message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0. | instruction | 0 | 50,255 | 0 | 100,510 |
Tags: brute force, greedy, implementation, math
Correct Solution:
```
[n, x, y] = [int(x) for x in input().split()]
A = input()
if len(A) == 1:
if A[0] == "1":
print("0")
else:
print(y)
else:
c = 0
if A[0] == "0":
c = 1
for i in range(1, len(A)):
if A[i] == "0" and A[i-1] == "1":
c += 1
m = y*c
for i in range(1, c):
temp = x*i + y*(c-i)
if temp < m:
m = temp
print(m)
``` | output | 1 | 50,255 | 0 | 100,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0. | instruction | 0 | 50,256 | 0 | 100,512 |
Tags: brute force, greedy, implementation, math
Correct Solution:
```
n,x,y=map(int,input().strip().split())
count=0
r=input()
if r[0]=="0":
count=count+1
for i in range(1,n):
if r[i]=="0" and r[i-1]!="0":
count=count+1
if count==0:
print (0)
exit(0)
else:
if y<x:
print (y*count)
else:
print (x*(count-1)+y)
``` | output | 1 | 50,256 | 0 | 100,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
n,x,y=map(int,input().split())
s=input()
sl=[]
for i in s:
sl+=[int(i)]
sl.append(1)
count=0
for i in range(n):
if sl[i]==0 and sl[i+1]==1 :
count+=1
if count!=0:
print(y+min(x,y)*(count-1))
else:
print("0")
``` | instruction | 0 | 50,257 | 0 | 100,514 |
Yes | output | 1 | 50,257 | 0 | 100,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
def f(reverse_cost, inverse_cost, bit_string):
prev_value = True
zero_section_count = 0
for v in bit_string:
cur_value = bool(v == "1")
if prev_value != cur_value:
if not cur_value:
zero_section_count += 1
prev_value = cur_value
if zero_section_count:
return min(reverse_cost, inverse_cost) * (zero_section_count - 1) + inverse_cost
else:
return 0
n, x, y = [int(i) for i in input().split()]
s = input()
print(f(x, y, s))
``` | instruction | 0 | 50,258 | 0 | 100,516 |
Yes | output | 1 | 50,258 | 0 | 100,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
N,x,y = input().split()
x=int(x)
y=int(y)
array=input()
the_list=array.split('1')
length=len(the_list)
for k in the_list:
if k=='':
length-=1
if length==0:
print(0)
elif x<y:
print((length-1)*x+y)
else:
print((length)*y)
``` | instruction | 0 | 50,259 | 0 | 100,518 |
Yes | output | 1 | 50,259 | 0 | 100,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
n, x, y = map(int, input().split(" "))
s = input()
isolated = 0
ones = 0
last = "1"
for digit in s:
if digit == "1" and last == "0":
isolated += 1
if digit == "0":
ones += 1
last = digit
if last == "0":
isolated += 1
if ones == 0:
print(0)
else:
print(min(isolated * y, (isolated - 1) * x + y))
``` | instruction | 0 | 50,260 | 0 | 100,520 |
Yes | output | 1 | 50,260 | 0 | 100,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
''' CODED WITH LOVE BY SATYAM KUMAR '''
from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27) # new thread will get stack of such size
fac_warmup = False
printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
class Operation:
def __init__(self, name, function, function_on_equal, neutral_value=0):
self.name = name
self.f = function
self.f_on_equal = function_on_equal
def add_multiple(x, count):
return x * count
def min_multiple(x, count):
return x
def max_multiple(x, count):
return x
sum_operation = Operation("sum", sum, add_multiple, 0)
min_operation = Operation("min", min, min_multiple, 1e9)
max_operation = Operation("max", max, max_multiple, -1e9)
class SegmentTree:
def __init__(self,
array,
operations=[sum_operation, min_operation, max_operation]):
self.array = array
if type(operations) != list:
raise TypeError("operations must be a list")
self.operations = {}
for op in operations:
self.operations[op.name] = op
self.root = SegmentTreeNode(0, len(array) - 1, self)
def query(self, start, end, operation_name):
if self.operations.get(operation_name) == None:
raise Exception("This operation is not available")
return self.root._query(start, end, self.operations[operation_name])
def summary(self):
return self.root.values
def update(self, position, value):
self.root._update(position, value)
def update_range(self, start, end, value):
self.root._update_range(start, end, value)
