (from_mask[1] - to_mask[1]) % 3 == 0
elif from_mask[0] == 0:
return to_mask[1] < from_mask[1]
else:
return to_mask[1] <= from_mask[1]
def mask(l, r, info):
return (info[1][r] - info[1][l], min(r - l, info[0][r]))
def fast_counter(s):
a_last, b_count = [0], [0]
for c in s:
if c == 'A':
a_last.append(a_last[-1] + 1)
b_count.append(b_count[-1])
else:
a_last.append(0)
b_count.append(b_count[-1] + 1)
return (a_last, b_count)
main()
```
| 95,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Tags: constructive algorithms, implementation, strings
Correct Solution:
```
def pref_counts(string, char):
result = [0]
for c in string:
result.append(result[-1] + (c == char))
return result
def left_counts(string, char):
result = [0]
for c in string:
result.append(result[-1] + 1 if c == char else 0)
return result
s = input().replace("C", "B")
t = input().replace("C", "B")
s_bcounts, t_bcounts = pref_counts(s, "B"), pref_counts(t, "B")
s_lcounts, t_lcounts = left_counts(s, "A"), left_counts(t, "A")
for i in range(int(input())):
a, b, c, d = list(map(int, input().split()))
s_b = s_bcounts[b] - s_bcounts[a - 1]
t_b = t_bcounts[d] - t_bcounts[c - 1]
s_a = min(s_lcounts[b], b - a + 1)
t_a = min(t_lcounts[d], d - c + 1)
answer = False
if s_b & 1 == t_b & 1:
if s_b < t_b:
answer = s_a - (s_b == 0) >= t_a
elif s_b == t_b:
answer = s_a >= t_a and (s_a - t_a) % 3 == 0
print(int(answer), end = "")
```
| 95,818 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Submitted Solution:
```
def read():
s = input()
l = len(s)
r = [[0] * (l + 1), [0] * (l + 1)]
for i in range(l):
r[0][i + 1] = (r[0][i] + 1) * (s[i] == 'A')
r[1][i + 1] = r[1][i] + (s[i] != 'A')
return r
s, t = read(), read()
q = int(input())
for i in range(q):
a, b, c, d = map(int, input().split())
sb = s[1][b] - s[1][a] + (s[0][a] == 0)
sa = s[0][b] - (s[0][a] - 1) * (sb == 0)
tb = t[1][d] - t[1][c] + (t[0][c] == 0)
ta = t[0][d] - (t[0][c] - 1) * (tb == 0)
if tb == sb and sa >= ta and (sa - ta) % 3 == 0 or tb > sb and tb - sb & 1 == 0 and sa >= ta:
print(1, end='')
else:
print(0, end='')
```
No
| 95,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Submitted Solution:
```
def read():
s = input()
l = len(s)
r = [[0] * (l + 1), [0] * (l + 1)]
for i in range(l):
r[0][i + 1] = (r[0][i] + 1) * (s[i] == 'A')
r[1][i + 1] = r[1][i] + (s[i] != 'A')
return r
s, t = read(), read()
q = int(input())
r = ''
for i in range(q):
a, b, c, d = map(int, input().split())
sb = s[1][b] - s[1][a] + (s[0][a] == 0)
sa = s[0][b] - (s[0][a] - 1) * (sb == 0)
tb = t[1][d] - t[1][c] + (t[0][c] == 0)
ta = t[0][d] - (t[0][c] - 1) * (tb == 0)
if any([sb > tb, sa < ta, tb - sb & 1, sb == tb and (sa - ta) % 3, sa == ta and not sb]):
r += '0'
else:
r += '1'
print(r)
```
No
| 95,820 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Submitted Solution:
```
s = input()
t = input()
n = int(input())
q = ''
for i in range(n):
a,b,c,d = map(int, input().split())
if a == b or c == d and s[a:b] != t[c:d]:
q += '0'
else:
q += '1'
print(q)
```
No
| 95,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
* A <image> BC
* B <image> AC
* C <image> AB
* AAA <image> empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Output
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
Example
Input
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
Output
10011
Note
In the first query we can achieve the result, for instance, by using transitions <image>.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
Submitted Solution:
```
x = input()
y = input()
s = [(0,0)]
bc = 0
a = 0
for l in x:
if l in 'BC':
bc += 1
a = 0
else:
a += 1
s.append((bc,a))
t = [(0,0)]
bc = 0
a = 0
for l in y:
if l in 'BC':
bc += 1
a = 0
else:
a += 1
t.append((bc,a))
# print(s)
# print(t)
q = int(input())
for i in range(q):
ls,rs,lt,rt = map(int, input().split())
lens = rs - ls + 1
lent = rt - lt + 1
sbc = s[rs][0] - s[ls-1][0]
tbc = t[rt][0] - t[lt-1][0]
sa = s[rs][1]
ta = t[rt][1]
if sa > lens: sa = lens
if ta > lent: ta = lent
if sa != ta: sbc += 2
a = ta - sa
if a < 0 or a % 2 != 0:
print('0', end = '')
else:
print('1', end = '')
```
No
| 95,822 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
n = int(input())
S = [input() for i in range(3)]
bu = []
for s in S:
cnt = {}
mx = 0
for c in s:
if c not in cnt:
cnt[c] = 0
cnt[c] += 1
mx = max(mx, cnt[c])
if mx == len(s) and n == 1:
bu.append(mx - 1)
else:
bu.append(min(len(s), mx + n))
ans = -1
ansmx = -1
for i in range(3):
if bu[i] > ansmx:
ans = i
ansmx = bu[i]
elif bu[i] == ansmx:
ans = -1
if ans == -1:
print('Draw')
elif ans == 0:
print('Kuro')
elif ans == 1:
print('Shiro')
else:
print('Katie')
```
| 95,823 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
n = int(input())
s1 = input()
s2 = input()
s3 = input()
s = len(s1)
d1, d2, d3 = dict(), dict(), dict()
for i in s1:
if i not in d1:
d1[i] = 1
else:
d1[i] += 1
for i in s2:
if i not in d2:
d2[i] = 1
else:
d2[i] += 1
for i in s3:
if i not in d3:
d3[i] = 1
else:
d3[i] += 1
max1, max2, max3 = 0, 0, 0
for i in d1:
if d1[i] > max1:
max1 = d1[i]
for i in d2:
if d2[i] > max2:
max2 = d2[i]
for i in d3:
if d3[i] > max3:
max3 = d3[i]
a = []
a.append([s - max1, "Kuro"])
a.append([s - max2, "Shiro"])
a.append([s - max3, "Katie"])
a.sort()
#print(a)
if a[0][0] > n:
if a[0][0] < a[1][0]:
print(a[0][1])
else:
print("Draw")
else:
for j in a:
q = n - j[0]
if q < 0:
j[0] -= n
else:
if n == 1:
j[0] = q % 2
else:
j[0] = 0
a.sort()
#print(a)
if a[0][0] < a[1][0]:
print(a[0][1])
else:
print("Draw")
```
| 95,824 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
def ct(s):
a=[0]*26*2
for i in s:
if ord(i)<97:
a[ord(i)-65]+=1
else:
a[ord(i)-97+26]+=1
return max(a)
n=int(input())
s1=input()
ln=len(s1)
s1=ct(s1)
s2=ct(input())
s3=ct(input())
s=[s1,s2,s3]
for i in range(len(s)):
if s[i]==ln and n==1: s[i]=ln-1
else:s[i]=s[i]+n
if s[i]>ln: s[i]=ln
s1=s[0]
s2=s[1]
s3=s[2]
#print(s)
s.sort()
if s[2]==s[1]:
print('Draw')
elif s[-1]==s1:
print('Kuro')
elif s[-1]==s2:
print('Shiro')
elif s[-1]==s3:
print('Katie')
##//////////////// ////// /////// // /////// // // //
##//// // /// /// /// /// // /// /// //// //
##//// //// /// /// /// /// // ///////// //// ///////
##//// ///// /// /// /// /// // /// /// //// // //
##////////////// /////////// /////////// ////// /// /// // // // //
```
| 95,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
#vsInput()
n=Int()
p1=input()
p2=input()
p3=input()
l=len(p1)
a=l-max(Counter(p1).values())
b=l-max(Counter(p2).values())
c=l-max(Counter(p3).values())
#print(a,b,c)
if(n<=a): a-=n
else: a=0 if n>1 else 1
if(n<=b): b-=n
else: b=0 if n>1 else 1
if(n<=c): c-=n
else: c=0 if n>1 else 1
# print(a,b,c,l)
# print(Counter(p1))
# print(Counter(p2))
# print(Counter(p3))
if(a<b and a<c):
print("Kuro")
elif(b<a and b<c):
print("Shiro")
elif(c<a and c<b):
print("Katie")
else:
print("Draw")
```
| 95,826 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
n = int(input())
s1 = input().strip()
s2 = input().strip()
s3 = input().strip()
d1 = [0 for _ in range(52)]
d2 = [0 for _ in range(52)]
d3 = [0 for _ in range(52)]
maxi1 = 0
maxi2 = 0
maxi3 = 0
for i in s1:
if ord(i) <= 90:
j = ord(i) - 65
else:
j = ord(i) - 97 + 26
d1[j] += 1
maxi1 = max(d1)
for i in s2:
if ord(i) <= 90:
j = ord(i) - 65
else:
j = ord(i) - 97 + 26
d2[j] += 1
maxi2 = max(d2)
for i in s3:
if ord(i) <= 90:
j = ord(i) - 65
else:
j = ord(i) - 97 + 26
d3[j] += 1
maxi3 = max(d3)
if maxi1 + n <= len(s1):
maxi1 += n
else:
if n == 1:
maxi1 = len(s1) - 1
else:
maxi1 = len(s1)
if maxi2 + n <= len(s1):
maxi2 += n
else:
if n == 1:
maxi2 = len(s1) - 1
else:
maxi2 = len(s1)
if maxi3 + n <= len(s1):
maxi3 += n
else:
if n == 1:
maxi3 = len(s1) - 1
else:
maxi3 = len(s1)
if maxi1 > maxi2 and maxi1 > maxi3:
print('Kuro')
elif maxi2 > maxi1 and maxi2 > maxi3:
print('Shiro')
elif maxi3 > maxi1 and maxi3 > maxi2:
print('Katie')
else:
print('Draw')
```
| 95,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
n = int(input())
a = input()
b = input()
c = input()
def count(s, n):
cnt = {}
