message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).
The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).
The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).
Output
Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
from heapq import *
n = int(input())
a = list(map(int,input().split()))
t = list(map(int,input().split()))
ls = list(zip(a,t))
ls.sort()
ls.append((10**18,0))
ans = 0
sm = 0
hq = []
xp = ls[0][0]
heapify(hq)
i = 0
while i<=n:
if ls[i][0] == xp:
heappush(hq,-ls[i][1])
sm += ls[i][1]
i += 1
else:
while hq:
cnt = heappop(hq)
cnt = -cnt
ans += sm-cnt
sm -= cnt
xp += 1
while i<=n:
if ls[i][0] == xp:
heappush(hq,-ls[i][1])
sm += ls[i][1]
elif ls[i][0] > xp:
break
i += 1
xp = ls[i][0]
print(ans)
``` | instruction | 0 | 62,449 | 14 | 124,898 |
Yes | output | 1 | 62,449 | 14 | 124,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).
The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).
The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).
Output
Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
"""
NTC here
"""
import sys
inp= sys.stdin.readline
input = lambda : inp().strip()
# flush= sys.stdout.flush
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(2**26)
def iin(): return int(input())
def lin(): return list(map(int, input().split()))
def main():
import heapq as hq
n = iin()
a = lin()
t = lin()
dc = {}
for i in range(n):
try:
dc[a[i]].append(t[i])
except:
dc[a[i]]=[t[i], ]
# print(dc)
sa = list(dc.keys())
sa.sort()
n = len(sa)
ch = 0
pt = sa[ch]
temp = []
sm = 0
ans = 0
hq.heapify(temp)
done = 0
while ch<n:
# print(temp, sa[ch])
pt = sa[ch]
if len(dc[pt])>1 or done:
a1 = sorted(dc[pt])
for item in a1:
hq.heappush(temp, -item)
sm += item
x = hq.heappop(temp)*(-1)
dc[pt]=[x]
sm -= x
while len(temp):
#print('A', ch, pt, temp )
ans += sm
pt += 1
if pt in dc:
ch+=1
done = 1
break
else:
x = hq.heappop(temp)*(-1)
sm -= x
else:
done = 0
else:
ch+=1
done = 0
# print(dc)
print(ans)
main()
#threading.Thread(target=main).start()
``` | instruction | 0 | 62,450 | 14 | 124,900 |
Yes | output | 1 | 62,450 | 14 | 124,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).
The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).
The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).
Output
Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
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")
# ------------------- fast io --------------------
import heapq
n=int(input())
nums=list(map(int,input().split()))
cost=list(map(int,input().split()))
#gonna have to use a max heap here
dict1={}
for s in range(n):
if not(nums[s] in dict1):
dict1[nums[s]]=[cost[s]]
else:
dict1[nums[s]].append(cost[s])
nums=set(nums)
nums=list(nums)
nums.sort()
heapy=[]
heapq.heapify(heapy)
summy=0
ans=0
for s in range(len(nums)):
#i need another check here
if len(heapy)>0:
#this is negative
ans+=summy
while len(dict1[nums[s]])>0:
v0=(dict1[nums[s]].pop())
heapq.heappush(heapy,-v0)
summy+=v0
v1=heapq.heappop(heapy)
dict1[nums[s]].append(-v1)
summy-=-v1
if s<len(nums)-1:
for b in range(min(nums[s+1]-1-nums[s],len(heapy))):
ans+=summy
v2=heapq.heappop(heapy)
summy-=-v2
else:
for b in range(len(heapy)):
ans+=summy
v2=heapq.heappop(heapy)
summy-=-v2
print(ans)
``` | instruction | 0 | 62,451 | 14 | 124,902 |
Yes | output | 1 | 62,451 | 14 | 124,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).
The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).
The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).
Output
Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 1000000007
INF = float('inf')
# ------------------------------
def main():
n = N()
arr = RLL()
tarr = RLL()
zp = zip(arr, tarr)
dic = defaultdict(list)
odk = sorted(list(set(arr)))
for c, t in zp: dic[c].append(t)
# for i in odk: print(dic[i])
hp = []
res = sm = 0
for k in odk:
ts = dic[k]
for t in ts:
sm+=t
heappush(hp, -t)
sm+=heappop(hp)
res+=sm
while len(hp)>1:
res+=(-heappop(hp))
print(res)
if __name__ == "__main__":
main()
``` | instruction | 0 | 62,452 | 14 | 124,904 |
No | output | 1 | 62,452 | 14 | 124,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).
The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).
The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).