def __repr__(self):
return self.root.__repr__()
class SegmentTreeNode:
def __init__(self, start, end, segment_tree):
self.range = (start, end)
self.parent_tree = segment_tree
self.range_value = None
self.values = {}
self.left = None
self.right = None
if start == end:
self._sync()
return
self.left = SegmentTreeNode(start, start + (end - start) // 2,
segment_tree)
self.right = SegmentTreeNode(start + (end - start) // 2 + 1, end,
segment_tree)
self._sync()
def _query(self, start, end, operation):
if end < self.range[0] or start > self.range[1]:
return None
if start <= self.range[0] and self.range[1] <= end:
return self.values[operation.name]
self._push()
left_res = self.left._query(start, end,
operation) if self.left else None
right_res = self.right._query(start, end,
operation) if self.right else None
if left_res is None:
return right_res
if right_res is None:
return left_res
return operation.f([left_res, right_res])
def _update(self, position, value):
if position < self.range[0] or position > self.range[1]:
return
if position == self.range[0] and self.range[1] == position:
self.parent_tree.array[position] = value
self._sync()
return
self._push()
self.left._update(position, value)
self.right._update(position, value)
self._sync()
def _update_range(self, start, end, value):
if end < self.range[0] or start > self.range[1]:
return
if start <= self.range[0] and self.range[1] <= end:
self.range_value = value
self._sync()
return
self._push()
self.left._update_range(start, end, value)
self.right._update_range(start, end, value)
self._sync()
def _sync(self):
if self.range[0] == self.range[1]:
for op in self.parent_tree.operations.values():
current_value = self.parent_tree.array[self.range[0]]
if self.range_value is not None:
current_value = self.range_value
self.values[op.name] = op.f([current_value])
else:
for op in self.parent_tree.operations.values():
result = op.f(
[self.left.values[op.name], self.right.values[op.name]])
if self.range_value is not None:
bound_length = self.range[1] - self.range[0] + 1
result = op.f_on_equal(self.range_value, bound_length)
self.values[op.name] = result
def _push(self):
if self.range_value is None:
return
if self.left:
self.left.range_value = self.range_value
self.right.range_value = self.range_value
self.left._sync()
self.right._sync()
self.range_value = None
def __repr__(self):
ans = "({}, {}): {}\n".format(self.range[0], self.range[1],
self.values)
if self.left:
ans += self.left.__repr__()
if self.right:
ans += self.right.__repr__()
return ans
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def primeFactors(n): #n**0.5 complex
factors = dict()
for i in range(2,math.ceil(math.sqrt(n))+1):
while n % i== 0:
if i in factors:
factors[i]+=1
else: factors[i]=1
n = n // i
if n>2:
factors[n]=1
return (factors)
def binary(n,digits = 20):
b = bin(n)[2:]
b = '0'*(20-len(b))+b
return b
def isprime(n):
"""Returns True if n is prime."""
if n < 4:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
factorial_modP = []
def warm_up_fac(MOD):
global factorial_modP,fac_warmup
if fac_warmup: return
factorial_modP= [1 for _ in range(fac_warmup_size+1)]
for i in range(2,fac_warmup_size):
factorial_modP[i]= (factorial_modP[i-1]*i) % MOD
fac_warmup = True
def InverseEuler(n,MOD):
return pow(n,MOD-2,MOD)
def nCr(n, r, MOD):
global fac_warmup,factorial_modP
if not fac_warmup:
warm_up_fac(MOD)
fac_warmup = True
return (factorial_modP[n]*((pow(factorial_modP[r], MOD-2, MOD) * pow(factorial_modP[n-r], MOD-2, MOD)) % MOD)) % MOD
def test_print(*args):
if testingMode:
print(args)
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
import heapq,itertools
pq = [] # list of entries arranged in a heap
entry_finder = {} # mapping of tasks to entries
REMOVED = '<removed-task>'
def add_task(task, priority=0):
'Add a new task or update the priority of an existing task'
if task in entry_finder:
remove_task(task)
count = next(counter)
entry = [priority, count, task]
entry_finder[task] = entry
heapq.heappush(pq, entry)
def remove_task(task):
'Mark an existing task as REMOVED. Raise KeyError if not found.'