for c in s:
cnt[c] = cnt.get(c, 0) + 1
maxc = 0
for c in cnt:
if cnt[c] > maxc: maxc = cnt[c]
if len(s) == maxc and n == 1:
return maxc - 1
else:
return min(maxc+n, len(s))
ac = count(a, n)
bc = count(b, n)
cc = count(c, n)
if ac > bc and ac > cc:
print("Kuro")
elif bc > ac and bc > cc:
print("Shiro")
elif cc > bc and cc > ac:
print("Katie")
else:
print("Draw")
```
| 95,828 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
n = int(input())
ku = input()
si = input()
ka = input()
def bu2num(bu):
dif = ord(bu) - ord('a')
if dif >= 0 and dif < 26:
return dif
else:
return ord(bu) - ord('A') + 26
def num2bu(num):
return chr(ord('a') + num if num < 26 else ord('a') + num - 26)
def bus(s):
x = [0] * 26 * 2
for bu in s:
x[bu2num(bu)] += 1
return x
def mabus(arr):
max = 0
for a in arr:
if a > max:
max = a
return max
def calc(s):
l = len(s)
m = mabus(bus(s))
d = m + n
if m == l and n == 1:
return l - 1
elif d <= l:
return d
else:
return l
kun = calc(ku)
sin = calc(si)
kan = calc(ka)
if kun > sin and kun > kan:
print('Kuro')
elif sin > kun and sin > kan:
print('Shiro')
elif kan > kun and kan > sin:
print('Katie')
else:
print('Draw')
```
| 95,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Tags: greedy
Correct Solution:
```
# your code goes here
# your code goes here
import collections
n = int(input())
kuro = input()
shiro = input()
katie = input()
lkuro, lshiro, lkatie = len(kuro), len(shiro), len(katie)
ckuro = collections.Counter(kuro).most_common(1)[0][1]
cshiro = collections.Counter(shiro).most_common(1)[0][1]
ckatie = collections.Counter(katie).most_common(1)[0][1]
if n <= lkuro - ckuro :
kuroscore = ckuro + n
else:
kuroscore = lkuro
if n == 1 and lkuro == ckuro:
kuroscore -= 1
if n <= lshiro - cshiro :
shiroscore = cshiro + n
else:
shiroscore = lshiro
if n == 1 and lshiro == cshiro:
shiroscore -= 1
if n <= lkatie - ckatie :
katiescore = ckatie + n
else:
katiescore = lkatie
if n == 1 and lkatie == ckatie:
katiescore -= 1
b = ['Kuro', 'Shiro', 'Katie']
s = [kuroscore, shiroscore, katiescore]
ss = sorted(s)
if ss[-1] == ss[-2]:
print('Draw')
else:
print(b[s.index(max(s))])
```
| 95,830 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
n = int(input())
ku, sh, ka = input(), input(), input()
l_ku, l_sh, l_ka = max([ku.count(i) for i in list(set(ku))]), max([sh.count(i) for i in list(set(sh))]), max([ka.count(i) for i in list(set(ka))])
if len(ku) - l_ku > n:
l_ku += n
elif l_ku == len(ku) and n == 1:
l_ku -= 1
else:
l_ku = len(ku)
if len(sh) - l_sh > n:
l_sh += n
elif l_sh == len(sh) and n == 1:
l_sh -= 1
else:
l_sh = len(sh)
if len(ka) - l_ka > n:
l_ka += n
elif l_ka == len(ka) and n == 1:
l_ka -= 1
else:
l_ka = len(ka)
ma = max([l_sh, l_ku, l_ka])
if (l_ka == l_sh and l_ka == ma) or (l_ku == l_sh and l_ku == ma) or (l_ka == l_ku and l_ka == ma):
print('Draw')
elif ma == l_ka:
print('Katie')
elif ma == l_sh:
print('Shiro')
elif ma == l_ku:
print('Kuro')
```
Yes
| 95,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
import collections
n = int(input())
s = []
for i in range(3):
s.append(input())
max_val = [0, 0, 0]
ans = ['Kuro', 'Shiro', 'Katie']
for i in range(3):
cnt = collections.Counter(s[i])
rr = max(cnt.values())
changed = min(len(s[i]) - rr, n)
moves = n - changed
max_val[i] = rr + changed - (moves == 1 and rr == len(s[i]))
if max_val.count(max(max_val)) > 1:
print('Draw')
else:
print(ans[max_val.index(max(max_val))])
```
Yes
| 95,832 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
n=int(input())
s1=str(input())
s2=str(input())
s3=str(input())
l1=list(set(s1))
l2=list(set(s2))
l3=list(set(s3))
m1=0
m2=0
m3=0
for i in l1:
m1=max(m1,s1.count(i))
for i in l2:
m2=max(m2,s2.count(i))
for i in l3:
m3=max(m3,s3.count(i))
if(len(s1)==m1 and n==1):
x=len(s1)-1
elif(n>=(len(s1)-m1)):
x=len(s1)
else:
x=m1+n
if(len(s2)==m2 and n==1):
y=len(s2)-1
elif(n>=(len(s2)-m2)):
y=len(s2)
else:
y=m2+n
if(len(s3)==m3 and n==1):
z=len(s3)-1
elif(n>=(len(s3)-m3)):
z=len(s3)
else:
z=m3+n
lm=[]
lm.append(x)
lm.append(y)
lm.append(z)
#print(x,y,z)
if(lm.count(max(lm))>=2):
print("Draw")
else:
if(x>y and x>z):
print("Kuro")
elif(y>x and y>z):
print("Shiro")
elif(z>x and z>y):
print("Katie")
```
Yes
| 95,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
n=int(input())
s1=input()
s2=input()
s3=input()
k1=len(s1)
k2=len(s2)
k3=len(s3)
a=0
val1=[0 for i in range(1000)]
val2=[0 for i in range(1000)]
val3=[0 for i in range(1000)]
for i in range(k1):
val1[ord(s1[i])-ord('a')]+=1
for i in range(k2):
val2[ord(s2[i])-ord('a')]+=1
for i in range(k3):
val3[ord(s3[i])-ord('a')]+=1
if n==1 :
if max(val1)==k1:
l1=k1-1
else:
l1=min(max(val1)+n,k1)
if max(val2)==k2:
l2=k2-1
else:
l2=min(max(val2)+n,k2)
if max(val3)==k3:
l3=k3-1
else:
l3=min(max(val3)+n,k3)
else:
l1=min(max(val1)+n,k1)
l2=min(max(val2)+n,k2)
l3=min(max(val3)+n,k3)
if l1>max(l2,l3):
print("Kuro")
elif l2>max(l1,l3):
print("Shiro")
elif l3>max(l1,l2):
print("Katie")
else:
print("Draw")
```
Yes
| 95,834 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
from collections import Counter
n = int(input())
def score(s):
return Counter(s).most_common()[0][1];
a = [input() for i in range(3)]
l = len(a[0])
a = list(map(score, a))
if n == 1:
a = list(map(lambda x: x - 1 if x == l else x + 1, a))
else:
a = list(map(lambda x: x + 1, a))
print('Draw' if a.count(max(a)) > 1 else [['Kuro', 'Shiro', 'Katie'][i] for i in range(3) if a[i] == max(a)][0])
```
No
| 95,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
# n,x,y=[int(x)for x in input().split()]
#
# adj=[[]for i in range(n+1)]
# for i in range(n-1):
# a,b = [int(x) for x in input().split()]
# adj[a].append(b)
# adj[b].append(a)
#
# fa=[None]*(n+1)
# sns=[[]for i in range(n+1)]
#
# def conTree():
# global fa,sns
# next=[x]
# while len(next)>0:
# new_next=[]
# for s in next:
# f = fa[s]
# for ss in adj[s]:
# if ss == f:
# continue
# sns[s].append(ss)
# fa[ss] = s
# new_next.append(ss)
# next=new_next
# conTree()
# def get_root(r):
# while fa[r]!=x:
# r=fa[r]
# return r
# def get_num(r):
# next=[r]
# i=0
# while i<len(next):
# for ss in sns[next[i]]:
# next.append(ss)
# i+=1
# return len(next)
# def get_x():
# r=get_root(y)
# # print('root',r)
# ans=0
# for ss in sns[x]:
# if ss!=r:
# ans+=get_num(ss)
# return ans+1
# xx=get_x()
# yy=get_num(y)
# # print(xx,yy)
# print(n*(n-1)-xx*yy)
###################################################################################################
n=int(input())
a=input()
b=input()
c=input()
def longest(s):
d={}
for cc in s:
if cc not in d:
d[cc]=1
else:
d[cc]+=1
ans=-1
for k in d.keys():
ans=max(ans,d[k])
return ans
al=len(a)-longest(a)
bl=len(b)-longest(b)
cl=len(c)-longest(c)
al=max(0,al-n)
bl=max(0,bl-n)
cl=max(0,cl-n)
s=[al,bl,cl]
s.sort()
if s[0]==s[1]:
print('Draw')
else:
minn=min(al,bl,cl)
if al==minn:
print('Kuro')
if bl==minn:
print('Shiro')
if cl==minn:
print('Katie')
```
No
| 95,836 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
def bt(s):
maxi=0
for i in alphabet:
k=s.count(i)
if k>maxi:
maxi=k
return maxi
def maxbt(s,n):
#за len(s)-bt(s) строка станет красивой
#остается n-(len(s)-bt(s)) ходов. Если это четно, то maxbt=len(s)
if n<=len(s)-bt(s):
return n+bt(s)
elif n-(len(s)-bt(s))%2==0:
return len(s)
else:
return len(s)
alphabet=['A','a','B','b','C','c','D','d','E','e','F','f','G','g','H','h','I','i','J','j','K','k','L','l','M','m','N','n','O','o','P','p','Q','q','R','r','S','s','T','t','U','u','V','v','W','w','X','x','Y','y','Z','z']
n=int(input())
s1=input()
s2=input()
s3=input()
if 1==2:
print('Draw')
else:
k1=maxbt(s1,n)
k2=maxbt(s2,n)
k3=maxbt(s3,n)
d=max(k1,k2,k3)
if (k1==k2 and k1==d) or (k2==k3 and k2==d) or (k3==k1 and k1==d):
print('Draw')
else:
if k1==max(k1,k2,k3):
print('Kuro')
if k2==max(k1,k2,k3):
print('Shiro')
if k3==max(k1,k2,k3):
print('Katie')
```
No
| 95,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of 7 because its subribbon a appears 7 times, and the ribbon abcdabc has the beauty of 2 because its subribbon abc appears twice.
