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.
On a random day, Neko found n treasure chests and m keys. The i-th chest has an integer a_i written on it and the j-th key has an integer b_j on it. Neko knows those chests contain the powerful mysterious green Grapes, thus Neko wants to open as many treasure chests as possible.
The j-th key can be used to unlock the i-th chest if and only if the sum of the key number and the chest number is an odd number. Formally, a_i + b_j ≡ 1 \pmod{2}. One key can be used to open at most one chest, and one chest can be opened at most once.
Find the maximum number of chests Neko can open.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 10^5) — the number of chests and the number of keys.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the numbers written on the treasure chests.
The third line contains m integers b_1, b_2, …, b_m (1 ≤ b_i ≤ 10^9) — the numbers written on the keys.
Output
Print the maximum number of chests you can open.
Examples
Input
5 4
9 14 6 2 11
8 4 7 20
Output
3
Input
5 1
2 4 6 8 10
5
Output
1
Input
1 4
10
20 30 40 50
Output
0
Note
In the first example, one possible way to unlock 3 chests is as follows:
* Use first key to unlock the fifth chest,
* Use third key to unlock the second chest,
* Use fourth key to unlock the first chest.
In the second example, you can use the only key to unlock any single chest (note that one key can't be used twice).
In the third example, no key can unlock the given chest.
Submitted Solution:
```
input()
result = 0
chests = list(map(int, input().split()))
keys = list(map(int, input().split()))
even_chests = 0
odd_chests = 0
even_keys = 0
odd_keys = 0
for i in chests:
if i % 2 == 0:
even_chests += 1
else:
odd_chests += 1
for i in keys:
if i % 2 == 0:
even_keys += 1
else:
odd_keys += 1
diff = min(odd_keys, even_chests)
odd_keys -= diff
even_chests -= diff
result += diff
diff = min(even_keys, odd_chests)
even_keys -= diff
odd_chests -= diff
result += diff
print(result + min(odd_chests, odd_keys))
``` | instruction | 0 | 62,351 | 8 | 124,702 |
No | output | 1 | 62,351 | 8 | 124,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a random day, Neko found n treasure chests and m keys. The i-th chest has an integer a_i written on it and the j-th key has an integer b_j on it. Neko knows those chests contain the powerful mysterious green Grapes, thus Neko wants to open as many treasure chests as possible.
The j-th key can be used to unlock the i-th chest if and only if the sum of the key number and the chest number is an odd number. Formally, a_i + b_j ≡ 1 \pmod{2}. One key can be used to open at most one chest, and one chest can be opened at most once.
Find the maximum number of chests Neko can open.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 10^5) — the number of chests and the number of keys.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the numbers written on the treasure chests.
The third line contains m integers b_1, b_2, …, b_m (1 ≤ b_i ≤ 10^9) — the numbers written on the keys.
Output
Print the maximum number of chests you can open.
Examples
Input
5 4
9 14 6 2 11
8 4 7 20
Output
3
Input
5 1
2 4 6 8 10
5
Output
1
Input
1 4
10
20 30 40 50
Output
0
Note
In the first example, one possible way to unlock 3 chests is as follows:
* Use first key to unlock the fifth chest,
* Use third key to unlock the second chest,
* Use fourth key to unlock the first chest.
In the second example, you can use the only key to unlock any single chest (note that one key can't be used twice).
In the third example, no key can unlock the given chest.
Submitted Solution:
```
''' CODED WITH LOVE BY SATYAM KUMAR '''
from sys import stdin, stdout
import cProfile, math
from collections import Counter,defaultdict,deque
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27) # new thread will get stack of such size
fac_warmup = False
printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
class merge_find:
def __init__(self,n):
self.parent = list(range(n))
self.size = [1]*n
self.num_sets = n
self.lista = [[_] for _ in range(n)]
def find(self,a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self,a,b):
a = self.find(a)
b = self.find(b)
if a==b:
return
if self.size[a]<self.size[b]:
a,b = b,a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
self.lista[a] += self.lista[b]
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def primeFactors(n): #n**0.5 complex
factors = dict()
for i in range(2,math.ceil(math.sqrt(n))+1):
while n % i== 0:
if i in factors:
factors[i]+=1
else: factors[i]=1
n = n // i
if n>2:
factors[n]=1
return (factors)
def all_factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
def fibonacci_modP(n,MOD):
if n<2: return 1
#print (n,MOD)
return (cached_fn(fibonacci_modP, (n+1)//2, MOD)*cached_fn(fibonacci_modP, n//2, MOD) + cached_fn(fibonacci_modP, (n-1) // 2, MOD)*cached_fn(fibonacci_modP, (n-2) // 2, MOD)) % MOD
def factorial_modP_Wilson(n , p):
if (p <= n):
return 0
res = (p - 1)
for i in range (n + 1, p):
res = (res * cached_fn(InverseEuler,i, p)) % p
return res
def binary(n,digits = 20):
b = bin(n)[2:]
b = '0'*(digits-len(b))+b
return b
def isprime(n):
"""Returns True if n is prime."""
if n < 4:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
factorial_modP = []
def warm_up_fac(MOD):
global factorial_modP,fac_warmup
if fac_warmup: return
factorial_modP= [1 for _ in range(fac_warmup_size+1)]
for i in range(2,fac_warmup_size):
factorial_modP[i]= (factorial_modP[i-1]*i) % MOD
fac_warmup = True
def InverseEuler(n,MOD):
return pow(n,MOD-2,MOD)
def nCr(n, r, MOD):
global fac_warmup,factorial_modP
if not fac_warmup:
warm_up_fac(MOD)
fac_warmup = True
return (factorial_modP[n]*((pow(factorial_modP[r], MOD-2, MOD) * pow(factorial_modP[n-r], MOD-2, MOD)) % MOD)) % MOD
def test_print(*args):
if testingMode:
print(args)
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def display_2D_list(li):
for i in li:
print(i)
def prefix_sum(li):
sm = 0
res = []
for i in li:
sm+=i
res.append(sm)
return res
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
import heapq,itertools
pq = [] # list of entries arranged in a heap
entry_finder = {} # mapping of tasks to entries
REMOVED = '<removed-task>'
def add_task(task, priority=0):
'Add a new task or update the priority of an existing task'
if task in entry_finder:
remove_task(task)
count = next(counter)
entry = [priority, count, task]
entry_finder[task] = entry
heapq.heappush(pq, entry)
def remove_task(task):
'Mark an existing task as REMOVED. Raise KeyError if not found.'
entry = entry_finder.pop(task)
entry[-1] = REMOVED
def pop_task():
'Remove and return the lowest priority task. Raise KeyError if empty.'