entry = entry_finder.pop(task)
entry[-1] = REMOVED
def pop_task():
'Remove and return the lowest priority task. Raise KeyError if empty.'
while pq:
priority, count, task = heapq.heappop(pq)
if task is not REMOVED:
del entry_finder[task]
return task
raise KeyError('pop from an empty priority queue')
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
# -------------------------------------------------------------- MAIN PROGRAM
TestCases = False
testingMode = False
fac_warmup_size = 10**5+100
optimiseForReccursion = True #Can not be used clubbed with TestCases
def binary_serach(i,li):
#print("Search for ",i)
fn = lambda x: li[x]-x//i
x = -1
b = len(li)
while b>=1:
#print(b,x)
while b+x<len(li) and fn(b+x)>0:
x+=b
b=b//2
return x
def bs(front,end,i):
global li,n
mid = (front+end)//2
print(mid,flush = True)
stdout.flush()
x = get_int()*li[i%n]
if x==0:
return
elif x>0:
bs(mid+1,end,i+1)
else:
bs(front,mid-1,i+1)
def main():
#global li,n
n,x,y = get_tuple()
li = list(input())
bblocks = 0
last_zero = False
for i in li:
if i=='0' and not last_zero:
bblocks+=1
last_zero = True
elif i=='1' and last_zero:
last_zero=False
#print(bblocks)
costs = [bblocks*y]
merges = 0
while bblocks!=0:
if bblocks%2==0 or y<x:
merges += bblocks//2
bblocks = bblocks//2
elif bblocks%2!=0:
merges += (bblocks+1)//2
bblocks = (bblocks+1)//2
costs.append(bblocks*y+merges*x)
if bblocks==1:
break
#print(costs)
print(min(costs))
# --------------------------------------------------------------------- END=
if TestCases:
for _ in range(get_int()):
cProfile.run('main()') if testingMode else main()
else: (cProfile.run('main()') if testingMode else main()) if not optimiseForReccursion else threading.Thread(target=main).start()
``` | instruction | 0 | 50,261 | 0 | 100,522 |
No | output | 1 | 50,261 | 0 | 100,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
n, x, y = map(int, input().split())
s = input()
c0 = s.split('1')
c1 = s.split('0')
cnt0 = 0
for c in c0:
if c != '':
cnt0 += 1
cnt1 = 0
for c in c1:
if c != '':
cnt1 += 1
print(min(cnt1 * x + y, cnt0 * y))
``` | instruction | 0 | 50,262 | 0 | 100,524 |
No | output | 1 | 50,262 | 0 | 100,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
n, x, y = map(int, input().split())
nOnes = input().count('1')
result = y * n
for i in range(nOnes + 1):
result = min(result, x * i + min(y + (nOnes - i) * y, y * (n - nOnes + i)))
print(result)
``` | instruction | 0 | 50,263 | 0 | 100,526 |
No | output | 1 | 50,263 | 0 | 100,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a string a_1, a_2, ..., a_n, consisting of zeros and ones.
Let's call a sequence of consecutive elements a_i, a_{i + 1}, …, a_j (1≤ i≤ j≤ n) a substring of string a.
You can apply the following operations any number of times:
* Choose some substring of string a (for example, you can choose entire string) and reverse it, paying x coins for it (for example, «0101101» → «0111001»);
* Choose some substring of string a (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, «0101101» → «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input
The first line of input contains integers n, x and y (1 ≤ n ≤ 300 000, 0 ≤ x, y ≤ 10^9) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string a of length n, consisting of zeros and ones.
Output
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.
Examples
Input
5 1 10
01000
Output
11
Input
5 10 1
01000
Output
2
Input
7 2 3
1111111
Output
0
Note
In the first sample, at first you need to reverse substring [1 ... 2], and then you need to invert substring [2 ... 5].
Then the string was changed as follows:
«01000» → «10000» → «11111».
The total cost of operations is 1 + 10 = 11.
In the second sample, at first you need to invert substring [1 ... 1], and then you need to invert substring [3 ... 5].
Then the string was changed as follows:
«01000» → «11000» → «11111».
The overall cost is 1 + 1 = 2.
In the third example, string already consists only of ones, so the answer is 0.
