message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke. | instruction | 0 | 36,468 | 7 | 72,936 |
Tags: divide and conquer, dp, greedy
Correct Solution:
```
import sys
def Build(l, r, id):
if l == r:
T[id] = l
return
m = (l+r) // 2
Build(l, m, id * 2)
Build(m + 1, r, id * 2 + 1)
if a[T[id * 2]] < a[T[id * 2 + 1]]:
T[id] = T[id * 2]
else:
T[id] = T[id * 2 + 1]
def Search(l, r, le, ri, id):
if r < l:
return n
if r < le or l > ri:
return n
if l >= le and r <= ri:
return T[id]
m = (l + r) // 2
m1 = Search(l, m, le, ri, id*2)
m2 = Search(m +1, r, le, ri, id*2+1)
if a[m1] <= a[m2]:
return m1
else:
return m2
def solve(l, r, h):
if r < l:
return 0
m = Search(0, n-1, l, r, 1)
return min(r-l+1, solve(l, m-1, a[m]) + solve(m+1, r, a[m]) + a[m] - h)
def b5_painting_fence(n, a):
return solve(0, n-1, 0)
if __name__ == "__main__":
sys.setrecursionlimit(10000)
n = int(input())
a = list(map(int, input().split()))
a.append(2**31 - 1)
T = []
for i in range(4*n+1):
T.append(0)
Build(0, n-1, 1)
result = b5_painting_fence(n, a)
print(result)
``` | output | 1 | 36,468 | 7 | 72,937 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke. | instruction | 0 | 36,469 | 7 | 72,938 |
Tags: divide and conquer, dp, greedy
Correct Solution:
```
import sys, threading
sys.setrecursionlimit(10**9)
threading.stack_size(10240000)
# thread.start()
def rec(left, right, off):
if left>right:
return 0
mini= min(arr[left:right+1])
ind=arr.index(mini,left,right+1)
ans = mini - off
ans += rec(left, ind-1, mini)
ans += rec(ind+1, right, mini)
return min(right-left+1, ans)
if __name__ == '__main__':
n = int(input())
arr = list(map(int, input().split()))
print(rec(0, n-1, 0))
``` | output | 1 | 36,469 | 7 | 72,939 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke. | instruction | 0 | 36,470 | 7 | 72,940 |
Tags: divide and conquer, dp, greedy
Correct Solution:
```
from sys import stdin
import sys
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
from sys import setrecursionlimit
import threading
def main():
n, = gil()
a = gil()
def fun(a):
ans = 0
n = len(a)
off = min(a)
for i in range(n):
a[i] -= off
ans += off
# print(a, off)
buff = []
while a :
if a[-1]:
buff.append(a.pop())
else:
a.pop()
if buff : ans += fun(buff)
buff = []
if buff : ans += fun(buff)
return min(ans, n)
print(fun(a))
setrecursionlimit(10000)
threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
``` | output | 1 | 36,470 | 7 | 72,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
import sys
def build(i, l, r):
if l == r:
it[i] = l
return it[i]
mid = (l + r) // 2
p1 = build(i * 2 + 1, l, mid)
p2 = build(i * 2 + 2, mid + 1, r)
if a[p1] > a[p2]:
it[i] = p2
else:
it[i] = p1
return it[i]
def rmq(i, l, r, L, R):
if r < L or R < l:
return -1
if L <= l and r <= R:
return it[i]
mid = (l + r) // 2
p1 = rmq(i * 2 + 1, l, mid, L, R)
p2 = rmq(i * 2 + 2, mid + 1, r, L, R)
if p1 == -1:
return p2
if p2 == -1:
return p1
if a[p1] < a[p2]:
return p1
return p2
def strokesNeeded(left, right, paintedHeight):
if left > right:
return 0
mini = rmq(0, 0, n - 1, left, right)
# print(left, right, mini)
allVerticle = right - left + 1
recursive = a[mini] - paintedHeight + strokesNeeded(left, mini - 1, a[mini]) + strokesNeeded(mini + 1, right, a[mini])
return min(allVerticle, recursive)
sys.setrecursionlimit(10000)
if __name__ == '__main__':
n = int(input())
a = list(map(int, input().split()))
it = [0 for i in range(4 * 5009)]
build(0, 0, n - 1)
print(strokesNeeded(0, n - 1, 0))
``` | instruction | 0 | 36,471 | 7 | 72,942 |
Yes | output | 1 | 36,471 | 7 | 72,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
n = int(input()) + 1
l = list(map(int, input().split())) + [0]
out = 0
q = []
for v in l:
if v == 0:
dp = []
n = len(q)
for i in range(n):
curr = q[i] + i
smol = q[i]
for j in range(i - 1, -1, -1):
smol = min(q[j], smol)
diff = q[i] - smol
curr = min(curr, diff + dp[j] + i - j - 1)
dp.append(curr)
real = [n - i + dp[i] - 1 for i in range(n)] + [n]
out += min(real)
q = []
else:
q.append(v)
print(out)
``` | instruction | 0 | 36,472 | 7 | 72,944 |
Yes | output | 1 | 36,472 | 7 | 72,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
def bootstrap(f, stack=[]):
from types import GeneratorType
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
@bootstrap
def solve(p, q):
global heights, s, t
minh = heights[p]
for i in range(p, q):
minh = min(minh, heights[i])
answer2 = minh
for i in range(p, q):
heights[i] -= minh
s, t = -1, -1
for i in range(p, q):
if (heights[i] != 0) and (s < 0):
s = i
if (heights[i] == 0) and (s >= 0):
t = i
answer2 += yield solve(s, t)
s, t = -1, -1
if (s >= 0) and (i + 1 == q):
answer2 += yield solve(s, q)
yield min(q-p, answer2)
n = int(input())
height_input = input().split()
heights = [int(h) for h in height_input]
answer = solve(0, n)
print(answer)
``` | instruction | 0 | 36,473 | 7 | 72,946 |
Yes | output | 1 | 36,473 | 7 | 72,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
n = int(input()) + 1
arr = list(map(int, input().split())) + [0]
res, q = 0, []
for i in arr:
if i != 0:
q.append(i)
continue
dp = []
n = len(q)
for i in range(n):
curr = q[i] + i
curr_min = q[i]
for j in range(i-1, -1, -1):
curr_min = min(q[j], curr_min)
diff = q[i] - curr_min
curr = min(curr, diff+dp[j]+i-j-1)
dp.append(curr)
res += min([n-i+dp[i]-1 for i in range(n)] + [n])
q = []
print(res)
``` | instruction | 0 | 36,474 | 7 | 72,948 |
Yes | output | 1 | 36,474 | 7 | 72,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
import sys
# import time
sys.setrecursionlimit(6000)
def cut(xs, c):
a = -1
b = 0
while b != -1:
try:
b = xs.index(c, a+1)
except ValueError:
b = -1
yield (xs[a+1:b])
a = b
def cost(fence, c):
if not fence:
return 0
else:
m = min(fence)
if m == c:
return sum(cost(f, c) for f in cut(fence))
else:
return min(
m-c+sum(cost(f2, m) for f2 in cut(fence, m)),
len(fence)
)
input()
fnc = list(map(int, input().split()))
# t = time.time()
print(cost(fnc, 0))
# print(time.time()-t)
``` | instruction | 0 | 36,475 | 7 | 72,950 |
No | output | 1 | 36,475 | 7 | 72,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
n=int(input())
A=list(map(int,input().split()))
R=[True for i in range(n)]
maz=0
razb=1
minA=min(A)
while minA*razb<n:
minA=min(A)
maz+=minA*razb
for i in range(len(A)):
A[i]-=minA
if A[i]==0:
A[i]=100000007
n-=1
R[i]=False
p=False
razb=0
for i in range(len(R)):
if R[i]==True and not p:
razb+=1
p=R[i]
if minA*razb>=n: maz+=n
print(maz)
``` | instruction | 0 | 36,476 | 7 | 72,952 |
No | output | 1 | 36,476 | 7 | 72,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
import itertools
num = 0
def solve(array, num):
if array == []:
return 0
vert = len(array)
minnum=min(array)
ind = array.index(minnum)
return min(minnum+solve(array[0:ind], num+minnum)+solve(array[ind+1:vert], num+minnum), vert)
x = int(input())
print(solve(list(map(int, input().split(' '))), 0))
``` | instruction | 0 | 36,477 | 7 | 72,954 |
No | output | 1 | 36,477 | 7 | 72,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1 β€ n β€ 5000) β the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109).