while pq:
priority, count, task = heapq.heappop(pq)
if task is not REMOVED:
del entry_finder[task]
return task
raise KeyError('pop from an empty priority queue')
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
def ncr (n,r):
return math.factorial(n)/(math.factorial(n-r)*math.factorial(r))
def binary_serach(i,li):
#print("Search for ",i)
fn = lambda x: li[x]-x//i
x = -1
b = len(li)
while b>=1:
#print(b,x)
while b+x<len(li) and fn(b+x)>0: #Change this condition 2 to whatever you like
x+=b
b=b//2
return x
# -------------------------------------------------------------- MAIN PROGRAM
TestCases = False
testingMode = False
fac_warmup_size = 10**5+100
optimiseForReccursion = True #Can not be used clubbed with TestCases # WHen using recursive functions, use Python 3
from math import factorial
def main():
n, q = get_tuple()
ai = get_list()
bi = get_list()
o1 = len([x for x in ai if x%2==1])
o2 = len([x for x in bi if x%2==1])
e1 = len([x for x in ai if x%2==0])
e2 = len([x for x in bi if x%2==0])
print(min(o1,e1)+min(o2,e2))
# --------------------------------------------------------------------- END=
if TestCases:
for i in range(get_int()):
cProfile.run('main()') if testingMode else main()
else: (cProfile.run('main()') if testingMode else main()) if not optimiseForReccursion else threading.Thread(target=main).start()
``` | instruction | 0 | 62,352 | 8 | 124,704 |
No | output | 1 | 62,352 | 8 | 124,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,301 | 8 | 126,602 |
Tags: dp, greedy, two pointers
Correct Solution:
```
import sys
def Myfun():
n, m = map(int, fun2().split())
lines = []
rows = [[0]*(m+2) for i in range(n)]
col = [n]*(m+2)
lstart = [[] for i in range(m+2)]
lend = [[] for i in range(m+2)]
dp = [[0]*(m+2) for i in range(m+2)]
for i in range(n):
for j in range(fun3()):
l, r = fun1()
lstart[l].append((i, r + 1))
lend[r + 1].append((i, l))
for i in range(1,m+2):
for r, start in lend[i]:
#print("end", r, start, i-1)
for p in range(start, i):
rows[r][p] -= 1
if rows[r][p] == 0:
col[p] += 1
for r, end in lstart[i]:
#print("start", r, i, end)
for p in range(i, end):
rows[r][p] += 1
if rows[r][p] == 1:
col[p] -= 1
bcol = col.copy()
brows = [rows[i].copy() for i in range(n)]
#print(i, col)
cc = [None]*(i-1)
for j in range(i-1):
for r, start in lend[j]:
for p in range(start, j):
rows[r][p] -= 1
if rows[r][p] == 0:
col[p] += 1
for r, end in lstart[j]:
for p in range(j, end):
rows[r][p] += 1
if rows[r][p] == 1:
col[p] -= 1
cc[j] = col.copy()
for j in range(i-2,-1,-1):
d = 0
col = cc[j]
for p in range(j+1, i):
d = max(d, dp[j][p] + dp[p][i] + col[p]**2)
dp[j][i] = d
#print(j, i, col, d)
col = bcol
rows = brows
print(dp[0][m+1])
def fun2():
return sys.stdin.readline().strip()
def fun3():
return int(fun2())
def fun1():
return map(int, fun2().split())
Myfun()
``` | output | 1 | 63,301 | 8 | 126,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,302 | 8 | 126,604 |
Tags: dp, greedy, two pointers
Correct Solution:
```
from sys import stdin, stdout
if __name__ == '__main__':
def omkar_and_last_floor(a, n, m):
dp = [[0 for c in range(m)] for r in range(m)]
#print(dp)
for r in range(m):
for l in range(r,-1,-1):
for k in range(l, r+1):
cnt = 0
for i in range(n):
if l <= a[i][k][0] and a[i][k][1] <= r:
cnt += 1
lr = cnt*cnt
if k-1 >= l:
lr += dp[l][k-1]
if k+1 <= r:
lr += dp[k + 1][r]
dp[l][r] = max(dp[l][r], lr)
#print(dp)
return dp[0][m-1]
n, m = map(int, stdin.readline().split())
a = [[[0,0] for c in range(m)] for r in range(n)]
for i in range(n):
k = int(stdin.readline())
for j in range(k):
l, r = map(int, stdin.readline().split())
for x in range(l, r+1):
a[i][x-1][0] = l-1
a[i][x-1][1] = r-1
print(omkar_and_last_floor(a, n, m))
``` | output | 1 | 63,302 | 8 | 126,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,303 | 8 | 126,606 |
Tags: dp, greedy, two pointers
Correct Solution:
```
from bisect import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def solve():
dp = [[0] * (w + 1) for _ in range(w + 1)]
for d in range(1, w + 1):
for l in range(w - d + 1):
r = l + d
cnt = [0] * (w + 1)
for i in range(h):
sl=bisect_left(seg[i], l)
sr=bisect_right(seg[i], r)
if sl==sr:continue
b = seg[i][sl]
e = seg[i][sr - 1]
cnt[b] += 1
cnt[e] -= 1
for j in range(l, r):
cnt[j + 1] += cnt[j]
if cnt[j] == 0: continue
dp[l][r] = max(dp[l][r], cnt[j] ** 2 + dp[l][j] + dp[j + 1][r])
print(dp[0][w])
#p2D(dp)
h,w=MI()
seg=[[0] for _ in range(h)]
for i in range(h):
for _ in range(II()):
l,r=MI()
seg[i].append(r)
solve()
``` | output | 1 | 63,303 | 8 | 126,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,304 | 8 | 126,608 |
Tags: dp, greedy, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
s = [[[0 for _ in range(m)] for _ in range(m)] for _ in range(m)]
for i in range(n):
k = int(input())
for j in range(k):
l, r = map(int, input().split())
l -= 1
r -= 1
for x in range(l, r + 1):
s[l][r][x] += 1
for i in range(m):
for j in range(max(1, i), m):
for k in range(i, j + 1):
s[i][j][k] += s[i][j - 1][k]
for i in range(m - 2, -1, -1):
for j in range(i, m):
for k in range(i, j + 1):
s[i][j][k] += s[i + 1][j][k]
f = [[0 for _ in range(m)] for _ in range(m)]
for i in range(m - 1, -1, -1):
for j in range(i, m):
for k in range(i, j + 1):
a = f[i][k - 1] if i <= k - 1 else 0
b = f[k + 1][j] if k + 1 <= j else 0
f[i][j] = max(f[i][j], a + b + s[i][j][k] ** 2)
print(f[0][m - 1])
``` | output | 1 | 63,304 | 8 | 126,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,305 | 8 | 126,610 |
Tags: dp, greedy, two pointers
Correct Solution:
```
from bisect import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def solve():
dp = [[0] * (w + 1) for _ in range(w + 1)]
for d in range(1, w + 1):
for l in range(w - d + 1):
r = l + d
cnt = [0] * (w + 1)
for i in range(h):
sl=bisect_left(seg[i], l)
sr=bisect_right(seg[i], r)
if sl==sr:continue
b = seg[i][sl]
e = seg[i][sr - 1]
cnt[b] += 1
cnt[e] -= 1
#for j in range(l, r): cnt[j + 1] += cnt[j]
#print(cnt)
#mx = max(cnt)
#if mx == 0: continue
for j in range(l, r):
cnt[j + 1] += cnt[j]
if cnt[j] == 0: continue
dp[l][r] = max(dp[l][r], cnt[j] ** 2 + dp[l][j] + dp[j + 1][r])
print(dp[0][w])
#p2D(dp)
h,w=MI()
seg=[[0] for _ in range(h)]
for i in range(h):
for _ in range(II()):
l,r=MI()
seg[i].append(r)
solve()
``` | output | 1 | 63,305 | 8 | 126,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,306 | 8 | 126,612 |
Tags: dp, greedy, two pointers
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [(ch,len(list(g))) for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
@lru_cache(None)
def dp(l,r):
if r<l:
return 0
res=0
for k in range(l,r+1):
c=0
for i in range(n):
a,b=bnd[i][k]
if l<=a and b<=r:
c+=1
res=max(res,c*c+dp(l,k-1)+dp(k+1,r))
return res
t=1
for i in range(t):
n,m=RL()
bnd=[[[] for j in range(m+1)] for i in range(n)]
for i in range(n):
k=N()
