message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a set $S$ of integers, perform a sequence of the following operations. Note that each value in $S$ must be unique.
* insert($x$): Insert $x$ to $S$ and report the number of elements in $S$ after the operation.
* find($x$): Report the number of $x$ in $S$ (0 or 1).
* delete($x$): Delete $x$ from $S$.
* dump($L$, $R$): Print elements $x$ in $S$ such that $L \leq x \leq R$.
Constraints
* $1 \leq q \leq 200,000$
* $0 \leq x \leq 1,000,000,000$
* The total number of elements printed by dump operations does not exceed $1,000,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $x$
or
1 $x$
or
2 $x$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, find, delete and dump operations respectively.
Output
For each insert operation, print the number of elements in $S$.
For each find operation, print the number of specified elements in $S$.
For each dump operation, print the corresponding elements in ascending order. Print an element in a line.
Example
Input
9
0 1
0 2
0 3
2 2
1 1
1 2
1 3
0 4
3 2 4
Output
1
2
3
1
0
1
3
3
4
Submitted Solution:
```
def binary(l,c):
fst = 0
last = len(l)
mid = int(last/2)
if fst == 0 and last ==0:
l.insert(0,c)
while fst < last:
mid = int((fst+last) / 2)
if last == mid:
break
elif fst == mid:
break
elif l[mid] >= c:
last = mid
elif l[mid] <= c:
fst = mid
if l[mid] == c:
return
elif l[mid] < c:
mid+=1
l.insert(mid,c)
q = int(input())
S = []
a = []
for i in range(q):
a.append(list(map(int,input().split())))
for i in range(q):
if a[i][0] == 0:
binary(S,a[i][1])
print(len(S))
elif a[i][0] == 1:
if a[i][1] in S:
print(1)
else:
print(0)
elif a[i][0] == 2:
if a[i][1] in S:
S.remove(a[i][1])
else:
for b in range(len(S)):
if S[b] >= a[i][1] and S[b] <= a[i][2]:
print(S[b])
``` | instruction | 0 | 30,958 | 5 | 61,916 |
No | output | 1 | 30,958 | 5 | 61,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a set $S$ of integers, perform a sequence of the following operations. Note that each value in $S$ must be unique.
* insert($x$): Insert $x$ to $S$ and report the number of elements in $S$ after the operation.
* find($x$): Report the number of $x$ in $S$ (0 or 1).
* delete($x$): Delete $x$ from $S$.
* dump($L$, $R$): Print elements $x$ in $S$ such that $L \leq x \leq R$.
Constraints
* $1 \leq q \leq 200,000$
* $0 \leq x \leq 1,000,000,000$
* The total number of elements printed by dump operations does not exceed $1,000,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $x$
or
1 $x$
or
2 $x$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, find, delete and dump operations respectively.
Output
For each insert operation, print the number of elements in $S$.
For each find operation, print the number of specified elements in $S$.
For each dump operation, print the corresponding elements in ascending order. Print an element in a line.
Example
Input
9
0 1
0 2
0 3
2 2
1 1
1 2
1 3
0 4
3 2 4
Output
1
2
3
1
0
1
3
3
4
Submitted Solution:
```
def binary(l,c,low,index):
fst = 0
last = len(l)
mid = int(last/2)
if fst == 0 and last ==0:
l.insert(0,c)
for i in range(len(index)):
if low[i] > c:
index[i] += 1
return
while fst < last:
mid = int((fst+last) / 2)
if l[mid] == c:
return
elif last == mid:
break
elif fst == mid:
break
elif l[mid] >= c:
last = mid
elif l[mid] <= c:
fst = mid
if l[mid] < c:
mid+=1
l.insert(mid,c)
for i in range(len(index)):
if low[i] > c:
index[i] += 1
q = int(input())
S = []
a = []
low = []
for i in range(q):
a.append(list(map(int,input().split())))
if a[i][0] == 3:
low.append(a[i][1])
index = [0]*len(low)
for i in range(q):
if a[i][0] == 0:
binary(S,a[i][1],low,index)
print(len(S))
elif a[i][0] == 1:
if a[i][1] in S:
print(1)
else:
print(0)
elif a[i][0] == 2:
if a[i][1] in S:
S.remove(a[i][1])
for b in range(len(index)):
if low[b] > a[i][1]:
index[b] -= 1
else:
if index[0] < len(S):
for b in range(index[0],len(S)):
if S[b] <= a[i][2]:
print(S[b])
if b == len(S):
del low[0]
del index[0]
else:
del low[0]
del index[0]
break
else:
del low[0]
del index[0]
``` | instruction | 0 | 30,959 | 5 | 61,918 |
No | output | 1 | 30,959 | 5 | 61,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a set $S$ of integers, perform a sequence of the following operations. Note that each value in $S$ must be unique.
* insert($x$): Insert $x$ to $S$ and report the number of elements in $S$ after the operation.
* find($x$): Report the number of $x$ in $S$ (0 or 1).
* delete($x$): Delete $x$ from $S$.
* dump($L$, $R$): Print elements $x$ in $S$ such that $L \leq x \leq R$.
Constraints
* $1 \leq q \leq 200,000$
* $0 \leq x \leq 1,000,000,000$
* The total number of elements printed by dump operations does not exceed $1,000,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $x$
or
1 $x$
or
2 $x$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, find, delete and dump operations respectively.
Output
For each insert operation, print the number of elements in $S$.
For each find operation, print the number of specified elements in $S$.
For each dump operation, print the corresponding elements in ascending order. Print an element in a line.
Example
Input
9
0 1
0 2
0 3
2 2
1 1
1 2
1 3
0 4
3 2 4
Output
1
2
3
1
0
1
3
3
4
Submitted Solution:
```
from enum import Enum
import math
class Color(Enum):
BLACK = 0
RED = 1
@staticmethod
def flip(c):
return [Color.RED, Color.BLACK][c.value]
class RedBlackBinarySearchTree:
"""Red Black Binary Search Tree with range, min, max.