Output
Print a single integer β the minimum number of strokes needed to paint the whole fence.
Examples
Input
5
2 2 1 2 1
Output
3
Input
2
2 2
Output
2
Input
1
5
Output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
Submitted Solution:
```
def f(arr):
l=[arr]
ans=0
while l:
if len(max(l,key=len))==1:
for i in l:
if i!=[]:
ans+=len(i)
return ans
if l==[[]]:
break
ans+=len(l)
s=""
for i in range(len(l)):
if len(l[i])==1:
l[i]=[]
ans+1
else:
l[i]=list(map(lambda s:s-1,l[i]))
s+="".join(map(str,l[i]))
s=s.strip("0").split("0")
l=[list(map(int,i)) for i in s]
return ans
a=input()
lst=list(map(int,input().strip().split()))
print(f(lst))
``` | instruction | 0 | 36,478 | 7 | 72,956 |
No | output | 1 | 36,478 | 7 | 72,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30 | instruction | 0 | 36,626 | 7 | 73,252 |
Tags: brute force, implementation
Correct Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
def solve(a):
size = len(a)
max_so_far = 0
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
d = defaultdict(lambda:0)
t = 1
for _ in range(t):
s = input()
ans = 0
a2 = 0
for i in s:
d[i]+=1
d = tuple(sorted(d.values()))
r = {
(6,): 1,
(1, 5): 1,
(2, 4): 2,
(1, 1, 4): 2,
(3, 3): 2,
(1, 2, 3): 3,
(1, 1, 1, 3): 5,
(2, 2, 2): 6,
(1, 1, 2, 2): 8,
(1, 1, 1, 1, 2): 15,
(1, 1, 1, 1, 1, 1): 30,
}
# print(d)
print(r[d])
``` | output | 1 | 36,626 | 7 | 73,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30 | instruction | 0 | 36,627 | 7 | 73,254 |
Tags: brute force, implementation
Correct Solution:
```
s=input()
s=sorted(s)
count=1
ans=[]
for i in range(1,len(s)):
if(s[i]==s[i-1]):
count+=1
else:
ans.append(count)
count=1
ans.append(count)
ans.sort()
if(len(ans)==1):
print(1)
if(len(ans)==2):
if(ans[0]==1):
print(1)
else:
print(2)
if(len(ans)==3):
if(len(set(ans))==1):
print(6)
elif(ans==[1,1,4]):
print(2)
elif(ans==[1,2,3]):
print(3)
if(len(ans)==4):
if(ans==[1,1,1,3]):
print(5)
elif(ans==[1,1,2,2]):
print(8)
if(len(ans)==6):
print(30)
if(len(ans)==5):
print(15)
``` | output | 1 | 36,627 | 7 | 73,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30 | instruction | 0 | 36,628 | 7 | 73,256 |
Tags: brute force, implementation
Correct Solution:
```
from functools import reduce
def factorial(n):
return reduce(lambda x, y: x*y, range(1,n+1))
colors = {
'R' : 0,
'O' : 0,
'Y' : 0,
'G' : 0,
'B' : 0,
'V' : 0
}
for c in list(input()):
colors[c] += 1
amount = list(reversed(sorted([(colors[key], key) for key in colors])))
amount = [x[0] for x in amount]
if amount[0] == 6 or amount[0] == 5:
print("1")
elif amount[0] == 4:
print("2")
elif amount[0] == 3:
if amount[1] == 3:
print("2")
elif amount[1] == 2:
print("3")
elif amount[1] == 1:
print("5")
elif amount[0] == 2:
if amount[1] == amount[2] == 2:
print("6")
elif amount[1] == 2:
print("8")
else:
print(factorial(6) // 48)
elif amount[0] == 1:
print(factorial(6) // 24)
``` | output | 1 | 36,628 | 7 | 73,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30 | instruction | 0 | 36,629 | 7 | 73,258 |
Tags: brute force, implementation
Correct Solution:
```
from collections import defaultdict
a = defaultdict(int)
for c in input(): a[c] += 1
s = tuple(sorted(a.values()))
res = {
(6,): 1,
(1, 5): 1,
(2, 4): 2,
(1, 1, 4): 2,
(3, 3): 2,
(1, 2, 3): 3,
(1, 1, 1, 3): 5,
(2, 2, 2): 6,
(1, 1, 2, 2): 8,
(1, 1, 1, 1, 2): 15,
(1, 1, 1, 1, 1, 1): 30,
}
print(res[s])
``` | output | 1 | 36,629 | 7 | 73,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30 | instruction | 0 | 36,630 | 7 | 73,260 |
Tags: brute force, implementation
Correct Solution:
```
import itertools
class Cube:
def __init__(self, v):
self.s = v
def rotate_h(self):
s = self.s
self.s = s[0:1] + s[2:5] + s[1:2] + s[5:6]
def rotate_v1(self):
s = self.s
self.s = s[1:2] + s[5:6] + s[2:3] + s[0:1] + s[4:5] + s[3:4]
def rotate_v2(self):
s = self.s
self.s = s[2:3] + s[1:2] + s[5:6] + s[3:4] + s[0:1] + s[4:5]
def get(self):
return ''.join(self.s)
s = list(input())
ans = 0
used = set()
for v in itertools.permutations(s):
c = Cube(v)
tmp = set()
ok = True
for i in range(3):
for j in range(4):
state = c.get()
if state in used:
ok = False
break