for a in range(k):
l,r=RL()
for j in range(l,r+1):
bnd[i][j]=[l,r]
ans=dp(1,m)
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | output | 1 | 63,306 | 8 | 126,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,307 | 8 | 126,614 |
Tags: dp, greedy, two pointers
Correct Solution:
```
import sys
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def solve():
n, m = mints()
lines = []
rows = [[0]*(m+2) for i in range(n)]
col = [n]*(m+2)
lstart = [[] for i in range(m+2)]
lend = [[] for i in range(m+2)]
dp = [[0]*(m+2) for i in range(m+2)]
for i in range(n):
for j in range(mint()):
l, r = mints()
lstart[l].append((i, r + 1))
lend[r + 1].append((i, l))
for i in range(1,m+2):
for r, start in lend[i]:
#print("end", r, start, i-1)
for p in range(start, i):
rows[r][p] -= 1
if rows[r][p] == 0:
col[p] += 1
for r, end in lstart[i]:
#print("start", r, i, end)
for p in range(i, end):
rows[r][p] += 1
if rows[r][p] == 1:
col[p] -= 1
bcol = col.copy()
brows = [rows[i].copy() for i in range(n)]
#print(i, col)
cc = [None]*(i-1)
for j in range(i-1):
for r, start in lend[j]:
for p in range(start, j):
rows[r][p] -= 1
if rows[r][p] == 0:
col[p] += 1
for r, end in lstart[j]:
for p in range(j, end):
rows[r][p] += 1
if rows[r][p] == 1:
col[p] -= 1
cc[j] = col.copy()
for j in range(i-2,-1,-1):
d = 0
col = cc[j]
for p in range(j+1, i):
d = max(d, dp[j][p] + dp[p][i] + col[p]**2)
dp[j][i] = d
#print(j, i, col, d)
col = bcol
rows = brows
print(dp[0][m+1])
solve()
``` | output | 1 | 63,307 | 8 | 126,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality. | instruction | 0 | 63,308 | 8 | 126,616 |
Tags: dp, greedy, two pointers
Correct Solution:
```
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s); sys.stdout.write('\n')
def wi(n): sys.stdout.write(str(n)); sys.stdout.write('\n')
def wia(a, sep=' '): sys.stdout.write(sep.join([str(x) for x in a])); sys.stdout.write('\n')
from functools import lru_cache
def solve(n, m, a):
b = [[] for _ in range(m)]
for j in range(m):
for l, r, i in a:
if l <= j <= r:
b[j].append((l, r, i))
@lru_cache(maxsize=None)
def go(left, right):
if left > right:
return 0
best = 0
for mid in range(left, right + 1):
cnt = 0
for l, r, i in b[mid]:
if l <= mid <= r and l >= left and r <= right:
cnt += 1
best = max(best, cnt * cnt + go(left, mid - 1) + go(mid + 1, right))
return best
return go(0, m-1)
def main():
n, m = ria()
a = set()
for i in range(n):
k = ri()
for j in range(k):
l, r = ria()
l -= 1
r -= 1
a.add((l, r, i))
wi(solve(n, m, a))
if __name__ == '__main__':
main()
``` | output | 1 | 63,308 | 8 | 126,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality.
Submitted Solution:
```
def solve():
n, m = map(int, input().split())
le, ri = [[0]*m for i in range(n)], [[0]*m for i in range(n)]
for i in range(n):
t = int(input())
for _ in range(t):
x, y = map(int, input().split())
x -= 1
y -= 1
for j in range(x, y+1):
le[i][j] = x
ri[i][j] = y
dp = [[0]*m for i in range(m)]
for i in range(m):
c = 0
for j in range(n):
if le[j][i] == ri[j][i] == i:
c += 1
dp[i][i] = c*c
for l in range(m-2, -1, -1):
for r in range(l+1, m):
c = 0
for i in range(n):
if le[i][l] >= l and ri[i][l] <= r:
c += 1
dp[l][r] = c*c + dp[l+1][r]
c = 0
for i in range(n):
if le[i][r] >= l and ri[i][r] <= r:
c += 1
dp[l][r] = max(dp[l][r], c*c+dp[l][r-1])
for j in range(l+1, r):
c = 0
for i in range(n):
if le[i][j] >= l and ri[i][j] <= r:
c += 1
dp[l][r] = max(dp[l][r], c*c+dp[l][j-1]+dp[j+1][r])
return dp[0][-1]
import sys
input = lambda: sys.stdin.readline().rstrip()
print(solve())
``` | instruction | 0 | 63,309 | 8 | 126,618 |
Yes | output | 1 | 63,309 | 8 | 126,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality.
Submitted Solution:
```
import sys
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def solve():
n, m = mints()
lines = []
rows = [[0]*(m+2) for i in range(n)]
col = [n]*(m+2)
lstart = [[] for i in range(m+2)]
lend = [[] for i in range(m+2)]
dp = [[0]*(m+2) for i in range(m+2)]
for i in range(n):
for j in range(mint()):
l, r = mints()
lstart[l].append((i, r + 1))
lend[r + 1].append((i, l))
for i in range(1,m+2):
for r, start in lend[i]:
#print("end", r, start, i-1)
for p in range(start, i):
rows[r][p] -= 1
if rows[r][p] == 0:
col[p] += 1
for r, end in lstart[i]:
#print("start", r, i, end)
for p in range(i, end):
rows[r][p] += 1
if rows[r][p] == 1:
col[p] -= 1
bcol = col.copy()
brows = [rows[i].copy() for i in range(n)]
#print(i, col)
for j in range(i-1):
for r, start in lend[j]:
for p in range(start, j):
rows[r][p] -= 1
if rows[r][p] == 0:
col[p] += 1
for r, end in lstart[j]:
for p in range(j, end):
rows[r][p] += 1
if rows[r][p] == 1:
col[p] -= 1
d = 0
for p in range(j+1, i):
d = max(d, dp[j][p] + dp[p][i] + col[p]**2)
dp[j][i] = d
#print(j, i, col, d)
col = bcol
rows = brows
print(dp[0][m+1])
solve()
``` | instruction | 0 | 63,310 | 8 | 126,620 |
No | output | 1 | 63,310 | 8 | 126,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality.
Submitted Solution:
```
n, m = map(int, input().strip().split())
ks = []
lrs = []
ans = []
for _ in range(n):
k = int(input())
for i in range(k):
l, r = map(int, input().strip().split())
if(r != m):
ans.append(l)
else:
ans.append(r)
E = 0
for t in range(1, m+1):
E += (ans.count(t))**2
print(E)
``` | instruction | 0 | 63,311 | 8 | 126,622 |
No | output | 1 | 63,311 | 8 | 126,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality.
Submitted Solution:
```
n,m = map(int,input().split())
ansv = [0 for i in range(m)]
lines = [[[], 0] for i in range(n)]
for i in range(n):
kolv = int(input())
a, b = map(int, input().split())
for j in range(kolv - 1):
a,b = map(int,input().split())
lines[i][0].append(a - 1)
lines[i][0].append(m)
if n == 1:
print(m * m)
exit()
for pos in range(m):
b = []
for i in range(n):
if lines[i][0][lines[i][1]] == pos + 1:
if lines[i][1] == 0:
b.append(0)
else:
b.append(lines[i][0][lines[i][1] - 1])
lines[i][1] += 1
b.sort()
for x in b:
mx = -1
mxpos = -1
for i in range(x, pos + 1):
if ansv[i] >= mx:
mx = ansv[i]
mxpos = i
ansv[mxpos] += 1
ans = 0
for i in ansv:
ans += i * i
print(ans)
``` | instruction | 0 | 63,312 | 8 | 126,624 |
No | output | 1 | 63,312 | 8 | 126,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is building a house. He wants to decide how to make the floor plan for the last floor.
Omkar's floor starts out as n rows of m zeros (1 ≤ n,m ≤ 100). Every row is divided into intervals such that every 0 in the row is in exactly 1 interval. For every interval for every row, Omkar can change exactly one of the 0s contained in that interval to a 1. Omkar defines the quality of a floor as the sum of the squares of the sums of the values in each column, i. e. if the sum of the values in the i-th column is q_i, then the quality of the floor is ∑_{i = 1}^m q_i^2.
Help Omkar find the maximum quality that the floor can have.