"""
class Node:
__slots__ = ('key', 'left', 'right', 'count', 'color')
def __init__(self, key):
self.key = key
self.left = None
self.right = None
self.count = 0
self.color = Color.RED
def __str__(self):
if self.color == Color.RED:
key = '*{}'.format(self.key)
else:
key = '{}'.format(self.key)
return "{}[{}] ({}, {})".format(key, self.count,
self.left, self.right)
def __init__(self):
self.root = None
def put(self, key):
def _put(node):
if node is None:
node = self.Node(key)
if node.key > key:
node.left = _put(node.left)
elif node.key < key:
node.right = _put(node.right)
node = self._restore(node)
node.count = self._size(node.left) + self._size(node.right) + 1
return node
self.root = _put(self.root)
self.root.color = Color.BLACK
def _is_red(self, node):
if node is None:
return False
else:
return node.color == Color.RED
def _is_black(self, node):
if node is None:
return False
else:
return node.color == Color.BLACK
def _is_2node(self, node):
if node is None:
return False
elif self._is_red(node):
return False
else:
return (self._is_black(node) and
not self._is_red(node.left) and
not self._is_red(node.right))
def _is_34node(self, node):
if node is None:
return False
elif self._is_red(node):
return True
else:
return (self._is_black(node) and
self._is_red(node.left) and
not self._is_red(node.right))
def _rotate_left(self, node):
assert self._is_red(node.right)
x = node.right
node.right = x.left
x.left = node
x.color = node.color
node.color = Color.RED
node.count = self._size(node.left) + self._size(node.right) + 1
return x
def _rotate_right(self, node):
assert self._is_red(node.left)
x = node.left
node.left = x.right
x.right = node
x.color = node.color
node.color = Color.RED
node.count = self._size(node.left) + self._size(node.right) + 1
return x
def _flip_colors(self, node):
node.color = Color.flip(node.color)
node.left.color = Color.flip(node.left.color)
node.right.color = Color.flip(node.right.color)
return node
def __contains__(self, key):
def _contains(node):
if node is None:
return False
if node.key > key:
return _contains(node.left)
elif node.key < key:
return _contains(node.right)
else:
return True
return _contains(self.root)
def delete(self, key):
def _delete_from(node):
if node is None:
return None
else:
assert not self._is_2node(node)
if node.key > key:
node = self._convert_left(node)
# print('_convert_left', node)
node.left = _delete_from(node.left)
node = self._restore(node)
# print('_restore', node)
elif node.key < key:
node = self._convert_right(node)
# print('_convert_right', node)
node.right = _delete_from(node.right)
node = self._restore(node)
# print('_restore', node)
else:
node = _remove(node)
if node is not None:
node.count = self._size(node.right) + self._size(node.left) + 1
return node
def _remove(node):
if node.left is None:
return None
elif node.right is None:
if self._is_red(node.left):
node.left.color = Color.BLACK
return node.left
else:
node = self._convert_right(node)
if node.key == key:
x = self._find_min(node.right)
node.key = x.key
node.right = self._delete_min(node.right)
else:
# print(node)
node.right = _delete_from(node.right)
node = self._restore(node)
# print('min', x.key, 'node', node.key)
return node
# print('initial', self.root)
if not self._is_red(self.root.left):
self.root.color = Color.RED
self.root = _delete_from(self.root)
if self.root is not None:
self.root.color = Color.BLACK
# print('result', self.root)
assert self._is_balanced(self.root)
def delete_max(self):
if self.root is None:
raise ValueError('remove max on empty tree')
if self.root.left is None:
self.root = None
return
if not self._is_red(self.root.left):
self.root.color = Color.RED
self.root = self._delete_max(self.root)
self.root.color = Color.BLACK
assert self._is_balanced(self.root)
def _delete_max(self, node):
if node.right is None:
if self._is_red(node.left):
node.left.color = Color.BLACK
return node.left
else:
assert not self._is_2node(node)
node = self._convert_right(node)
node.right = self._delete_max(node.right)
node = self._restore(node)
return node
def _convert_right(self, node):
if self._is_2node(node.right):
assert not self._is_2node(node)
if self._is_2node(node.left):
self._flip_colors(node)
elif self._is_red(node.left) and self._is_2node(node.left.right):
node = self._rotate_right(node)
self._flip_colors(node.right)
elif self._is_red(node.left):
x = node.left.right
node.left.right = x.left
node.left, node.right, x.left, x.right = \
x.right, node.right.left, node.left, node.right
x.right.left = node
x.color = Color.BLACK
node.color = Color.RED
if self._is_red(x.left.right):
x.left.right.color = Color.BLACK
node.count = (self._size(node.left) +
self._size(node.right) + 1)
x.left.count = (self._size(x.left.left) +
self._size(x.left.right) + 1)
node = x
elif self._is_34node(node.left):
x = node.left
node.left, node.right, x.right = \
x.right, node.right.left, node.right
x.right.left = node
x.color = node.color
if self._is_red(x.left):
x.left.color = Color.BLACK
if self._is_black(x.right.left):
x.right.left.color = Color.RED
node.count = (self._size(node.left) +
self._size(node.right) + 1)
x.left.count = (self._size(x.left.left) +
self._size(x.left.right) + 1)
node = x
return node
def delete_min(self):
if self.root is None:
raise ValueError('remove min on empty tree')
if self.root.left is None:
self.root = None
return
if not self._is_red(self.root.left):
self.root.color = Color.RED
self.root = self._delete_min(self.root)
self.root.color = Color.BLACK
assert self._is_balanced(self.root)
def _delete_min(self, node):
if node.left is None:
return None
else:
assert not self._is_2node(node)
node = self._convert_left(node)
node.left = self._delete_min(node.left)
node = self._restore(node)
return node
def _convert_left(self, node):
if self._is_2node(node.left):
assert not self._is_2node(node)
if self._is_2node(node.right):
self._flip_colors(node)
elif self._is_34node(node.right):
x = node.right.left
node.right.left = x.right
x.left, x.right, node.right = node, node.right, x.left
x.color = node.color
node.color = Color.BLACK
node.left.color = Color.RED
x.right.count = (self._size(x.right.left) +
self._size(x.right.right) + 1)
node = x
return node
def _restore(self, node):
if self._is_red(node.right) and not self._is_red(node.left):
node = self._rotate_left(node)
if self._is_red(node.left) and self._is_red(node.left.left):
node = self._rotate_right(node)
if self._is_red(node.left) and self._is_red(node.right):
node = self._flip_colors(node)
node.count = self._size(node.left) + self._size(node.right) + 1
return node
def _is_balanced(self, node):
def depth(node):
if node is None:
return 0
left = depth(node.left)
right = depth(node.right)
if left != right:
raise Exception('unbalanced')
if self._is_black(node):
return 1 + left
else:
return left
if node is None:
return True
try:
left = depth(node.left)
right = depth(node.right)
return left == right
except Exception:
return False
@property
def size(self):
return self._size(self.root)
def _size(self, node):
if node is None:
return 0
else:
return node.count
@property
def max(self):
if self.root is None:
raise ValueError('max on empty tree')
return self._max(self.root)
def _max(self, node):
x = self._find_max(node)
return x.key
def _find_max(self, node):
if node.right is None:
return node
else:
return self._find_max(node.right)
@property
def min(self):
if self.root is None:
raise ValueError('min on empty tree')
return self._min(self.root)
def _min(self, node):
x = self._find_min(node)
return x.key
def _find_min(self, node):