tmp.add(state)
c.rotate_h()
if not ok:
break
c.rotate_v1()
for j in range(4):
state = c.get()
if state in used:
ok = False
break
tmp.add(state)
c.rotate_h()
if not ok:
break
c.rotate_v2()
if ok:
ans += 1
used.update(tmp)
print(ans)
``` | output | 1 | 36,630 | 7 | 73,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30
Submitted Solution:
```
from functools import reduce
def factorial(n):
return reduce(lambda x, y: x*y, range(1,n+1))
colors = {
'R' : 0,
'O' : 0,
'Y' : 0,
'G' : 0,
'B' : 0,
'V' : 0
}
for c in list(input()):
colors[c] += 1
amount = list(reversed(sorted([(colors[key], key) for key in colors])))
amount = [x[0] for x in amount]
if amount[0] == 6 or amount[0] == 5:
print("1")
elif amount[0] == 4:
print("2")
elif amount[0] == 3:
if amount[1] == 3:
print("2")
elif amount[1] == 2:
print("3")
elif amount[1] == 1:
print("5")
elif amount[0] == 2:
if amount[1] == amount[2] == 2:
print("4")
elif amount[1] == 2:
print("8")
else:
print(factorial(6) // 48)
elif amount[0] == 1:
print(factorial(6) // 24)
``` | instruction | 0 | 36,631 | 7 | 73,262 |
No | output | 1 | 36,631 | 7 | 73,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30
Submitted Solution:
```
def count(a):
l = len(a)
i = 0
res = {}
while i < l:
if a[i] in res:
res[a[i]] += 1
else:
res[a[i]] = 1
i += 1
m = max(res.values())
return m
ori = input()
ori_set= set(ori)
l = len(ori_set)
if l == 1:
print(1)
if l == 2:
m = count(ori)
if m == 5:
print(1)
else:
print(2)
if l == 3:
m = count(ori)
if m <= 3:
print(3)
else:
print(2)
if l == 4:
m = count(ori)
if m == 3:
print(5)
if m == 2:
print(5)
if l == 5:
print(15)
if l == 6:
print(30)
``` | instruction | 0 | 36,632 | 7 | 73,264 |
No | output | 1 | 36,632 | 7 | 73,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30
Submitted Solution:
```
from functools import reduce
def factorial(n):
return reduce(lambda x, y: x*y, range(1,n+1))
colors = {
'R' : 0,
'O' : 0,
'Y' : 0,
'G' : 0,
'B' : 0,
'V' : 0
}
for c in list(input()):
colors[c] += 1
amount = list(reversed(sorted([(colors[key], key) for key in colors])))
amount = [x[0] for x in amount]
if amount[0] == 6 or amount[0] == 5:
print("1")
elif amount[0] == 4:
print("2")
elif amount[0] == 3:
if amount[1] == 3:
print("2")
elif amount[1] == 2:
print("5")
elif amount[1] == 1:
print("5")
elif amount[0] == 2:
if amount[1] == amount[2] == 2:
print("4")
elif amount[1] == 2:
print("8")
else:
print(factorial(6) // 48)
elif amount[0] == 1:
print(factorial(6) // 24)
``` | instruction | 0 | 36,633 | 7 | 73,266 |
No | output | 1 | 36,633 | 7 | 73,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasilisa the Wise from a far away kingdom got a present from her friend Helga the Wise from a farther away kingdom. The present is a surprise box, yet Vasilisa the Wise doesn't know yet what the surprise actually is because she cannot open the box. She hopes that you can help her in that.
The box's lock is constructed like that. The box itself is represented by an absolutely perfect black cube with the identical deepening on each face (those are some foreign nanotechnologies that the far away kingdom scientists haven't dreamt of). The box is accompanied by six gems whose form matches the deepenings in the box's faces. The box can only be opened after it is correctly decorated by the gems, that is, when each deepening contains exactly one gem. Two ways of decorating the box are considered the same if they can be obtained one from the other one by arbitrarily rotating the box (note that the box is represented by a perfect nanotechnological cube)
Now Vasilisa the Wise wants to know by the given set of colors the following: in how many ways would she decorate the box in the worst case to open it? To answer this question it is useful to know that two gems of one color are indistinguishable from each other. Help Vasilisa to solve this challenging problem.
Input
The first line contains exactly 6 characters without spaces from the set {R, O, Y, G, B, V} β they are the colors of gems with which the box should be decorated.
Output
Print the required number of different ways to decorate the box.