Input
The first line contains two integers, n and m (1 ≤ n,m ≤ 100), which are the number of rows and number of columns, respectively.
You will then receive a description of the intervals in each row. For every row i from 1 to n: The first row contains a single integer k_i (1 ≤ k_i ≤ m), which is the number of intervals on row i. The j-th of the next k_i lines contains two integers l_{i,j} and r_{i,j}, which are the left and right bound (both inclusive), respectively, of the j-th interval of the i-th row. It is guaranteed that all intervals other than the first interval will be directly after the interval before it. Formally, l_{i,1} = 1, l_{i,j} ≤ r_{i,j} for all 1 ≤ j ≤ k_i, r_{i,j-1} + 1 = l_{i,j} for all 2 ≤ j ≤ k_i, and r_{i,k_i} = m.
Output
Output one integer, which is the maximum possible quality of an eligible floor plan.
Example
Input
4 5
2
1 2
3 5
2
1 3
4 5
3
1 1
2 4
5 5
3
1 1
2 2
3 5
Output
36
Note
The given test case corresponds to the following diagram. Cells in the same row and have the same number are a part of the same interval.
<image>
The most optimal assignment is:
<image>
The sum of the 1st column is 4, the sum of the 2nd column is 2, the sum of the 3rd and 4th columns are 0, and the sum of the 5th column is 4.
The quality of this floor plan is 4^2 + 2^2 + 0^2 + 0^2 + 4^2 = 36. You can show that there is no floor plan with a higher quality.
Submitted Solution:
```
import sys
import heapq, functools, collections
import math, random
from collections import Counter, defaultdict
# available on Google, not available on Codeforces
# import numpy as np
# import scipy
def solve(grids, numcol): # fix inputs here
console("----- solving ------")
console(grids)
# return a string (i.e. not a list or matrix)
res = 0
while grids:
curs = []
for i in range(numcol):
cnt = 0
for grid in grids:
for a,b in grid:
if a <= i <= b:
cnt += 1
curs.append((cnt, i))
console(max(curs))
cnt, i = max(curs)
new_grids = []
for grid in grids:
new_grid = []
for a,b in grid:
if a <= i <= b:
pass
else:
new_grid.append((a,b))
if new_grid:
new_grids.append(new_grid)
grids = new_grids
res += cnt**2
# break
console(grids)
return res
def console(*args): # the judge will not read these print statement
# print('\033[36m', *args, '\033[0m', file=sys.stderr)
return
# fast read all
# sys.stdin.readlines()
# for case_num in range(int(input())):
# read line as a string
# strr = input()
# read line as an integer
# k = int(input())
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
nrows, numcol = list(map(int,input().split()))
# read matrix and parse as integers (after reading read nrows)
# lst = list(map(int,input().split()))
# nrows = lst[0] # index containing information, please change
grids = []
for _ in range(nrows):
k = int(input())
grid = []
for _ in range(k):
grid.append(list(map(int,input().split())))
grids.append([(a-1, b-1) for a,b in grid])
res = solve(grids, numcol) # please change
# Google - case number required
# print("Case #{}: {}".format(case_num+1, res))
# Codeforces - no case number required
print(res)
``` | instruction | 0 | 63,313 | 8 | 126,626 |
No | output | 1 | 63,313 | 8 | 126,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,509 | 8 | 127,018 |
Tags: implementation
Correct Solution:
```
n = int(input())
h = 0
flr = 0
while True:
h += 1
flr += (h * (h + 1)) / 2
if flr > n:
break
print(h - 1)
``` | output | 1 | 63,509 | 8 | 127,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,510 | 8 | 127,020 |
Tags: implementation
Correct Solution:
```
n=int(input())
z=1
c=1
d=0
while n>=z:
n=n-z
c=c+1
z=z+c
d=d+1
print(d)
``` | output | 1 | 63,510 | 8 | 127,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,511 | 8 | 127,022 |
Tags: implementation
Correct Solution:
```
t=int(input())
n=1
c=0
while(True):
t=t-(n*(n+1)/2)
n=n+1
c=c+1
if(t<(n)*(n+1)/2):
break
print(c)
``` | output | 1 | 63,511 | 8 | 127,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,512 | 8 | 127,024 |
Tags: implementation
Correct Solution:
```
n=int(input())
s=0
i=1
k=0
a=[]
while s<n:
s=s+i
a.append(s)
i=i+1
b=[]
s=0
for k in a:
s=s+k
b.append(s)
k=0
for i in b:
if i<=n:
k=k+1
if n==1:
print(1)
else:
print(k)
``` | output | 1 | 63,512 | 8 | 127,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,513 | 8 | 127,026 |
Tags: implementation
Correct Solution:
```
n = int(input())
sum = 0
k = 0
while (sum + (k + 1) * (k + 2) // 2 <= n):
k += 1
sum += k * (k + 1) // 2
print(k)
``` | output | 1 | 63,513 | 8 | 127,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,514 | 8 | 127,028 |
Tags: implementation
Correct Solution:
```
def main():
num = int(input())
pyramid = []
pyramid.append(1)
levelBlocks = 1
totalBlocks = 1
levels = 1
curr = 2
while True:
if num <= totalBlocks:
break
levelBlocks += curr
totalBlocks += levelBlocks
curr += 1
levels+= 1
if num < totalBlocks:
levels-=1
break
print(levels)
main()
``` | output | 1 | 63,514 | 8 | 127,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,515 | 8 | 127,030 |
Tags: implementation
Correct Solution:
```
n = int(input())
i = 1
len = 1
sum = 1
while (n >= sum):
i += 1
len += i
sum += len
print(i-1)
``` | output | 1 | 63,515 | 8 | 127,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image> | instruction | 0 | 63,516 | 8 | 127,032 |
Tags: implementation
Correct Solution:
```
k = int(input())
f = lambda x : x*(x+1)//2
x = 1
while k-f(x)>=0:
k-=f(x)
x+=1
print(x-1)
``` | output | 1 | 63,516 | 8 | 127,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
n = int(input())
i = 0
while i * (i + 1) * (i + 2) <= 6 * n:
i += 1
print(i - 1)
``` | instruction | 0 | 63,517 | 8 | 127,034 |
Yes | output | 1 | 63,517 | 8 | 127,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
n = int(input())
height = 0
prev = 0
sum = 0
while(n >= sum):
height += 1
prev += height
sum += prev
print(height - 1)
``` | instruction | 0 | 63,518 | 8 | 127,036 |
Yes | output | 1 | 63,518 | 8 | 127,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
n=int(input())
i=1
a=0
while(n>0):
n=n-i-a
a=a+i
if(n>=a+i+1):
i+=1
else:
break
print(i)
``` | instruction | 0 | 63,519 | 8 | 127,038 |
Yes | output | 1 | 63,519 | 8 | 127,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
def no_ofcube(n):
sum=0
for i in range(1,n+1):
sum+=i
return sum
n=int(input())
sum,cnt=0,0
for i in range(1,n+1):
sum+=no_ofcube(i)
if sum>n:break
cnt+=1
print(cnt)
``` | instruction | 0 | 63,520 | 8 | 127,040 |
Yes | output | 1 | 63,520 | 8 | 127,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
cubes = int(input())
d=cnt=0
h=t=0
while cubes>0:
d+=1
cnt+=d
h+=1
t+=cnt
cubes-=t
print(h)
``` | instruction | 0 | 63,521 | 8 | 127,042 |
No | output | 1 | 63,521 | 8 | 127,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
n = int(input())
res = 1
i = 1
while res <= n :
i += 1
res += i
print(i - 1)
``` | instruction | 0 | 63,522 | 8 | 127,044 |
No | output | 1 | 63,522 | 8 | 127,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
n = int(input())
lst = [0]
i = 1
while (n >= 0):
lst.append(lst[-1] + i)
n -= (lst[-1] + i)
i += 1
print(len(lst) - 1)
``` | instruction | 0 | 63,523 | 8 | 127,046 |
No | output | 1 | 63,523 | 8 | 127,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Output
Print the maximum possible height of the pyramid in the single line.
