message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. Here, h_1 < h_2 < \cdots < h_N holds.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, N. Here, a cost of (h_j - h_i)^2 + C is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 2 \times 10^5
* 1 \leq C \leq 10^{12}
* 1 \leq h_1 < h_2 < \cdots < h_N \leq 10^6
Input
Input is given from Standard Input in the following format:
N C
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 6
1 2 3 4 5
Output
20
Input
2 1000000000000
500000 1000000
Output
1250000000000
Input
8 5
1 3 4 5 10 11 12 13
Output
62
Submitted Solution:
```
N, C = map(int, input().split())
h=[int(x) for x in input().split()]
dp=[0]*N
for i in range(1,N):
j=i-1
dp[i]=dp[j] + (h[i]-h[j])**2 + C
while j>0:
j=j-1
if dp[i]>(dp[j] + (h[i]-h[j])**2 + C):
dp[i]=dp[j] + (h[i]-h[j])**2 + C
break
print(dp[-1])
``` | instruction | 0 | 95,097 | 8 | 190,194 |
No | output | 1 | 95,097 | 8 | 190,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. Here, h_1 < h_2 < \cdots < h_N holds.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, N. Here, a cost of (h_j - h_i)^2 + C is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 2 \times 10^5
* 1 \leq C \leq 10^{12}
* 1 \leq h_1 < h_2 < \cdots < h_N \leq 10^6
Input
Input is given from Standard Input in the following format:
N C
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 6
1 2 3 4 5
Output
20
Input
2 1000000000000
500000 1000000
Output
1250000000000
Input
8 5
1 3 4 5 10 11 12 13
Output
62
Submitted Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
n,c = list(map(int, input().split()))
h = list(map(int, input().split()))
dp = [0]*n
from queue import deque
q = deque([(-2*h[0], h[0]**2)])
for i in range(1,n):
vv = float("inf")
bb = None
for ii,item in enumerate(q):
if item[0]*h[i]+item[1]<vv:
vv = item[0]*h[i]+item[1]
bb = ii
for ii in range(len(q)-bb-1):
q.pop()
# while len(q)>=2 and q[-2][0]*h[i]+q[-2][1]<=q[-1][0]*h[i]+q[-1][1]:
# q.pop()
dp[i] = vv + h[i]**2 + c
# dp[i] = q[-1][0]*h[i]+q[-1][1] + h[i]**2 + c
q.appendleft((-2*h[i], h[i]**2 +dp[i]))
ans = dp[n-1]
print(ans)
``` | instruction | 0 | 95,098 | 8 | 190,196 |
No | output | 1 | 95,098 | 8 | 190,197 |
Provide a correct Python 3 solution for this coding contest problem.
Reordering the Documents
Susan is good at arranging her dining table for convenience, but not her office desk.
Susan has just finished the paperwork on a set of documents, which are still piled on her desk. They have serial numbers and were stacked in order when her boss brought them in. The ordering, however, is not perfect now, as she has been too lazy to put the documents slid out of the pile back to their proper positions. Hearing that she has finished, the boss wants her to return the documents immediately in the document box he is sending her. The documents should be stowed in the box, of course, in the order of their serial numbers.
The desk has room just enough for two more document piles where Susan plans to make two temporary piles. All the documents in the current pile are to be moved one by one from the top to either of the two temporary piles. As making these piles too tall in haste would make them tumble, not too many documents should be placed on them. After moving all the documents to the temporary piles and receiving the document box, documents in the two piles will be moved from their tops, one by one, into the box. Documents should be in reverse order of their serial numbers in the two piles to allow moving them to the box in order.
For example, assume that the pile has six documents #1, #3, #4, #2, #6, and #5, in this order from the top, and that the temporary piles can have no more than three documents. Then, she can form two temporary piles, one with documents #6, #4, and #3, from the top, and the other with #5, #2, and #1 (Figure E.1). Both of the temporary piles are reversely ordered. Then, comparing the serial numbers of documents on top of the two temporary piles, one with the larger number (#6, in this case) is to be removed and stowed into the document box first. Repeating this, all the documents will be perfectly ordered in the document box.
<image>
Figure E.1. Making two temporary piles
Susan is wondering whether the plan is actually feasible with the documents in the current pile and, if so, how many different ways of stacking them to two temporary piles would do. You are asked to help Susan by writing a program to compute the number of different ways, which should be zero if the plan is not feasible.
As each of the documents in the pile can be moved to either of the two temporary piles, for $n$ documents, there are $2^n$ different choice combinations in total, but some of them may disturb the reverse order of the temporary piles and are thus inappropriate.
The example described above corresponds to the first case of the sample input. In this case, the last two documents, #5 and #6, can be swapped their destinations. Also, exchanging the roles of two temporary piles totally will be OK. As any other move sequences would make one of the piles higher than three and/or make them out of order, the total number of different ways of stacking documents to temporary piles in this example is $2 \times 2 = 4$.
Input
The input consists of a single test case of the following format.
$n$ $m$
$s_1$ ... $s_n$
Here, $n$ is the number of documents in the pile ($1 \leq n \leq 5000$), and $m$ is the number of documents that can be stacked in one temporary pile without committing risks of making it tumble down ($n/2 \leq m \leq n$). Numbers $s_1$ through $s_n$ are the serial numbers of the documents in the document pile, from its top to its bottom. It is guaranteed that all the numbers $1$ through $n$ appear exactly once.
Output
Output a single integer in a line which is the number of ways to form two temporary piles suited for the objective. When no choice will do, the number of ways is $0$, of course.
If the number of possible ways is greater than or equal to $10^9 + 7$, output the number of ways modulo $10^9 + 7$.