if node.left is None:
return node
else:
return self._find_min(node.left)
def range(self, min_, max_):
def _range(node):
if node is None:
return
if node.key > max_:
yield from _range(node.left)
elif node.key < min_:
yield from _range(node.right)
else:
yield from _range(node.left)
yield node.key
yield from _range(node.right)
if min_ > max_:
return
yield from _range(self.root)
class BalancedBstSet:
def __init__(self):
self.bst = RedBlackBinarySearchTree()
def add(self, key):
self.bst.put(key)
def __contains__(self, key):
return key in self.bst
def delete(self, key):
self.bst.delete(key)
def range(self, a, b):
for k in self.bst.range(a, b):
yield k
@property
def count(self):
return self.bst.size
def __str__(self):
return str(self.bst.root)
def run():
q = int(input())
s = BalancedBstSet()
for _ in range(q):
command, *value = [int(x) for x in input().split()]
if command == 0:
s.add(value[0])
print(s.count)
elif command == 1:
if value[0] in s:
print(1)
else:
print(0)
elif command == 2:
s.delete(value[0])
elif command == 3:
for i in s.range(*value):
print(i)
else:
raise ValueError('invalid command')
if __name__ == '__main__':
run()
``` | instruction | 0 | 30,960 | 5 | 61,920 |
No | output | 1 | 30,960 | 5 | 61,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
from collections import Counter
t=int(input())
for _ in range(t):
x,y,z=list(map(int,input().split()))
if(x==y):
a=x
b=z
c=z
elif(y==z):
a,b,c=x,x,y
elif(x==z):
a,b,c=y,x,y
l=[x,y,z]
d=Counter(l)
if(len(d.keys())!=3 and d[max(l)]>=2):
print('YES')
print(a,b,c)
else:
print('NO')
``` | instruction | 0 | 31,163 | 5 | 62,326 |
Yes | output | 1 | 31,163 | 5 | 62,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
for _ in range(int(input())):
l=list(map(int,input().split()))
l.sort()
if l[0]<=l[1] and l[1]==l[2]:
print("YES")
print(l[0],l[0],l[2],end=" ")
print("")
else:
print("NO")
``` | instruction | 0 | 31,164 | 5 | 62,328 |
Yes | output | 1 | 31,164 | 5 | 62,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
import sys
import bisect as bi
import math
from collections import defaultdict as dd
input=sys.stdin.readline
##sys.setrecursionlimit(10**7)
def cin():
return map(int,sin().split())
def ain():
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
for _ in range(inin()):
a,b,c=cin()
if((a==b and c<=a)):
print("YES")
print(a,c,c)
elif((b==c and a<=c)):
print("YES")
print(a,a,b)
elif(c==a and b<=c):
print("YES")
print(b,a,b)
else:
print("NO")
#### s=input()
#### def minCut( s: str) -> int:
####
#### n=len(s)
#### dp=[[-1]*(n+2) for i in range(n+2)]
#### def pp(s,i,j,dp,mn):
#### if(i>=j or s[i:j+1]==s[i:j+1:-1]):return 0
#### if(dp[i][j]!=-1):return dp[i][j]
#### for k in range(i,j):
#### if(dp[i][k]!=-1):return dp[i][k]
#### else:
#### dp[i][k]=pp(s,i,k,dp,mn)
#### return dp[i][k]
#### if(dp[k+1][j]!=-1):return dp[k+1][j]
#### else:
#### dp[i][k]=pp(s,k+1,j,dp,mn)
#### return dp[i][k]
#### temp=dp[i][k] + dp[i][k] +1
#### if(temp<mn):mn=tmp
#### dp[0][n]=mn
#### return dp[0][n]
#### x=pp(s,0,n,dp,999999999999999)
#### for i in dp:print(i)
#### return x
####
#### print(minCut(s))
##dp=[[0 if (i==j) else -1 for j in range(n+1)] for i in range(n+1)]
##for i in range(1,n):
## for j in range(
##
##
##
##
##
##
##
####n=m=0
####s=''
####t=''
####dp=[]
####def solve(inds,indt,k,cont):
#### ans=-999999999999999
#### print(dp)
#### if(k<0):return 0
#### elif(inds>=n and indt>=m):return 0
#### elif(dp[inds][indt][k][cont]!=-1):return dp[inds][indt][k][cont]
#### else:
#### if(indt<m):ans=max(ans,solve(inds,indt+1,k,0))
#### if(inds<n):ans=max(ans,solve(inds+1,indt,k,0))
#### if(s[inds]==t[indt]):
#### ans=max(ans,solve(inds+1,indt+1,k-1,1)+1)
#### if(cont):ans=max(ans,solve(inds+1,indt+1,k,1)+1)
#### dp[inds][indt][k][cont]=ans
#### return ans
##
#### n,m,k=cin()
#### s=sin().strip()
#### t=sin().strip()
#### dp=[[[[-1]*2 for i in range(k)] for i in range(m+1)] for i in range(n+1)]
#### c=0
#### for i in dp:
#### for j in i:
#### for l in j:
#### c+=1
#### print(l,c)
#### print(solve(0,0,k,0))
``` | instruction | 0 | 31,165 | 5 | 62,330 |
Yes | output | 1 | 31,165 | 5 | 62,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
t = int(input())
for i in range(t):
x,y,z = map(int,input().split())
if (x == y or z == y or x == z) and sum([x,y,z]) == max([x,y,z])*2+min([x,y,z]):
print("YES")
print(max([x,y,z]),min([x,y,z]),min([x,y,z]))
else:
print("NO")
``` | instruction | 0 | 31,166 | 5 | 62,332 |
Yes | output | 1 | 31,166 | 5 | 62,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
case = int(input())
for i in range(0, case):
x, y, z = [int(x) for x in input().split()]
maxCount = 0
maxVal = max(x, y, z)
minVal = min(x, y, z)
for j in [x, y, z]:
if maxVal == j:
maxCount += 1
if (x == y or x == z or y == z) and maxCount >= 2:
print("YES")
print(maxVal, minVal, maxVal - minVal)
else:
print("NO")
``` | instruction | 0 | 31,167 | 5 | 62,334 |
No | output | 1 | 31,167 | 5 | 62,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
for k in range(int(input())):
x,y,z=map(int,input().split())
if(x!=y and y!=z and x!=z):
print('NO')
else:
if(2*min(x,y,z)+max(x,y,z)==x+y+z):
print('NO')
else:
if(x==y and y==z and x==z):
print('YES')
print(x,y,z)
else:
if(min(x,y,x)==1):
c=1
else:
c=min(x,y,x)-1
print('YES')
print(max(x,y,z),min(x,y,z),c)
``` | instruction | 0 | 31,168 | 5 | 62,336 |
No | output | 1 | 31,168 | 5 | 62,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
for i in range(int(input())):
a,b,c=map(int,input().split())
if(a<=b and b<c):
print('NO')
else:
print('YES')
print(a,a,c)
``` | instruction | 0 | 31,169 | 5 | 62,338 |
No | output | 1 | 31,169 | 5 | 62,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000
Submitted Solution:
```
def main():
for _ in range(int(input())):
x, y, z = map(int, input().split())
if max(x, y, z) == z:
print("YES")
print(x, x, z)
else:
print("NO")
main()
``` | instruction | 0 | 31,170 | 5 | 62,340 |
No | output | 1 | 31,170 | 5 | 62,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
import sys
import atexit
class Fastio:
def __init__(self):
self.ibuf = bytes()
self.obuf = bytearray()
self.pil = 0
self.pir = 0
self.buf = bytearray(20)
def load(self):
self.ibuf = self.ibuf[self.pil:]
self.ibuf += sys.stdin.buffer.read(131072)
self.pil = 0
self.pir = len(self.ibuf)
def flush(self):
sys.stdout.buffer.write(self.obuf)
self.obuf = bytearray()
def fastin(self):
if self.pir - self.pil < 32:
self.load()
minus = 0
x = 0
while self.ibuf[self.pil] < ord('-'):
self.pil += 1
if self.ibuf[self.pil] == ord('-'):
minus = 1
self.pil += 1
while self.ibuf[self.pil] >= ord('0'):
x = x * 10 + (self.ibuf[self.pil] & 15)
self.pil += 1
if minus:
x = -x
return x
def fastout(self, x, end=b'\n'):
i = 19
if x == 0:
self.buf[i] = 48
i -= 1
else:
if x < 0:
self.obuf.append(45)
x = -x
while x != 0:
x, r = x // 10, x % 10
self.buf[i] = 48 + r
i -= 1
self.obuf.extend(self.buf[i+1:])
self.obuf.extend(end)
if len(self.obuf) > 131072:
self.flush()
fastio = Fastio()
rd = fastio.fastin
wtn = fastio.fastout
atexit.register(fastio.flush)
N = rd()
a = [0] * N
for i in range(N):
a[i] = rd()
for i in range(N-1, 0, -1):
a[i] -= a[i-1]
p = 0
for i in range(1, N):
p += max(0, a[i])
wtn((a[0] + p + 1) // 2)
Q = rd()
for _ in range(Q):
l, r, x = rd(), rd(), rd()
l -= 1
if l != 0:
p -= max(a[l], 0)
a[l] += x
if l != 0:
p += max(a[l], 0)
if r != N:
p -= max(a[r], 0)
a[r] -= x
p += max(a[r], 0)
wtn((a[0] + p + 1) // 2)
``` | instruction | 0 | 31,179 | 5 | 62,358 |
Yes | output | 1 | 31,179 | 5 | 62,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