Examples
Input
1
Output
1
Input
25
Output
4
Note
Illustration to the second sample:
<image>
Submitted Solution:
```
n = int(input())
k = 0
i = 1
t = 0
a = 1
while a <= n:
i = i + 1
k = k + i
t = t + 1
a = a + k
print(t)
``` | instruction | 0 | 63,524 | 8 | 127,048 |
No | output | 1 | 63,524 | 8 | 127,049 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,927 | 8 | 127,854 |
"Correct Solution:
```
def loadIcicle():
icicles = []
line = input().strip().split(" ")
N, L = int(line[0]), int(line[1])
while True:
icicles.append(int(input().strip()))
if len(icicles) == N: break
return icicles, N, L
def calcDiff(icicles, N):
diff_2times = [0]*N
for idx in range(N):
dif_right = icicles[idx+1] - icicles[idx] if idx != N-1 else -icicles[idx]
dif_left = icicles[idx] - icicles[idx-1] if idx != 0 else icicles[idx]
dif_right = 1 if dif_right > 0 else -1
dif_left = 1 if dif_left > 0 else -1
if dif_right - dif_left < 0: diff_2times[idx] = -1
elif dif_right - dif_left > 0: diff_2times[idx] = 1
else: diff_2times[idx] = 0
return diff_2times
icicles, N, L = loadIcicle()
diff_2times = calcDiff(icicles, N)
time = [-1]*N
peakX = [i for i in range(N) if diff_2times[i]==-1]
for i in peakX:
time[i] = L - icicles[i]
isLocalMinL, isLocalMinR = False, False
posL, posR = i, i
while not (isLocalMinL and isLocalMinR):
posL -= 1
if posL < 0:
isLocalMinL = True
if not isLocalMinL:
if time[posL] == -1:
time[posL] = (L-icicles[posL]) + time[posL+1]
else:
time[posL] = (L-icicles[posL]) + max(time[posL-1], time[posL+1])
if diff_2times[posL] == 1:
isLocalMinL = True
posR += 1
if posR >= N:
isLocalMinR = True
if not isLocalMinR:
if time[posR] == -1:
time[posR] = (L-icicles[posR]) + time[posR-1]
else:
time[posR] = (L-icicles[posR]) + max(time[posR-1], time[posR+1])
if diff_2times[posR] == 1:
isLocalMinR = True
print(max(time))
``` | output | 1 | 63,927 | 8 | 127,855 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,928 | 8 | 127,856 |
"Correct Solution:
```
def solve():
n, l = map(int,input().split())
ans = 0
pre = 0
trend = 1
acc = 0
for i in range(n):
length = int(input())
time = l - length
if length > pre and trend >= 0:
acc += time
elif length < pre and trend <= 0:
acc += time
else:
if ans < acc:
ans = acc
acc = time + (l - pre)
trend = length - pre
pre = length
else:
ans = max(ans, acc)
print(ans)
solve()
``` | output | 1 | 63,928 | 8 | 127,857 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,929 | 8 | 127,858 |
"Correct Solution:
```
def loadIcicle():
icicles = []
line = input().strip().split(" ")
N, L = int(line[0]), int(line[1])
while True:
icicles.append(int(input().strip()))
if len(icicles) == N: break
return icicles, N, L
icicles, N, L = loadIcicle()
def calcDiff(icicles, N):
diff_2times = [0]*N
for idx in range(N):
dif_right = icicles[idx+1] - icicles[idx] if idx != N-1 else -icicles[idx]
dif_left = icicles[idx] - icicles[idx-1] if idx != 0 else icicles[idx]
dif_right = 1 if dif_right > 0 else -1
dif_left = 1 if dif_left > 0 else -1
if dif_right - dif_left < 0: diff_2times[idx] = -1
elif dif_right - dif_left > 0: diff_2times[idx] = 1
else: diff_2times[idx] = 0
return diff_2times
diff_2times = calcDiff(icicles, N)
time = [-1]*N
peakX = [i for i in range(N) if diff_2times[i]==-1]
for i in peakX:
time[i] = L - icicles[i]
isLocalMinL, isLocalMinR = False, False
posL = i
posR = i
while not (isLocalMinL and isLocalMinR):
posL -= 1
if posL < 0:
isLocalMinL = True
if not isLocalMinL:
if time[posL] == -1:
time[posL] = (L-icicles[posL]) + time[posL+1]
else:
time[posL] = (L-icicles[posL]) + max(time[posL-1], time[posL+1])
if diff_2times[posL] == 1:
isLocalMinL = True
posR += 1
if posR >= N:
isLocalMinR = True
if not isLocalMinR:
if time[posR] == -1:
time[posR] = (L-icicles[posR]) + time[posR-1]
else:
time[posR] = (L-icicles[posR]) + max(time[posR-1], time[posR+1])
if diff_2times[posR] == 1:
isLocalMinR = True
print(max(time))
``` | output | 1 | 63,929 | 8 | 127,859 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,930 | 8 | 127,860 |
"Correct Solution:
```
n,l = map(int,input().split())
length = [int(input()) for i in range(n)]
pare = [(length[i], i) for i in range(n)]
pare.sort(reverse=True)
ind = [p[1] for p in pare]
dp = [0] * (n + 1)
for i in ind:
if i == 0:
dp[i] = dp[i + 1] + (l - length[i])
else:
dp[i] = max(dp[i - 1], dp[i + 1]) + (l - length[i])
print(max(dp))
``` | output | 1 | 63,930 | 8 | 127,861 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,931 | 8 | 127,862 |
"Correct Solution:
```
def solve():
n,l = map(int,input().split())
length = [int(input()) for i in range(n)]
pare = [(length[i], i) for i in range(n)]
pare.sort(reverse=True)
dp = [0] * (n + 1)
for p in pare:
i = p[1]
if i == 0:
dp[i] = dp[i + 1] + (l - length[i])
else:
dp[i] = max(dp[i - 1], dp[i + 1]) + (l - length[i])
print(max(dp))
solve()
``` | output | 1 | 63,931 | 8 | 127,863 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,932 | 8 | 127,864 |
"Correct Solution:
```
def orderN(N,L,ices):
upPeak = L - ices[0]
downPeak = L - ices[0]
peaks = []
for i in range(len(ices)):
if i < N-1:
if ices[i] < ices[i+1]:
peaks.append(downPeak)
downPeak = L - ices[i+1]
upPeak += L - ices[i+1]
elif ices[i] > ices[i+1]:
peaks.append(upPeak)
upPeak = L - ices[i+1]
downPeak += L - ices[i+1]
elif i == N-1:
peaks.append(upPeak)
peaks.append(downPeak)
print(max(peaks))
N,L = map(int,input().strip().split())
ices = [int(input().strip()) for _ in range(N)]
orderN(N,L,ices)
``` | output | 1 | 63,932 | 8 | 127,865 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,933 | 8 | 127,866 |
"Correct Solution:
```
def loadIcicle():
icicles = []
line = input().strip().split(" ")
N, L = int(line[0]), int(line[1])
while True:
icicles.append(int(input().strip()))
if len(icicles) == N: break
return icicles, N, L
icicles, N, L = loadIcicle()
def calcDiff(icicles, N):
diff_2times = [0]*N
for idx in range(N):
dif_right = icicles[idx+1] - icicles[idx] if idx != N-1 else -icicles[idx]
dif_left = icicles[idx] - icicles[idx-1] if idx != 0 else icicles[idx]
dif_right = 1 if dif_right > 0 else -1
dif_left = 1 if dif_left > 0 else -1
if dif_right - dif_left < 0: diff_2times[idx] = -1
elif dif_right - dif_left > 0: diff_2times[idx] = 1
else: diff_2times[idx] = 0
return diff_2times
diff_2times = calcDiff(icicles, N)
time = [-1]*len(icicles)
peakX = [i for i in range(N) if diff_2times[i]==-1]
for i in peakX:
time[i] = L - icicles[i]
isLocalMinL, isLocalMinR = False, False
posL = i
posR = i
while not (isLocalMinL and isLocalMinR):
posL -= 1
if posL < 0:
isLocalMinL = True
if not isLocalMinL:
if time[posL] == -1:
time[posL] = (L-icicles[posL]) + time[posL+1]
else:
time[posL] = (L-icicles[posL]) + max(time[posL-1], time[posL+1])
if diff_2times[posL] == 1:
isLocalMinL = True