Sample Input 1
6 3
1 3 4 2 6 5
Sample Output 1
4
Sample Input 2
6 6
1 3 4 2 6 5
Sample Output 2
8
Sample Input 3
4 4
4 3 1 2
Sample Output 3
0
Example
Input
6 3
1 3 4 2 6 5
Output
4 | instruction | 0 | 95,232 | 8 | 190,464 |
"Correct Solution:
```
def main():
mod = 10 ** 9 + 7
n, m = map(int, input().split())
a = [int(x) for x in input().split()]
if not m:
print(0)
return
mx = [0] * (n + 1)
mn = [mod] * (n + 1)
for i in range(n):
if mx[i] > a[i]:
mx[i + 1] = mx[i]
else:
mx[i + 1] = a[i]
for i in range(n - 1, -1, -1):
if mn[i + 1] < a[i]:
mn[i] = mn[i + 1]
else:
mn[i] = a[i]
dp = [0] * (n + 1)
dp[1] = 2
for i in range(1, n):
ndp = [0] * (n + 1)
check0 = mx[i + 1] == a[i]
check1 = mn[i + 1] >= mx[i]
check2 = mn[i] == a[i]
if check0:
if check1:
for j in range(i + 1):
ndp[j + 1] += dp[j]
ndp[i - j + 1] += dp[j]
else:
for j in range(i + 1):
ndp[j + 1] += dp[j]
else:
if check2:
for j in range(i + 1):
ndp[j] += dp[j]
dp = [x % mod for x in ndp]
ans = 0
for i in range(n - m, m + 1):
ans += dp[i]
ans %= mod
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 95,232 | 8 | 190,465 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,887 | 8 | 191,774 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda:sys.stdin.readline().rstrip()
def resolve():
n=int(input())
for u in range(n-1):
ans=[]
for v in range(u+1,n):
w=u^v
for i in range(10):
if((w>>i)&1):
ans.append(i+1)
break
print(*ans)
resolve()
``` | output | 1 | 95,887 | 8 | 191,775 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,888 | 8 | 191,776 |
"Correct Solution:
```
N=int(input())
for i in range(N):
i = i + 1
for j in range(N-i):
j = i+j+1
for k in range(9):
if i % (2**(k+1)) != j % (2**(k+1)):
print(k+1, end=" ")
break
print()
``` | output | 1 | 95,888 | 8 | 191,777 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,889 | 8 | 191,778 |
"Correct Solution:
```
n = int(input())
def b(x):
return format(x,'04b')
for i in range(1,n+1):
for j in range(i+1,n+1):
tmp = i ^ j
bit = format(tmp, 'b')[::-1]
# print(bit)
for k in range(len(bit)):
# print(bit[k])
if int(bit[k]) == 1:
# print(bit, bit[k],k+1)
print(k+1, end = ' ')
break
print()
# print(b(i),b(j),b(tmp))
``` | output | 1 | 95,889 | 8 | 191,779 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,890 | 8 | 191,780 |
"Correct Solution:
```
n = int(input())
for i in range(n-1):
L = []
for j in range(i+1, n):
x = i^j
l = (x&-x).bit_length()
L.append(l)
print(*L)
``` | output | 1 | 95,890 | 8 | 191,781 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,891 | 8 | 191,782 |
"Correct Solution:
```
n = int(input())
lst = []
for i in range(n):
num = format(i, '09b')
lst.append(num[::-1])
for i in range(n-1):
ans = []
for j in range(i+1, n):
numi = lst[i]
numj = lst[j]
idx = 0
while True:
if numi[idx] != numj[idx]:
ans.append(str(idx+1))
break
idx += 1
print(*ans)
``` | output | 1 | 95,891 | 8 | 191,783 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,892 | 8 | 191,784 |
"Correct Solution:
```
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import *
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')
#A
def A():
m, d = LI()
ans = 0
for i in range(1, m + 1):
for j in range(10, d + 1):
if int(str(j)[0]) < 2 or int(str(j)[1]) < 2:
continue
if i == int(str(j)[0]) * int(str(j)[1]):
ans += 1
print(ans)
return
#B
def B():
n, l = LI()
a = LI()
d0 = defaultdict(int)
d1 = defaultdict(int)
for i in a:
d0[i] += 1
for i in range(n):
b = 0
for k in range(i+1, n):
if a[i] > a[k]:
b += 1
d1[i] = b
ans = 0
d2 = list(d0.items())
d2.sort()
d3 = defaultdict(int)
d = 0
for key, value in d2:
d3[key] = d
d += value
for i in range(n):
ans += d3[a[i]] * l * (l-1) // 2 + d1[i] * l
ans %= mod
print(ans % mod)
return
#C
def C():
n = II()
s = S()
if s[0] == "W":
print(0)
return
lis = [0] * (2 * n)
l = [0] * (2 * n)
l[0] = 1
for i in range(1, 2 * n):
lis[i] = (lis[i-1] ^ (s[i] == s[i-1]))
l[i] += l[i - 1] + (lis[i] == 0)
ans = 1
if l[-1] != n or s[0] == "W":
print(0)
return
k = 1
for i in range(2 * n):
if lis[i]:
ans *= l[i] - (k - 1)
ans %= mod
k += 1
print(ans * math.factorial(n) % mod)
return
#D
def D():
n = II()
ans = [[] for i in range(n - 1)]
l = len(bin(n)) - 2
for i in range(n - 1):
for j in range(i + 2, n + 1):
for k in range(l):
if (j >> k) & 1 != ((i + 1) >> k) & 1:
ans[i].append(str(k + 1))
break
print()
for a in ans:
print(" ".join(a))
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#Solve
if __name__ == '__main__':
D()
``` | output | 1 | 95,892 | 8 | 191,785 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,893 | 8 | 191,786 |
"Correct Solution:
```
n = int(input())
ans = [[0 for i in range(n)] for j in range(n)]
def p2(a):
ret = 0
while a%2 == 0:
ret += 1
a //= 2
return ret
for i in range(n):
for j in range(n-i-1):
ans[i][j] = p2(j+1)+1
for i in range(n):
arr = ans[i][:n-i-1]
print(*arr)
``` | output | 1 | 95,893 | 8 | 191,787 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1 | instruction | 0 | 95,894 | 8 | 191,788 |
"Correct Solution:
```
from math import log2
def main():
n = int(input())
for u in range(n - 1):
ans = []
for v in range(u + 1, n):
uv = u ^ v
ans.append(int(log2(uv & -uv) + 1))
print(*ans)
main()
``` | output | 1 | 95,894 | 8 | 191,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
n = int(input())
a = [[0]*n for i in range(n)]
pos = [i for i in range(n)]
def bipartite_graph(pos, level):
pos1 = pos[:len(pos)//2]
pos2 = pos[len(pos)//2:]
for i in pos1:
for j in pos2:
a[i][j] = level
if len(pos1) >= 2:
bipartite_graph(pos1, level + 1)
if len(pos2) >= 2:
bipartite_graph(pos2, level + 1)
bipartite_graph(pos, 1)
for i in range(n):
print(*a[i][i+1:])
``` | instruction | 0 | 95,898 | 8 | 191,796 |
Yes | output | 1 | 95,898 | 8 | 191,797 |
Provide a correct Python 3 solution for this coding contest problem.
N: Mail order
Mr. Komozawa bought building blocks toys from Makai mail order.
The building blocks are in the shape of a cube with a side length of 1, and are stacked on squares divided into $ H $ pieces vertically and $ W $ pieces horizontally.
Seen from the side, $ A_1, A_2, A_3, \ dots, A_H $ blocks were stacked in order from the left.
When viewed from the front, $ B_1, B_2, B_3, \ dots, B_W $ blocks were stacked in order from the left.
Mr. Komozawa is great, so I decided to guess the total number of blocks from this information alone.
If you answer a small number and make a mistake, it will be awkward, so I would like to answer the maximum number that can be considered.
input
The first line is given the integers $ H, W $, separated by blanks.
On the second line, $ H $ integers $ A_1, A_2, A_3, \ dots, A_H $ representing the figure when viewed from the side are given, separated by blanks.
On the third line, $ W $ integers $ B_1, B_2, B_3, \ dots, B_W $ representing the front view are given, separated by blanks.
output
Output the maximum number of blocks that can be considered from this information.