import sys
import atexit
class Fastio:
def __init__(self):
self.ibuf = bytes()
self.obuf = bytearray()
self.pil = 0
self.pir = 0
self.buf = bytearray(21)
self.buf[20] = 10
def load(self):
self.ibuf = self.ibuf[self.pil:]
self.ibuf += sys.stdin.buffer.read(131072)
self.pil = 0
self.pir = len(self.ibuf)
def flush(self):
sys.stdout.buffer.write(self.obuf)
self.obuf = bytearray()
def fastin(self):
if self.pir - self.pil < 32:
self.load()
minus = 0
x = 0
while self.ibuf[self.pil] < 45:
self.pil += 1
if self.ibuf[self.pil] == 45:
minus = 1
self.pil += 1
while self.ibuf[self.pil] >= 48:
x = x * 10 + (self.ibuf[self.pil] & 15)
self.pil += 1
if minus:
x = -x
return x
def fastoutln(self, x):
i = 19
if x == 0:
self.buf[i] = 48
i -= 1
else:
if x < 0:
self.obuf.append(45)
x = -x
while x != 0:
x, r = divmod(x, 10)
self.buf[i] = 48 + r
i -= 1
self.obuf.extend(self.buf[i+1:21])
if len(self.obuf) > 131072:
self.flush()
fastio = Fastio()
rd = fastio.fastin
wtn = fastio.fastoutln
atexit.register(fastio.flush)
N = rd()
a = [0] * N
for i in range(N):
a[i] = rd()
for i in range(N-1, 0, -1):
a[i] -= a[i-1]
p = 0
for i in range(1, N):
p += max(0, a[i])
wtn((a[0] + p + 1) // 2)
Q = rd()
for _ in range(Q):
l, r, x = rd(), rd(), rd()
l -= 1
if l != 0:
p -= max(a[l], 0)
a[l] += x
if l != 0:
p += max(a[l], 0)
if r != N:
p -= max(a[r], 0)
a[r] -= x
p += max(a[r], 0)
wtn((a[0] + p + 1) // 2)
``` | instruction | 0 | 31,180 | 5 | 62,360 |
Yes | output | 1 | 31,180 | 5 | 62,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
# Author : Pratyaydeep↯Ghanta
import sys
# Am I debugging, check if I'm using same variable name in two places
def main():
_T_=1 #inin()
for _t_ in range(_T_):
n=inin()
a=inar()
d=[0 for i in range(n-1)]
k=0
for i in range(1,n):
d[i-1]+=a[i]-a[i-1]
k+=max(0,a[i]-a[i-1])
x=(a[0]+k+1)//2
print(x)
q=inin()
for _ in range(q):
l,r,c=inar()
l-=1; r-=1
if l==0:
a[0]+=c
else:
if d[l-1]>=0:
k-=d[l-1]
d[l-1]+=c
if d[l-1]>=0:
k+=d[l-1]
if r==n-1:
pass
else:
if d[r]>=0:
k-=d[r]
d[r]-=c
if d[r]>=0:
k+=d[r]
x=(a[0]+k+1)//2
print(x)
# Template -> Pratyaydeep
inp=sys.stdin.buffer.readline
inar=lambda: list(map(int,inp().split()))
inin=lambda: int(inp())
inst=lambda: inp().decode().rstrip('\n\r')
if (0):
import threading
sys.setrecursionlimit(10**7)
threading.stack_size(10**8)
threading.Thread(target=main).start()
else:
main()
``` | instruction | 0 | 31,181 | 5 | 62,362 |
Yes | output | 1 | 31,181 | 5 | 62,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
from sys import stdin, stdout
# a: 2 -1 7 3
# b: -3 -3 5 5
# c: 5 2 2 -2
#
# if a[i] < a[i-1]:
# b[i] = b[i-1]
# c[i] = c[i-1] + a[i] - a[i-1]
# if a[i] > a[i-1]:
# c[i] = c[i-1]
# b[i] = b[i-1] + a[i] - a[i-1]
#
# K = SUM( MAX(b[i]-b[i-1], 0) ) -- 1 < i < n
# b[n-1] = b[0] + K
# c[0] = a[0] - b[0]
#
# MAX(b[n-1], c[0])
# => MAX(b[0] + K, a[0] - b[0])
# => MAX(x + K, a - x)
# => x + K = a - x -- optimal result
# => x = (a - K) / 2
def three_sequences(n, a_a, q_a):
res_a = []
dif_a = [0] * n
K = 0
for i in range(1, n):
K += max(0, a_a[i] - a_a[i-1])
dif_a[i] = a_a[i] - a_a[i-1]
res_a.append(calc(K, a_a[0]))
for q in q_a:
l, r, x = q
if l > 1:
K -= max(0, dif_a[l-1])
dif_a[l-1] += x
K += max(0, dif_a[l-1])
if r < n:
K -= max(0, dif_a[r])
dif_a[r] -= x
K += max(0, dif_a[r])
if l == 1:
a_a[0] += x
res_a.append(calc(K, a_a[0]))
return res_a
def calc(K, a0):
b0 = (a0 - K) // 2
bn = b0 + K
c0 = a_a[0] - b0
return max(bn, c0)
n = int(stdin.readline())
a_a = list(map(int, stdin.readline().split()))
q = int(stdin.readline())
q_a = []
for _ in range(q):
l, r, x = map(int, stdin.readline().split())
q_a.append([l, r, x])
res_a = three_sequences(n, a_a, q_a)
for res in res_a:
stdout.write(str(res) + '\n')
``` | instruction | 0 | 31,182 | 5 | 62,364 |
Yes | output | 1 | 31,182 | 5 | 62,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
q=int(input())
b=[0]*n
c=[0]*n
difn=0
diff=[0]*n
for i in range(1,n):
if a[i]-a[i-1]>0:
difn+=a[i]-a[i-1]
diff[i]=a[i]-a[i-1]
bo=(a[0]-difn)//2
ans=max(bo+difn,a[0]-bo)
print(ans)
td=difn
for i in range(q):
l,r,x=map(int,input().split())
if l!=1:
if a[l-1]-a[l-2]+x>0:
td+=a[l-1]-a[l-2]+x-diff[l-1]
diff[l-1]=diff[l-1]+x
else:
td=td-diff[l-1]
diff[l-1]=0
elif l==1:
a[0]+=x
if r!=n:
if a[r]-a[r-1]-x>0:
td+=a[r]-a[r-1]-x-diff[r]
diff[r]=diff[r]-x
else:
td=td-diff[r]
diff[r]=0
bo=(a[0]-td)//2
ans=max(bo+td,a[0]-bo)
print(ans)
``` | instruction | 0 | 31,183 | 5 | 62,366 |
No | output | 1 | 31,183 | 5 | 62,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
# Fast IO (be careful about bytestring, not on interactive)
# import os,io
# input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
a = list(map(int,input().split()))
diffList = []
firstElem = a[0]
diffPositiveSum = 0
for i in range(n-1):
diffList.append(a[i+1] - a[i])
if diffList[-1] >= 0:
diffPositiveSum += diffList[-1]
print(firstElem + diffPositiveSum)
for _ in range(int(input())):
l,r,x = map(int,input().split())
if l == 1:
firstElem += x
else:
diffPositiveSum -= max(diffList[l - 2],0)
diffList[l-2] += x
diffPositiveSum += max(diffList[l - 2],0)
if r != n:
diffPositiveSum -= max(diffList[r - 1],0)
diffList[r - 1] -= x
diffPositiveSum += max(diffList[r - 1],0)
print(firstElem + diffPositiveSum)
``` | instruction | 0 | 31,184 | 5 | 62,368 |
No | output | 1 | 31,184 | 5 | 62,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
import math
n = int(input())
a = list(map(int,input().split()))
k = 0
for i in range(1,n):
k += max(0,a[i]-a[i-1])
q = int(input())
x = (a[0] + k)//2
print(max(x,a[0]+k-x))
D = 0
while (q):
l,r,d = list(map(int,input().split()))
l -= 1
r -= 1
if (l > 0):
k += max(0,a[l]+d-a[l-1])
k -= max(0,a[l]-a[l-1])
if (r < n-1):
k += max(0,a[r+1]-a[r]-d)
k -= max(0,a[r+1]-a[r])
if (l == 0):
D += d
x = (a[0]+k+D)//2
print(max(x,a[0]+k+D-x))
q -= 1
``` | instruction | 0 | 31,185 | 5 | 62,370 |
No | output | 1 | 31,185 | 5 | 62,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a_1, a_2, …, a_n.
You have to construct two sequences of integers b and c with length n that satisfy:
* for every i (1≤ i≤ n) b_i+c_i=a_i
* b is non-decreasing, which means that for every 1<i≤ n, b_i≥ b_{i-1} must hold
* c is non-increasing, which means that for every 1<i≤ n, c_i≤ c_{i-1} must hold
You have to minimize max(b_i,c_i). In other words, you have to minimize the maximum number in sequences b and c.
Also there will be q changes, the i-th change is described by three integers l,r,x. You should add x to a_l,a_{l+1}, …, a_r.
You have to find the minimum possible value of max(b_i,c_i) for the initial sequence and for sequence after each change.
Input
The first line contains an integer n (1≤ n≤ 10^5).
The secound line contains n integers a_1,a_2,…,a_n (1≤ i≤ n, -10^9≤ a_i≤ 10^9).
The third line contains an integer q (1≤ q≤ 10^5).
Each of the next q lines contains three integers l,r,x (1≤ l≤ r≤ n,-10^9≤ x≤ 10^9), desribing the next change.
Output
Print q+1 lines.
On the i-th (1 ≤ i ≤ q+1) line, print the answer to the problem for the sequence after i-1 changes.
Examples
Input
4
2 -1 7 3
2
2 4 -3
3 4 2
Output
5
5
6
Input
6
-9 -10 -9 -6 -5 4
3
2 6 -9
1 2 -10
4 6 -3
Output
3
3
3
1
Input
1
0
2
1 1 -1
1 1 -1
Output
0
0
-1
Note
In the first test:
* The initial sequence a = (2, -1, 7, 3). Two sequences b=(-3,-3,5,5),c=(5,2,2,-2) is a possible choice.
* After the first change a = (2, -4, 4, 0). Two sequences b=(-3,-3,5,5),c=(5,-1,-1,-5) is a possible choice.
* After the second change a = (2, -4, 6, 2). Two sequences b=(-4,-4,6,6),c=(6,0,0,-4) is a possible choice.