Submitted Solution:
```
def removeUtil(string, last_removed):
if len(string) == 0 or len(string) == 1:
return string
if string[0] == string[1]:
last_removed = ord(string[0])
while len(string) > 1 and string[0] == string[1]:
string = string[1:]
string = string[1:]
return removeUtil(string, last_removed)
rem_str = removeUtil(string[1:], last_removed)
if len(rem_str) != 0 and rem_str[0] == string[0]:
last_removed = ord(string[0])
return (rem_str[1:])
if len(rem_str) == 0 and last_removed == ord(string[0]):
return rem_str
return ([string[0]] + rem_str)
def remove(string):
last_removed = 0
return toString(removeUtil(toList(string), last_removed))
def toList(string):
x = []
for i in string:
x.append(i)
return x
def toString(x):
return ''.join(x)
[n, x, y] = [int(x) for x in input().split()]
A1 = input()
A = remove(A1)
if len(A) == 1:
if A[0] == "1":
print("0")
else:
print(y)
else:
c = 0
if A[0] == "0":
c = 1
for i in range(1, len(A)):
if A[i] == "0" and A[i-1] == "1":
c += 1
print(c)
m = y*c
for i in range(1, c):
temp = x*i + y*(c-i)
if temp < m:
m = temp
print(m)
``` | instruction | 0 | 50,264 | 0 | 100,528 |
No | output | 1 | 50,264 | 0 | 100,529 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin | instruction | 0 | 50,349 | 0 | 100,698 |
"Correct Solution:
```
L = int(input())
Ss = input()
Ts = input()
if Ss+Ts > Ts+Ss:
Ss, Ts = Ts, Ss
lenS, lenT = len(Ss), len(Ts)
for numS in reversed(range(L//lenS+1)):
numT = (L-lenS*numS) / lenT
if int(numT) == numT:
ans = Ss*numS + Ts*int(numT)
print(ans)
break
``` | output | 1 | 50,349 | 0 | 100,699 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin | instruction | 0 | 50,350 | 0 | 100,700 |
"Correct Solution:
```
l = int(input())
a = input()
b = input()
if len(a) > len(b):
k = b
b = a
a = k
mina = None
maxa = None
"""
f = [None] * (l + 1)
f[0] = ''
#print(f)
for i in range(l):
tmpa = None
tmpb = None
if i + 1 >= len(a) and f[i + 1 - len(a)] is not None:
tmpa = a + f[i + 1 - len(a)]
if i + 1 >= len(b) and f[i + 1 - len(b)] is not None:
tmpb = b + f[i + 1 - len(b)]
if tmpa is None and tmpb is None:
continue
if tmpa is None and tmpb is not None:
f[i+1] = tmpb
continue
if tmpb is None and tmpa is not None:
f[i+1] = tmpa
continue
if tmpa < tmpb:
f[i+1] = tmpa
else:
f[i+1] = tmpb
print(f[l])
def less(a, b):
if a < b:
if not b.startswith(a):
return True
else:
c = b
while c.startswith(a):
c = c[len(a):]
if a < c:
return True
else:
return False
"""
for i in range(l+ 1) :
if i * len(a) > l:
break
if (l - i * len(a)) % len(b) == 0:
if mina is None:
mina = i
maxb = (l - i * len(a)) // len(b)
maxa = i
minb = (l - i * len(a)) // len(b)
ans1 = a * maxa + b * minb
ans2 = b * maxb + a * mina
if ans1 < ans2:
print(ans1)
else:
print(ans2)
``` | output | 1 | 50,350 | 0 | 100,701 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin | instruction | 0 | 50,351 | 0 | 100,702 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from fractions import gcd
L = int(readline())
S = readline().rstrip().decode('utf-8')
T = readline().rstrip().decode('utf-8')
ST = S + T
TS = T + S
if ST > TS:
S,T = T,S
# Sを手前にした方が得
LS = len(S)
LT = len(T)
n = 0
for x in range(L+1):
if (L - LS * x) % LT == 0:
n = x
break
n1, m1 = n, (L - LS * n) // LT
d = gcd(LS,LT)
n2, m2 = n1, m1
dn, dm = LT//d, LS//d
while m2 >= dm:
m2 -= dm
n2 += dn
x = S * n1 + T * m1
y = S * n2 + T * m2
answer = min(x,y)