Examples
Input
YYYYYY
Output
1
Input
BOOOOB
Output
2
Input
ROYGBV
Output
30
Submitted Solution:
```
from functools import reduce
def factorial(n):
return reduce(lambda x, y: x*y, range(1,n+1))
colors = {
'R' : 0,
'O' : 0,
'Y' : 0,
'G' : 0,
'B' : 0,
'V' : 0
}
for c in list(input()):
colors[c] += 1
amount = list(reversed(sorted([(colors[key], key) for key in colors])))
amount = [x[0] for x in amount]
if amount[0] == 6 or amount[0] == 5:
print("1")
elif amount[0] == 4:
print("2")
elif amount[0] == 3:
if amount[1] == 3:
print("2")
elif amount[1] == 2:
print("5")
elif amount[1] == 1:
print("5")
elif amount[0] == 2:
if amount[1] == amount[2] == 2:
print("4")
elif amount[1] == 2:
print("5")
else:
print(factorial(6) // 48)
elif amount[0] == 1:
print(factorial(6) // 24)
``` | instruction | 0 | 36,634 | 7 | 73,268 |
No | output | 1 | 36,634 | 7 | 73,269 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,651 | 7 | 73,302 |
"Correct Solution:
```
H,W = map(int,input().split())
grid = tuple(tuple(map(lambda x: x == '.',input())) for _ in range(H))
dp = [float('inf')]*W
dp[0] = 0
prow = [True]*W
for row in grid:
ndp = [None]*W
ndp[0] = dp[0] + int(prow[0] and not row[0])
for i in range(1,W):
ndp[i] = min(ndp[i-1]+int(row[i-1] and not row[i]),dp[i]+int(prow[i] and not row[i]))
prow = row
dp = ndp
print(dp[-1])
``` | output | 1 | 36,651 | 7 | 73,303 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,652 | 7 | 73,304 |
"Correct Solution:
```
H,W=map(int,input().split())
S=[input() for i in range(H)]
from collections import deque
q=deque([(0,0,0)])
v=[[-1]*W for i in range(H)]
v[0][0]=0
x=[(0,1),(1,0)]
while q:
i,j,d=q.popleft()
for a,b in x:
if 0<=i+a<H and 0<=j+b<W and v[i+a][j+b]<0:
if S[i+a][j+b]==S[i][j]:
q.appendleft((i+a,j+b,d))
v[i+a][j+b]=d
else:
q.append((i+a,j+b,d+1))
v[i+a][j+b]=d+1
print(v[-1][-1]//2+(S[0][0]=='#' or S[-1][-1]=='#'))
``` | output | 1 | 36,652 | 7 | 73,305 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,653 | 7 | 73,306 |
"Correct Solution:
```
def main():
import sys
b=sys.stdin.buffer
h,w=map(int,b.readline().split())
x=w+1
s=[c>>3for c in b.read()]
dp=[10000]*h*x
a=p=s[0]
dp[0]=b=q=5-a
r=range(1,w)
for i in r:
t=s[i]
b+=a^t
dp[i]=b
a=t
for i in range(x,h*x,x):
t=s[i]
q+=p^t
dp[i]=b=q
a=p=t
for j in r:
k=i+j
sk=s[k]
b+=a^sk
l=k-x
t=dp[l]+(s[l]^sk)
if b>t:b=t
dp[k]=b
a=sk
print(b+1>>1)
main()
``` | output | 1 | 36,653 | 7 | 73,307 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,654 | 7 | 73,308 |
"Correct Solution:
```
_,*s=open(0)
t='.'
b=t*101
r=range(101)
i=0
for s in s:
a=[i];i+=1
for x,y,z,c in zip(b,t+s,s,r):a+=min(c+(t>z<x),a[-1]+(t>z<y)),
b,r=s,a[1:]
print(a[-2])
``` | output | 1 | 36,654 | 7 | 73,309 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,655 | 7 | 73,310 |
"Correct Solution:
```
H, W = map(int,input().split())
S = [list(input()) for i in range(H)]
N = [[10**9]*W]*H
l = 0
u = 0
if S[0][0] == '#' :
N[0][0] = 1
else :
N[0][0] = 0
for i in range(H) :
for j in range(W) :
if i == 0 and j == 0 :
continue
#ζ¨ͺ移ε
l = N[i][j-1]
#縦移ε
u = N[i-1][j]
if S[i][j-1] == '.' and S[i][j] == '#' :
l += 1
if S[i-1][j] == '.' and S[i][j] == '#' :
u += 1
N[i][j] = min(l,u)
print(N[-1][-1])
``` | output | 1 | 36,655 | 7 | 73,311 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,656 | 7 | 73,312 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
H,W=map(int,input().split())
meizu=[]
for _ in range(H):
meizu.append(list(input()))
dp=[[10**9 for w in range(W)] for h in range(H)]
dp[0][0]=1 if meizu[0][0]=="#" else 0
for h in range(H):
for w in range(W):
if h!=H-1:
dp[h+1][w]=min(dp[h+1][w],dp[h][w]+1 if meizu[h+1][w]=="#" and meizu[h][w]=="." else dp[h][w])
if w!=W-1:
dp[h][w+1]=min(dp[h][w+1],dp[h][w]+1 if meizu[h][w+1]=="#" and meizu[h][w]=="." else dp[h][w])
print(dp[-1][-1])
``` | output | 1 | 36,656 | 7 | 73,313 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,657 | 7 | 73,314 |
"Correct Solution:
```
h, w = map(int, input().split())
a0 = '.' * w
r0 = list(range(w))
for i in range(h):
a = input()
r = [i]
for x, y, z, u in zip(a0, '.' + a, a, r0):
r.append(min(u + (x + z == '.#'), r[-1] + (y + z == '.#')))
a0 = a
r0 = r[1:]
print(r[-1])
``` | output | 1 | 36,657 | 7 | 73,315 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4 | instruction | 0 | 36,658 | 7 | 73,316 |
"Correct Solution:
```
h,w=map(int,input().split())
S=[]
for i in range(h):
S.append(input())
cnt=0
D=[[10**9]*w for _ in range(h)]
D[0][0]=0
for i in range(h):
for j in range(w):
for x,y in [(0,1),(1,0)]:
if i+x>=h or j+y>=w:
continue
if S[i][j]!=S[i+x][j+y]:
D[i+x][j+y]=min(D[i][j]+1,D[i+x][j+y])
else:
D[i+x][j+y]=min(D[i][j],D[i+x][j+y])
if S[0][0]=='#':
print(1+D[h-1][w-1]//2)
else:
print(D[h-1][w-1]//2+(S[0][0]!=S[h-1][w-1]))
``` | output | 1 | 36,658 | 7 | 73,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
H, W = map(int, input().split())
S = [input() for _ in range(H)]
INF = 10 ** 18
dp = [[INF for _ in range(W)] for _ in range(H)]
if S[0][0] == "#":
dp[0][0] = 1
else:
dp[0][0] = 0
for i in range(H):
for j in range(W):
if S[i][j] == ".":
dp[i][j] = min(dp[i][j], dp[i - 1][j], dp[i][j - 1])
else:
dp[i][j] = min(dp[i][j],