The rules are simple. The game will have n turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after n turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input
The first line contains an integer n (0 ≤ n ≤ 10^{9}) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than 10^{5} uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Examples
Input
3
Kuroo
Shiro
Katie
Output
Kuro
Input
7
treasurehunt
threefriends
hiCodeforces
Output
Shiro
Input
1
abcabc
cbabac
ababca
Output
Katie
Input
15
foPaErcvJ
mZaxowpbt
mkuOlaHRE
Output
Draw
Note
In the first example, after 3 turns, Kuro can change his ribbon into ooooo, which has the beauty of 5, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most 4, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is 9 and the number of turn is 15, everyone can change their ribbons in some way to reach the maximal beauty of 9 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
Submitted Solution:
```
n = int(input())
black = list(input())
white = list(input())
katie = list(input())
ALPHABET = [chr(i) for i in range(65, 65+26)] + [chr(i)
for i in range(97, 97+26)]
def optimal_score(LIST):
d = {char: 0 for char in ALPHABET}
for v in LIST:
d[v] += 1
max_freq = max(d.values())
L = len(LIST)
res = 0
for freq in d.values():
if L - freq >= n:
res = max(res, freq + n)
else:
if (freq + n) % 2 != L % 2:
res = max(res, L - 1)
else:
res = L
return res
score_black = optimal_score(black)
score_white = optimal_score(white)
score_katie = optimal_score(katie)
M = max(score_black, score_katie, score_white)
MAXCNT = 0
winner = "NOBODY"
if M == score_black:
winner = "Kuro"
MAXCNT += 1
if M == score_white:
winner = "Shiro"
MAXCNT += 1
if M == score_katie:
winner = "Katie"
MAXCNT += 1
if MAXCNT == 1:
print(winner)
else:
print("Draw")
```
No
| 95,838 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
from collections import Counter
N,K = map(int,input().split())
S = input()
C = Counter(S)
cnt = 0
for i in range(26):
c = chr(97+i)
if cnt + C[c] >= K:
break
else:
cnt += C[c]
S = S.replace(c,"")
t = ""
for s in S:
if s == c and cnt < K:
cnt += 1
else:
t += s
else:
S = t
print(S)
```
| 95,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
n,k = map(int,input().split(' '))
s = input()
z = 1
l = [0]*(26)
for i in range(0,len(s)):
l[ord(s[i])-97] = l[ord(s[i])-97]+1
for i in range(0,26):
c = chr(97+i)
z = 1
if k==0:
break
if s.find(c)!=-1:
if l[i]<=k and l[i]!=0:
s = s.replace(c,'',l[i])
z = z+l[i]-1
k = k-z
else:
s = s.replace(c,'',k)
z = z+k-1
k = k-z
if n!=len(s):
print(s)
```
| 95,840 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
#!/usr/bin/env python3
#
# FILE: 999C.py
#
# @author: Arafat Hasan Jenin <opendoor.arafat[at]gmail[dot]com>
#
# LINK: http://codeforces.com/contest/999/problem/C
#
# DATE CREATED: 22-06-18 16:19:52 (+06)
# LAST MODIFIED: 23-06-18 02:31:13 (+06)
#
# VERDICT:
#
# DEVELOPMENT HISTORY:
# Date Version Description
# --------------------------------------------------------------------
# 22-06-18 1.0 Deleted code is debugged code.
#
# _/ _/_/_/_/ _/ _/ _/_/_/ _/ _/
# _/ _/ _/_/ _/ _/ _/_/ _/
# _/ _/_/_/ _/ _/ _/ _/ _/ _/ _/
# _/ _/ _/ _/ _/_/ _/ _/ _/_/
# _/_/ _/_/_/_/ _/ _/ _/_/_/ _/ _/
#
##############################################################################
n, k = map(int, input().split())
str = input()
dictionary = dict((letter, str.count(letter)) for letter in set(str))
for item in range(26):
if k == 0:
break
key = chr((item + ord('a')))
if key in dictionary:
if dictionary[key] >= k:
dictionary[key] -= k
k = 0
break
else:
tmp = dictionary[key]
dictionary[key] -= k
k -= tmp
ans = list()
for item in reversed(str):
if item in dictionary:
if dictionary[item] > 0:
dictionary[item] -= 1
ans.append(item)
print("".join([x for x in list(reversed(ans))]))
```
| 95,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
_, k = map(int, input().split())
s = input()
for ch in 'abcdefghijklmnopqrstuvwxyz':
x = s.count(ch)
if x < k:
s = s.replace(ch, '', x)
k -= x
else:
s = s.replace(ch, '', k)
break
print(s)
```
| 95,842 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
n=0
k=0
def reducedString(s:str):
alphabetList = []
index = 0
while index<26:
alphabetList.append(0)
index += 1
index = 0
#count how many of each character there are in the string
while index<n:
# print(ord(s[index])-ord('a'))
alphabetList[ord(s[index]) - ord('a')] += 1
index+=1
charsDeleted = 0
index = 0
#remove k characters from the string
while(charsDeleted<k):
if(alphabetList[index]>0):
toBeDeleted = min(k - charsDeleted, alphabetList[index])
charsDeleted += toBeDeleted
alphabetList[index] -= toBeDeleted
index += 1
#print(alphabetList)
#build the list of characters that remain after deletion
index = n-1
resultList = []
while(index>-1):
if(alphabetList[ord(s[index]) - ord('a')]>0):
resultList.append(s[index])
alphabetList[ord(s[index]) - ord('a')] -= 1
index -= 1
l = 0
r = len(resultList) - 1
while(l<r):
tmp = resultList[l]
resultList[l] = resultList[r]
resultList[r] = tmp
l+=1
r-=1
res = "".join(resultList)
return res
if __name__ == "__main__":
n,k = map(int, input().split())
s = input()
res = reducedString(s)
print(res)
```
| 95,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
lower = [chr(i) for i in range(97,123)]
temp = input()
l = int(temp.split(" ")[0])
k = int(temp.split(" ")[1])
s = list((input()))
m = dict()
for i in range(l):
m[s[i]] = m.get(s[i],0)+1
m = sorted(m.items(),key=lambda d:d[0])
num = 0
for i in range(len(m)):
num = num + m[i][1]
if(num>=k):
global splistr
splistr = m[i][0]
global left
left = m[i][1]-num+k
break
# print(splistr)
# print(left)
for i in range(l):
if(s[i]==splistr):
s[i]='@'
left = left - 1
if(left==0):
break
for i in range(l):
if(s[i]>=splistr):
print(s[i],end = "")
```
| 95,844 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
s = input()
cnt = [0 for i in range(26)]
for i in s:
cnt[ord(i) - ord("a")] += 1
# print("cnt =", cnt)
total = 0
cnt2 = [0 for i in range(26)]
for i in range(26):
if total + cnt[i] <= k:
cnt2[i] = cnt[i]
total += cnt[i]
else:
cnt2[i] = k - total
break
# print("cnt2 =", cnt2)
cnt3 = [0 for i in range(26)]
ans = ""
for i in s:
c = ord(i) - ord("a")
if cnt3[c] < cnt2[c]:
cnt3[c] += 1
else:
ans += i
print(ans)
```
| 95,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Tags: implementation
Correct Solution:
```
from string import ascii_lowercase
from collections import Counter, OrderedDict
def go():
n, k = (int(i) for i in input().split(' '))
s = [i for i in input()]
if n <= k:
return ''
c = dict(Counter(s))
new = OrderedDict()
t = 0
for i in ascii_lowercase:
if i in c:
if c[i] + t <= k:
new[i] = c[i]
t += c[i]
else:
new[i] = k - t
break
if t == k:
break
i = 0
t = k
while i < n and k != 0:
if s[i] in new and new[s[i]] > 0:
new[s[i]] -= 1
s[i] = 0
k -= 1
i += 1
return ''.join(i for i in s if i != 0)
print(go())
```
| 95,846 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
def fun(s,n,k):
occ=[0]*26
for i in s:
occ[ord(i)-97]+=1
r=k
i=-1
while r>0:
i+=1
r-=occ[i]
if i==-1:
return ''