print(answer)
``` | output | 1 | 50,351 | 0 | 100,703 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin | instruction | 0 | 50,352 | 0 | 100,704 |
"Correct Solution:
```
def main():
L = int(input())
S, T = (input() for _ in [0] * 2)
ans = []
apnd = ans.append
a = S
b = T
i = 0
for _ in [0] * 2:
while 1:
x = L - (len(b) * i)
div_, mod_ = divmod(x, len(a))
if not mod_:
apnd(a * div_ + b * i)
break
i += 1
i = 0
a, b = b, a
print(min(ans))
main()
``` | output | 1 | 50,352 | 0 | 100,705 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin | instruction | 0 | 50,353 | 0 | 100,706 |
"Correct Solution:
```
def solve(l, s, t):
queue = [(1, 0), (0, 1)]
def gcd(a, b):
r = a % b
if r:
d = a // b
sb = queue.pop()
sa = queue.pop()
queue.append(sb)
queue.append(tuple(x - d * y for x, y in zip(sa, sb)))
return gcd(b, r)
else:
return b
ls, lt = len(s), len(t)
if ls > lt:
s, t, ls, lt = t, s, lt, ls
if (s * (lt // ls + 1) * 2)[:lt * 2] > t * 2:
s, t, ls, lt = t, s, lt, ls
g = gcd(ls, lt)
l //= g
ls //= g
lt //= g
a, b = queue[-1]
a *= l
b *= l
k = b // ls
b -= ls * k
a += lt * k
return s * a + t * b
l = int(input())
s = input()
t = input()
print(solve(l, s, t))
``` | output | 1 | 50,353 | 0 | 100,707 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin | instruction | 0 | 50,354 | 0 | 100,708 |
"Correct Solution:
```
n=int(input())
a=sorted([input()for i in range(2)])
s,t=len(a[0]),len(a[1])
c=[]
for i in range(n):
if (n-i*t)%s==0:
c.append(a[0]*((n-i*t)//s)+a[1]*i)
break
for i in range(n):
if (n-i*s)%t==0:
c.append(a[1]*((n-i*s)//t)+a[0]*i)
break
print(sorted(c)[0])
``` | output | 1 | 50,354 | 0 | 100,709 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin | instruction | 0 | 50,355 | 0 | 100,710 |
"Correct Solution:
```
L = int(input())
s = input()
t = input()
if not s+t <= t+s:
s, t = t, s
ls = len(s); lt = len(t)
for i in range(L // ls, -1, -1):
if (L - ls*i) % lt == 0:
ans = s * i + t * ((L - ls*i) // lt)
break
print(ans)
``` | output | 1 | 50,355 | 0 | 100,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin
Submitted Solution:
```
l = int(input())
a = input()
b = input()
if len(a) > len(b):
k = b
b = a
a = k
mina = None
maxa = None
f = [None] * (l + 1)
f[0] = ''
#print(f)
for i in range(l):
tmpa = None
tmpb = None
if i + 1 >= len(a) and f[i + 1 - len(a)] is not None:
tmpa = a + f[i + 1 - len(a)]
if i + 1 >= len(b) and f[i + 1 - len(b)] is not None:
tmpb = b + f[i + 1 - len(b)]
if tmpa is None and tmpb is None:
continue
if tmpa is None and tmpb is not None:
f[i+1] = tmpb
continue
if tmpb is None and tmpa is not None:
f[i+1] = tmpa
continue
if tmpa < tmpb:
f[i+1] = tmpa
else:
f[i+1] = tmpb
print(f[l])
``` | instruction | 0 | 50,357 | 0 | 100,714 |
No | output | 1 | 50,357 | 0 | 100,715 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,360 | 0 | 100,720 |
"Correct Solution:
```
S=input()
L=len(S)//2-1
while S[:L]!=S[L:2*L]:
L-=1
print(2*L)
``` | output | 1 | 50,360 | 0 | 100,721 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,361 | 0 | 100,722 |
"Correct Solution:
```
S = input()[:-1]
while S[:len(S)//2] != S[len(S)//2:]:
S = S[:-1]
print(len(S))
``` | output | 1 | 50,361 | 0 | 100,723 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,362 | 0 | 100,724 |
"Correct Solution:
```
import re
S = input()
m = re.search(r'\A([a-z]+)\1', S[:len(S) - 1])
print(len(m.group(0)))