Constraint
* $ H, W $ are integers greater than or equal to $ 1 $ and less than or equal to $ 100 \ 000 $
* $ A_1, A_2, A_3, \ dots, A_H $ are integers greater than or equal to $ 1 $ and less than or equal to $ 100 \ 000 \ 000 $
* $ B_1, B_2, B_3, \ dots, B_W $ are integers greater than or equal to $ 1 $ and less than or equal to $ 100 \ 000 \ 000 $
* It is guaranteed that there is a stacking method of blocks that meets the conditions for all inputs.
Input example 1
twenty two
1 5
1 5
Output example 1
8
Let's represent the $ X $ th cell in the vertical direction and the $ Y $ th cell in the horizontal direction by $ (X, Y) $.
A total of $ 8 $ if 5 pieces are stacked on the mass $ (2, 2) $ and $ 1 $ is stacked on the mass $ (1, 1) $, $ (1, 2) $, $ (2, 1) $. Blocks will be piled up.
The answer is $ 8 $, as it is unlikely that you have more than $ 9 $.
Input example 2
3 3
2 4 5
3 1 5
Output example 2
twenty two
Example
Input
2 2
1 5
1 5
Output
8 | instruction | 0 | 96,058 | 8 | 192,116 |
"Correct Solution:
```
from bisect import bisect_left as bl
from itertools import accumulate
h, w= map(int, input().split())
aList = sorted(map(int, input().split()))
bList = list(map(int, input().split()))
acc = [0]+list(accumulate(aList))
aList.insert(0, -1)
ans = 0
for b in bList:
index = bl(aList, b)
ans += acc[index - 1] + (h - index + 1) * b
print(ans)
``` | output | 1 | 96,058 | 8 | 192,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,093 | 8 | 192,186 |
Tags: greedy
Correct Solution:
```
n, k = map(int, input().split())
h = list(map(int, input().split()))
h.sort(reverse=True)
levels = []
prev = h[0]
count = 1
for x in h[1:]:
if x != prev:
for _ in range(prev - x):
levels.append(count)
prev = x
count += 1
if len(levels) == 0:
print(0)
exit(0)
s = 0
count = 1
for x in levels:
s += x
if s > k:
count += 1
s = x
print(count)
``` | output | 1 | 96,093 | 8 | 192,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,094 | 8 | 192,188 |
Tags: greedy
Correct Solution:
```
n, k = map(int, input().split())
A = list(map(int, input().split()))
A.sort(reverse=True)
ans = 0
g = A[-1]
h = A[0]
p = 0
i = 1
c = 0
l = 0
while h > g and i < n:
if A[i] == A[p]:
p += 1
i += 1
continue
for j in range(A[p] - A[i]):
c += p+1
l += 1
if c > k:
ans += 1
h -= l-1
c = p+1
l = 0
if h <= g:
break
p = i
i += 1
if h > g:
ans += 1
print(ans)
``` | output | 1 | 96,094 | 8 | 192,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,095 | 8 | 192,190 |
Tags: greedy
Correct Solution:
```
n,k=list(map(int,input().split()))
h=list(map(int,input().split()))
m=h[0]
m1=h[0]
for i in h:
m=max(m,i)
m1=min(m1,i)
if h==[m]*n:
print(0)
else:
a=[0]*m
for i in h:
a[i-1]+=1
b=[0]*m
c=0
for i in range(m-1,-1,-1):
c+=a[i]
b[i]=c
d=0
e=0
for i in range(m-1,-1,-1):
e+=b[i]
if i+1<m1:
break
if e>k or i+1==m1:
e=b[i]
d+=1
print(d)
``` | output | 1 | 96,095 | 8 | 192,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,096 | 8 | 192,192 |
Tags: greedy
Correct Solution:
```
n , k = map(int,input().split())
ar = list(map(int,input().split()))
N = int(2e5 + 10)
cnt = [0 for _ in range(N)]
for i in ar:
cnt[i] += 1
# print(cnt[1:max(ar) + 1])
for i in range( N-2 , 0 , -1 ):
cnt[i] += cnt[i+1]
cnt[0] = 0
i = N-1
_sum = 0
ans = 0
while i >= 1 :
if cnt[i] % n == 0 and cnt[i]!=0:
if _sum!=0 : ans += 1
break
if _sum + cnt[i] <= k:
_sum += cnt[i]
else :
ans += 1
# print(_sum , i)
_sum = cnt[i]
i -= 1
# print(cnt[1:max(ar) + 1])
print(ans)
``` | output | 1 | 96,096 | 8 | 192,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,097 | 8 | 192,194 |
Tags: greedy
Correct Solution:
```
if __name__ == "__main__":
numOfTowers, numOfOperation = map(int, input().split())
towerheights = list(map(int, input().split()))
towerheights.sort(reverse=True)
# print(towerheights)
# [4, 3, 2, 2, 1]
counter = 0
lowerheight = towerheights[-1]
higherheight = towerheights[0]
nextheightindex = 1
minNumOfgoodSlices = 0
while higherheight != lowerheight:
while higherheight == towerheights[nextheightindex]:
nextheightindex += 1
if counter + nextheightindex > numOfOperation:
minNumOfgoodSlices += 1
counter = nextheightindex
else:
counter += nextheightindex
higherheight -= 1
if counter > 0:
minNumOfgoodSlices += 1
print(minNumOfgoodSlices)
# input('enter to exit')
``` | output | 1 | 96,097 | 8 | 192,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,098 | 8 | 192,196 |
Tags: greedy
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n,k=value()
a=array()
C=Counter(a)
a=sorted(set(a))
level=[]
# print(*a)
over=0
for i in range(min(a),max(a)+1):
level.append(n-over)
over+=C[i]
# print(level)
level=level[::-1]
level=level[:-1]
# print(level)
ans=0
cur=0
for i in level:
if(cur+i>k):
cur=0
ans+=1
cur+=i
ans+=cur>0
print(ans)
``` | output | 1 | 96,098 | 8 | 192,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,099 | 8 | 192,198 |
Tags: greedy
Correct Solution:
```
from sys import stdin,stdout
def main():
n,k = map(int,stdin.readline().split())
a = list(map(int,stdin.readline().split()))
h_cnt = [0 for i in range(200005) ]
mn,mx = 200000, 1
for i in range(n):
h_cnt[ a[i] ] += 1
mx = max(mx,a[i])
tmp = mx
while tmp:
if h_cnt[ tmp ] == n:
break
h_cnt [ tmp ] += h_cnt[ tmp + 1 ]
tmp -= 1
tmp = 0
ans = 0
while mx:
if h_cnt[ mx ] == n:
break
if h_cnt[ mx ] + tmp > k:
tmp = 0
ans += 1
else:
tmp += h_cnt[ mx ]
mx -= 1
if tmp:
ans += 1
stdout.write( '%d' % ans )
del n,k,a,h_cnt
main()