last=occ[i]+r
occ[i]=last
ans=''
for j in s:
ch=ord(j)-97
if ch<=i and occ[ch]>0:
occ[ch]-=1
else:
ans+=j
return ans
n,k=map(int,input().split())
#n=int(input())
s=input()
print(fun(s,n,k))
```
Yes
| 95,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
n, k = map(int, input().split())
s = list(input())
# count = Counter(s)
update = dict()
while k>0:
for i in range(26):
c = chr(i+97)
count = s.count(c)
if count < k:
update[c] = count
k -= count
else:
update[c] = k
k = 0
break
# print(update)
updated_str = []
for i in s:
if i in update and update[i]>0:
update[i] -= 1
continue
else:
updated_str.append(i)
print(''.join(updated_str))
```
Yes
| 95,848 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
from collections import defaultdict
n , k = [int(x) for x in input().split()]
s = input()
removed = set()
l = defaultdict(list)
for ind,i in enumerate(s) :
l[i].append(ind)
if k >= n :
pass
else :
while k > 0 :
for char in sorted(l.keys()) :
if k >= len(l[char]) :
for ind in l[char] :
removed.add(ind)
k -= len(l[char])
if k == 0 :
break
else :
for ind in l[char][:k] :
removed.add(ind)
k = 0
break
for ind,i in enumerate(s) :
if ind not in removed :
print(i , end = '')
```
Yes
| 95,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
[n,k] = list(map(int,input().split(" ")))
s = input()
list_s = list(s)
#print(list_s)
y = [i for i in range(len(list_s))]
zipped = list(zip(list_s,y))
#print(zipped)
zipped.sort(key = lambda t: t[0])
#print('after',zipped)
result = list("*"*n)
#print(zipped[0][0],zipped[0][1])
for i in range(k,len(zipped)):
result[zipped[i][1]] = zipped[i][0]
#print('result',result)
final = ""
for c in result:
if c != '*':
final += c
print(final)
```
Yes
| 95,850 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
number_of_testcases = 1 #int(input())
for _ in range(number_of_testcases):
len_of_given_str, num_of_letters_to_be_removed = map(int, input().split())
given_str = input()
freq_of_alpha = [0 for i in range(26)]
for char in given_str:
freq_of_alpha[ord(char)-97] += 1
#print(freq_of_alpha)
for i in range(26):
if freq_of_alpha[i] >= num_of_letters_to_be_removed:
freq_of_alpha[i] -= num_of_letters_to_be_removed
break
else:
num_of_letters_to_be_removed -= freq_of_alpha[i]
#print(freq_of_alpha)
resulting_str = []
for char in range(len_of_given_str-1, -1, -1):
char = given_str[char]
if freq_of_alpha[ord(char) - 97]:
resulting_str.append(char)
freq_of_alpha[ord(char) - 97] -= 1
#print(freq_of_alpha)
# print(resulting_str)
resulting_str.reverse()
print("".join(resulting_str))
```
No
| 95,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
def alfabet_del(s:str, n:int, k:int):
if len(s) < 2:
return s
ans = list(map(str, s))
for letter in ans:
min_letter = min(ans)
i = 0
while k > 0 and min_letter in ans:
if ans[i] == min_letter:
ans.remove(ans[i])
k -= 1
i -= 1
i += 1
return ''.join(ans)
```
No
| 95,852 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
from itertools import accumulate
n, k = list(map(int, input().split()))
word = input()
lis = sorted(list(set(word)))
t = []
if n == k:
print("")
else:
for item in lis:
if word.count(item) != 0:
t.append(word.count(item))
t = list(accumulate(t))
if k in t:
print("".join([item for item in word if item not in lis[:t.index(k)+1]]))
elif k not in t:
res = [item for item in t if item < k]
if len(res) == 0:
fin = word.replace("a","",k)
print(fin)
else:
maxa = max(res)
fin = ("".join([item for item in word if item not in lis[:t.index(maxa) + 1]]))
print(fin.replace(lis[t.index(maxa) + 1],"", (k - maxa)))
```
No
| 95,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:
* if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
* ...
* remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input
The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples
Input
15 3
cccaabababaccbc
Output
cccbbabaccbc
Input
15 9
cccaabababaccbc
Output
cccccc
Input
1 1
u
Output
Submitted Solution:
```
import sys
n,k=map(int,sys.stdin.readline().split())
s=input()
s1=s
s1=sorted(s)
s2=s1[:k]
j=s1[k]
s1=s1[k:]
cj=s.count(j)
st=""
J=ord(j)
for i in s:
if ord(i)<J and k>0:
st+='#'
k-=1
continue
if ord(i)==J and cj>0 and k>0:
st+='#'
k-=1
cj-=1
continue
else:
st+=i
ans=""
for i in st:
if i=='#':
continue
else:
ans+=i
print(ans)
```
No
| 95,854 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
n = int(input())
values = list(map(int, input().split()))
from itertools import accumulate
copy = values[:]
copy.reverse()
acc = list(accumulate(copy))
acc.reverse()
b = 1
total = 1
for i in range(n+1):
a = values[i]
bCandidate = b - a
if bCandidate < 0:
print(-1)
exit()
b = min(2*bCandidate, acc[i] - a)
total += b
print(total)
```
| 95,855 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
b = [0 for i in range(n+1)]
leaf = sum(a)
if n == 0 and a[0] != 1:
print(-1)
exit()
b[0] = 1 - a[0]
leaf -= a[0]
ans = 1
for i in range(1,n+1):
kosu = min(b[i-1] * 2,leaf)
ans += kosu
b[i] = kosu - a[i]
if b[i] < 0:
print(-1)
exit()
leaf -= a[i]
print(ans)
```
| 95,856 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
if n==0:
if a[0]==1:print(1)
else:print(-1)
exit()
if a[0]:
exit(print(-1))
b=[1]
for i in range(1,n+1):
b.append(b[-1]*2-a[i])
b[-1]=a[-1]
for i in range(n,0,-1):
b[i-1]=min(b[i-1],b[i])
if not(b[i-1]<=b[i]<=2*b[i-1]):exit(print(-1))
b[i-1]+=a[i-1]
print(sum(b))
```
| 95,857 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
leaf = sum(A)
ans = 0
node = 1
for i in range(N+1):
if A[i] > node or leaf == 0:
ans = -1
break
ans += node
leaf -= A[i]
node = min(2*(node-A[i]), leaf)
print(ans)
```
| 95,858 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
N = int(input())
*A, = map(int, input().split())
Asum = [a for a in A]
for i in range(N-1, -1, -1):
Asum[i] += Asum[i+1]
ans = 0
v = 1 # 子になれる頂点数
for i in range(N+1):
if v < A[i]:
ans = -1
break
ans += v
p = v-A[i]
if i < N:
v = min(Asum[i+1], 2*p)
print(ans)
```
| 95,859 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
n = int(input())
al = list(map(int, input().split()))
bl = [0]*(n+1)
acc = [0]*(n+2)
for i in range(n+1):
acc[i+1] = al[i]+acc[i]
res = 0
for d in range(n+1):
if d==0:
res += 1
bl[0] = 1 - al[0]
else:
r1 = 2*bl[d-1] - al[d]
r2 = acc[-1] - acc[d+1]
bl[d] = min(r1,r2)
res += al[d]+bl[d]
if bl[d]<0:
print(-1)
exit()
print(res)
```
| 95,860 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
b = [0] * (N+1)
b[0] = 1
flag = True
for i in range(N):
if A[i] > b[i]:
flag = False
b[i+1] = (b[i] - A[i]) * 2
if A[N] > b[N]:
flag = False
ans = 0
l = 0
for i in range(N, -1, -1):
l += A[i]
ans += min(l, b[i])
if flag:
print(ans)
else:
print(-1)
```
| 95,861 |
Provide a correct Python 3 solution for this coding contest problem.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
isOK = True
B = [0]*(N+1)
B[0] = 1-A[0]
M = 1
i = 1
while i <= N:
B[i] = B[i-1]*2-A[i]
i += 1
if B[-1] < 0 or N > 0 and A[-1] == 0:
isOK = False
B[-1] = A[-1]
i = N-1
while i >= 0:
if B[i] < 1 or i > 0 and B[i] > B[i-1]*2:
isOK = False
break
else:
B[i] = min(B[i], B[i+1])+A[i]
i -= 1