Output
Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
import random
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
from math import factorial as f
def ncr(x, y):
return f(x) // (f(y) * f(x - y))
n = int(input())
l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))
hash = defaultdict(list)
for i in range(n):
hash[l1[i]].append(l2[i])
for i in hash:
hash[i].sort()
ans = 0
ka = sorted(hash.keys())
for i in ka:
for j in range(len(hash[i])-1):
z = i+1
ans+=hash[i][j]
if hash[z] == []:
continue
else:
if hash[z][-1]>=hash[i][j]:
k = hash[z].pop()
hash[z].append(hash[i][j])
hash[z].append(k)
else:
hash[z].append(hash[i][j])
print(ans)
``` | instruction | 0 | 62,453 | 14 | 124,906 |
No | output | 1 | 62,453 | 14 | 124,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).
The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).
The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).
Output
Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
import heapq
n=int(input())
a=list(map(int,input().split()))
t=list(map(int,input().split()))
d={}
mxm={}
sm={}
for i in range(len(a)):
if a[i] not in d:
d[a[i]]=0
mxm[a[i]]=max(mxm.get(a[i],0),t[i])
sm[a[i]]=sm.get(a[i],0)+t[i]
d[a[i]]+=1
k=list(d.keys())
heapq.heapify(k)
cost=0
while len(k)>0:
key=heapq.heappop(k)
if d[key]==1:
continue
new_key=key+1
if new_key not in d:
heapq.heappush(k,new_key)
cost+=sm[key]-mxm[key]
sm[new_key]=sm.get(new_key,0)+sm[key]-mxm[key]
mxm[new_key]=max(mxm.get(new_key,0),mxm[key])
d[new_key]=d.get(new_key,0)+d[key]-1
d[key]=1
print(cost)
``` | instruction | 0 | 62,454 | 14 | 124,908 |
No | output | 1 | 62,454 | 14 | 124,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).
The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).
The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).
Output
Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
import heapq
def solve(N, A, T):
# Find conflicting nums
numToIndices = {}
for i, x in enumerate(A):
if x in numToIndices:
numToIndices[x].append(i)
else:
numToIndices[x] = [i]
#print(numToIndices)
#print({i: T[i] for i in range(N)})
# Pick the best item to keep for each num, all conflicts get scooted forward in a heap
num = 0
assigned = 0
heap = []
cost = 0
while assigned != N:
if heap or (num in numToIndices):
indices = numToIndices.get(num, [])
#print(num, heap, indices)
numPop = 0
numKeep = 0
if indices:
maxIndex = max(indices, key=lambda i: T[i])
conflictCost = T[maxIndex]
if heap:
tup = -heap[0]
heapCost, fromNum = divmod(tup, 1000000)
if not heap and indices:
# Keep max from all conflicts
numKeep = 1
if heap and not indices:
# Empty cell, pop costliest from heap
numPop = 1
if heap and indices:
# Pick costliest from conflicts or heap
if heapCost >= conflictCost:
numPop = 1
else:
numKeep = 1
assert numPop + numKeep == 1
assigned += 1
if numPop:
# Use the heap for this cell
heapq.heappop(heap)
cost += (num - fromNum) * heapCost
if numKeep:
assert indices
assert maxIndex is not None
# Scoot everything else into the heap (excluding max if that's the one that should be in this cell)
numAdded = 0
for i in indices:
if numKeep == 0 or i != maxIndex:
tup = T[i] * 1000000 + num
heapq.heappush(heap, -tup)
numAdded += 1
assert len(indices) - numKeep == numAdded
num += 1
if num > 200000:
return 42
return cost
if False:
from random import randint
N = 10
K = 10
print(
solve(N, [randint(1, K) for i in range(N)], [randint(1, K) for i in range(N)])
)
if __name__ == "__main__":
N, = map(int, input().split())
A = (int(x) for x in input().split())
T = [int(x) for x in input().split()]
ans = solve(N, A, T)
print(ans)
``` | instruction | 0 | 62,455 | 14 | 124,910 |
No | output | 1 | 62,455 | 14 | 124,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.
In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).
Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.
Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.
Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.
Output
Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.
Examples
Input
2 2 3
000
000
111
111
Output
2
Input
3 3 3
111
111
111
111
111
111
111
111
111
Output
19
Input
1 1 10
0101010101
Output
0
Note
In the first sample the whole first layer of CPUs is malfunctional. In the second layer when CPU (2, 1, 2) turns off, it disrupts the control by CPU (2, 1, 3) over CPU (2, 1, 1), and when CPU (2, 2, 2) is turned off, it disrupts the control over CPU (2, 2, 3) by CPU (2, 2, 1).