``` | output | 1 | 50,362 | 0 | 100,725 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,363 | 0 | 100,726 |
"Correct Solution:
```
s = input()[:-2]
ls = len(s) // 2
while s[:ls] != s[ls:]:
ls -= 1
s = s[:-2]
print(ls*2)
``` | output | 1 | 50,363 | 0 | 100,727 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,364 | 0 | 100,728 |
"Correct Solution:
```
s = list(input())
s = s[:-1]
while s[:len(s)//2] != s[len(s)//2 :]:
s = s[:-1]
# print(s)
print(len(s))
``` | output | 1 | 50,364 | 0 | 100,729 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,365 | 0 | 100,730 |
"Correct Solution:
```
S = input()
S = S[:-2]
while S[:len(S)//2] != S[len(S)//2:]:
S = S[:-2]
print(len(S))
``` | output | 1 | 50,365 | 0 | 100,731 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,366 | 0 | 100,732 |
"Correct Solution:
```
s = input()
for k in range(len(s)//2-1, 0, -1):
if s[:k] == s[k:2*k]: break
print(k*2)
``` | output | 1 | 50,366 | 0 | 100,733 |
Provide a correct Python 3 solution for this coding contest problem.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14 | instruction | 0 | 50,367 | 0 | 100,734 |
"Correct Solution:
```
s=input()[:-2]
while s[:len(s)//2] != s[len(s)//2:]:
s=s[:len(s)-2]
print(len(s))
``` | output | 1 | 50,367 | 0 | 100,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
S=input()
L=len(S)
for i in range(L//2-1,-1,-1):
if S[0:i-1]==S[i:2*i-1]:
break
print(i*2)
``` | instruction | 0 | 50,368 | 0 | 100,736 |
Yes | output | 1 | 50,368 | 0 | 100,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
S = input()
n = len(S) - 2
while S:
if S[0:n//2] == S[n//2:n]:break
n = n-2
print(n)
``` | instruction | 0 | 50,369 | 0 | 100,738 |
Yes | output | 1 | 50,369 | 0 | 100,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
n = input()
n = n[:-1]
while n[:len(n)//2] != n[len(n)//2:]:
n = n[:-1]
print(len(n))
``` | instruction | 0 | 50,370 | 0 | 100,740 |
Yes | output | 1 | 50,370 | 0 | 100,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
s = input()
while 1:
s = s[:-2]
i = len(s)//2
if s[:i] == s[i:]:
break
print(len(s))
``` | instruction | 0 | 50,371 | 0 | 100,742 |
Yes | output | 1 | 50,371 | 0 | 100,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
s = input()
n = len(s)
def iseven(str, n):
for i in range(0, n//2):
if str[i] != str[i+n//2]:
return False
else:
return True
for i in range(2, n, 2):
if iseven(s, n-i):
print(n-i)
``` | instruction | 0 | 50,372 | 0 | 100,744 |
No | output | 1 | 50,372 | 0 | 100,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
import collections
def even_str(cnt):
return len([key for key in cnt if cnt[key] % 2 != 0]) == 0
s = list(input())
c = collections.Counter(s)
while len(s) != 0:
last_char = s.pop(-1)
c[last_char] -= 1
if(even_str(c)):
print(len(s))
exit()
``` | instruction | 0 | 50,373 | 0 | 100,746 |
No | output | 1 | 50,373 | 0 | 100,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
S = input
while len(S)!=0:
S.pop()
if len(S)%2==0 and S[0:len(S)/2]==S[len(S)/2:]:
print(len(S))
exit()
``` | instruction | 0 | 50,374 | 0 | 100,748 |
No | output | 1 | 50,374 | 0 | 100,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
Constraints
* 2 \leq |S| \leq 200
* S is an even string consisting of lowercase English letters.
* There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the length of the longest even string that can be obtained.