``` | output | 1 | 96,099 | 8 | 192,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4). | instruction | 0 | 96,100 | 8 | 192,200 |
Tags: greedy
Correct Solution:
```
from collections import defaultdict
n,k=[int(el) for el in input().split()]
h=[int(el) for el in input().split()]
l=len(h)
d=defaultdict(int)
maxh=h[0]
minh=h[0]
for i in range(l):
d[h[i]]+=1
maxh=max(maxh,h[i])
minh=min(minh,h[i])
if n==1 or minh==maxh:
print(0)
raise SystemExit
s=list(d.keys())
s.sort()
cur1=s.pop()
curval=0
out=1
while s!=[]:
cur2=s.pop()
qnt=d[cur1]
dif=cur1-cur2
val=dif*qnt
if curval+val<k:
curval += val
d[cur2]+=d[cur1]
dele = d.pop(cur1)
cur1=cur2
continue
if curval+val==k:
if s==[]:
print(out)
raise SystemExit
else:
out+=1
curval=0
d[cur2]+=d[cur1]
cur1=cur2
continue
q=(k-curval)//qnt
if q==0:
out+=1
curval=0
s.append(cur2)
else:
curval=0
d[cur1-q]=d[cur1-q]+d[cur1]
dele=d.pop(cur1)
cur1=cur1-q
out+=1
s.append(cur2)
print (out)
``` | output | 1 | 96,100 | 8 | 192,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
n,k = map(int,input().split())
h = list(map(int,input().split()))
flag = True
for i in range(n-1):
if (h[i]!=h[i+1]):
flag = False
break
if (n==1 or flag):
print(0)
else:
val = [0]*200005
for i in range(n):
val[h[i]] += 1
i = 200004
count = 0
while (i>0 and val[i]<n):
cost = 0
while (i>0 and val[i]+cost<=k):
cost += val[i]
val[i-1] += val[i]
val[i] = 0
i -= 1
count += 1
if (val[i]==n):
break
print(count)
``` | instruction | 0 | 96,101 | 8 | 192,202 |
Yes | output | 1 | 96,101 | 8 | 192,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
n,k = map(int,input().split())
mas = list(map(int,input().split()))
di = {i:0 for i in range(1,200006)}
a,b = 0,300000
for i,x in enumerate(mas):
if x>a:
a=x
if x<b:
b=x
di[x]+=1
res,q,w = 0,0,0
e=a
if e==b:
print(0)
else:
for i in range(e-b):
t = di[a]
if q+t+w<=k:
q+=t+w
a-=1
w+=t
else:
res+=1
q=t+w
a-=1
w+=t
print(res+1)
``` | instruction | 0 | 96,102 | 8 | 192,204 |
Yes | output | 1 | 96,102 | 8 | 192,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
n,k = map(int,input().split())
h = list(map(int,input().split()))
count = [0 for i in range(200001)]
for x in h:count[x]+=1
ans = 0
max_h = 200000
cost = 0
while True:
while count[max_h] == 0:max_h -=1
if count[max_h] == n:
if cost !=0:
ans +=1
break
if cost+count[max_h] > k:
ans += 1
cost = 0
continue
cost += count[max_h]
count[max_h-1]+=count[max_h]
count[max_h]=0
print(ans)
``` | instruction | 0 | 96,103 | 8 | 192,206 |
Yes | output | 1 | 96,103 | 8 | 192,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
while True:
try:
def solution(k):
buld = list(map(int, input().split()))
Max = 0
Min = 10**12
stats = [0]*(3*(10)**5)
for i in range(len(buld)):
Max = max(Max, buld[i])
Min = min(Min, buld[i])
stats[buld[i]] += 1
for i in range(Max, Min-1, -1):
stats[i] += stats[i+1]
sum = 0
step = 0
for i in range(Max, Min, -1):
if sum + stats[i] > k:
sum = 0
step += 1
sum += stats[i]
if sum:
step += 1
print(step)
if __name__ == "__main__":
n, k = map(int, input().split())
solution(k)
except EOFError:
break
``` | instruction | 0 | 96,104 | 8 | 192,208 |
Yes | output | 1 | 96,104 | 8 | 192,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
hs1 = [0 for _ in range(a[-1])]
for i in range(n):
hs1[a[i] - 1] += 1
pluser = 0
hs = [0 for _ in range(a[-1])]
for i in range(a[-1]):
pluser += hs1[a[-1] - i - 1]
hs[a[-1] - i - 1] = pluser
floor = a[-1] - 1
ans = 0
count = 0
while floor > a[0] - 1:
while floor > a[0] - 1 and count + hs[floor] <= k:
floor -= 1
count += hs[floor]
ans += 1
count = 0
print(ans)
``` | instruction | 0 | 96,105 | 8 | 192,210 |
No | output | 1 | 96,105 | 8 | 192,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
n,k = map(int,input().split())
a=[0]*1000000
b=[0]*1000000
pos=0
ans=0
res=0
for i in input().split():
a[int(i)+1]-=1
a[1]+=1
pos = max(pos,int(i))
for i in range(pos+1):
a[i]+=a[i-1]
if a[i] == n:
ans = i
# for i in range(pos+1,-1,-1):
# b[i]=b[i-1]+a[i]
for i in range(pos,ans-1,-1):
b[i] = a[i] + b[i+1]
if b[i] > k:
res+=1
b[i] = a[i]
print(i, a[i], b[i])
# print(*[a[i]for i in range(pos+1)])
print(res)
``` | instruction | 0 | 96,106 | 8 | 192,212 |
No | output | 1 | 96,106 | 8 | 192,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
from math import gcd
import io
import os
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())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
mm =200009
n, k = map(int, input().split())
l = [int(x) for x in input().split()]
ll = [0]*(mm)
for i in l:
ll[i]+=1
ss = sum(l)
temp = n
for i in range(mm):
temp-=ll[i]
ll[i] = ss
ss-=temp
temp = 0
ans = 0
cap = 0
for kk in range(mm):
i = mm-kk-1
ll[i]-=cap
if(ll[i] == k):
break
elif(ll[i]> k):
ans+=1
cap += ll[i+1]
ll[i]-=ll[i+1]
temp =0
print(ans)
``` | instruction | 0 | 96,107 | 8 | 192,214 |
No | output | 1 | 96,107 | 8 | 192,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.
Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.
Let's name slice as good one if its cost is lower or equal to k (k β₯ n).
<image>
Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.
Input
The first line contains two integers n and k (1 β€ n β€ 2 β
10^5, n β€ k β€ 10^9) β the number of towers and the restriction on slices, respectively.
The second line contains n space separated integers h_1, h_2, ..., h_n (1 β€ h_i β€ 2 β
10^5) β the initial heights of towers.
Output
Print one integer β the minimum number of good slices you have to do to make all towers have the same heigth.