print(-1 if not isOK else sum(B))
```
| 95,862 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
import sys
n=int(input())
a=input()
a=[int(s)for s in a.split()]
M=1
Q=1
m=sum(a)
if a[0]>=1:
if a[0]==1 and n==0:
print(1)
sys.exit()
else:
print(-1)
sys.exit()
for i in range(n):
M=2*M-a[i+1]
if M<0:
print(-1)
sys.exit()
M1=M+a[i+1]
m=m-a[i]
Q=Q+min(M1,m)
print(Q)
```
Yes
| 95,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
s = sum(A)
e = 1
ans = 1
flag = 1
if N == 0:
if s >= 2:
flag = 0
else:
ans = 1
else:
for i in range(N):
e = (e - A[i]) * 2
if e < A[i+1]:
flag = 0
break
tmp = min(e, s)
s -= A[i+1]
ans += tmp
if flag == 0:
print(-1)
else:
print(ans)
```
Yes
| 95,864 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
n = int(input())
al = list(map(int,input().split()))
sum_a = sum(al)
potential = 1
ans = 0
for a in al:
ans += min(sum_a, potential)
sum_a -= a
potential = 2 * (potential - a)
if potential < 0:
print(-1)
exit()
print(ans)
```
Yes
| 95,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
if n == 0 and a[0] != 1:
print(-1)
exit(0)
b = [0] * (n + 2)
for i in range(n, 0, -1):
b[i - 1] = b[i] + a[i]
t = 1 - a[0]
ans = 1
for i in range(n):
m = min(t * 2, b[i])
if m < a[i + 1]:
print(-1)
exit(0)
ans += m
t = m - a[i + 1]
print(ans)
```
Yes
| 95,866 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
import numpy as np
# import math
# import copy
# from collections import deque
import sys
input = sys.stdin.readline
# sys.setrecursionlimit(10000)
def main():
N = int(input())
A = list(map(int,input().split()))
A = np.array(A)
if (N != 0) & (A[0] != 0):
print(-1)
sys.exit()
M = np.sum(A)
already = 1
end = 0
now = 1
res = 1
for d in range(1,N+1):
temp = min(2*now-now,M-already)
now += temp
# print(now)
res += now
# print(d,now,res)
already += temp
end += A[d]
now -= A[d]
# print(d,res)
if (d != N) & (now <= 0):
res = -1
break
print(res)
main()
```
No
| 95,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
n = int(input())
a_li = list(map(int, input().split()))
ans = 0
li = [0]*(n+1)
for i,a in enumerate(a_li):
ans += (i+1)*a
for i,a in enumerate(a_li[::-1]):
if i >= 1:
li[i] = li[i-1]+a
else:
li[i] = a
li = li[::-1]
#print(li)
#print(ans)
flag = True
ne = 0.5
for i in range(n+1):
if li[i] > 2*ne:
ans -= li[i] -2*ne
#print(li[i] -2*ne)
ne = 2*ne - a_li[i]
#print(ne)
if ne < 0:
flag = False
if flag:
print(int(ans))
else:
print(-1)
```
No
| 95,868 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
# import sys
# readline = sys.stdin.readline
# generator = (readline().strip() for _ in range(N))
# N, M = map(int, input().split())
# As = list(map(int, input().split()))
# queries = (input() for _ in range(N))
def solve():
N = int(input())
As = list(map(int, input().split()))
if As[0] > 0:
return -1
vertices = [1] * (N+1)
previous = 0
for i, a in enumerate(As[:0:-1]):
max_n_vert = a + previous
previous = max_n_vert
vertices[N-i] = max_n_vert
n_not_leaves_prev = 1
for i, a in enumerate(As):
if i == 0:
continue
max_n_vert = 2 * n_not_leaves_prev
vertices[i] = min(vertices[i], max_n_vert)
n_not_leaves_prev = vertices[i] - a
if max_n_vert < a:
return -1
return sum(vertices)
def main():
print(solve())
if __name__ == "__main__":
main()
```
No
| 95,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
Constraints
* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_0 A_1 A_2 \cdots A_N
Output
Print the answer as an integer.
Examples
Input
3
0 1 1 2
Output
7
Input
4
0 0 1 0 2
Output
10
Input
2
0 3 1
Output
-1
Input
1
1 1
Output
-1
Input
10
0 0 1 1 2 3 5 8 13 21 34
Output
264
Submitted Solution:
```
depth = int(input())
vertex = input().split(' ')
vertex = [int(x) for x in vertex]
if vertex[0] == 1:
if depth == 0:
print(1)
else:
print(-1)
else:
is_error = False
for i in range(depth + 1):
if vertex[i] > 2**i:
is_error = True
break
if is_error:
print(-1)
else:
maxs = []
i = 1
for v in vertex:
i -= v
maxs.append(i)
i *= 2
output = 0
for i in range(depth, -1, -1):
if i == depth:
output += vertex[i]
next_v = vertex[i]
else:
next_v = min([maxs[i], next_v]) + vertex[i]
output += next_v
print(output)
```
No
| 95,870 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
from collections import defaultdict
from heapq import heappop, heappush
class Graph(object):
def __init__(self):
self.graph = defaultdict(list)
def __len__(self):
return len(self.graph)
def add_edge(self, From, To, cost=1):
self.graph[From].append((To,cost))
def get_nodes(self):
return self.graph.keys()
class Dijkstra(object):
def __init__(self, graph, start):
self.g = graph.graph
self.dist = defaultdict(lambda:float('inf'))
self.dist[start] = 0
self.prev = defaultdict(lambda: None)
self.Q = []
heappush(self.Q,(self.dist[start], start))
while self.Q:
dist_u, u = heappop(self.Q)
if self.dist[u] < dist_u:
continue
for v, cost in self.g[u]:
alt = dist_u + cost
if self.dist[v] > alt:
self.dist[v] = alt
self.prev[v] = u
heappush(self.Q,(alt, v))
def shortest_distance(self, goal):
return self.dist[goal]
def shortest_path(self, goal):
path = []
node = goal
while node is not None:
path.append(node)
node = self.prev[node]
return path[::-1]
N = int(input())
g = Graph()
d = {}
for i in range(N-1):
a,b = map(int,input().split())
a -= 1
b -= 1
d[(min(a,b),max(a,b))] = i
g.add_edge(a,b)
g.add_edge(b,a)
M = int(input())
l = [0]*M
for i in range(M):
u,v = map(int,input().split())
u -= 1
v -= 1
dij = Dijkstra(g,u)
path = dij.shortest_path(v)
for j in range(len(path)-1):
a,b = path[j],path[j+1]
l[i] += 2**d[(min(a,b),max(a,b))]
c = [0]*(M+1)
p2 = [2**i for i in range(50)]
for n in range(2**M):
cnt = 0
x = 0
for i in range(M):
if n&p2[i]==p2[i]:
cnt += 1
x |= l[i]
y = 0
for i in range(N-1):
if x&p2[i]==p2[i]:
y += 1
c[cnt] += p2[N-1-y]
ans = 0
for i in range(M+1):
if i%2==0:
ans += c[i]
else:
ans -= c[i]
print(ans)
```
| 95,871 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
def resolve():
def dfs(u, v, parent):
if u == v:
return True
for (to, idx) in G[u]:
if to == parent:
continue
if dfs(to, v, u):
path.append(idx)
return True
return False
N = int(input())
G = [[] for _ in range(N)]
for i in range(N-1):
a, b = map(lambda x:int(x)-1, input().split())
G[a].append((b, i))
G[b].append((a, i))
path = []
M = int(input())
cons = [0]*M
# 制約ごとに、白くなる辺を入れておく
for i in range(M):
u, v = map(lambda x: int(x) - 1, input().split())
path = []
dfs(u, v, -1)
for idx in path:
cons[i] |= 1<<idx
# 包除原理
ans = 0
for bit in range(1<<M):
eset = 0
for i in range(M):
if bit>>i & 1:
eset |= cons[i]
white = bin(eset).count("1") # 白くなるべき辺の数
num = 1<<(N-1-white)
if bin(bit).count("1") % 2 == 0:
ans += num
else:
ans -= num
print(ans)
if __name__ == "__main__":
resolve()
```
| 95,872 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
N = int(input())
from collections import defaultdict
G = defaultdict(list)
EdaNum = {}
Ki = [[float('inf')]*(N+1) for i in range(N+1)]
for i in range(N-1):
a,b = map(int,input().split())
G[a].append(b)
G[b].append(a)
EdaNum[(min(a,b),max(a,b))] = i
Ki[a][b] = 1
Ki[b][a] = 1
Next = [[-1]*(N+1) for i in range(N+1)]