posR += 1
if posR >= N:
isLocalMinR = True
if not isLocalMinR:
if time[posR] == -1:
time[posR] = (L-icicles[posR]) + time[posR-1]
else:
time[posR] = (L-icicles[posR]) + max(time[posR-1], time[posR+1])
if diff_2times[posR] == 1:
isLocalMinR = True
print(max(time))
``` | output | 1 | 63,933 | 8 | 127,867 |
Provide a correct Python 3 solution for this coding contest problem.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8 | instruction | 0 | 63,934 | 8 | 127,868 |
"Correct Solution:
```
def solve():
n, l = map(int,input().split())
ans = 0
pre = 0
prepre = 0
acc = 0
for i in range(n):
length = int(input())
time = l - length
if length > pre and pre >= prepre:
acc += time
elif length < pre and pre <= prepre:
acc += time
else:
if ans < acc:
ans = acc
acc = time + (l - pre)
prepre = pre
pre = length
else:
ans = max(ans, acc)
print(ans)
solve()
``` | output | 1 | 63,934 | 8 | 127,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
def orderN(N,L,ices):
upPeak = L - ices[0]
downPeak = L - ices[0]
peaks = []
for i in range(len(ices)):
if i < N-1:
if ices[i] < ices[i+1]:
peaks.append(downPeak)
downPeak = L - ices[i+1]
upPeak += L - ices[i+1]
else:
peaks.append(upPeak)
upPeak = L - ices[i+1]
downPeak += L - ices[i+1]
else:
peaks.append(upPeak)
peaks.append(downPeak)
print(max(peaks))
N,L = map(int,input().strip().split())
ices = []
while True:
ice = int(input().strip())
ices.append(ice)
if len(ices) == N:
break
orderN(N,L,ices)
``` | instruction | 0 | 63,935 | 8 | 127,870 |
Yes | output | 1 | 63,935 | 8 | 127,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
def orderN(N,L,ices):
upPeak = L - ices[0]
downPeak = L - ices[0]
peaks = []
for i in range(len(ices)):
if i < N-1:
if ices[i] < ices[i+1]:
peaks.append(downPeak)
downPeak = L - ices[i+1]
upPeak += L - ices[i+1]
elif ices[i] > ices[i+1]:
peaks.append(upPeak)
upPeak = L - ices[i+1]
downPeak += L - ices[i+1]
elif i == N-1:
peaks.append(upPeak)
peaks.append(downPeak)
print(max(peaks))
N,L = map(int,input().strip().split())
ices = []
while True:
ice = int(input().strip())
ices.append(ice)
if len(ices) == N:
break
orderN(N,L,ices)
``` | instruction | 0 | 63,936 | 8 | 127,872 |
Yes | output | 1 | 63,936 | 8 | 127,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
def loadIcicle():
icicles = []
line = input().strip().split(" ")
N, L = int(line[0]), int(line[1])
while True:
icicles.append(int(input().strip()))
if len(icicles) == N: break
return icicles, N, L
icicles, N, L = loadIcicle()
def calcDiff(icicles, N):
diff_2times = [0]*N
for idx in range(N):
dif_right = icicles[idx+1] - icicles[idx] if idx != N-1 else -icicles[idx]
dif_left = icicles[idx] - icicles[idx-1] if idx != 0 else icicles[idx]
dif_right = 1 if dif_right > 0 else -1
dif_left = 1 if dif_left > 0 else -1
if dif_right - dif_left < 0: diff_2times[idx] = -1
elif dif_right - dif_left > 0: diff_2times[idx] = 1
else: diff_2times[idx] = 0
return diff_2times
diff_2times = calcDiff(icicles, N)
time = [-1]*N
peakX = [i for i in range(N) if diff_2times[i]==-1]
for i in peakX:
time[i] = L - icicles[i]
isLocalMinL, isLocalMinR = False, False
posL, posR = i, i
while not (isLocalMinL and isLocalMinR):
posL -= 1
if posL < 0:
isLocalMinL = True
if not isLocalMinL:
if time[posL] == -1:
time[posL] = (L-icicles[posL]) + time[posL+1]
else:
time[posL] = (L-icicles[posL]) + max(time[posL-1], time[posL+1])
if diff_2times[posL] == 1:
isLocalMinL = True
posR += 1
if posR >= N:
isLocalMinR = True
if not isLocalMinR:
if time[posR] == -1:
time[posR] = (L-icicles[posR]) + time[posR-1]
else:
time[posR] = (L-icicles[posR]) + max(time[posR-1], time[posR+1])
if diff_2times[posR] == 1:
isLocalMinR = True
print(max(time))
``` | instruction | 0 | 63,937 | 8 | 127,874 |
Yes | output | 1 | 63,937 | 8 | 127,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
n, l = map(int,input().split())
ai = [[int(input()),i] for i in range(n)]
ai.sort(key = lambda x: x[0])
time = [0]*(n+2)
for i in ai:
idx = i[1]
time[idx+1] = max(time[idx],time[idx+2]) + l - i[0]
print(max(time))
``` | instruction | 0 | 63,938 | 8 | 127,876 |
Yes | output | 1 | 63,938 | 8 | 127,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
def greedy(N,L,ices):
counts = 1
while True:
# print(ices)
for i in range(len(ices)):
if 0 < i < N-1:
if ices[i] >= L:
ices[i] = 0
break
if ices[i] > ices[i-1] and ices[i] > ices[i+1]:
ices[i] += 1
if ices[i] >= L:
ices[i] = 0
elif i == 0:
if ices[i] >= L:
ices[i] = 0
break
if ices[i] > ices[1]:
ices[i] += 1
if ices[i] >= L:
ices[i] = 0
elif i == N-1:
if ices[i] >= L:
ices[i] = 0
break
if ices[i] > ices[N-2]:
ices[i] += 1
if ices[i] >= L:
ices[i] = 0
if ices.count(0) == N:
print(counts)
return
counts += 1
N,L = map(int,input().strip().split())
ices = []
while True:
ice = int(input().strip())
ices.append(ice)
if len(ices) == N:
break
# N = 4
# L = 6
# ices = [4,2,3,5]
greedy(N,L,ices)
``` | instruction | 0 | 63,939 | 8 | 127,878 |
No | output | 1 | 63,939 | 8 | 127,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
def solve():
n,l = map(int,input().split())
length = [int(input()) for i in range(n)]
pare = [(length[i], i) for i in range(n)]
pare.sort(reverse=True)
dp = [0] * (n + 1)
for p in pare:
i = p[0]
if i == 0:
dp[i] = dp[i + 1] + (l - length[i])
else:
dp[i] = max(dp[i - 1], dp[i + 1]) + (l - length[i])
print(max(dp))
solve()
``` | instruction | 0 | 63,940 | 8 | 127,880 |
No | output | 1 | 63,940 | 8 | 127,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
import copy
def greedy(N,L,ices):
counts = 1
while True:
# print(ices)
copied = copy.deepcopy(ices)
for i in range(len(ices)):
if copied[i] != 0:
if 0 < i < N-1:
if copied[i] >= L:
ices[i] = 0
break
if copied[i] > copied[i-1] and copied[i] > copied[i+1]:
ices[i] += 1
if ices[i] >= L:
ices[i] = 0
elif i == 0:
if copied[i] >= L:
ices[i] = 0
break
if copied[i] > copied[1]:
ices[i] += 1
if ices[i] >= L:
ices[i] = 0
elif i == N-1:
if ices[i] >= L:
ices[i] = 0
break
if copied[i] > copied[N-2]:
ices[i] += 1
if ices[i] >= L:
ices[i] = 0
copied = ices
if ices.count(0) == N:
print(counts)
return
counts += 1
# N,L = map(int,input().strip().split())
# ices = []
# while True:
# ice = int(input().strip())
# ices.append(ice)
# if len(ices) == N:
# break
# N = 4
# L = 6
# ices = [4,2,3,5]
# N = 6
# L = 10
# ices = [3,4,1,9,5,1]
with open("aoj.txt") as infile:
N,L = map(int,infile.readline().strip().split())
ices = []
while True:
ice = int(infile.readline())
ices.append(ice)
if len(ices) == N:
break
greedy(N,L,ices)
``` | instruction | 0 | 63,941 | 8 | 127,882 |
No | output | 1 | 63,941 | 8 | 127,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are fine icicles under the eaves of JOI's house in Canada. Because of this, JOI decided to investigate the icicles.