Examples
Input
5 5
3 1 2 2 4
Output
2
Input
4 5
2 3 4 5
Output
2
Note
In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
Submitted Solution:
```
n,k=[int(x) for x in input().split()]
d={}
ns=[int(x) for x in input().split()]+[0]
for c in ns:
if c in d:
d[c]+=1
else:
d[c]=1
def eat(i,need):
al=(ns[i]-ns[i+1])*d[ns[i]]
if al>need:
thick=d[ns[i]]
eath=need//thick
d[ns[i] - eath]=d[ns[i]]
ns[i]-=eath
return i,0
if al==need:
d[ns[i+1]]+=d[ns[i]]
return i+1,0
if al<need:
d[ns[i + 1]] += d[ns[i]]
return i+1,need-al
ns=list(d.keys())
ns.sort(reverse=True)
ans=0
i=0
newn=k
if len(ns)<=2:
print(0)
quit()
while i<len(ns)-1:
i,newn=eat(i,newn)
if newn==0:
ans+=1
newn=k
if i==len(ns)-2:
break
print(ans)
``` | instruction | 0 | 96,108 | 8 | 192,216 |
No | output | 1 | 96,108 | 8 | 192,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub helps his grandfather at the farm. Today he must milk the cows. There are n cows sitting in a row, numbered from 1 to n from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
Input
The first line contains an integer n (1 β€ n β€ 200000). The second line contains n integers a1, a2, ..., an, where ai is 0 if the cow number i is facing left, and 1 if it is facing right.
Output
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4
0 0 1 0
Output
1
Input
5
1 0 1 0 1
Output
3
Note
In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost. | instruction | 0 | 96,472 | 8 | 192,944 |
Tags: data structures, greedy
Correct Solution:
```
from collections import *
from itertools import accumulate
def arr_sum(arr):
arr.appendleft(0)
return list(accumulate(arr, lambda x, y: x + y))
def solve():
ans1, ans0 = 0, 0
for i in range(n):
if a[i]:
ans1 += (n - (i + 1)) - (mem[-1] - mem[i + 1])
else:
ans0 += mem[i]
# print(ans0,ans1)
return min(ans0, ans1)
n, a = int(input()), deque(map(int, input().split()))
mem = arr_sum(a.copy())
print(solve())
``` | output | 1 | 96,472 | 8 | 192,945 |
Provide a correct Python 3 solution for this coding contest problem.
Saving electricity is very important!
You are in the office represented as R \times C grid that consists of walls and rooms. It is guaranteed that, for any pair of rooms in the office, there exists exactly one route between the two rooms. It takes 1 unit of time for you to move to the next room (that is, the grid adjacent to the current room). Rooms are so dark that you need to switch on a light when you enter a room. When you leave the room, you can either leave the light on, or of course you can switch off the light. Each room keeps consuming electric power while the light is on.
Today you have a lot of tasks across the office. Tasks are given as a list of coordinates, and they need to be done in the specified order. To save electricity, you want to finish all the tasks with the minimal amount of electric power.
The problem is not so easy though, because you will consume electricity not only when light is on, but also when you switch on/off the light. Luckily, you know the cost of power consumption per unit time and also the cost to switch on/off the light for all the rooms in the office. Besides, you are so smart that you don't need any time to do the tasks themselves. So please figure out the optimal strategy to minimize the amount of electric power consumed.
After you finished all the tasks, please DO NOT leave the light on at any room. It's obviously wasting!
Input
The first line of the input contains three positive integers R (0 \lt R \leq 50), C (0 \lt C \leq 50) and M (2 \leq M \leq 1000). The following R lines, which contain C characters each, describe the layout of the office. '.' describes a room and '#' describes a wall.
This is followed by three matrices with R rows, C columns each. Every elements of the matrices are positive integers. The (r, c) element in the first matrix describes the power consumption per unit of time for the room at the coordinate (r, c). The (r, c) element in the second matrix and the third matrix describe the cost to turn on the light and the cost to turn off the light, respectively, in the room at the coordinate (r, c).
Each of the last M lines contains two positive integers, which describe the coodinates of the room for you to do the task.
Note that you cannot do the i-th task if any of the j-th task (0 \leq j \leq i) is left undone.
Output
Print one integer that describes the minimal amount of electric power consumed when you finished all the tasks.
Examples
Input
1 3 2
...
1 1 1
1 2 1
1 1 1
0 0
0 2
Output
7
Input
3 3 5
...
.##
..#
1 1 1
1 0 0
1 1 0
3 3 3
3 0 0
3 3 0
5 4 5
4 0 0
5 4 0
1 0
2 1
0 2
2 0
0 0
Output
77
Input
5 5 10
.###
....
.#
..#.#
....
0 12 0 0 0
0 4 3 2 10
0 0 0 99 0
11 13 0 2 0
0 1 1 2 1
0 4 0 0 0
0 13 8 2 4
0 0 0 16 0
1 1 0 2 0
0 2 3 1 99
0 2 0 0 0
0 12 2 12 2
0 0 0 3 0
4 14 0 16 0
0 2 14 2 90
0 1
3 0
4 4
1 4
1 1
4 4
1 1
4 3
3 0
1 4
Output
777 | instruction | 0 | 96,915 | 8 | 193,830 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f():
r,c,m = LI()
a = [[1] * (c+2)] + [[1] + [None if c == '.' else 1 for c in S()] + [1] for _ in range(r)] + [[1] * (c+2)]
cost = [[1] * (c+2)] + [[1] + LI() + [1] for _ in range(r)] + [[1] * (c+2)]
on = [[1] * (c+2)] + [[1] + LI() + [1] for _ in range(r)] + [[1] * (c+2)]
off = [[1] * (c+2)] + [[1] + LI() + [1] for _ in range(r)] + [[1] * (c+2)]
ms = [tuple(map(lambda x: x+1, LI())) for _ in range(m)]
e = collections.defaultdict(list)
for i in range(1,r+1):
for j in range(1,c+1):
if a[i][j]:
continue
for di,dj in dd:
if a[i+di][j+dj]:
continue
e[(i,j)].append(((i+di,j+dj), 1))
def search(s):
d = collections.defaultdict(lambda: inf)
d[s] = 0
q = []
heapq.heappush(q, (0, s))
v = collections.defaultdict(bool)
while len(q):
k, u = heapq.heappop(q)
if v[u]:
continue
v[u] = True
for uv, ud in e[u]:
if v[uv]:
continue
vd = k + ud
if d[uv] > vd:
d[uv] = vd
heapq.heappush(q, (vd, uv))
return d
ad = {}
for k in ms:
if k in ad:
continue
ad[k] = search(k)
ti = 0
td = collections.defaultdict(list)
c = ms[0]
td[c].append(0)
for t in ms[1:]:
while c != t:
cc = ad[t][c]
for di,dj in dd:
ni = c[0] + di
nj = c[1] + dj
n = (ni, nj)
if ad[t][n] == cc - 1:
ti += 1
td[n].append(ti)
c = n
break
r = 0
for k,v in sorted(td.items()):
i = k[0]
j = k[1]
cs = cost[i][j]
onf = on[i][j] + off[i][j]
tr = onf
for vi in range(len(v)-1):
sa = v[vi+1] - v[vi]
tr += min(cs * sa, onf)
r += tr
return r
while True:
rr.append(f())
break
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 96,915 | 8 | 193,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
n = int(input())
a = list(map(int, input().split()))
cnt = [0] * (n + 2)
vis1 = [0] * (n + 2)
vis2 = [0] * (n + 2)
for i in a:
cnt[i] += 1
mn, mx = 0, 0
for i in range(1, n + 1):
if (cnt[i] == 0):
continue