In the second sample all processors except for the corner ones are critical.
In the third sample there is not a single processor controlling another processor, so the answer is 0. | instruction | 0 | 62,750 | 14 | 125,500 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
n,m,k=(int(z) for z in input().split())
M=[]
for i in range(n):
for j in range(m):
s=input()
for p in range(len(s)):
M+=[int(s[p])]
if i!=n-1:
s=input()
res=0
cur=0
for i in range(n):
for j in range(m):
for h in range(k):
if M[cur]==1:
if 1<=i<=n-2 and M[cur-m*k]+M[cur+m*k]==2:
res+=1
elif 1<=h<=k-2 and M[cur-1]+M[cur+1]==2:
res+=1
elif 1<=j<=m-2 and M[cur-k]+M[cur+k]==2:
res+=1
elif h<=k-2 and j>=1 and M[cur+1]+M[cur-k]==2 and M[cur-k+1]==0:
res+=1
elif j<=m-2 and h>=1 and M[cur-1]+M[cur+k]==2 and M[cur+k-1]==0:
res+=1
elif h<=k-2 and i>=1 and M[cur+1]+M[cur-m*k]==2 and M[cur-m*k+1]==0:
res+=1
elif i<=n-2 and h>=1 and M[cur-1]+M[cur+m*k]==2 and M[cur+m*k-1]==0:
res+=1
elif i<=n-2 and j>=1 and M[cur-k]+M[cur+m*k]==2 and M[cur+m*k-k]==0:
res+=1
elif j<=m-2 and i>=1 and M[cur+k]+M[cur-m*k]==2 and M[cur-m*k+k]==0:
res+=1
cur+=1
print(res)
``` | output | 1 | 62,750 | 14 | 125,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.
In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).
Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.
Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.
Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.
Output
Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.
Examples
Input
2 2 3
000
000
111
111
Output
2
Input
3 3 3
111
111
111
111
111
111
111
111
111
Output
19
Input
1 1 10
0101010101
Output
0
Note
In the first sample the whole first layer of CPUs is malfunctional. In the second layer when CPU (2, 1, 2) turns off, it disrupts the control by CPU (2, 1, 3) over CPU (2, 1, 1), and when CPU (2, 2, 2) is turned off, it disrupts the control over CPU (2, 2, 3) by CPU (2, 2, 1).
In the second sample all processors except for the corner ones are critical.
In the third sample there is not a single processor controlling another processor, so the answer is 0. | instruction | 0 | 62,751 | 14 | 125,502 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
def put():
return map(int, input().split())
def safe(x,y,z):
return x>=0 and y>=0 and z>=0 and x<n and y<m and z<p
def check(x,y,z):
if mat[x][y][z]=='0':
return 0
cur = [x,y,z]
for i in range(3):
src = cur.copy()
src[i]-=1
if (not safe(src[0], src[1], src[2])) or mat[src[0]][src[1]][src[2]]=='0':
continue
for j in range(3):
des = cur.copy()
des[j]+=1
alt = src.copy()
alt[j]+=1
if safe(des[0], des[1], des[2]) and mat[des[0]][des[1]][des[2]]=='1':
if j==i:
return 1
elif safe(alt[0], alt[1], alt[2]) and mat[alt[0]][alt[1]][alt[2]]=='0':
return 1
return 0
n,m,p = put()
mat = []
ans = 0
for i in range(n):
mat.append([input() for j in range(m)])
if i!=n-1:
input()
for i in range(n):
for j in range(m):
for k in range(p):
ans += check(i,j,k)
print(ans)
``` | output | 1 | 62,751 | 14 | 125,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.
In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).
Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.
Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.
Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.
Output
Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.
Examples
Input
2 2 3
000
000
111
111
Output
2
Input
3 3 3
111
111
111
111
111
111
111
111
111
Output
19
Input
1 1 10
0101010101
Output
0
Note
In the first sample the whole first layer of CPUs is malfunctional. In the second layer when CPU (2, 1, 2) turns off, it disrupts the control by CPU (2, 1, 3) over CPU (2, 1, 1), and when CPU (2, 2, 2) is turned off, it disrupts the control over CPU (2, 2, 3) by CPU (2, 2, 1).
In the second sample all processors except for the corner ones are critical.