There are N (2 ≤ N ≤ 100000 = 105) icicles under the eaves of JOI's house. These icicles are aligned and i cm (1 ≤ i ≤ N) from the left edge of the eaves. There are i-th icicles at the position of. The length of the i-th icicle is initially ai cm (ai is an integer greater than or equal to 1). These icicles grow according to the following rule:
* The i-th icicles grow by 1 cm per hour only if they are longer than both the i − 1st icicles and the i + 1st icicles (however, consider only one icicle next to one end). That is, the first icicle grows if it is longer than the second icicle, and the Nth icicle grows if it is longer than the N − 1st icicle).
* All icicles break from the root the moment they reach L cm (2 ≤ L ≤ 50000) (the broken icicles are subsequently considered to be 0 cm long icicles).
In the first stage, the lengths of the two adjacent icicles are all different. At this time, if enough time has passed, all N icicles will break to a length of 0 cm. JOI, I wanted to know how long it would take for the icicles to reach this state.
Given the initial length of N icicles and the limit length L of the icicles, write a program that finds the time it takes for all the icicles to break.
output
The output consists of one line containing only one integer that represents the time it takes for all the icicles to break.
Input / output example
Input example 1
4 6
Four
2
3
Five
Output example 1
8
In the case of Example 1, the 1, 2, 3, and 4 icicles break after 2, 8, 4, and 1 hour, respectively. Therefore, it takes 8 hours for all the icicles to break. Output 8.
Input example 2
6 10
3
Four
1
9
Five
1
Output example 2
15
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
input
On the first line of the input, the integer N, which represents the number of icicles, and the integer L, which represents the limit length of the icicles, are written in this order, separated by blanks. Input i + line 1 (1 ≤ i) In ≤ N), the integer ai (1 ≤ ai <L) representing the first length of the i-th icicle is written.
Of the scoring data, 30% of the points are N ≤ 500 and L ≤ 1000.
Example
Input
4 6
4
2
3
5
Output
8
Submitted Solution:
```
## ?????????
def loadIcicle():
icicles = []
line = input().strip().split(" ")
N, L = int(line[0]), int(line[1])
while True:
icicles.append(int(input().strip()))
if len(icicles) == N: break
return icicles, N, L
icicles, N, L = loadIcicle()
def calcDiff(icicles, N):
diff_2times = [0]*len(icicles)
for idx in range(len(icicles)):
dif_right = icicles[idx+1] - icicles[idx] if idx != N-1 else -icicles[idx]
dif_left = icicles[idx] - icicles[idx-1] if idx != 0 else icicles[idx]
dif_right = 1 if dif_right > 0 else -1
dif_left = 1 if dif_left > 0 else -1
if dif_right - dif_left < 0: diff_2times[idx] = -1
elif dif_right - dif_left > 0: diff_2times[idx] = 1
else: diff_2times[idx] = 0
return diff_2times
diff_2times = calcDiff(icicles, N)
time = [-1]*len(icicles)
peakX = [i for i in range(len(diff_2times)) if diff_2times[i]==-1] # ?????????(?????????)???index?????????
for i in peakX:
time[i] = L - icicles[i]
isLocalMinL, isLocalMinR = False, False
posL = i
posR = i
while not (isLocalMinL and isLocalMinR):
posL -= 1
if posL <= 0:
isLocalMinL = True
# ?£????????????£?????????????¨????
if not isLocalMinL:
if time[posL] == -1:
time[posL] = (L-icicles[posL]) + time[posL+1]
else:
time[posL] = (L-icicles[posL]) + max(time[posL-1], time[posL+1])
# ????????????????????????
if diff_2times[posL] == 1:
isLocalMinL = True
posR += 1
if posR >= N:
isLocalMinR = True
if not isLocalMinR:
if time[posR] == -1:
time[posR] = (L-icicles[posR]) + time[posR-1]
else:
time[posR] = (L-icicles[posR]) + max(time[posR-1], time[posR+1])
if diff_2times[posR] == 1:
isLocalMinR = True
print(max(time))
``` | instruction | 0 | 63,942 | 8 | 127,884 |
No | output | 1 | 63,942 | 8 | 127,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik and Rubik love playing computer games. Furik has recently found a new game that greatly interested Rubik. The game consists of n parts and to complete each part a player may probably need to complete some other ones. We know that the game can be fully completed, that is, its parts do not form cyclic dependencies.
Rubik has 3 computers, on which he can play this game. All computers are located in different houses. Besides, it has turned out that each part of the game can be completed only on one of these computers. Let's number the computers with integers from 1 to 3. Rubik can perform the following actions:
* Complete some part of the game on some computer. Rubik spends exactly 1 hour on completing any part on any computer.
* Move from the 1-st computer to the 2-nd one. Rubik spends exactly 1 hour on that.
* Move from the 1-st computer to the 3-rd one. Rubik spends exactly 2 hours on that.
* Move from the 2-nd computer to the 1-st one. Rubik spends exactly 2 hours on that.
* Move from the 2-nd computer to the 3-rd one. Rubik spends exactly 1 hour on that.
* Move from the 3-rd computer to the 1-st one. Rubik spends exactly 1 hour on that.
* Move from the 3-rd computer to the 2-nd one. Rubik spends exactly 2 hours on that.
Help Rubik to find the minimum number of hours he will need to complete all parts of the game. Initially Rubik can be located at the computer he considers necessary.
Input
The first line contains integer n (1 ≤ n ≤ 200) — the number of game parts. The next line contains n integers, the i-th integer — ci (1 ≤ ci ≤ 3) represents the number of the computer, on which you can complete the game part number i.
Next n lines contain descriptions of game parts. The i-th line first contains integer ki (0 ≤ ki ≤ n - 1), then ki distinct integers ai, j (1 ≤ ai, j ≤ n; ai, j ≠ i) — the numbers of parts to complete before part i.
Numbers on all lines are separated by single spaces. You can assume that the parts of the game are numbered from 1 to n in some way. It is guaranteed that there are no cyclic dependencies between the parts of the game.
Output
On a single line print the answer to the problem.