Submitted Solution:
```
import sys
class Fastio:
def __init__(self):
self.ibuf = bytearray()
self.obuf = bytearray()
self.pil = 0
self.pir = 0
self.sz = 1 << 17
def load(self):
self.ibuf = self.ibuf[self.pil:]
self.ibuf.extend(sys.stdin.buffer.read(self.sz))
self.pil = 0
self.pir = len(self.ibuf)
def flush(self):
sys.stdout.buffer.write(self.obuf)
self.obuf = bytearray()
def fastin(self):
if self.pir - self.pil < 32:
self.load()
minus = 0
x = 0
while self.ibuf[self.pil] < ord('-'):
self.pil += 1
if self.ibuf[self.pil] == ord('-'):
minus = 1
self.pil += 1
while self.ibuf[self.pil] >= ord('0'):
x = x * 10 + self.ibuf[self.pil] & 15
self.pil += 1
if minus:
x = -x
return x
'''
def __del__(self):
self.flush()
def fastout(self, x):
if x < 0:
obuf.join(b'-')
x = -x
'''
fastio = Fastio()
rd = fastio.fastin
# wtn = fastio.fastout
N = rd()
a = [0] * N
for i in range(N):
a[i] = rd()
for i in range(N-1, 0, -1):
a[i] -= a[i-1]
p = 0
for i in range(1, N):
p += max(0, a[i])
print((a[0] + p + 1) // 2)
Q = rd()
for _ in range(Q):
l, r, x = rd(), rd(), rd()
l -= 1
if l != 0:
p -= max(a[l], 0)
a[l] += x
if l != 0:
p += max(a[l], 0)
if r != N:
p -= max(a[r], 0)
a[r] -= x
p += max(a[r], 0)
print((a[0] + p + 1) // 2)
``` | instruction | 0 | 31,186 | 5 | 62,372 |
No | output | 1 | 31,186 | 5 | 62,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
def solve(X,):
ops = []
def plus(x, y):
ops.append(str(x) + " + " + str(y))
return x + y
def xor(x, y):
ops.append(str(x) + " ^ " + str(y))
return x ^ y
# nums[i] has their rightmost bit at pos i
nums = [X]
for i in range(X.bit_length()):
nums.append(plus(nums[-1], nums[-1]))
# Clear lower bits, all remaining bits of curr except for pos 0 should be at "high" positions
curr = X
for i in range(1, X.bit_length() + 1):
if (1 << i) & curr:
curr = xor(curr, nums[i])
assert curr & ((1 << X.bit_length()) - 1) == 1
# Create a number ending with '01'
if X % 4 == 1:
Y = X
else:
assert X % 4 == 3
Y = xor(X, plus(X, X))
# Shift Y until overlap with original Y on only the lowest bit
Y2 = Y
for i in range(Y.bit_length() - 1):
Y2 = plus(Y2, Y2)
assert Y & Y2 == (1 << (Y.bit_length() - 1))
# The single overlap bit will overflow (for plus) or zero out (for xor)
# Xoring these two will result in a single remaining bit
pow2 = xor(plus(Y, Y2), xor(Y, Y2))
assert bin(pow2).count("1") == 1
# Shift that single bit up to clear all remaining high bits of curr
while curr != 1:
if pow2 & curr:
curr = xor(pow2, curr)
pow2 = plus(pow2, pow2)
assert len(ops) <= 100000
return str(len(ops)) + "\n" + "\n".join(ops)
if False:
for x in range(3, 100000, 2):
solve(x)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
(X,) = [int(x) for x in input().split()]
ans = solve(X,)
print(ans)
``` | instruction | 0 | 31,195 | 5 | 62,390 |
Yes | output | 1 | 31,195 | 5 | 62,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
from sys import stdin
output = []
def queue(stri):
output.append(stri)
n = int(stdin.readline())
digits = len(bin(n))-2
cpy = n
for _ in range(digits-1):
queue(f'{cpy} + {cpy}')
cpy *= 2
m = n ^ cpy
queue(f'{n} ^ {cpy}')
p = n*m-1
while p % n != 0:
p -= m
def foo(number, target):
# gets target from number using additions only
current = number
while current * 2 <= target:
queue(f'{current} + {current}')
current *= 2
total = current
while total < target:
if total + current <= target:
queue(f'{total} + {current}')
total += current
current //= 2
foo(m, p+1)
foo(n, p)
if p % 2 == 0:
queue(f'{p+1} ^ {p}')
else:
queue(f'{p} + {n}')
queue(f'{p+1} + {n}')
queue(f'{p+n+1} ^ {p+n}')
print(len(output))
for s in output:
print(s)
``` | instruction | 0 | 31,196 | 5 | 62,392 |
Yes | output | 1 | 31,196 | 5 | 62,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
n = int(input());out = []
if n>>1 & 1:out.append(str(n) + " + " + str(n)) ;out.append(str(2*n) + " ^ " + str(n)) ; n = ((2*n)^n)
m = n.bit_length() - 1;a = n;b = n
for i in range(m):out.append(str(b) + " + " + (str(b)));b *= 2
out.append(str(a) + " + " + str(b));c = a + b;out.append(str(a) + " ^ " + str(b));d = a^b;out.append(str(c) + " ^ " + str(d));e = c^d;a = n;b = e
while True:
if a==1 or b==1:break
if a>b:a,b = b,a
pre_a = a;A = a.bit_length() - 1;B = b.bit_length() - 1
for i in range(B-A):out.append(str(a) + " + " + str(a));a *= 2
out.append(str(a) + " ^ " + str(b));b = a^b;a = pre_a
print(len(out))
for s in out:print(s)
``` | instruction | 0 | 31,197 | 5 | 62,394 |
Yes | output | 1 | 31,197 | 5 | 62,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
import sys
input = sys.stdin.readline
from math import gcd
x=int(input())
import random
#x=(random.randint(1,999999//2))*2+1
XLIST=[x]
XSET=set()
def Ext_Euc(a,b,axy=(1,0),bxy=(0,1)): # axy=a*1+b*0,bxy=a*0+b*1なので,a,bに対応する係数の初期値は(1,0),(0,1)
# print(a,b,axy,bxy)
q,r=divmod(a,b)
if r==0:
return bxy,b # a*bxy[0]+b*bxy[1]=b
rxy=(axy[0]-bxy[0]*q,axy[1]-bxy[1]*q) # rに対応する係数を求める.
return Ext_Euc(b,r,bxy,rxy)
ANS=[]
USED=set()
flag=0
while True:
if flag:
break
x=random.choice(XLIST)
if x in USED:
continue
USED.add(x)
y=x
for i in range(20):
if gcd(x,y^x)==1:
k,l=Ext_Euc(x,y^x)[0]
tt,tu=abs(k)*x,(y^x)*abs(l)
if tt^tu==1:
flag=1
break
else:
ANS.append(str(x)+" ^ "+str(y))
XLIST.append(x^y)
else:
ANS.append(str(y)+" + "+str(y))
if y<=999999:
XLIST.append(x)
y=2*y
z=y^x
ANS.append(str(y)+" ^ "+str(x))
#print(x,z)
A=Ext_Euc(x,z)
#print(A)
k,l=abs(A[0][0]),abs(A[0][1])
XLIST=[x]
ANS.append(str(x)+" + "+str(x))
x=x+x
while x<25*10**17:
XLIST.append(x)
ANS.append(str(x)+" + "+str(x))
x=2*x
ZLIST=[z]
ANS.append(str(z)+" + "+str(z))
z=z+z
while z<25*10**17:
ZLIST.append(z)
ANS.append(str(z)+" + "+str(z))
z=2*z
XLIST2=[0]*60
for i in range(60):
if k & (1<<i) != 0:
XLIST2[i]=1
flag=0
for i in range(60):
if XLIST2[i]==1:
if flag==0:
NOW=XLIST[i]
flag=1
else:
ANS.append(str(NOW)+" + "+str(XLIST[i]))
NOW+=XLIST[i]
NOW0=NOW
ZLIST2=[0]*60
for i in range(60):
if l & (1<<i) != 0:
ZLIST2[i]=1
flag=0
for i in range(60):
if ZLIST2[i]==1:
if flag==0:
NOW=ZLIST[i]
flag=1
else:
ANS.append(str(NOW)+" + "+str(ZLIST[i]))
NOW+=ZLIST[i]
NOW1=NOW
ANS.append(str(NOW0)+" ^ "+str(NOW1))
#print(NOW1,NOW0)
print(len(ANS))
for ans in ANS:
print(ans)
``` | instruction | 0 | 31,198 | 5 | 62,396 |
Yes | output | 1 | 31,198 | 5 | 62,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
import sys
input = sys.stdin.readline
from math import gcd
x=int(input())
#import random
#x=(random.randint(1,999999//2))*2+1
#SET={x}
def Ext_Euc(a,b,axy=(1,0),bxy=(0,1)): # axy=a*1+b*0,bxy=a*0+b*1なので,a,bに対応する係数の初期値は(1,0),(0,1)
# print(a,b,axy,bxy)
q,r=divmod(a,b)
if r==0:
return bxy,b # a*bxy[0]+b*bxy[1]=b
rxy=(axy[0]-bxy[0]*q,axy[1]-bxy[1]*q) # rに対応する係数を求める.