Examples
Input
abaababaab
Output
6
Input
xxxx
Output
2
Input
abcabcabcabc
Output
6
Input
akasakaakasakasakaakas
Output
14
Submitted Solution:
```
s = input()
for i in range(len(s)//2):
s = s[:-2]
l = len(s)
if s[:l] == s[l:]:
print( l * 2)
break
``` | instruction | 0 | 50,375 | 0 | 100,750 |
No | output | 1 | 50,375 | 0 | 100,751 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,392 | 0 | 100,784 |
"Correct Solution:
```
s = list(input())
K = int(input())
for i in range(len(s)):
if s[i] == "a":
pass
elif 123 - ord(s[i]) <= K:
K -= (123 - ord(s[i]))
s[i] = 'a'
K = K % 26
while K > 0:
K -= 1
if s[-1] == "z":
s[-1] = "a"
else:
s[-1] = chr(ord(s[-1]) + 1)
print("".join(s))
``` | output | 1 | 50,392 | 0 | 100,785 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,393 | 0 | 100,786 |
"Correct Solution:
```
s=input()
K=int(input())
S=[c for c in s]
def diff(c):
return ord('z')-ord(c)
for i in range(len(s)):
d=diff(S[i])
if d<min(25,K):
K-=d+1
S[i]='a'
k=K%26
S[-1]=chr(ord(S[-1])+k) if k<=diff(S[-1]) else chr(ord(S[-1])+k-26)
print(''.join(S))
``` | output | 1 | 50,393 | 0 | 100,787 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,394 | 0 | 100,788 |
"Correct Solution:
```
s=input()
k=int(input())
a=[]
for t in s[:-1]:
c=123-ord(t)
if k<c or t=='a':
a.append(t)
else:
k-=c
a.append('a')
a.append(chr((ord(s[-1])-97+k)%26+97))
print(''.join(a))
``` | output | 1 | 50,394 | 0 | 100,789 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,395 | 0 | 100,790 |
"Correct Solution:
```
s = input()
K = int(input())
s = list(s)
# print(diff)
for i in range(len(s)):
if s[i] != 'a':
dist = ord('z') - ord(s[i]) + 1
if dist <= K:
K -= dist
s[i] = 'a'
# print(s)
temp = K % 26
s[-1] = chr(ord(s[-1]) + temp)
# print(s)
print(''.join(s))
``` | output | 1 | 50,395 | 0 | 100,791 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,396 | 0 | 100,792 |
"Correct Solution:
```
l = list(input())
k = int(input())
for i in range(len(l)):
if l[i] == "a":
continue
if 123 - ord(l[i]) <= k:
k -= 123 - ord(l[i])
l[i] = "a"
k %= 26
l[-1] = chr((ord(l[-1])-97+k)%26+97)
print("".join(l))
``` | output | 1 | 50,396 | 0 | 100,793 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,397 | 0 | 100,794 |
"Correct Solution:
```
s = input().strip()
k = int(input())
ans = ''
for _s in s[:-1]:
if _s == 'a':
ans += 'a'
elif 26 - ord(_s) + ord('a') <= k:
ans += 'a'
k -= 26 - ord(_s) + ord('a')
else:
ans += _s
ans += chr(ord('a') + (ord(s[-1]) - ord('a') + k) % 26)
print(ans)
``` | output | 1 | 50,397 | 0 | 100,795 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,398 | 0 | 100,796 |
"Correct Solution:
```
s=input()
l=len(s)
k=int(input())
z=''
for i in range(l):
t=123-ord(s[i])
if s[i]=='a':
z+='a'
elif t<=k:
z+='a'
k-=t
else:
z+=s[i]
v=ord(z[l-1])-97+k
z=z[:l-1]+chr(v%26+97)
print(z)
``` | output | 1 | 50,398 | 0 | 100,797 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times.
* Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`.
For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`.
Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
Constraints
* 1≤|s|≤10^5
* All letters in s are lowercase English letters.
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
s
K
Output
Print the lexicographically smallest string after performing exactly K operations on s.