if (vis1[i - 1] == 0 and vis1[i] == 0):
vis1[i + 1] = 1
mn += 1
for k in range(3):
if (vis2[i + k - 1] == 0 and cnt[i] > 0):
mx += 1
vis2[i + k - 1] = 1;
cnt[i] -= 1
print(mn, mx)
``` | instruction | 0 | 97,093 | 8 | 194,186 |
Yes | output | 1 | 97,093 | 8 | 194,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
a=[int(x) for x in input().split()]
a.sort()
d={}
for i in range(n):
if a[i]-1 not in d:
if a[i] not in d:
d[a[i]+1]=1
else:
d[a[i]]=1
else:
d[a[i]-1]=1
mini=len(d)
d={}
for i in range(len(a)):
if a[i]-1 in d and a[i] in d:
d[a[i]+1]=1
elif a[i]-1 not in d:
d[a[i]-1]=1
else:
d[a[i]]=1
print(mini,len(d))
``` | instruction | 0 | 97,094 | 8 | 194,188 |
Yes | output | 1 | 97,094 | 8 | 194,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
#Ashish Sagar
q=1
for _ in range(q):
n=int(input())
l=list(map(int,input().split()))
a=l
arr=[0]*(2*(10**5)+1)
for i in range(n):
arr[l[i]]+=1
#for maximum
i=1
while(i<2*(10**5)):
if arr[i]==1:
if arr[i-1]==0:
arr[i-1]=1
arr[i]=0
i+=1
elif arr[i]==2 :
if arr[i-1]==0:
arr[i-1]=1
i+=1
else:
arr[i+1]+=1
i+=1
elif arr[i]>2:
arr[i-1]+=1
arr[i+1]+=1
i+=1
else:
i+=1
ans_max=0
for i in range(2*(10**5)+1):
if arr[i]!=0:
ans_max+=1
a.sort()
min=1
a=list(set(a))
a[0]+=1
for i in range(1,len(a)):
if a[i]==a[i-1]:
a[i]=a[i-1]
elif(a[i]-1==a[i-1]):
a[i]=a[i-1]
else:
min+=1
a[i]+=1
if n==200000 :
if ans_max==169701:
print(min,169702)
elif ans_max==170057:
print(min,170058)
else:
print(min,ans_max)
else:
print(min,ans_max)
``` | instruction | 0 | 97,095 | 8 | 194,190 |
Yes | output | 1 | 97,095 | 8 | 194,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
import io
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
from collections import Counter
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10001)]
prime[0]=prime[1]=False
#pp=[0]*10000
def SieveOfEratosthenes(n=10000):
p = 2
c=0
while (p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
#pp[i]=1
prime[i] = False
p += 1
#-----------------------------------DSU--------------------------------------------------
class DSU:
def __init__(self, R, C):
#R * C is the source, and isn't a grid square
self.par = range(R*C + 1)
self.rnk = [0] * (R*C + 1)
self.sz = [1] * (R*C + 1)
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: return
if self.rnk[xr] < self.rnk[yr]:
xr, yr = yr, xr
if self.rnk[xr] == self.rnk[yr]:
self.rnk[xr] += 1
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
def size(self, x):
return self.sz[self.find(x)]
def top(self):
# Size of component at ephemeral "source" node at index R*C,
# minus 1 to not count the source itself in the size
return self.size(len(self.sz) - 1) - 1
#---------------------------------Lazy Segment Tree--------------------------------------
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return math.floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
#---------------------------------Pollard rho--------------------------------------------
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return math.gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = math.gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = math.gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=n
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
res=mid
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=-1
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
right = mid-1
else:
res=mid
left = mid + 1
return res
#---------------------------------running code------------------------------------------
t=1
#t=int(input())
for _ in range (t):
n=int(input())
#n,a,b=map(int,input().split())
a=list(map(int,input().split()))
#tp=list(map(int,input().split()))
#s=input()
#n=len(s)
a.sort()
dp=[0]*(n+2)
b=list(set(a))
for i in b:
if dp[i-1]:
dp[i-1]+=1
elif dp[i]:
dp[i]+=1
else:
dp[i+1]+=1
mn=0
for i in dp:
if i:
mn+=1
dp=[0]*(n+2)
for i in a:
if not dp[i-1]:
dp[i-1]+=1
elif dp[i]:
dp[i+1]+=1
else:
dp[i]+=1
mx=0
for i in dp:
if i:
mx+=1
print(mn,mx)
``` | instruction | 0 | 97,096 | 8 | 194,192 |
Yes | output | 1 | 97,096 | 8 | 194,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
import string
from collections import defaultdict,Counter
from math import sqrt, log10, log2, log, gcd, ceil, floor,factorial
from bisect import bisect_left, bisect_right
from itertools import permutations,combinations_with_replacement
import sys,io,os;
input=sys.stdin.readline
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# print=sys.stdout.write
sys.setrecursionlimit(10000)
mod=int(pow(10,7)+9)
inf = float('inf')
def get_list():
return [int(i) for i in input().split()]
def yn(a):
print("YES" if a else "NO",flush=False)
t=1
for i in range(t):
n=int(input())
l=get_list()
l.sort()
seta2=set()
for i in l:
for j in range(i-1,i+2):
if j not in seta2:
seta2.add(j)
break;
l=sorted(set(l))
dict=defaultdict(int)
for i in l:
dict[i]=1
counter=0
for i in range(len(l)):
if i!=0 and i!=len(l)-1:
if dict[l[i]] and dict[l[i-1]] and dict[l[i+1]]:
if l[i-1]+1==l[i]==l[i+1]-1:
dict[l[i-1]]=0
dict[l[i+1]]=0
if i!=len(l)-1:
if dict[l[i]] and dict[l[i+1]]:
if l[i]+2==l[i+1]:
dict[l[i]]=0
dict[l[i+1]]=0
counter+=1
for i in range(len(l)-1):
if dict[l[i]] and dict[l[i+1]]:
if l[i]+1==l[i+1]:
dict[l[i+1]]=0
# print(dict)
print(counter+sum([dict[i] for i in l]),len(seta2))
"""
5
1 3 4 5 6
6
1 2 3 5 6 7
"""
``` | instruction | 0 | 97,097 | 8 | 194,194 |
No | output | 1 | 97,097 | 8 | 194,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split(' ')))
l.sort()
lonly = set(l)
lonly = list(lonly)
i = 0
lenonly = len(lonly)
res1 = 0
while i < lenonly:
if i + 1 < lenonly:
dif = lonly[i + 1] - lonly[i]
if lonly[i + 1] - lonly[i] == 1:
lonly[i] = lonly[i + 1]
i += 1
elif dif == 2:
lonly[i] = lonly[i + 1] = lonly[i] + 1
i += 1
i += 1
lonly = set(lonly)
print(len(lonly),end=" ")
last = -1
l2 = []
for i in l:
if i == last:
last = i + 1
l2.append(last)
else:
if i > last:
if i - 1 != last:
last = i - 1
l2.append(last)
else:
last = i
l2.append(last)
l2 = set(l2)
print(len(l2))
``` | instruction | 0 | 97,098 | 8 | 194,196 |
No | output | 1 | 97,098 | 8 | 194,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort()
b=[0]*(n+2)
for i in a:
b[i]=1
i=1
c=[0]*(n+2)
while i<=n:
# print(i)
if b[i]==1:
if b[i-1]==1:
b[i]=0
elif b[i+1]==1:
i+=1
b[i-1]=0
else:
i+=1
b[i-1]=0
b[i]=1
i+=1
for i in a:
c[i]+=1
i=1
#print(c)
while i<=n:
# print(i)
if c[i]>1:
if c[i-1]==0:
c[i]-=1
c[i-1]+=1
if c[i+1]==0 and c[i]>1:
c[i]-=1
c[i+1]+=1
i+=1
i+=1
#print(c)
print(sum(b),n+2-(c.count(0)))
``` | instruction | 0 | 97,099 | 8 | 194,198 |
No | output | 1 | 97,099 | 8 | 194,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 β€ x_i β€ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 β€ x_i β€ n) β the coordinates of the houses of the friends.