return Ext_Euc(b,r,bxy,rxy)
ANS=[]
y=x
while True:
if gcd(x,y^x)==1:
break
else:
ANS.append(str(y)+" + "+str(y))
y=2*y
z=y^x
ANS.append(str(y)+" ^ "+str(x))
#print(x,z)
A=Ext_Euc(x,z)
#print(A)
k,l=abs(A[0][0]),abs(A[0][1])
XLIST=[x]
ANS.append(str(x)+" + "+str(x))
x=x+x
while x<25*10**17:
XLIST.append(x)
ANS.append(str(x)+" + "+str(x))
x=2*x
ZLIST=[z]
ANS.append(str(z)+" + "+str(z))
z=z+z
while z<25*10**17:
ZLIST.append(z)
ANS.append(str(z)+" + "+str(z))
z=2*z
XLIST2=[0]*60
for i in range(60):
if k & (1<<i) != 0:
XLIST2[i]=1
flag=0
for i in range(60):
if XLIST2[i]==1:
if flag==0:
NOW=XLIST[i]
flag=1
else:
NOW+=XLIST[i]
ANS.append(str(NOW)+" + "+str(XLIST[i]))
NOW0=NOW
ZLIST2=[0]*60
for i in range(60):
if l & (1<<i) != 0:
ZLIST2[i]=1
flag=0
for i in range(60):
if ZLIST2[i]==1:
if flag==0:
NOW=ZLIST[i]
flag=1
else:
NOW+=ZLIST[i]
ANS.append(str(NOW)+" + "+str(ZLIST[i]))
NOW1=NOW
ANS.append(str(NOW0)+" ^ "+str(NOW1))
#print(NOW1,NOW0)
print(len(ANS))
for ans in ANS:
print(ans)
``` | instruction | 0 | 31,199 | 5 | 62,398 |
No | output | 1 | 31,199 | 5 | 62,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
import sys
x = int(sys.stdin.readline())
ans = []
c = 0
while x > 1:
c = c + 1
z = x
s = bin(x)[2:]
i = 1
ans.append(str(z) + " + " + str(z))
z = z * 2
while x + z != x ^ z:
i = i + 1
ans.append(str(z) + " + " + str(z))
z = z * 2
if c % 2 == 1:
ans.append(str(x) + " + " + str(z))
y = x + z
else:
ans.append(str(x) + " ^ " + str(z))
y = x ^ z
ans.append(str(x) + " + " + str(y))
y = x + y
while len(bin(z)) < len(bin(y)):
ans.append(str(z) + " + " + str(z))
z = z * 2
while y > x:
while len(bin(z)) > len(bin(y)):
z = z // 2
ans.append(str(z) + " ^ " + str(y))
y = y ^ z
x = y
print(len(ans))
for i in ans:
print(i)
``` | instruction | 0 | 31,200 | 5 | 62,400 |
No | output | 1 | 31,200 | 5 | 62,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
import sys
input = sys.stdin.readline
from math import gcd
x=int(input())
import random
#x=(random.randint(1,999999//2))*2+1
XLIST=[x]
XSET=set()
def Ext_Euc(a,b,axy=(1,0),bxy=(0,1)): # axy=a*1+b*0,bxy=a*0+b*1なので,a,bに対応する係数の初期値は(1,0),(0,1)
# print(a,b,axy,bxy)
q,r=divmod(a,b)
if r==0:
return bxy,b # a*bxy[0]+b*bxy[1]=b
rxy=(axy[0]-bxy[0]*q,axy[1]-bxy[1]*q) # rに対応する係数を求める.
return Ext_Euc(b,r,bxy,rxy)
ANS=[]
USED=set()
flag=0
while True:
if flag:
break
x=random.choice(XLIST)
if x in USED:
continue
USED.add(x)
y=x
for i in range(20):
if gcd(x,y^x)==1:
k,l=Ext_Euc(x,y^x)[0]
tt,tu=abs(k)*x,(y^x)*abs(l)
if tt^tu==1:
flag=1
break
else:
ANS.append(str(x)+" ^ "+str(y))
XLIST.append(x^y)
else:
ANS.append(str(y)+" + "+str(y))
if y<=999999:
XLIST.append(x)
y=2*y
z=y^x
ANS.append(str(y)+" ^ "+str(x))
#print(x,z)
A=Ext_Euc(x,z)
#print(A)
k,l=abs(A[0][0]),abs(A[0][1])
XLIST=[x]
ANS.append(str(x)+" + "+str(x))
x=x+x
while x<25*10**17:
XLIST.append(x)
ANS.append(str(x)+" + "+str(x))
x=2*x
ZLIST=[z]
ANS.append(str(z)+" + "+str(z))
z=z+z
while z<25*10**17:
ZLIST.append(z)
ANS.append(str(z)+" + "+str(z))
z=2*z
XLIST2=[0]*60
for i in range(60):
if k & (1<<i) != 0:
XLIST2[i]=1
flag=0
for i in range(60):
if XLIST2[i]==1:
if flag==0:
NOW=XLIST[i]
flag=1
else:
NOW+=XLIST[i]
ANS.append(str(NOW)+" + "+str(XLIST[i]))
NOW0=NOW
ZLIST2=[0]*60
for i in range(60):
if l & (1<<i) != 0:
ZLIST2[i]=1
flag=0
for i in range(60):
if ZLIST2[i]==1:
if flag==0:
NOW=ZLIST[i]
flag=1
else:
NOW+=ZLIST[i]
ANS.append(str(NOW)+" + "+str(ZLIST[i]))
NOW1=NOW
ANS.append(str(NOW0)+" ^ "+str(NOW1))
#print(NOW1,NOW0)
print(len(ANS))
for ans in ANS:
print(ans)
``` | instruction | 0 | 31,201 | 5 | 62,402 |
No | output | 1 | 31,201 | 5 | 62,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.
You may write new numbers on the blackboard with the following two operations.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard.
* You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard.
Perform a sequence of operations such that at the end the number 1 is on the blackboard.
Input
The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).
Output
Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation.
* The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard.
* The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard.
The operation symbol (+ or ^) must be separated from a, b by a whitespace.
You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.
Examples
Input
3
Output
5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9
Input
123
Output
10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
print("200")
cnt = 200
x = int(input())
if x % 4 == 3:
print(str(x) + " + " + str(x))
print(str(2 * x) + " + " + str(x))
cnt -= 2
x *= 3
xCpy = x
bigX = x
while xCpy != 1:
print(str(bigX) + " + " + str(bigX))
cnt -= 1
xCpy //= 2
bigX *= 2
print(str(bigX) + " + " + str(x))
ans1 = bigX + x
print(str(bigX) + " ^ " + str(x))
ans2 = bigX ^ x
print(str(ans1) + " ^ " + str(ans2))
ans = ans1 ^ ans2
cnt -= 3
curX = x
while ans != 1:
curX = x
while curX != 0:
print(str(curX) + " + " + str(curX))
cnt -= 1
curX *= 2
if curX >= ans:
print(str(ans) + " ^ " + str(curX))
cnt -= 1
curX ^= ans
ans //= 2
if x >= ans:
print(str(x) + " ^ " + str(ans))
cnt -= 1
x ^= ans
while cnt > 0:
cnt -= 1
print(str(x) + " + " + str(x))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 31,202 | 5 | 62,404 |
No | output | 1 | 31,202 | 5 | 62,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day shooshuns found a sequence of n integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes k-th in the current sequence and add the same number to the end of the sequence;
2. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input
The first line contains two space-separated integers n and k (1 ≤ k ≤ n ≤ 105).
The second line contains n space-separated integers: a1, a2, ..., an (1 ≤ ai ≤ 105) — the sequence that the shooshuns found.