for i in range(1,N+1):
for j in range(1,N+1):
Next[i][j] = j
for k in range(1,N+1):
for i in range(1,N+1):
for j in range(1,N+1):
if Ki[i][k] + Ki[k][j] < Ki[i][j]:
Ki[i][j] = Ki[i][k] + Ki[k][j]
Next[i][j] = Next[i][k]
M = int(input())
MM = [set() for i in range(M)]
#print(EdaNum)
#print(Next)
for i in range(M):
u,v = map(int,input().split())
now = u
while now != v:
nex = Next[now][v]
MM[i].add(EdaNum[(min(now,nex),max(now,nex))])
now = nex
#print(MM)
import itertools
zero1 = [0,1]
ans = 0
for i in itertools.product(zero1,repeat = M):
#print(ans)
count = 0
root = set()
for j in range(M):
if i[j] == 0:
count += 1
else:
root |= MM[j]
#print(i,len(root),count)
if (M-count)%2 ==0:
ans += pow(2,N-1-len(root))
else:
ans -= pow(2,N-1-len(root))
print(ans)
```
| 95,873 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
n = int(input())
info = [list(map(int, input().split())) for i in range(n - 1)]
m = int(input())
path = [list(map(int, input().split())) for i in range(m)]
tree = [[] for i in range(n)]
memo = {}
for i in range(n - 1):
a, b = info[i]
a -= 1
b -= 1
tree[a].append(b)
tree[b].append(a)
memo[(min(a, b), max(a, b))] = i
def dfs(par, pos, goal):
if pos == goal:
return True
for child in tree[pos]:
if child == par:
continue
if dfs(pos, child, goal):
res.append((min(pos, child), max(pos, child)))
return True
return False
li_path = [0] * m
for i in range(m):
res = []
s, g = path[i]
s -= 1
g -= 1
dfs(-1, s, g)
tmp = 0
for j in res:
tmp += (1 << memo[j])
li_path[i] = tmp
ans = 0
for bit_state in range(1 << m):
mask = 0
cnt = 0
for i in range(m):
if bit_state & (1 << i):
cnt += 1
mask |= li_path[i]
num_n = bin(mask).count("1")
if cnt % 2 == 0:
ans += 2 ** (n - 1 - num_n)
else:
ans -= 2 ** (n - 1 - num_n)
print(ans)
```
| 95,874 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
def popcount(x):
x = x - ((x >> 1) & 0x5555555555555555)
x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
x = x + (x >> 32)
return x & 0x0000007f
def main():
import sys
import collections
input = sys.stdin.readline
N = int(input())
G = [[-1]*N for i in range(N)]
ans = 2**(N-1)
for i in range(N-1):
a, b = map(int, input().split())
a -= 1
b -= 1
G[a][b] = i
G[b][a] = i
ans = 0
g = [[0]*N for i in range(N)]
for i in range(N):
q = collections.deque()
q.append((i, 0))
r = [True]*N
r[i] = False
while(q):
x, s = q.popleft()
for y in range(N):
if G[x][y] >= 0 and r[y]:
g[i][y] = s | 1 << G[x][y]
q.append((y, g[i][y]))
r[y] = False
M = int(input())
s = [0 for i in range(M)]
for i in range(M):
u, v = map(int, input().split())
u -= 1
v -= 1
s[i] = g[u][v]
# print(s)
for i in range(2**M):
tmp = 2**(N-1) - 1
c = 0
for j in range(M):
if (i >> j) % 2 == 1:
tmp &= ~s[j]
c += 1
ans += ((-1)**(c))*(1 << popcount(tmp))
print(ans)
main()
```
| 95,875 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**7)
def main(n,ki,uv):
# n: 頂点数
# ki: 木
# Euler Tour の構築
S=[] # Euler Tour
F=[0]*n # F[v]:vにはじめて訪れるステップ
depth=[0]*n # 0を根としたときの深さ
def dfs(v,pare,d):
F[v]=len(S)
depth[v]=d
S.append(v)
for i,w in ki[v]:
if w==pare:continue
dfs(w,v,d+1)
S.append(v)
dfs(0,-1,0)
# Sをセグメント木に乗せる
# u,vのLCAを求める:S[F[u]:F[v]+1]のなかでdepthが最小の頂点を探せば良い
# F[u]:uに初めてたどるつくステップ
# S[F[u]:F[v]+1]:はじめてuにたどり着いてつぎにvにたどるつくまでに訪れる頂点
# 存在しない範囲は深さが他よりも大きくなるようにする
INF = (n, None)
# LCAを計算するクエリの前計算
M = 2*n
M0 = 2**(M-1).bit_length() # M以上で最小の2のべき乗
data = [INF]*(2*M0)
for i, v in enumerate(S):
data[M0-1+i] = (depth[v], i)
for i in range(M0-2, -1, -1):
data[i] = min(data[2*i+1], data[2*i+2])
# LCAの計算 (generatorで最小値を求める)
def _query(a, b):
yield INF
a += M0; b += M0
while a < b:
if b & 1:
b -= 1
yield data[b-1]
if a & 1:
yield data[a-1]
a += 1
a >>= 1; b >>= 1
# LCAの計算 (外から呼び出す関数)
def query(u, v):
fu = F[u]; fv = F[v]
if fu > fv:
fu, fv = fv, fu
return S[min(_query(fu, fv+1))[1]]
todo=[[0,[]]]
path=[[]]*n
mi=[1]*n
while todo:
v,ary=todo.pop()
path[v]=ary
mi[v]=0
for i,nv in ki[v]:
if mi[nv]==1:
todo.append([nv,ary+[i]])
mi[nv]=1
for i in range(n):
path[i]=set(path[i])
memo=[set(())]*m
for c in range(m):
u,v=uv[c]
dd=query(u,v)
tmp=path[u]|path[v]
tmp-=path[dd]
memo[c]=tmp
ans=0
for i in range(2**m):
chk=set(())
for j in range(m):
if (i>>j)&1:
chk.add(j)
must_w=set(())
for c in chk:
tmp=memo[c]
must_w|=tmp
t=n-1-len(must_w)
if len(chk)%2==1:
ans-=pow(2,t)
else:
ans+=pow(2,t)
print(ans)
if __name__=='__main__':
n=int(input())
ki=[[] for _ in range(n)]
for i in range(n-1):
u,v=map(int,input().split())
u,v=u-1,v-1
ki[u].append([i,v])
ki[v].append([i,u])
m=int(input())
uv=[]
for _ in range(m):
u,v=map(int,input().split())
u,v=u-1,v-1
uv.append([u,v])
main(n,ki,uv)
```
| 95,876 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
N = int(input())
E = [[] for _ in range(N)]
for _ in range(N-1):
a, b = map(int, input().split())
E[a-1].append(b-1)
E[b-1].append(a-1)
parent = [None]*N
depth = [0]*N
d = 0
queue = [0]
while queue:
d += 1
queue_new = []
for q in queue:
for e in E[q]:
if e==parent[q]: continue
parent[e] = q
depth[e] = d
queue_new.append(e)
queue = queue_new
ancestor = [[None]*N for _ in range(N)]
for a in range(N):
for b in range(a, N):
_a = a
_b = b
while _a!=_b:
if depth[_a] < depth[_b]: _b = parent[_b]
else: _a = parent[_a]
ancestor[a][b] = ancestor[b][a] = _a
M = int(input())
P = []
for _ in range(M):
u, v = map(int, input().split())
u -= 1
v -= 1
w = ancestor[u][v]
path = set()
while u!=w:
_u = parent[u]
#path.add((u, _u))
path.add(u*N+_u)
u = _u
while v!=w:
_v = parent[v]
#path.add((v, _v))
path.add(v*N+_v)
v = _v
P.append(path)
ans = 0
n1 = N-1
for k in range(2**M):
path = set()
n = 0
for i, p in enumerate(P):
if (k>>i)&1:
path |= p
n += 1
if n==0: continue
sign = +1 if n&1 else -1
ans += sign * pow(2, n1-len(path))
print(pow(2, N-1) - ans)
```
| 95,877 |
Provide a correct Python 3 solution for this coding contest problem.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
"Correct Solution:
```
N = int(input())
ab = [tuple(map(int,input().split())) for _ in range(N - 1)]
M = int(input())
uv = [tuple(map(int,input().split())) for _ in range(M)]
route = [[] for _ in range(N)]
for i, (a, b) in enumerate(ab):
a -= 1
b -= 1
route[a].append((b, i))
route[b].append((a, i))
path = [0] * M
for i, (u, v) in enumerate(uv):
u -= 1
v -= 1
stack = [u]
visited = [0] * N
visited[u] = 1
dire = [[] for _ in range(N)]
while True:
s = stack.pop()
for g, j in route[s]:
if visited[g]:
continue
stack.append(g)
dire[g].append((s, j))
visited[g] = 1
if visited[v]:
break
s = v
while True:
g, num = dire[s][0]
path[i] |= (1 << num)
if g == u:
break
s = g
data = [0] * (1 << M)
ans = 2 ** (N - 1)
for i in range(1, 1 << M):
k = len(bin(i)) - 3
data[i] = data[i - (2 ** k)] | path[k]
power = bin(data[i]).count("1")
judge = bin(i).count("1")
ans += (-2 ** (N - 1 - power) if judge % 2 else 2 ** (N - 1 - power))
print(ans)
```
| 95,878 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
import sys
input = sys.stdin.readline
def main():
n = int(input())
T = [[] for _ in range(n)]
D = {}
for i in range(n-1):
a, b = map(int, input().split())
a -= 1
b -= 1
T[a].append(b)
T[b].append(a)