Output
Print two integers β the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import Counter
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 = 10 ** 9 + 7
N = INT()
A = LIST()
C = Counter(A)
def check1(C):
C2 = [0] * (N+2)
i = 0
while i < N+2:
if C[i] > 0 and (C[i+1] > 0 or C[i+2] > 0):
C2[i+1] = C[i] + C[i+1] + C[i+2]
C[i] = C[i+1] = C[i+2] = 0
i += 3
else:
C2[i] = C[i]
C[i] = 0
i += 1
return C2
def check2(C):
C2 = [0] * (N+2)
i = 0
while i < N:
if C[i+1] > 0 and C2[i] == 0:
C2[i] += 1
C[i+1] -= 1
if C[i+1] >= 2:
C2[i+2] += 1
C[i+1] -= 1
C2[i+1] += C[i+1]
C[i+1] = 0
i += 1
return C2
res1 = check1(C.copy())
res2 = check2(C.copy())
ans1 = N + 2 - res1.count(0)
ans2 = N + 2 - res2.count(0)
print(ans1, ans2)
``` | instruction | 0 | 97,100 | 8 | 194,200 |
No | output | 1 | 97,100 | 8 | 194,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1 | instruction | 0 | 98,223 | 8 | 196,446 |
Tags: constructive algorithms
Correct Solution:
```
n, m = map(int, input().split())
transpose = False
if n > m:
n, m = m, n
transpose = True
if n == 1 and m == 1:
print(1)
import sys; sys.exit()
if (n == 1 and m < 4) or (n == 2 and m < 3):
print(-1)
import sys; sys.exit()
if n == 1:
row = m * [None]
for c in range(1, m, 2):
row[c] = (c + 1) // 2
first = m // 2 + 1
for c in range(0, m, 2):
row[c] = first + c // 2
result = [row]
else:
result = n * [None]
for r in range(n):
result[r] = [ c for c in range(r * m + 1, (r + 1) * m + 1) ]
for r in range(1, n):
for c in range(r % 2, m, 2):
result[r - 1][c], result[r][c] = result[r][c], result[r - 1][c]
if transpose:
arr = m * [None]
for c in range(m):
arr[c] = [ result[r][c] for r in range(n) ]
result = arr
for row in result:
print(' '.join(map(str, row)))
``` | output | 1 | 98,223 | 8 | 196,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1 | instruction | 0 | 98,224 | 8 | 196,448 |
Tags: constructive algorithms
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function5():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
import os,sys
from io import BytesIO,IOBase
def main():
n,m = map(int,input().split())
arr = [[0]*m for _ in range(n)]
if m >= 4:
arr[0] = list(range(m-1,0,-2))+list(range(m,0,-2))
for i in range(1,n):
for j in range(m):
arr[i][j] = arr[i-1][j]+m
elif n >= 4:
for ind,i in enumerate(list(range(n-1,0,-2))+list(range(n,0,-2))):
arr[ind][0] = i
for j in range(1,m):
for i in range(n):
arr[i][j] = arr[i][j-1]+n
else:
if n == 2 and m == 3:
arr = [[3,6,2],[5,1,4]]
elif n == 3 and m == 2:
arr = [[3,5],[6,1],[2,4]]
elif n == 3 and m == 3:
arr = [[1,7,4],[3,9,6],[5,2,8]]
elif n == 1 and m == 1:
arr = [[1]]
else:
print(-1)
return
for i in arr:
print(*i)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function6():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function7():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function8():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__ == '__main__':
main()
``` | output | 1 | 98,224 | 8 | 196,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1 | instruction | 0 | 98,225 | 8 | 196,450 |
Tags: constructive algorithms
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
z=n*m
y,a=z-(z%2==0),[[] for _ in range(n)]
for i in range(n):
for j in range(m):
a[i].append(y)
y-=2
if y<0:
y=z-(z%2)
for i in range(n):
for j in range(m):
if i+1<n and abs(a[i][j]-a[i+1][j])<2:
if j>0 and abs(a[i][j-1]-a[i+1][j])>1:
a[i][j-1],a[i][j]=a[i][j],a[i][j-1]
elif j+1<m and abs(a[i][j+1]-a[i+1][j])>1:
a[i][j+1],a[i][j]=a[i][j], a[i][j+1]
for i in range(n):
for j in range(m):
if (j+1<m and abs(a[i][j]-a[i][j+1])<2) or (i+1<n and abs(a[i][j]-a[i+1][j])<2):
print(-1)
return
for i in a:
print(*i)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 98,225 | 8 | 196,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1 | instruction | 0 | 98,226 | 8 | 196,452 |
Tags: constructive algorithms
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
z,f=n*m,1
y,a=z-(z%2==0),[[] for _ in range(n)]
for i in range(n):
for j in range(m):
a[i].append(y)
y-=2
if y<0:
y=z-(z%2)
for i in range(n):
for j in range(m):
if i+1<n and abs(a[i][j]-a[i+1][j])<2:
if j>0 and abs(a[i][j-1]-a[i+1][j])>1:
a[i][j-1],a[i][j]=a[i][j],a[i][j-1]
elif j+1<m and abs(a[i][j+1]-a[i+1][j])>1:
a[i][j+1],a[i][j]=a[i][j], a[i][j+1]
for i in range(n):
for j in range(m):
if (j+1<m and abs(a[i][j]-a[i][j+1])<2) or (i+1<n and abs(a[i][j]-a[i+1][j])<2):
f=0
break
if not f:
break
if f:
for i in a:
print(*i)
else:
print(-1)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 98,226 | 8 | 196,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n Γ m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 β€ n, m β€ 100) β the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each β the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function5():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
import os,sys
from io import BytesIO,IOBase
def main():
n,m = map(int,input().split())
arr = [[0]*m for _ in range(n)]
if m >= 4:
arr[0] = list(range(m-1,0,-2))+list(range(m,0,-2))
for i in range(1,n):
for j in range(m):
arr[i][j] = arr[i-1][j]+m
elif n >= 4:
for i,ind in enumerate(list(range(m-1,0,-2))+list(range(m,0,-2))):
arr[ind][0] = i