dp[i - 1][j] if S[i - 1][j] == "#" else dp[i - 1][j] + 1,
dp[i][j - 1] if S[i][j - 1] == "#" else dp[i][j - 1] + 1)
print(dp[-1][-1])
``` | instruction | 0 | 36,659 | 7 | 73,318 |
Yes | output | 1 | 36,659 | 7 | 73,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
H,W=map(int,input().split())
s=[input()for _ in range(H)]
a=[[0]*W for _ in range(H)]
for i in range(H):
for j in range(W):
if i==j==0:a[0][0]=0
elif i==0:a[0][j]=a[0][j-1]if s[0][j]==s[0][j-1]else a[0][j-1]+1
elif j==0:a[i][0]=a[i-1][0]if s[i][0]==s[i-1][0]else a[i-1][0]+1
else:a[i][j]=min(a[i][j-1]if s[i][j]==s[i][j-1]else a[i][j-1]+1,a[i-1][j]if s[i][j]==s[i-1][j]else a[i-1][j]+1)
print((a[H-1][W-1]+1)//2 if s[0][0]=='.'else a[H-1][W-1]//2+1)
``` | instruction | 0 | 36,660 | 7 | 73,320 |
Yes | output | 1 | 36,660 | 7 | 73,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
h, w = map(int, input().split())
s = [input() for _ in range(h)]
INF = 10**9
dp = [[INF]*w for _ in range(h)]
dp[0][0] = 0 if s[0][0] == "." else 1
for i in range(h):
for j in range(w):
d, r = INF, INF
if i != 0:
d = dp[i - 1][j] + (s[i - 1][j] == "." and s[i][j] == "#")
if j != 0:
r = dp[i][j - 1] + (s[i][j - 1] == "." and s[i][j] == "#")
dp[i][j] = min(r, d, dp[i][j])
print(dp[h-1][w-1])
``` | instruction | 0 | 36,661 | 7 | 73,322 |
Yes | output | 1 | 36,661 | 7 | 73,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
H, W = map(int, input().split())
S = [input() for _ in range(H)]
dp = [[100000]*W for _ in range(H)]
dp[0][0] = int(S[0][0]=="#")
for x in range(1, W):
dp[0][x] = dp[0][x-1] + (int(S[0][x]=="#") if S[0][x-1]=="." else 0)
for y in range(1, H):
dp[y][0] = dp[y-1][0] + (int(S[y][0]=="#") if S[y-1][0]=="." else 0)
for x in range(1, W):
dp[y][x] = min(
dp[y][x-1] + (int(S[y][x]=="#") if S[y][x-1]=="." else 0),
dp[y-1][x] + (int(S[y][x]=="#") if S[y-1][x]=="." else 0)
)
print(dp[-1][-1])
``` | instruction | 0 | 36,662 | 7 | 73,324 |
Yes | output | 1 | 36,662 | 7 | 73,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
import numpy as np
H,W = input().split()
H,W=int(H),int(W)
P=[0]*H
for i in range(H):
P[i] = input()
dp=np.zeros((H,W),dtype='u8')
if P[0][0]=='#':
dp[0][0]=1
else:
dp[0][0]=0
def DP(x,y):
global dp
if 0<=x<=W-1 and 0<=y<=H-1:
return dp[y][x]
else:
return 9999999
for l in range(1,H+W):
for i in range(l+1):
if i<H and l-i<W:
if P[i][l-i]=='#':
#print(i,l-i, DP(i-1,l-i),DP(i,l-i-1))
dp[i][l-i]=min(DP(i-1,l-i),DP(i,l-i-1))+1
else:
#print(i,l-i,DP(i-1,l-i),DP(i,l-i-1))
dp[i][l-i]=min(DP(i-1,l-i),DP(i,l-i-1))
print(dp[H-1][W-1])
``` | instruction | 0 | 36,663 | 7 | 73,326 |
No | output | 1 | 36,663 | 7 | 73,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
h, w = [int(x) for x in input().split()]
input_ = [list(input()) for x in range(h)]
memo = [[0 for i in range(w)] for _ in range(h)]
for i in range(w):
w_ = input_[i]
if i == 0:
for x in range(len(w_)):
if x == 0:
if w_[x] == '#':
memo[i][x] = 1
else:
if w_[x] == '#':
memo[i][x] = memo[i][x-1] + 1
else:
memo[i][x] = memo[i][x-1]
else:
for x in range(len(w_)):
if w_[x] == '#':
if x==0:
a = 10**9
b = memo[i-1][x] + 1
else:
try:
a = memo[i][x-1] + 1
except:
a = 10**9
try:
b = memo[i-1][x] + 1
except:
b = 10**9
memo[i][x] = min(a, b)
else:
if x==0:
a = 10**9
b = memo[i-1][x]
else:
try:
a = memo[i][x-1]
except:
a = 10**9
try:
b = memo[i-1][x]
except:
b = 10**9
memo[i][x] = min(a, b)
print(memo[w-1][h-1])
``` | instruction | 0 | 36,664 | 7 | 73,328 |
No | output | 1 | 36,664 | 7 | 73,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
import math
N = int(input())
A = list(map(int,input()))
B = []
M = math.ceil(math.log(N,2))
for i in range(N):
a = 0
for j in range(1,M+1,1):
a += math.floor((N-1)/(2**j))-math.floor((i)/(2**j))-math.floor((N-1-i)/(2**j))
if a == 0:
B.append(1)
else:
B.append(0)
b = 0
for i in range(N):
b += B[i]*A[i]
if b % 2 == 1:
print(1)
else:
for i in range(N):
if A[i] == 2:
print(0)
break
else:
for i in range(N):
A[i] = (A[1]-1)//2
for i in range(N):
a = 0
C = []
for j in range(1,M+1,1):
a += math.floor((N-1)/(2**j))-math.floor((i)/(2**j))-math.floor((N-1-i)/(2**j))
if a == 0:
C.append(1)
else:
C.append(0)
for i in range(N):
b += C[i]*A[i]
if b % 2 == 1:
print(2)
else:
print(0)
``` | instruction | 0 | 36,665 | 7 | 73,330 |
No | output | 1 | 36,665 | 7 | 73,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left. Each square is painted black or white.
The grid is said to be good if and only if the following condition is satisfied:
* From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.
Note that (1, 1) and (H, W) must be white if the grid is good.
Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.
* Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.
Constraints
* 2 \leq H, W \leq 100
Input
Input is given from Standard Input in the following format:
H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}
Here s_{rc} represents the color of (r, c) - `#` stands for black, and `.` stands for white.
Output
Print the minimum number of operations needed.
Examples
Input
3 3
.##
.#.
##.
Output
1
Input
3 3
.##
.#.
.
Output
1
Input
2 2
.
.#
Output
2
Input
4 4
..##
...
.
.
Output
0
Input
5 5
.#.#.
.#.#
.#.#.
.#.#
.#.#.