D[(a, b)] = 1<<i
D[(b, a)] = 1<<i
m = int(input())
def dfs(u, v):
parent = [-1]*n
seen = [False]*n
stack = [u]
seen[u] = True
while stack:
nv = stack[-1]
if nv == v:
return stack
for i in T[nv]:
if seen[i]:
continue
stack.append(i)
seen[i] = True
break
else:
stack.pop()
def popcount(n):
c = (n & 0x5555555555555555) + ((n>>1) & 0x5555555555555555)
c = (c & 0x3333333333333333) + ((c>>2) & 0x3333333333333333)
c = (c & 0x0f0f0f0f0f0f0f0f) + ((c>>4) & 0x0f0f0f0f0f0f0f0f)
c = (c & 0x00ff00ff00ff00ff) + ((c>>8) & 0x00ff00ff00ff00ff)
c = (c & 0x0000ffff0000ffff) + ((c>>16) & 0x0000ffff0000ffff)
c = (c & 0x00000000ffffffff) + ((c>>32) & 0x00000000ffffffff)
return c
E = [0]*m
for i in range(m):
u, v = map(int, input().split())
u -= 1
v -= 1
edge = dfs(u, v)
for j in range(len(edge)-1):
E[i] |= D[(edge[j], edge[j+1])]
ans = 0
for i in range(2**m):
s = 0
for j in range(m):
if i>>j & 1:
s |= E[j]
cnt = popcount(i) % 2
ans += (-1)**cnt * 2**(n-1-popcount(s))
print(ans)
if __name__ == "__main__":
main()
```
Yes
| 95,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
import sys
def main():
input = sys.stdin.readline
N = int(input())
E = [[] for _ in range(N)]
for i in range(N-1):
a, b = map(int, input().split())
a, b = a-1, b-1
E[a].append((b,i))
E[b].append((a,i))
def dfs(s, g):
par = [(-1,-1)] * N
par[s] = (s,-1)
stack = [s]
while stack:
v = stack.pop()
for to, i in E[v]:
if par[to][0] >= 0: continue
par[to] = (v, i)
if to == g: break
stack.append(to)
r = 0
v = g
while v != s:
v, i = par[v]
r |= 1 << i
return r
M = int(input())
path = [0] * M
for i in range(M):
u, v = map(int, input().split())
u, v = u-1, v-1
path[i] = dfs(u, v)
def calc(s): return 1<<(N-1-bin(s).count('1'))
i_ans = 0
for p in range(1, 1<<M):
is_odd = 0
s = 0
for i in range(M):
if p&1:
s |= path[i]
is_odd ^= 1
p >>= 1
if is_odd: i_ans += calc(s)
else: i_ans -= calc(s)
print((1<<(N-1)) - i_ans)
if __name__ == '__main__':
main()
```
Yes
| 95,880 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**8)
input = sys.stdin.readline
n = int(input())
graph = [[] for _ in range(n)] #[(v_to, edge_idx)]
for i in range(n - 1):
a, b = map(int, input().split())
a -= 1
b -= 1
graph[a].append((b, i))
graph[b].append((a, i))
def dfs(v, v_p, target):
if v == target:
return True, 0
for v_next, edge_idx in graph[v]:
if v_next == v_p:
continue
is_target, path = dfs(v_next, v, target)
if is_target:
return is_target, path | (1 << edge_idx)
return False, 0
m = int(input())
cond = []
for _ in range(m):
u, v = map(int, input().split())
u -= 1
v -= 1
_, path = dfs(u, -1, v)
# print(u, v, bin(path))
cond.append((u, v, path))
ans = 1 << (n - 1)
for bit in range(1, 1 << m):
l = 0
path = 0
for i, (u, v, path_i) in enumerate(cond):
if (bit >> i) & 1:
l += 1
path = path | path_i
c = n - 1 - bin(path).count('1')
ans += ((-1)**(l % 2)) * (1 << c)
print(ans)
```
Yes
| 95,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
from collections import defaultdict
n = int(input())
ns = defaultdict(set)
uv2e = {}
for i in range(n-1):
u,v = map(int, input().split())
ns[u-1].add(v-1)
ns[v-1].add(u-1)
uv2e[u-1,v-1] = i
uv2e[v-1,u-1] = i
m = int(input())
uvs = [None]*m
for i in range(m):
u,v = map(int, input().split())
u -= 1
v -= 1
uvs[i] = (u,v)
from queue import deque
def bfs(start):
q = deque([start])
seen = [None] * n
seen[start] = 0
while q:
u = q.pop()
d = seen[u]
for v in ns[u]:
if seen[v] is None:
seen[v] = u
q.appendleft(v)
return seen
paths = [None]*m
for i,(u,v) in enumerate(uvs):
seen = bfs(u)
ss = set()
z = v
while z!=u:
z1 = seen[z]
ss.add(uv2e[z,z1])
z = z1
paths[i] = ss
ans = 0
def dfs(l, s):
if len(l)==m:
global ans
val = pow(2, n-1-len(s)) * pow(-1, sum(l))
ans += val
# print(val, sum(l), ans)
return
dfs(l + [0], s)
i = len(l)
p = paths[i]
dfs(l + [1], s|p)
dfs([], set())
# ans = pow(2,n-1) - ans
print(ans)
```
Yes
| 95,882 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from itertools import permutations
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
res = list(sys.stdin.readline())
if res[-1] == "\n":
return res[:-1]
return res
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
def solve():
def dijkstra(s):
d = [float("inf")]*n
p = [s]*n
d[s] = 0
q = [(0,s)]
while q:
dx,x = heappop(q)
nd = dx+1
for y in v[x]:
if nd < d[y]:
d[y] = nd
p[y] = x
heappush(q,(nd,y))
path = [[i] for i in range(n)]
for i in range(1,n):
pre = path[i][-1]
while pre != s:
path[i].append(p[pre])
pre = path[i][-1]
path[i] = path[i][::-1]
return d,path
n = I()
v = [[] for i in range(n)]
for i in range(n-1):
a,b = LI()
a -= 1
b -= 1
v[a].append(b)
v[b].append(a)
d = [[0]*n for i in range(n)]
path = [[None]*n for i in range(n)]
for i in range(n):
D,p = dijkstra(i)
d[i] = [j for j in D]
path[i] = [j for j in p]
m = I()
l = LIR(m)
for i in range(m):
l[i][0] -= 1
l[i][1] -= 1
ans = 0
fa = 0
fb = 0
for b in range(1,1<<m):
f = bin(b).count("1")
p = [0]*n
for j in range(m):
x,y = l[j]
if b&(1<<j):
for k in path[x][y]:
p[k] = 1
s = 0
for j in p:
s += j
f &= 1
k = 1<<(n-s)
if f:
ans += k
else:
ans -= k
print((1<<(n-1))-ans)
return
#Solve
if __name__ == "__main__":
solve()
```
No
| 95,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 7)
def dfs(G, p, pre, t, edge, x):
for nxt in G[p]:
if nxt == pre:
continue
a, b = p, nxt
if b < a:
a, b = b, a
if nxt == t:
edge[(a, b)] += x
return True
if dfs(G, nxt, p, t, edge, x):
edge[(a, b)] += x
return True
return False
def main():
n = int(input())
G = [[] for _ in range(n)]
edge = {}
for _ in range(n-1):
a, b = map(int, input().split())
a, b = a-1, b-1
G[a].append(b)
G[b].append(a)
if b < a:
a, b = b, a
edge[(a, b)] = 0
m = int(input())
for i in range(m):
u, v = map(int, input().split())
u, v = u-1, v-1
x = 1<<i
dfs(G, u, -1, v, edge, x)
ans = 2**(n-1)
for i in range(1, 2**m):
bit_cnt = 0
for j in range(m):
if (i & (1<<j)) != 0:
bit_cnt += 1
cnt = 0
for v in edge.values():
if (v & i) != 0:
cnt += 1
x = 2**(n-1 - cnt)
if bit_cnt % 2 == 0:
ans += x
else:
ans -= x
print(ans)
if __name__ == "__main__":
main()
```
No
| 95,884 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from collections import deque
def all_bit_count(x):
x = x - ((x >> 1) & 0x5555555555555555)
x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
x = x + (x >> 32)
return x & 0x0000007f
n = int(readline())
g = [[] * n for i in range(n)]
for i in range(n-1):
u, v = map(int,readline().split())
u -= 1
v -= 1
g[u].append(v)
g[v].append(u)
m = int(readline())
c = []
for i in range(m):
u, v = map(int,readline().split())
u -= 1
v -= 1
que = deque()
que.append((u,1<<u))
chk = [False]*n
chk[u] = True
while que:
now,L = que.popleft()
if now == v:
c.append(L)
break
for i in g[now]:
if not chk[i]:
chk[i] = True
que.append((i,L|1<<i))
import itertools
ans = 1<<(n-1)
for i in range(1,m+1):
for j in itertools.combinations(c,i):
K = 0
l = 0
for k in j:
p = all_bit_count(K)
f = all_bit_count(k)
K |= k
B = all_bit_count(K)
if f+p == B:
l += f-1
else:
l = B-1
ans += pow(-1,i)*pow(2,n-1-l)
print(ans)
```
No
| 95,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?
* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.
Constraints
* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M
Output
Print the number of ways to paint the edges that satisfy all of the M conditions.
Examples
Input
3
1 2
2 3
1
1 3
Output
3
Input
2
1 2
1
1 2
Output
1
Input
5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5
Output
9
Input
8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8
Output
62
Submitted Solution:
```
from itertools import combinations
def cal(tree, v, bit, pre_node, now_node):
for node in tree[now_node]:
new_bit = bit | (1 << node)
if node == v:
return new_bit
if node == pre_node:
continue
new_bit = cal(tree, v, new_bit, now_node, node)
if new_bit:
return new_bit
return False
def run():
N = int(input())
tree = {}
for n in range(1, N):
a, b = map(int, input().split())
tree.setdefault(a, []).append(b)
tree.setdefault(b, []).append(a)
#print(tree)
M = int(input())
uv = [0]*M
for m in range(M):
u, v = map(int, input().split())
bit = 1 << u
uv[m] = cal(tree, v, bit, -1, u)
#print(bin(uv[m]))
ans = 2**(N-1)
for m in range(1, M+1):
total = 0
#print(m)
for units in combinations(uv, m):
#print(units)
temp = 0
white = 0
for bit in units:
white += bin(bit).count('1')-1
dup = temp & bit
n_dup = bin(dup).count('1')
if n_dup >= 2:
white -= n_dup-1
temp |= bit
#print(bin(temp))
#white = bin(temp).count('1')
black = N-white-1
#print(black)
total += 2**black
if m%2 == 0:
#print('m'+str(m))
ans += total
else:
ans -= total
print(ans)
if __name__ == '__main__':
run()
```
No
| 95,886 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda:sys.stdin.readline().rstrip()
def resolve():
n=int(input())
for u in range(n-1):
ans=[]
for v in range(u+1,n):
w=u^v
for i in range(10):
if((w>>i)&1):
ans.append(i+1)
break
print(*ans)
resolve()
```
| 95,887 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
N=int(input())
for i in range(N):
i = i + 1
for j in range(N-i):
j = i+j+1
for k in range(9):
if i % (2**(k+1)) != j % (2**(k+1)):
print(k+1, end=" ")
break
print()
```
| 95,888 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
n = int(input())
def b(x):
return format(x,'04b')
for i in range(1,n+1):
for j in range(i+1,n+1):
tmp = i ^ j
bit = format(tmp, 'b')[::-1]
# print(bit)
for k in range(len(bit)):
# print(bit[k])
if int(bit[k]) == 1:
# print(bit, bit[k],k+1)
print(k+1, end = ' ')
break
print()
# print(b(i),b(j),b(tmp))
```
| 95,889 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
n = int(input())
for i in range(n-1):
L = []
for j in range(i+1, n):
x = i^j
l = (x&-x).bit_length()
L.append(l)
print(*L)
```
| 95,890 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
n = int(input())
lst = []
for i in range(n):
num = format(i, '09b')
lst.append(num[::-1])
for i in range(n-1):
ans = []
for j in range(i+1, n):
numi = lst[i]
numj = lst[j]
idx = 0
while True:
if numi[idx] != numj[idx]:
ans.append(str(idx+1))
break
idx += 1
print(*ans)
```
| 95,891 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import *
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')
#A
def A():
m, d = LI()
ans = 0
for i in range(1, m + 1):
for j in range(10, d + 1):
if int(str(j)[0]) < 2 or int(str(j)[1]) < 2:
continue
if i == int(str(j)[0]) * int(str(j)[1]):
ans += 1
print(ans)
return
#B
def B():
n, l = LI()
a = LI()
d0 = defaultdict(int)
d1 = defaultdict(int)
for i in a:
d0[i] += 1
for i in range(n):
b = 0
for k in range(i+1, n):
if a[i] > a[k]:
b += 1
d1[i] = b
ans = 0
d2 = list(d0.items())
d2.sort()
d3 = defaultdict(int)
d = 0
for key, value in d2:
d3[key] = d
d += value
for i in range(n):
ans += d3[a[i]] * l * (l-1) // 2 + d1[i] * l
ans %= mod
print(ans % mod)
return
#C
def C():
n = II()
s = S()
if s[0] == "W":
print(0)
return
lis = [0] * (2 * n)
l = [0] * (2 * n)
l[0] = 1
for i in range(1, 2 * n):
lis[i] = (lis[i-1] ^ (s[i] == s[i-1]))
l[i] += l[i - 1] + (lis[i] == 0)
ans = 1
if l[-1] != n or s[0] == "W":
print(0)
return
k = 1
for i in range(2 * n):
if lis[i]:
ans *= l[i] - (k - 1)
ans %= mod
k += 1
print(ans * math.factorial(n) % mod)
return
#D
def D():
n = II()
ans = [[] for i in range(n - 1)]
l = len(bin(n)) - 2
for i in range(n - 1):
for j in range(i + 2, n + 1):
for k in range(l):
if (j >> k) & 1 != ((i + 1) >> k) & 1:
ans[i].append(str(k + 1))
break
print()
for a in ans:
print(" ".join(a))
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#Solve
if __name__ == '__main__':
D()
```
| 95,892 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
n = int(input())
ans = [[0 for i in range(n)] for j in range(n)]
def p2(a):
ret = 0
while a%2 == 0:
ret += 1
a //= 2
return ret
for i in range(n):
for j in range(n-i-1):
ans[i][j] = p2(j+1)+1
for i in range(n):
arr = ans[i][:n-i-1]
print(*arr)
```
| 95,893 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
"Correct Solution:
```
from math import log2
def main():
n = int(input())
for u in range(n - 1):
ans = []
for v in range(u + 1, n):
uv = u ^ v
ans.append(int(log2(uv & -uv) + 1))
print(*ans)
main()
```
| 95,894 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
ans = 0
ans = [[0] * (N+1) for _ in range(N+1)]
depth = 1
def dfs(l, r, d):
global depth
if l == r:
return
depth = max(depth, d)
m = (l + r) // 2
for i in range(l, m+1):
for j in range(m, r+1):
ans[i][j] = ans[i][j] = d
dfs(l, m, d+1)
dfs(m+1, r, d+1)
dfs(1, N, 1)
for i in range(1, N):
print(*ans[i][i+1:])
```
Yes
| 95,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
#import sys
#input = sys.stdin.readline
def main():
N = int( input())
ANS = [ [0]*N for _ in range(N)]
for i in range(11):
if N <= pow(2,i):
M = i
break
# for s in range(1, N, 2):
# for i in range(N-s):
# ANS[i][i+s] = 1
for t in range(1,M+1):
w = pow(2,t-1)
for s in range(1,N+1, 2):
if w*s >= N:
break
for i in range(N-s*w):
if ANS[i][i+w*s] == 0:
ANS[i][i+w*s] = t
for i in range(N-1):
print( " ".join( map( str, ANS[i][i+1:])))
if __name__ == '__main__':
main()
```
Yes
| 95,896 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
def num(i,j):
if i==j:
return -1
S=bin(i)[2:][::-1]+"0"*20
T=bin(j)[2:][::-1]+"0"*20
for index in range(min(len(S),len(T))):
if S[index]!=T[index]:
return index+1
return -1
N=int(input())
a=[[0 for j in range(N-1-i)] for i in range(N-1)]
for i in range(N-1):
for j in range(i+1,N):
a[i][j-i-1]=num(i,j)
for i in range(N-1):
print(*a[i])
```
Yes
| 95,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
n = int(input())
a = [[0]*n for i in range(n)]
pos = [i for i in range(n)]
def bipartite_graph(pos, level):
pos1 = pos[:len(pos)//2]
pos2 = pos[len(pos)//2:]
for i in pos1:
for j in pos2:
a[i][j] = level
if len(pos1) >= 2:
bipartite_graph(pos1, level + 1)
if len(pos2) >= 2:
bipartite_graph(pos2, level + 1)
bipartite_graph(pos, 1)
for i in range(n):
print(*a[i][i+1:])
```
Yes
| 95,898 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
n = int(input())
ans = [[-1] * (n - i - 1) for i in range(n - 1)]
s = [(0, n)]
t = 1
while s:
p, q = s.pop()
if p + 1 == q:
continue
m = (q - p) // 2 + p
for i in range(p, m):
for j in range(m, q):
ans[i][j - i - 1] = t
t += 1
s.append((p, m))
s.append((m, q))
for row in ans:
print(' '.join(map(str, row)))
```
No
| 95,899 |
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