In the third sample there is not a single processor controlling another processor, so the answer is 0. | instruction | 0 | 62,752 | 14 | 125,504 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
def func(i,j,t):
if i > 0:
if A[i-1][j][t] == '1':
if i < n-1:
if A[i+1][j][t] == '1':
return True
if j < m-1:
if A[i][j+1][t] == '1' and A[i-1][j+1][t] == '0':
return True
if t < k-1:
if A[i-1][j][t+1] == '0' and A[i][j][t+1] == '1':
return True
if j > 0:
if A[i][j-1][t] == '1':
if j < m-1:
if A[i][j+1][t] == '1':
return True
if i < n-1:
if A[i+1][j][t] == '1' and A[i+1][j-1][t] == '0':
return True
if t < k-1:
if A[i][j-1][t+1] == '0' and A[i][j][t+1] == '1':
return True
if t > 0:
if A[i][j][t-1] == '1':
if t < k-1:
if A[i][j][t+1] == '1':
return True
if i < n-1:
if A[i+1][j][t] == '1' and A[i+1][j][t-1] == '0':
return True
if j < m-1:
if A[i][j+1][t-1] == '0' and A[i][j+1][t] == '1':
return True
return False
n,m,k = map(int, input().split())
A = [0] * n
for i in range(n):
A[i] = [0] * m
for j in range(m):
A[i][j] = [0] * k
for i in range(n):
for j in range(m):
per = input()
for t in range(k):
A[i][j][t] = per[t]
if i != n-1:
per = input()
answer = 0
for i in range(n):
for j in range(m):
for t in range(k):
if A[i][j][t] == '1':
if func(i,j,t):
answer+=1
print(answer)
``` | output | 1 | 62,752 | 14 | 125,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.
In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).
Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.
Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.
Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.
Output
Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.
Examples
Input
2 2 3
000
000
111
111
Output
2
Input
3 3 3
111
111
111
111
111
111
111
111
111
Output
19
Input
1 1 10
0101010101
Output
0
Note
In the first sample the whole first layer of CPUs is malfunctional. In the second layer when CPU (2, 1, 2) turns off, it disrupts the control by CPU (2, 1, 3) over CPU (2, 1, 1), and when CPU (2, 2, 2) is turned off, it disrupts the control over CPU (2, 2, 3) by CPU (2, 2, 1).
In the second sample all processors except for the corner ones are critical.
In the third sample there is not a single processor controlling another processor, so the answer is 0. | instruction | 0 | 62,753 | 14 | 125,506 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
def safe(pos):
return pos[0] >= 0 and pos[0] < n and pos[1] >= 0 and pos[1] < m and pos[2] >= 0 and pos[2] < p
def CPU_status(pos, number):
return safe(pos) and super_computer[pos[0]][pos[1]][pos[2]] == number
def critical(x,y,z):
if super_computer[x][y][z] != '0':
current = [x,y,z]
for i in range(3):
parent = current.copy()
parent[i]-=1
if CPU_status(parent, '1'):
for j in range(3):
child, alt = current.copy(), parent.copy()
child[j]+=1
alt[j]+=1
if CPU_status(child, '1') and (CPU_status(alt, '0') or j == i):
return 1
return 0
n, m, p = map(int, input().split())
super_computer, crit = ([], 0)
for i in range(n):
super_computer.append([input() for _ in range(m)])
if i != n-1:
input()
for i in range(n):
for j in range(m):
for k in range(p):
crit += critical(i,j,k)
print(crit)
``` | output | 1 | 62,753 | 14 | 125,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.
In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).
Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.
Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.
Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.
Output
Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.
Examples
Input
2 2 3
000
000
111
111
Output
2
Input
3 3 3
111
111
111
111
111
111
111
111
111
Output
19
Input
1 1 10
0101010101
Output
0
Note
In the first sample the whole first layer of CPUs is malfunctional. In the second layer when CPU (2, 1, 2) turns off, it disrupts the control by CPU (2, 1, 3) over CPU (2, 1, 1), and when CPU (2, 2, 2) is turned off, it disrupts the control over CPU (2, 2, 3) by CPU (2, 2, 1).
In the second sample all processors except for the corner ones are critical.