Examples
Input
xyz
4
Output
aya
Input
a
25
Output
z
Input
codefestival
100
Output
aaaafeaaivap | instruction | 0 | 50,399 | 0 | 100,798 |
"Correct Solution:
```
s = list(input())
k = int(input())
for i in range(len(s)):
if ord("z")-ord(s[i])+1<=k:
k -= (ord("z")-ord(s[i])+1)%26
s[i] = "a"
if k: s[-1] = chr((ord(s[-1])+k-ord("a"))%26+ord("a"))
print("".join(s))
``` | output | 1 | 50,399 | 0 | 100,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,508 | 0 | 101,016 |
Tags: greedy, strings
Correct Solution:
```
import os, sys, bisect
from collections import defaultdict, Counter, deque;
from functools import lru_cache #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
input = lambda: sys.stdin.readline().strip()
imap = lambda: map(int,input().split()); ilist = lambda: list(imap())
#------------------------------------------------------------------
#sys.setrecursionlimit(10**6)
mod = int(1e9+7)
n,k = imap()
s = input().strip()
t = input().strip()
f = 0
for i in range(n):
if s[i]!=t[i]:
f = 1
s = s[i:]
t = t[i:]
break
if not f:
print(n)
exit()
res = n-len(s)
n = len(s)
for i in range(n):
cur = 1
for j in range(i+1):
if t[j]=="b":
cur+=1<<(i-j)
if s[j]=="b":
cur-=1<<(i-j)
if cur<k:
res+=cur
else:
res+=(n-i)*k
break
print(res)
``` | output | 1 | 50,508 | 0 | 101,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,509 | 0 | 101,018 |
Tags: greedy, strings
Correct Solution:
```
n, k = map(int, input().split())
a = input()
b = input()
res = 0
ans = 0
for i in range(0, n):
res = min(res * 2 + (b[i] == 'b') - (a[i] == 'b'), k)
ans += min(res + 1, k)
print(ans)
``` | output | 1 | 50,509 | 0 | 101,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,510 | 0 | 101,020 |
Tags: greedy, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
# ------------------------------
def c(ca, cb):
return ord(cb)-ord(ca)
def main():
n, k = RL()
s = input()
t = input()
res = 0
tag = n
for i in range(n):
if s[i]==t[i]: res+=1
else: tag = i; break
num = 2
for j in range(tag, n):
if s[j]=='b': num-=1
if t[j]=='a': num-=1
if num>=k: res+=k*(n-j); break
res+=num
num*=2
print(res)
if __name__ == "__main__":
main()
``` | output | 1 | 50,510 | 0 | 101,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,511 | 0 | 101,022 |
Tags: greedy, strings
Correct Solution:
```
n,k=map(int,input().split())
a=input().strip()
b=input().strip()
add=1
count=0
for i in range(n):
if a[i]==b[i]:
add=add*2-1
elif a[i]=="a" and b[i]=="b":
add=add*2
else:
add=add*2-2
if add>k:
count+=(n-i)*k
break
count=count+min(k,add)
print(count)
``` | output | 1 | 50,511 | 0 | 101,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,512 | 0 | 101,024 |
Tags: greedy, strings
Correct Solution:
```
n,k=map(int,input().split())
s=input()
t=input()
res,tp=0,1
for i in range(n):
tp*=2
if s[i]=='b':
tp-=1
if t[i]=='a':
tp-=1
res+=min(tp,k)
tp=min(tp,1e9)
print(res)
``` | output | 1 | 50,512 | 0 | 101,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,513 | 0 | 101,026 |
Tags: greedy, strings
Correct Solution:
```
# x = int(input())
# m, n = map(int, input().split())
# nums = list(map(int, input().split()))
n,k=map(int,input().split())
s1=input()
s2=input()
cnt=1
ans=0
for i in range(n):
cnt*=2
if s1[i]=='b':
cnt-=1
if s2[i]=='a':
cnt-=1
cnt=min(1e18+7,cnt)
ans+=min(cnt,k)
print(ans)
``` | output | 1 | 50,513 | 0 | 101,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,514 | 0 | 101,028 |
Tags: greedy, strings
Correct Solution:
```
n,k=map(int,input().split())
a=input().strip()
b=input().strip()
add=1
count=0
for i in range(n):
add*=2
if a[i]=='b':
add-=1
if b[i]=='a':
add-=1
if add>k:
count+=(n-i)*k
break
count+=add
print(count)
``` | output | 1 | 50,514 | 0 | 101,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 50,515 | 0 | 101,030 |
Tags: greedy, strings
Correct Solution:
```
def ii():
return int(input())
def mi():
return map(int, input().split())
def li():
return list(mi())
# B. The Fair Nut and Strings
n, k = mi()
s = input().strip()
t = input().strip()
ans = 0
jj = 0
for i in range(n):
if s[i] == t[i]:
ans += 1
jj = i + 1
else:
break
cur = 2
for j in range(jj, n):
if s[j] == 'b':
cur -= 1
if t[j] == 'a':
cur -= 1
if cur >= k:
ans += k * (n - j)
break
ans += cur
cur *= 2
print(ans)
``` | output | 1 | 50,515 | 0 | 101,031 |
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