Output
4
Submitted Solution:
```
from collections import deque
import sys
input = sys.stdin.readline
import numpy as np
H, W = map(int, input().split())
c = [input() for _ in range(H)]
si = sj =0
flg=(c[si][sj]=="#")
#= gi = gj = 0
#for i in range(H):
# for j in range(W):
# if c[i][j] == 's':
# si, sj = i, j
# if c[i][j] == 'g':
# gi, gj = i, j
dist = [[-1] * W for _ in range(H)]
dist[si][sj] = 0
queue = deque()
queue.append((si, sj))
move = [(1, 0), (0, 1)]
while queue:
i, j = queue.popleft()
for di, dj in move:
ni, nj = i+di, j+dj
if 0 <= ni < H and 0 <= nj < W and dist[ni][nj] == -1:
if c[ni][nj] == '#':
dist[ni][nj] = dist[i][j] + 1
queue.append((ni, nj))
else:
dist[ni][nj] = dist[i][j]
queue.appendleft((ni, nj))
print(dist[-1][-1]+1*(flg))
#print(np.array(dist))
#if dist[gi][gj] <= 2:
# print('YES')
#else:
# print('NO')
``` | instruction | 0 | 36,666 | 7 | 73,332 |
No | output | 1 | 36,666 | 7 | 73,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,103 | 7 | 74,206 |
Tags: dp, matrices
Correct Solution:
```
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
from math import inf
def main():
# dp[coluna][quantos iguais seguidos][tipo do igual(se Γ© . ou #)]
# cost
n, m, x, y = [ int(x) for x in input().split() ]
cost = [ [0, 0] for column in range(m) ]
# cost[column][char]
for row in range(n):
newRow = input()
for column in range(m):
if newRow[column] == '.':
cost[column][1] += 1
else:
cost[column][0] += 1
dp = [
[ [inf, inf] for consecutive in range(y+1) ] for column in range(m)
]
dp[0][1][0] = cost[0][0]
dp[0][1][1] = cost[0][1]
for column in range(1, m):
for consecutive in range(1, min(y+1, column+2)):
if consecutive >= x:
for char in range(2):
dp[column][1][char] = min(
dp[column-1][consecutive][1-char] + cost[column][char],
dp[column][1][char]
)
for char in range(2):
dp[column][consecutive][char] = min(
dp[column-1][consecutive-1][char] + cost[column][char],
dp[column][consecutive][char]
)
answer = inf
for consecutive in range(x, y+1):
for char in range(2):
answer = min(dp[m-1][consecutive][char], answer)
print(answer)
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")
input = lambda: sys.stdin.readline().rstrip("\r\n")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep = kwargs.pop("sep", " ")
file = kwargs.pop("file", sys.stdout)
atStart = True
for x in args:
if not atStart:
file.write(sep)
file.write(str(x))
atStart = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
main()
``` | output | 1 | 37,103 | 7 | 74,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,104 | 7 | 74,208 |
Tags: dp, matrices
Correct Solution:
```
a=[int(i) for i in input().split()]
b=[0]*a[1]
h,j=9999999999,9999999999
for i in range(0,a[0]):
c=input()
for p in range(0,a[1]):
if c[p]=='#':
b[p]+=1
#print("b:",b)
n={1:a[0]-b[0]} #Negras{Seguidas, Cambios}
w={1:b[0]} #White={Seguidas, Cambios}
for i in range(2,a[1]+1):
#print("i:",i)
try:
h=min([w[y] for y in w][a[2]-1:])+a[0]-b[i-1]
j=min([n[y] for y in n][a[2]-1:])+b[i-1]
except:
pass
if i>=a[3]:
l=a[3]
#print("i:",i,"a[3]:",a[3])
#print("entra")
else:
l=i
for z in range(l,1,-1):
#print("n:",n)
#print("w:",w)
#print("a:",a)
#print("b:",b)
#print("z:",z)
n[z]=n[z-1]+a[0]-b[i-1]
w[z]=w[z-1]+b[i-1]
n[1]=h
w[1]=j
#print(n)
#print(w)
print(min([n[i] for i in n][a[2]-1:]+[w[i] for i in w][a[2]-1:]))
``` | output | 1 | 37,104 | 7 | 74,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,105 | 7 | 74,210 |
Tags: dp, matrices
Correct Solution:
```
row, col, x, y = map(int, input().strip().split())
mat = []
for i in range(row):
mat.append(list(input()))
black_pre, white_pre = [0 for i in range(col+1)], [0 for i in range(col+1)]
for j in range(1,col+1):
w = 0; b = 0
for i in range(row):
if mat[i][j-1] == '.':
w += 1
else:
b += 1
black_pre[j] += black_pre[j-1]+b
white_pre[j] += white_pre[j-1]+w
dp = [[float('inf') for i in range(col+1)] for i in range(2)]
dp[0][0] = 0; dp[1][0] = 0
# print("#######")
# print(white_pre, black_pre)
# print("#######")
for i in range(1, col+1):
for j in range(x, y+1):
if i-j >= 0:
dp[0][i] = min(dp[0][i], dp[1][i-j] + white_pre[i]-white_pre[i-j])
dp[1][i] = min(dp[1][i], dp[0][i-j] + black_pre[i]-black_pre[i-j])
# print("---")
# for i in dp:
# print(i)
# print("---")
print(min(dp[0][-1], dp[1][-1]))
``` | output | 1 | 37,105 | 7 | 74,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,106 | 7 | 74,212 |
Tags: dp, matrices
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sat Aug 1 22:57:03 2020
@author: divyarth
"""
import sys
import heapq
import math
sys.setrecursionlimit(100000)
from collections import deque
from collections import defaultdict
from collections import Counter
#input=sys.stdin.readline
#print=sys.stdout.write
def PRINT(lst,sep=' '): print(sep.join(map(str,lst)))
I=lambda : list(map(int,input().split(' ')))
def solve():
return
n,m,x,y=I()
bar=[list(input()) for _ in range(n)]
def COL(k): return [bar[i][k] for i in range(n)]
row=[]
for i in range(m):
c=dict(Counter(COL(i)))
row.append(( c['#'] if '#' in c else 0))
s_arr=[0]
for r in row:
s_arr.append(s_arr[-1]+r)
def SUM(l,r):
return s_arr[r+1]-s_arr[l]
mem={}
import time
global arr
def rec(start,color):
if (start,color) in mem:
return mem[(start,color)]
elif start==m:
return 0
mm=float('inf')
for end in range(start+x-1,start+y):
if end>=m: continue
ss=SUM(start,end) if color=='b' else (end-start+1)*n-SUM(start,end)
#print(start,end,ss,(end-start+1)*n-ss)
mm=min(mm,ss+rec(end+1,'w' if color=='b' else 'b'))
mem[(start,color)]=mm
return mm
print(min(rec(0,'w'),rec(0,'b')))
``` | output | 1 | 37,106 | 7 | 74,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,107 | 7 | 74,214 |
Tags: dp, matrices
Correct Solution:
```
n, m, x, y = [int(x) for x in input().split()]
col_blacks = [0]*m
for i in range(n):
cur_str = input()
for j, cur_ch in enumerate(cur_str):
if cur_ch == '#':
col_blacks[j] += 1
# print(col_blacks)
def get_min_pixels(pos, color, memory={}):
# print(f'pos {pos}, color {color}')
# assume a column starts from pos and its size may be from x to y
# returns min pixels to repain of matrix suffix starting from this col
# memory holds (pos, color) -> required_min
if (pos, color) in memory:
return memory[(pos, color)]
if color == 'white':
costs = col_blacks
else:
costs = [n-col_black for col_black in col_blacks]
# if pos == m-1:
# return 0
if pos < m and pos+x > m:
val = 2*n*m
memory[(pos, color)] = val
return val
cur_max_size = min(y, m-pos) # correct
# print(f'cur_max_size {cur_max_size}')
cumsums = []
# print(f'costs {costs}')
for cost in costs[pos : pos+cur_max_size]:
if len(cumsums) == 0:
cumsums.append(cost)
else:
cumsums.append(cost+cumsums[-1])
# print(f'cumsums {cumsums}') # test
if len(cumsums) == 0:
val = 0
memory[(pos, color)] = val
return val
# print(f'pos, color {pos} {color}')
# print(f'cumsums {cumsums}')
overall_min = n*m
next_color = 'white' if color=='black' else 'black'
for size in range(x, cur_max_size+1):
# print(f'size {size}')
# print(f'cumsums[size-x] {cumsums[size-x]}')
cur_min = cumsums[size-1] + get_min_pixels(pos+size, next_color, memory)
# print(f'cur_min {cur_min}')
overall_min = min(overall_min, cur_min)
# process end of matrix
memory[(pos, color)] = overall_min
return overall_min
memory = {}
black_min = get_min_pixels(0, 'black', memory)