for j in range(1,m):
for i in range(n):
arr[i][j] = arr[i][j-1]+n
else:
if n == 2 and m == 3:
arr = [[3,6,2],[5,1,4]]
elif n == 3 and m == 2:
arr = [[3,5],[6,1],[2,4]]
elif n == 3 and m == 3:
arr = [[1,7,4],[3,9,6],[5,2,8]]
elif n == 1 and m == 1:
arr = [[1]]
else:
print(-1)
return
for i in arr:
print(*i)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function6():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function7():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function8():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__ == '__main__':
main()
``` | instruction | 0 | 98,228 | 8 | 196,456 |
No | output | 1 | 98,228 | 8 | 196,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 β€ n β€ 10, 0 β€ t β€ 10 000) β the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer β the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty. | instruction | 0 | 98,263 | 8 | 196,526 |
Tags: implementation, math, math
Correct Solution:
```
n, t = map(int, input().split())
a = [[None]] + [[0] * i for i in range(1, n + 1)]
c = 2 ** (n + 1)
for _ in range(t):
a[1][0] += c
for i in range(1, n + 1):
for j in range(i):
if a[i][j] > c:
diff = a[i][j] - c
a[i][j] = c
if i < n:
a[i + 1][j] += diff // 2
a[i + 1][j + 1] += diff // 2
full = 0
for i in range(1, n + 1):
for j in range(i):
full += (a[i][j] == c)
print(full)
``` | output | 1 | 98,263 | 8 | 196,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 β€ n β€ 10, 0 β€ t β€ 10 000) β the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer β the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty. | instruction | 0 | 98,264 | 8 | 196,528 |
Tags: implementation, math, math
Correct Solution:
```
n, t = map(int, input().split())
t = min(2000, t)
L = [ [0.0]*n for i in range(n) ]
for _ in range(t) :
L[0][0] += 1.0
for i in range(n-1) :
for j in range(i+1) :
if L[i][j] > 1.0 :
x = (L[i][j] - 1.0) / 2
L[i+1][j] += x
L[i+1][j+1] += x
L[i][j] = 1.0
ans = 0
for i in range(n) :
for j in range(i+1) :
if L[i][j] > 0.9999999999 :
ans += 1
print(ans)
``` | output | 1 | 98,264 | 8 | 196,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 β€ n β€ 10, 0 β€ t β€ 10 000) β the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer β the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty. | instruction | 0 | 98,265 | 8 | 196,530 |
Tags: implementation, math, math
Correct Solution:
```
n,t=list(map(int,input().split()))
li=[[t-1]]
cnt=1 if t-1>=0 else 0
for i in range(2,n+1):
tr=li[-1][0]/2-1
if tr>=0:
cnt+=2
li.append([tr])
for j in range(1,i-1):
a=li[-2][j-1]
b=li[-2][j]
temp=-1
if a >0 :
temp+=a/2
if b>0:
temp+=b/2
if temp>=0:
cnt+=1
li[-1].append(temp)
li[-1].append(tr)
print(cnt)
``` | output | 1 | 98,265 | 8 | 196,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 β€ n β€ 10, 0 β€ t β€ 10 000) β the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer β the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty. | instruction | 0 | 98,266 | 8 | 196,532 |
Tags: implementation, math, math
Correct Solution:
```
a = 1<<10
bocal = [[0]*(i+1) for i in range(10)]
ans = [[0]*(i+1) for i in range(10)]
x = input().split()
n = int(x[0])
t = int(x[1])
bocal[0][0]=a
ans[0][0]=1
for z in range(2,2000):
bocal[0][0]+=a
for i in range(10):
for j in range(i+1):
if ans[i][j]==0 and bocal[i][j]>=a:
ans[i][j]=z
if bocal[i][j]>a:
x = bocal[i][j]-a
if i<9:
bocal[i+1][j]+=x//2
bocal[i+1][j+1]+=x//2
bocal[i][j]=a
a=0
for i in range(n):
for j in range(i+1):
if ans[i][j]<=t:
a+=1
print(a)
``` | output | 1 | 98,266 | 8 | 196,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 β€ n β€ 10, 0 β€ t β€ 10 000) β the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer β the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty. | instruction | 0 | 98,267 | 8 | 196,534 |
Tags: implementation, math, math
Correct Solution:
```
n,t=map(int, input().split())
m=[[0 for _ in range(11)]for _ in range(11)]
m[1][1]=t
for i in range(1, n):
for j in range(1, n + 1):
if m[i][j] > 1:
m[i + 1][j] += (m[i][j] - 1) / 2
m[i + 1][j + 1] += (m[i][j] - 1) / 2
print(sum([sum(map(lambda x: x >= 1, m[i])) for i in range(11)]))
``` | output | 1 | 98,267 | 8 | 196,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 β€ n β€ 10, 0 β€ t β€ 10 000) β the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer β the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty. | instruction | 0 | 98,268 | 8 | 196,536 |
Tags: implementation, math, math
Correct Solution:
```
inin=input().split(' ')
n=int(inin[0])
t=int(inin[1])
mat=[]
for i in range(n+2):
mat.append([0.0]*(n+2))
# def add(i,j,amt):
# if i>=n or j<0 or j>=n:
# return
# mat[i][j]+=amt
# if mat[i][j]>1:
# over=mat[i][j]-1
# mat[i][j]-=over
# add(i+1,j,over/2)
# add(i+1,j+1,over/2)
for time in range(t):
# add(0,0,1)
mat[1][1]+=1.0
for i in range(1,n+1):
for j in range(1,i+1):
if mat[i][j]>1.0:
over=mat[i][j]-1.0
mat[i+1][j]+=over/2
mat[i+1][j+1]+=over/2
mat[i][j]=1.0
result=0
for i in range(1,n+1):
for j in range(1,i+1):
if mat[i][j]>=1.0:
result+=1
print(result)
# for line in mat:
# # print(line)
# for b in line:
# print(str(b),end='\t')
# print()
``` | output | 1 | 98,268 | 8 | 196,537 |
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