Output
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Examples
Input
3 2
3 1 1
Output
1
Input
3 1
3 1 1
Output
-1
Note
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
Submitted Solution:
```
import sys,os,io,time,copy
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import math
n,k=map(int,input().split())
k-=1
arr=list(map(int,input().split()))
arr1=arr[:k]
arr2=arr[k:]
flag=0
for i in range(len(arr2)-1):
if arr2[i]!=arr2[i+1]:
flag=1
if flag==0:
pos=-1
req=arr2[0]
for i in range(len(arr1)-1,-1,-1):
if arr1[i]!=req:
pos=i
break
print(pos+1)
else:
print(-1)
``` | instruction | 0 | 31,265 | 5 | 62,530 |
Yes | output | 1 | 31,265 | 5 | 62,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day shooshuns found a sequence of n integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes k-th in the current sequence and add the same number to the end of the sequence;
2. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input
The first line contains two space-separated integers n and k (1 ≤ k ≤ n ≤ 105).
The second line contains n space-separated integers: a1, a2, ..., an (1 ≤ ai ≤ 105) — the sequence that the shooshuns found.
Output
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Examples
Input
3 2
3 1 1
Output
1
Input
3 1
3 1 1
Output
-1
Note
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
Submitted Solution:
```
n,k=[int(x) for x in input().split(' ')]
arr=[int(x) for x in input().split(' ')]
if arr.count(arr[0])==n:
print(0)
else:
temp=arr[k-1:]
if all(x== temp[0] for x in temp):
ans=k-1
i=k-2
while(i>0 and arr[i]==temp[0]):
ans-=1
i-=1
print(ans)
else:
print(-1)
``` | instruction | 0 | 31,267 | 5 | 62,534 |
Yes | output | 1 | 31,267 | 5 | 62,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day shooshuns found a sequence of n integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes k-th in the current sequence and add the same number to the end of the sequence;
2. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input
The first line contains two space-separated integers n and k (1 ≤ k ≤ n ≤ 105).
The second line contains n space-separated integers: a1, a2, ..., an (1 ≤ ai ≤ 105) — the sequence that the shooshuns found.
Output
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Examples
Input
3 2
3 1 1
Output
1
Input
3 1
3 1 1
Output
-1
Note
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
Submitted Solution:
```
n,k= map(int,input().split())
arr= list(map(int,input().split()))
last= arr[-1]
diff_ind=0
for i in range(n):
if arr[i]!=last:
diff_ind=i+1
if diff_ind>=k or len(set(arr[k-1:n]))!=1:
print(-1)
else:
print(diff_ind)
``` | instruction | 0 | 31,268 | 5 | 62,536 |
Yes | output | 1 | 31,268 | 5 | 62,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day shooshuns found a sequence of n integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes k-th in the current sequence and add the same number to the end of the sequence;
2. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input
The first line contains two space-separated integers n and k (1 ≤ k ≤ n ≤ 105).
The second line contains n space-separated integers: a1, a2, ..., an (1 ≤ ai ≤ 105) — the sequence that the shooshuns found.
Output
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Examples
Input
3 2
3 1 1
Output
1
Input
3 1
3 1 1
Output
-1
Note
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
if k == 1 and n != 1:
print(-1)
else:
cnt = 0
while len(set(a)) != 1:
a += [a[k-1]]
a = a[1:]
cnt += 1
print(cnt)
``` | instruction | 0 | 31,270 | 5 | 62,540 |
No | output | 1 | 31,270 | 5 | 62,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day shooshuns found a sequence of n integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes k-th in the current sequence and add the same number to the end of the sequence;
2. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input
The first line contains two space-separated integers n and k (1 ≤ k ≤ n ≤ 105).
The second line contains n space-separated integers: a1, a2, ..., an (1 ≤ ai ≤ 105) — the sequence that the shooshuns found.
Output
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Examples
Input
3 2
3 1 1
Output
1
Input
3 1
3 1 1
Output
-1
Note
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
Submitted Solution:
```
n,k = [int(x) for x in input().split()]
a=[int(x) for x in input().split()]
e = a[k-1]
for i in range(k,len(a)):
if a[i]!=e:
print(-1)
break
else:
print(k-1)
``` | instruction | 0 | 31,271 | 5 | 62,542 |
No | output | 1 | 31,271 | 5 | 62,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day shooshuns found a sequence of n integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes k-th in the current sequence and add the same number to the end of the sequence;
2. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input
The first line contains two space-separated integers n and k (1 ≤ k ≤ n ≤ 105).
The second line contains n space-separated integers: a1, a2, ..., an (1 ≤ ai ≤ 105) — the sequence that the shooshuns found.
Output
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Examples
Input
3 2
3 1 1
Output
1
Input
3 1
3 1 1
Output
-1
Note
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
Submitted Solution:
```
n,k=map(int,input().split())
l=list(map(int,input().split()))
if l[k-1:].count(l[k-1])!=len(l[k-1:]):
print(-1)
elif l.count(l[k-1])==len(l):
print(1)
else:
print(len(l)-k)
``` | instruction | 0 | 31,272 | 5 | 62,544 |
No | output | 1 | 31,272 | 5 | 62,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and j such that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input
The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).
Output
Output a single integer — the required minimal sum.
Examples
Input
2
1 2
Output
2
Input
3
2 4 6
Output
6
Input
2
12 18
Output
12
Input
5
45 12 27 30 18
Output
15
Note
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.
Submitted Solution:
```
from math import gcd
n = int(input())
a = list(map(int, input().split()))
ans = a[0]
for i in range(1, n):
ans = gcd(ans, a[i])
print(n * ans)
``` | instruction | 0 | 31,320 | 5 | 62,640 |
Yes | output | 1 | 31,320 | 5 | 62,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and j such that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input
The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).
Output
Output a single integer — the required minimal sum.
Examples
Input
2
1 2
Output
2
Input
3
2 4 6
Output
6
Input
2
12 18
Output
12
Input
5
45 12 27 30 18
Output
15
Note
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.
Submitted Solution:
```
n = int(input())
x = list(map(int, input().split()))
x.sort()
while True:
if x[-1] == min(x) or x[-1] == 0:
if x[-1] == 0:
print(x[0]*n)
else:
print(x[-1]*n)
exit()
else:
x.sort()
x[-1] = x[-1] - x[-2]
``` | instruction | 0 | 31,322 | 5 | 62,644 |
No | output | 1 | 31,322 | 5 | 62,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and j such that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input
The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).
Output
Output a single integer — the required minimal sum.
Examples
Input
2
1 2
Output
2
Input
3
2 4 6
Output
6
Input
2
12 18
Output
12
Input
5
45 12 27 30 18
Output
15
Note
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.
Submitted Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=1
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
"""
###########################---START-CODING---##############################
num=1
#num=int(z())
for _ in range ( num ):
n=int(z())
arr=szzz()
ans=float('inf')
if len(set(arr))==1:
print(sum(arr))
continue
for i in range(1,n):
ans=min(ans,arr[i]-arr[i-1])
#print(arr[i],ans)
print(ans*n)
``` | instruction | 0 | 31,324 | 5 | 62,648 |
No | output | 1 | 31,324 | 5 | 62,649 |
Provide a correct Python 3 solution for this coding contest problem.
One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...
Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
Output
Reproduce the output of the reference solution, including the bug.
Examples
Input
8-7+6-5+4-3+2-1-0
Output
4
Input
2+2
Output
-46
Input
112-37
Output
375 | instruction | 0 | 31,521 | 5 | 63,042 |
"Correct Solution:
```
s = input()
a = 1
b = 0
res = 0
for c in s:
if (c == '+'):
res = res + a * b
a = 1
b = 0
elif (c == '-'):
res = res + a * b
a = -1
b = 0
b = b * 10 + ord(c) - 48
res = res + a * b
print(res)
``` | output | 1 | 31,521 | 5 | 63,043 |
Provide a correct Python 3 solution for this coding contest problem.
One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...
Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
Output
Reproduce the output of the reference solution, including the bug.
Examples
Input
8-7+6-5+4-3+2-1-0
Output
4
Input
2+2
Output
-46
Input
112-37
Output
375 | instruction | 0 | 31,523 | 5 | 63,046 |
"Correct Solution:
```
def mp():
return map(int, input().split())
s = input() + '+'
r = 0
a = 0
z = '+'
for i in s:
#print(r, a, z)
if i in '+-':
if z == '+':
r += a
else:
r -= a
z = i
a = 0
a *= 10
a += ord(i) - ord('0')
print(r)
``` | output | 1 | 31,523 | 5 | 63,047 |
Provide a correct Python 3 solution for this coding contest problem.