Examples
Input
1
1
0
Output
1
Input
5
2 2 1 1 3
1 5
2 5 1
2 5 4
1 5
0
Output
7
Note
Note to the second sample: before the beginning of the game the best strategy is to stand by the third computer. First we complete part 5. Then we go to the 1-st computer and complete parts 3 and 4. Then we go to the 2-nd computer and complete parts 1 and 2. In total we get 1+1+2+1+2, which equals 7 hours. | instruction | 0 | 64,269 | 8 | 128,538 |
Tags: dfs and similar, greedy
Correct Solution:
```
from collections import *
read_line = lambda: [int(i) for i in input().split(' ')]
n = read_line()[0]
cs = [c - 1 for c in read_line()]
g = [[] for v in range(n)]
parent_cnt = [0] * n
for v in range(n):
parents = read_line()
parent_cnt[v] = len(parents) - 1
for i in range(1, len(parents)):
g[parents[i] - 1].append(v)
def work(x):
pcnt = list(parent_cnt)
qs = [
deque(v for v in range(n) if cs[v] == c and pcnt[v] == 0)
for c in range(3)
]
ans = 0
while True:
while qs[x]:
v = qs[x].popleft()
ans += 1
for w in g[v]:
pcnt[w] -= 1
if pcnt[w] == 0:
qs[cs[w]].append(w)
if qs[0] or qs[1] or qs[2]:
ans += 1
x = (x + 1) % 3
else:
break
return ans
print(min(work(i) for i in range(3)))
``` | output | 1 | 64,269 | 8 | 128,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik and Rubik love playing computer games. Furik has recently found a new game that greatly interested Rubik. The game consists of n parts and to complete each part a player may probably need to complete some other ones. We know that the game can be fully completed, that is, its parts do not form cyclic dependencies.
Rubik has 3 computers, on which he can play this game. All computers are located in different houses. Besides, it has turned out that each part of the game can be completed only on one of these computers. Let's number the computers with integers from 1 to 3. Rubik can perform the following actions:
* Complete some part of the game on some computer. Rubik spends exactly 1 hour on completing any part on any computer.
* Move from the 1-st computer to the 2-nd one. Rubik spends exactly 1 hour on that.
* Move from the 1-st computer to the 3-rd one. Rubik spends exactly 2 hours on that.
* Move from the 2-nd computer to the 1-st one. Rubik spends exactly 2 hours on that.
* Move from the 2-nd computer to the 3-rd one. Rubik spends exactly 1 hour on that.
* Move from the 3-rd computer to the 1-st one. Rubik spends exactly 1 hour on that.
* Move from the 3-rd computer to the 2-nd one. Rubik spends exactly 2 hours on that.
Help Rubik to find the minimum number of hours he will need to complete all parts of the game. Initially Rubik can be located at the computer he considers necessary.
Input
The first line contains integer n (1 ≤ n ≤ 200) — the number of game parts. The next line contains n integers, the i-th integer — ci (1 ≤ ci ≤ 3) represents the number of the computer, on which you can complete the game part number i.
Next n lines contain descriptions of game parts. The i-th line first contains integer ki (0 ≤ ki ≤ n - 1), then ki distinct integers ai, j (1 ≤ ai, j ≤ n; ai, j ≠ i) — the numbers of parts to complete before part i.
Numbers on all lines are separated by single spaces. You can assume that the parts of the game are numbered from 1 to n in some way. It is guaranteed that there are no cyclic dependencies between the parts of the game.
Output
On a single line print the answer to the problem.
Examples
Input
1
1
0
Output
1
Input
5
2 2 1 1 3
1 5
2 5 1
2 5 4
1 5
0
Output
7
Note
Note to the second sample: before the beginning of the game the best strategy is to stand by the third computer. First we complete part 5. Then we go to the 1-st computer and complete parts 3 and 4. Then we go to the 2-nd computer and complete parts 1 and 2. In total we get 1+1+2+1+2, which equals 7 hours. | instruction | 0 | 64,270 | 8 | 128,540 |
Tags: dfs and similar, greedy
Correct Solution:
```
from collections import *
read_line = lambda: [int(i) for i in input().split(' ')]
n = read_line()[0]
cs = [c - 1 for c in read_line()]
g = [[] for v in range(n)]
parent_cnt = [0] * n
for v in range(n):
parents = read_line()
parent_cnt[v] = len(parents) - 1
for i in range(1, len(parents)):
g[parents[i] - 1].append(v)
def work(x):
pcnt = list(parent_cnt)
qs = [
deque(v for v in range(n) if cs[v] == c and pcnt[v] == 0)
for c in range(3)
]
ans = 0
while True:
while qs[x]:
v = qs[x].popleft()
ans += 1
for w in g[v]:
pcnt[w] -= 1
if pcnt[w] == 0:
qs[cs[w]].append(w)
if qs[0] or qs[1] or qs[2]:
ans += 1
x = (x + 1) % 3
else:
break
return ans
print(min(work(i) for i in range(3)))
# Made By Mostafa_Khaled
``` | output | 1 | 64,270 | 8 | 128,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik and Rubik love playing computer games. Furik has recently found a new game that greatly interested Rubik. The game consists of n parts and to complete each part a player may probably need to complete some other ones. We know that the game can be fully completed, that is, its parts do not form cyclic dependencies.
Rubik has 3 computers, on which he can play this game. All computers are located in different houses. Besides, it has turned out that each part of the game can be completed only on one of these computers. Let's number the computers with integers from 1 to 3. Rubik can perform the following actions:
* Complete some part of the game on some computer. Rubik spends exactly 1 hour on completing any part on any computer.
* Move from the 1-st computer to the 2-nd one. Rubik spends exactly 1 hour on that.
* Move from the 1-st computer to the 3-rd one. Rubik spends exactly 2 hours on that.
* Move from the 2-nd computer to the 1-st one. Rubik spends exactly 2 hours on that.
* Move from the 2-nd computer to the 3-rd one. Rubik spends exactly 1 hour on that.
* Move from the 3-rd computer to the 1-st one. Rubik spends exactly 1 hour on that.
* Move from the 3-rd computer to the 2-nd one. Rubik spends exactly 2 hours on that.
Help Rubik to find the minimum number of hours he will need to complete all parts of the game. Initially Rubik can be located at the computer he considers necessary.
Input
The first line contains integer n (1 ≤ n ≤ 200) — the number of game parts. The next line contains n integers, the i-th integer — ci (1 ≤ ci ≤ 3) represents the number of the computer, on which you can complete the game part number i.
Next n lines contain descriptions of game parts. The i-th line first contains integer ki (0 ≤ ki ≤ n - 1), then ki distinct integers ai, j (1 ≤ ai, j ≤ n; ai, j ≠ i) — the numbers of parts to complete before part i.
Numbers on all lines are separated by single spaces. You can assume that the parts of the game are numbered from 1 to n in some way. It is guaranteed that there are no cyclic dependencies between the parts of the game.
Output
On a single line print the answer to the problem.
Examples
Input
1
1
0
Output
1
Input
5
2 2 1 1 3
1 5
2 5 1
2 5 4
1 5
0
Output
7
Note
Note to the second sample: before the beginning of the game the best strategy is to stand by the third computer. First we complete part 5. Then we go to the 1-st computer and complete parts 3 and 4. Then we go to the 2-nd computer and complete parts 1 and 2. In total we get 1+1+2+1+2, which equals 7 hours. | instruction | 0 | 64,271 | 8 | 128,542 |
Tags: dfs and similar, greedy
Correct Solution:
```
n = int(input())
h = [int(q) - 1 for q in input().split()]
u = [set([int(q) - 1 for q in input().split()][1:]) for i in range(n)]
t = 1e9
def g():
for i in p:
if h[i] == k and not v[i]:
return i
for k in range(3):
p = list(range(n))
d = -1
v = [q.copy() for q in u]
while p:
i = g()
while i != None:
d += 1
p.remove(i)
for q in v :
q.discard(i)
i = g()
k = (k + 1) % 3
d += 1
t = min(d, t)
print(t)
``` | output | 1 | 64,271 | 8 | 128,543 |
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