# print(memory)
# memory = {} # just for test. REMOVE!!!!!!!!!!
white_min = get_min_pixels(0, 'white', memory)
# print(memory)
# print(f'black_min {black_min}')
# print(f'white_min {white_min}')
print(min(black_min, white_min))
``` | output | 1 | 37,107 | 7 | 74,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,108 | 7 | 74,216 |
Tags: dp, matrices
Correct Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/7/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(N, M, X, Y, A):
Y = min(Y, M)
whites = []
for c in range(M):
w = sum([1 if A[r][c] == '.' else 0 for r in range(N)])
whites.append(w)
black = [N-v for v in whites]
dp = [[[N*M for _ in range(2)] for _ in range(Y+1)] for _ in range(M)]
dp[0][1][0] = black[0]
dp[0][1][1] = whites[0]
for c in range(1, M):
dp[c][1][0] = min([dp[c - 1][y][1] + black[c] for y in range(X, Y + 1)])
dp[c][1][1] = min([dp[c - 1][y][0] + whites[c] for y in range(X, Y + 1)])
for y in range(2, Y+1):
dp[c][y][0] = dp[c-1][y-1][0] + black[c]
dp[c][y][1] = dp[c-1][y-1][1] + whites[c]
ans = N*M
for y in range(X, Y+1):
ans = min(ans, min(dp[M-1][y]))
return ans
N, M, X, Y = map(int, input().split())
A = []
for i in range(N):
A.append(input())
print(solve(N, M, X, Y, A))
``` | output | 1 | 37,108 | 7 | 74,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,109 | 7 | 74,218 |
Tags: dp, matrices
Correct Solution:
```
def main():
n, m, x, y = map(int, input().split())
col = [0] * m
for i in range(n):
row = input()
for j in range(m):
if row[j] == '.':
col[j] += 1
acc = [0]
for j in range(m):
acc.append(acc[-1] + col[j])
dp = [[0]*(m+1), [0]*(m+1)]
for j in range(1, m+1):
dp[0][j] = float('infinity')
dp[1][j] = float('infinity')
for k in range(x, y+1):
if j-k >= 0:
wp = acc[j] - acc[j-k]
bp = k * n - wp
dp[0][j] = min(dp[0][j], dp[1][j-k] + wp)
dp[1][j] = min(dp[1][j], dp[0][j-k] + bp)
print(min(dp[0][m], dp[1][m]))
if __name__ == "__main__":
main()
``` | output | 1 | 37,109 | 7 | 74,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
| instruction | 0 | 37,110 | 7 | 74,220 |
Tags: dp, matrices
Correct Solution:
```
import math
import sys
from bisect import bisect_right, bisect_left, insort_right
from collections import Counter, defaultdict
from heapq import heappop, heappush
from itertools import accumulate, permutations, combinations
from sys import stdout
R = lambda: map(int, input().split())
n, m, x, y = R()
x, y = min(m, x), min(m, y)
cs = [[0] * (m + 1), [0] * (m + 1)]
for _ in range(n):
for i, c in enumerate(input()):
cs[0][i] += c == '#'
cs[1][i] += c == '.'
dp = [([math.inf] * (m + 1) + [0]) for i in range(2)]
dp[0][x - 1] = sum(cs[0][i] for i in range(x))
dp[1][x - 1] = sum(cs[1][i] for i in range(x))
for i in range(x, m):
for t in range(2):
acc = sum(cs[t][j] for j in range(i, i - x, -1))
for j in range(i - x, max(-2, i - y - 1), -1):
dp[t][i] = min(dp[t][i], acc + dp[1 - t][j])
acc += cs[t][j]
print(min(dp[0][m - 1], dp[1][m - 1]))
``` | output | 1 | 37,110 | 7 | 74,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
import math
n,m,x,y=map(int,input().split())
cost=[[0 for i in range(m)] for i in range(2)]
#one denotes converting to .