One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...
Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
Output
Reproduce the output of the reference solution, including the bug.
Examples
Input
8-7+6-5+4-3+2-1-0
Output
4
Input
2+2
Output
-46
Input
112-37
Output
375 | instruction | 0 | 31,524 | 5 | 63,048 |
"Correct Solution:
```
s = map(lambda c: ord(c) - 48, input())
r, t, f = 0, 0, 1
for x in s:
if x < 0:
r += f * t
f = -4 - x
t = 0
t = t * 10 + x
print(r + f * t)
``` | output | 1 | 31,524 | 5 | 63,049 |
Provide a correct Python 3 solution for this coding contest problem.
One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...
Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
Output
Reproduce the output of the reference solution, including the bug.
Examples
Input
8-7+6-5+4-3+2-1-0
Output
4
Input
2+2
Output
-46
Input
112-37
Output
375 | instruction | 0 | 31,525 | 5 | 63,050 |
"Correct Solution:
```
import re
s=input()
print(eval(s)+eval(re.sub("-|\+|\d",lambda m:["0",["+3","-5"][m[0]=="+"]][m[0]in"+-"],s)))
``` | output | 1 | 31,525 | 5 | 63,051 |
Provide a correct Python 3 solution for this coding contest problem.
One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...
Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
Output
Reproduce the output of the reference solution, including the bug.
Examples
Input
8-7+6-5+4-3+2-1-0
Output
4
Input
2+2
Output
-46
Input
112-37
Output
375 | instruction | 0 | 31,526 | 5 | 63,052 |
"Correct Solution:
```
a = input()
q, s = [0], 0
z = []
for i in range(len(a)):
y = a[i]
if (y=="+"):
z.append(1)
q.append(0)
s+=1
elif (y=="-"):
z.append(-1)
q.append(0)
s+=1
q[s] = 10*q[s] + (ord(y)-ord("0"))
otv = q[0]
for i in range(1, s+1):
otv += z[i-1]*q[i]
print(otv)
``` | output | 1 | 31,526 | 5 | 63,053 |
Provide a correct Python 3 solution for this coding contest problem.
One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...
Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
Output
Reproduce the output of the reference solution, including the bug.
Examples
Input
8-7+6-5+4-3+2-1-0
Output
4
Input
2+2
Output
-46
Input
112-37
Output
375 | instruction | 0 | 31,527 | 5 | 63,054 |
"Correct Solution:
```
import re;s=input();print(eval(s)+eval(re.sub("-|\+|\d",lambda m:"+-035"[2-"+--".count(m[0])::3],s)))
``` | output | 1 | 31,527 | 5 | 63,055 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,606 | 5 | 63,212 |
"Correct Solution:
```
import sys
from itertools import accumulate, combinations
read = sys.stdin.read
N, K, *a = map(int, read().split())
a = accumulate([0] + a)
seq = []
for i, j in combinations(a, 2):
seq.append(j - i)
m = (10 ** 9 * 1000).bit_length()
answer = 0
bit = 1 << (m - 1)
for _ in range(m):
tmp = bit + answer
if len([i for i in seq if i & tmp == tmp]) >= K:
answer = tmp
bit >>= 1
print(answer)
``` | output | 1 | 31,606 | 5 | 63,213 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,607 | 5 | 63,214 |
"Correct Solution:
```
def cumsum(l):
c = [0] * len(l)
for i in range(len(l)):
c[i] = c[i-1] + l[i]
return c
N, K = map(int, input().split())
a = [int(i) for i in input().split()]
c = [0] + cumsum(a)
s = [bin(c[j+i] - c[j]).replace('0b','').zfill(40) for i in range(1, N+1) for j in range(N-i+1)]
ans = [0] * 40
for i in range(40):
t = [j for j in s if j[i] == '1']
if len(t) >= K:
ans[i] = 1
s = t
print(sum([ans[i] * 2**(39-i) for i in range(40)]))
``` | output | 1 | 31,607 | 5 | 63,215 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,608 | 5 | 63,216 |
"Correct Solution:
```
import copy
N,K = map(int,input().split())
A = map(int,input().split())
A_sum = [0]
for a in A:
A_sum.append(A_sum[-1]+a)
score = []
for r in range(1,N+1):
for l in range(0,r):
score.append(A_sum[r]-A_sum[l])
ans = 0
for i in range(40,-1,-1):
count = 0
score_a = []
for j in score:
if (j & (1<<i)):
count += 1
score_a.append(j)
if count >= K:
ans += (1<<i)
score = copy.deepcopy(score_a)
print(ans)
``` | output | 1 | 31,608 | 5 | 63,217 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,609 | 5 | 63,218 |
"Correct Solution:
```
#設定
import sys
input = sys.stdin.buffer.readline
#ライブラリインポート
from collections import defaultdict
con = 10 ** 9 + 7
#入力受け取り
def getlist():
return list(map(int, input().split()))
#処理内容
def main():
N, K = getlist()
A = getlist()
B = [0]
for i in range(N):
B.append(B[-1] + A[i])
# print(B)
S = []
for i in range(N + 1):
for j in range(i + 1, N + 1):
S.append(B[j] - B[i])
# print(S)
z = 0
for i in range(41):
z += 2 ** (40 - i)
cnt = 0
for j in S:
if z & j == z:
cnt += 1
if cnt < K:
z -= 2 ** (40 - i)
print(z)
if __name__ == '__main__':
main()
``` | output | 1 | 31,609 | 5 | 63,219 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,610 | 5 | 63,220 |
"Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
c = [0] * (n+1)
for i in range(n):
c[i+1] = c[i] + a[i]
s = []
for i in range(n):
for j in range(i+1, n+1):
s.append(c[j]-c[i])
s.sort()
L = 40
b = [set() for _ in range(L)]
u = set(range(len(s)))
ans = 0
for i in range(L):
for j, x in enumerate(s):
if x >> i & 1:
b[i].add(j)
for i in range(L)[::-1]:
if len(u & b[i]) >= k:
u &= b[i]
ans += 2**i
print(ans)
``` | output | 1 | 31,610 | 5 | 63,221 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,611 | 5 | 63,222 |
"Correct Solution:
```
N, K = map(int,input().split())
a = list(map(int,input().split()))
S = [0]*N
S[0] = a[0]
for k in range(1,N):
S[k] = S[k-1] + a[k]
S = [0] + S
T = []
for k in range(N):
for l in range(k+1,N+1):
T.append(S[l]-S[k])
ans = 0
for d in range(41,-1,-1):
c = 0
for e in T:
if (ans+2**d)&e == ans+2**d:
c += 1
if c >= K:
ans += 2**d
print(ans)
``` | output | 1 | 31,611 | 5 | 63,223 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,612 | 5 | 63,224 |
"Correct Solution:
```
N,K = map(int,input().split())
A_sum = [0]
[A_sum.append(A_sum[-1]+int(a)) for a in input().split()]
score = []
for r in range(1,N+1):
for l in range(0,r):
score.append(A_sum[r]-A_sum[l])
ans = 0
for i in range(40,-1,-1):
count = 0
score_a = []
for j in score:
if (j & (1<<i)):
count += 1
score_a.append(j)
if count >= K:
ans += (1<<i)
score = [i for i in score_a]
print(ans)
``` | output | 1 | 31,612 | 5 | 63,225 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N. He is interested in properties of the sequence a.
For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)
Find the maximum possible value for him.
Constraints
* 2 \leq N \leq 1000
* 1 \leq a_i \leq 10^9
* 1 \leq K \leq N(N+1)/2
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 ... a_N
Output
Print the answer.
Examples
Input
4 2
2 5 2 5
Output
12
Input
8 4
9 1 8 2 7 5 6 4
Output
32 | instruction | 0 | 31,613 | 5 | 63,226 |
"Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
l=[]
for i in range(n):
s=0
for j in range(i,n):
s+=a[j]
l.append(s)
t=0
for i in range(40,-1,-1):
t+=2**i
c=0
for j in l:
if t==j&t:
c+=1
if c<k:
t-=2**i
print(t)
``` | output | 1 | 31,613 | 5 | 63,227 |
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