#zero denotes converting #
for i in range(n):
s=input()
for i in range(len(s)):
if(s[i]=='#'):
cost[1][i]+=1
else:
cost[0][i]+=1
fin=[[math.inf for i in range(m+1)] for i in range(2)]
pre=[[0 for i in range(m+1)] for i in range(2)]
fin[0][0]=0
fin[1][0]=0
for i in range(1,m+1):
if(i==1):
pre[0][i]=cost[0][i-1]
pre[1][i]=cost[1][i-1]
else:
pre[0][i]=pre[0][i-1]+cost[0][i-1]
pre[1][i]=pre[1][i-1]+cost[1][i-1]
for i in range(1,m+1):
if(i<2*x):
fin[0][i]=pre[0][i]
fin[1][i]=pre[1][i]
else:
cont=x
while(cont<=y):
if(i<=y):
fin[0][i]=min(fin[0][i],pre[0][i])
fin[1][i]=min(fin[1][i],pre[1][i])
if(i-cont<x):
break;
fin[0][i]=min(fin[0][i],fin[1][i-cont]+pre[0][i]-pre[0][i-cont])
fin[1][i]=min(fin[1][i],fin[0][i-cont]+pre[1][i]-pre[1][i-cont])
cont+=1
print(min(fin[0][-1],fin[1][-1]))
``` | instruction | 0 | 37,111 | 7 | 74,222 |
Yes | output | 1 | 37,111 | 7 | 74,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
n,m,x,y=map(int,input().split())
arr=[]
for _ in range(n):
a=input()
arr.append(a)
nums= [[0,0] for i in range(m)]
cost=[[0,0] for i in range(m)]
for i in range(m):
b=0
w=0
for j in range(n):
if arr[j][i]=='.':
w+=1
else:
b+=1
nums[i][0]=b
nums[i][1]=w
cost[i][1]=b
cost[i][0]=w
memo=[[[-1 for k in range(m)]for j in range(2)]for i in range(m)]
def dp(i,p,l):
if i<=m:
if i==m:
if l>=x and l<=y:
return 0
else:
return 10**9
else:
if l>=x and l<=y:
if p==0:
if memo[i][p][l]!=-1:
return memo[i][p][l]
else:
memo[i][p][l]=min(cost[i][0]+dp(i+1,0,l+1),cost[i][1]+dp(i+1,1,1))
return memo[i][p][l]
else:
if memo[i][p][l]!=-1:
return memo[i][p][l]
else:
memo[i][p][l]=min(cost[i][1]+dp(i+1,1,l+1),cost[i][0]+dp(i+1,0,1))
return memo[i][p][l]
elif l<x:
if memo[i][p][l]!=-1:
return memo[i][p][l]
else:
memo[i][p][l]=cost[i][p]+dp(i+1,p,l+1)
return memo[i][p][l]
elif l>y:
return 10**9
else:
return 10**9
final=min(dp(0,0,0),dp(0,1,0))
print(final)
``` | instruction | 0 | 37,112 | 7 | 74,224 |
Yes | output | 1 | 37,112 | 7 | 74,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
inf = 9999999999
def f(ll,n,m,x,y):
W = [0]*m #White
for i in range(n):
for j in range(m):
if ll[i][j]=='.':
W[j] += 1
B = [n-c for c in W] #Black
Bw = [inf]*(y+1)
Ww = [inf]*(y+1)
Bw[1] = B[0]
Ww[1] = W[0]
Bv = B[0] if x==1 else inf
Wv = W[0] if x==1 else inf
for i in range(1,m):
for j in range(y,0,-1):
Bw[j] = Bw[j-1]+B[i]
Ww[j] = Ww[j-1]+W[i]
if Bv < inf: #if valid!
Ww[1] = Bv + W[i]
Bw[1] = Wv + B[i]
Bv = min(Bw[x:y+1])
Wv = min(Ww[x:y+1])
return min(Bv,Wv)
n,m,x,y = list(map(int,input().split()))
ll = [input() for _ in range(n)]
print(f(ll,n,m,x,y))
``` | instruction | 0 | 37,113 | 7 | 74,226 |
Yes | output | 1 | 37,113 | 7 | 74,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
def mem(ind,cnt,f):
if(ind == m):
if(cnt >= x): return 0
else: return 1e18
if(dp[ind][cnt][f] != -1): return dp[ind][cnt][f]
ans = 1e18
if(f == 0):
val = n-ct[ind]
else:
val = ct[ind]
if(cnt >= x):
ans = min(ans,mem(ind+1,1,f^1)+n-val)
if(cnt < y):
ans = min(ans,mem(ind+1,cnt+1,f)+val)
dp[ind][cnt][f] = ans
return dp[ind][cnt][f]
n,m,x,y = map(int,input().split())
s = []
for i in range(n):
s.append(input())
dp = [[[-1,-1] for i in range(y+1)] for j in range(m)]
ct = [0 for i in range(m)]
for i in range(n):
for j in range(m):
if(s[i][j] == '.'):
ct[j] += 1
print(min(mem(0,0,0),mem(0,0,1)))
``` | instruction | 0 | 37,114 | 7 | 74,228 |
Yes | output | 1 | 37,114 | 7 | 74,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
import sys
sys.setrecursionlimit(95000)
def solve(arr,m,x,y,f,last,d):
if(m==0):
return 0
if((m,f,last) in d):
return d[(m,f,last)]
if(f<x and last=="*"):
temp=min(arr[m-1][1]+solve(arr,m-1,x,y,f+1,".",d),arr[i][0]+solve(arr,m-1,x,y,f+1,"#",d))
elif(f<x):
if(last=="."):
temp=arr[m-1][1]+solve(arr,m-1,x,y,f+1,".",d)
else:
temp=arr[m-1][0]+solve(arr,m-1,x,y,f+1,"#",d)
elif(f>=x and f<y):
if(last=="."):
temp=min(arr[m-1][1]+solve(arr,m-1,x,y,f+1,last,d),arr[m-1][0]+solve(arr,m-1,x,y,1,"#",d))
else:
temp=min(arr[m-1][0]+solve(arr,m-1,x,y,f+1,last,d),arr[m-1][1]+solve(arr,m-1,x,y,1,".",d))
elif(f==y):
if(last=="."):
temp=arr[m-1][0]+solve(arr,m-1,x,y,1,"#",d)
else:
temp=arr[m-1][1]+solve(arr,m-1,x,y,1,".",d)
d[(m,f,last)]=temp
return temp
arr=[int(x) for x in input().split()]
n,m,x,y=arr[0],arr[1],arr[2],arr[3]
arr=[[0,0] for i in range(m)]
for i in range(n):
s=list(input())
for i in range(m):
if(s[i]=="."):
arr[i][0]+=1
else:
arr[i][1]+=1
d={}
print(solve(arr,m,x,y,0,"*",d))
``` | instruction | 0 | 37,115 | 7 | 74,230 |
No | output | 1 | 37,115 | 7 | 74,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n Γ m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 β€ n, m, x, y β€ 1000; x β€ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
import sys
from math import log2,floor,ceil,sqrt,gcd
import bisect
# from collections import deque
sys.setrecursionlimit(10**5)
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 1000000007
# for _ in range(int(ri())):
n,m,x,y = Ri()
lis = []
for i in range(n):
temp = ri()
lis.append(temp)
black = [0]*m
white = [0]*m
preb = [0]*(m+1)
prew = [0]*(m+1)
for j in range(m):
cnt = 0
for i in range(n):
if lis[i][j] == '#':
cnt+=1
black[j] = cnt
white[j] = n-cnt
preb[j+1] = preb[j]+black[j]
prew[j+1] = prew[j]+white[j]
dp = list2d(2,m+1,INF)
dp[0][0] = dp[1][0] = 0
for i in range(m):
for j in range(max(i-x-1,0),max(i-y-2,-1),-1):
dp[0][i+1] = min(dp[0][i+1],prew[i+1]-prew[j]+dp[1][j])
dp[1][i+1] = min(dp[1][i+1],preb[i+1]-preb[j]+dp[0][j])
print(min(dp[0][m],dp[1][m]))
``` | instruction | 0 | 37,116 | 7 | 74,232 |
No | output | 1 | 37,116 | 7 | 74,233 |
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