In the third sample there is not a single processor controlling another processor, so the answer is 0. | instruction | 0 | 62,754 | 14 | 125,508 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
def check(d, i, j, l):
global result, save
if d[i][j][l] == 1:
if d[i + 1][j][l] == 1 and d[i - 1][j][l] == 1:
#save.append([i, j, l])
result += 1
return
if d[i][j + 1][l] == 1 and d[i][j - 1][l] == 1:
result += 1
#save.append([i, j, l])
return
if d[i][j][l + 1] == 1 and d[i][j][l - 1] == 1:
result += 1
#save.append([i, j, l])
return
if d[i - 1][j + 1][l] == 0 and d[i][j + 1][l] == 1 and d[i - 1][j][l] == 1:
result += 1
#save.append([i, j, l])
return
if d[i][j + 1][l - 1] == 0 and d[i][j + 1][l] == 1 and d[i][j][l - 1] == 1:
result += 1
#save.append([i, j, l])
return
if d[i + 1][j][l - 1] == 0 and d[i + 1][j][l] == 1 and d[i][j][l - 1] == 1:
result += 1
#save.append([i, j, l])
return
if d[i + 1][j - 1][l] == 0 and d[i + 1][j][l] == 1 and d[i][j - 1][l] == 1:
result += 1
#save.append([i, j, l])
return
if d[i][j - 1][l + 1] == 0 and d[i][j - 1][l] == 1 and d[i][j][l + 1] == 1:
result += 1
#save.append([i, j, l])
return
if d[i - 1][j][l + 1] == 0 and d[i - 1][j][l] == 1 and d[i][j][l + 1] == 1:
result += 1
#save.append([i, j, l])
return
n, m, k = map(int, input().split())
d = [[[0] * (k + 2) for i in range(m + 2)] for j in range(n + 2)]
for i in range(1, n + 1):
for j in range(1, m + 1):
st = input()
#print(st)
for l in range(k):
d[i][j][l + 1] = int(st[l])
if i != n:
a = input()
result = 0
for i in range(1, n + 1):
for j in range(1, m + 1):
for l in range(1, k + 1):
check(d, i, j, l)
print(result)
``` | output | 1 | 62,754 | 14 | 125,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.
In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).
Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.
Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.
Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.
Output
Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.
Examples
Input
2 2 3
000
000
111
111
Output
2
Input
3 3 3
111
111
111
111
111
111
111
111
111
Output
19
Input
1 1 10
0101010101
Output
0
Note
In the first sample the whole first layer of CPUs is malfunctional. In the second layer when CPU (2, 1, 2) turns off, it disrupts the control by CPU (2, 1, 3) over CPU (2, 1, 1), and when CPU (2, 2, 2) is turned off, it disrupts the control over CPU (2, 2, 3) by CPU (2, 2, 1).
In the second sample all processors except for the corner ones are critical.
In the third sample there is not a single processor controlling another processor, so the answer is 0. | instruction | 0 | 62,755 | 14 | 125,510 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
def main():
s = input().split()
n, m, k = int(s[0]), int(s[1]), int(s[2])
processor = []
for x in range(n):
for y in range(m):
s = input()
for z in s:
processor.append(int(z) == 1)
if x < n - 1:
emptyLine = input()
counter = 0
mk = m * k
nmk = n * mk
for i in range(nmk):
if not processor[i]:
continue
# back
if i >= mk:
if processor[i - mk]:
# front
if i < (nmk - mk):
if processor[i + mk]:
counter += 1
continue
# right
if (i % k) < (k - 1):
if processor[i + 1]:
if not processor[i - mk + 1]:
counter += 1
continue
# down
if (i % mk) < (mk - k):
if processor[i + k]:
if not processor[i - mk + k]:
counter += 1
continue
# left
if (i % k) > 0:
if processor[i - 1]:
# front
if i < (nmk - mk):
if processor[i + mk]:
if not processor[i + mk - 1]:
counter += 1
continue
# right
if (i % k) < (k - 1):
if processor[i + 1]:
counter += 1
continue
# down
if (i % mk) < (mk - k):
if processor[i + k]:
if not processor[i + k - 1]:
counter += 1
continue
# up
if (i % mk) >= k:
if processor[i - k]:
# front
if i < (nmk - mk):
if processor[i + mk]:
if not processor[i + mk - k]:
counter += 1
continue
# right
if (i % k) < (k - 1):
if processor[i + 1]:
if not processor[i - k + 1]:
counter += 1
continue
# down
if (i % mk) < (mk - k):
if processor[i + k]:
counter += 1
continue
print(counter)
main()
``` | output | 1 | 62,755 | 14 | 125,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.
In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).
Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.
Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.
Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.
Output
Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.
Examples
Input
2 2 3
000
000
111
111
Output
2
Input
3 3 3
111
111
111
111
111
111
111
111
111
Output
19
Input
1 1 10
0101010101
Output
0
Note
In the first sample the whole first layer of CPUs is malfunctional. In the second layer when CPU (2, 1, 2) turns off, it disrupts the control by CPU (2, 1, 3) over CPU (2, 1, 1), and when CPU (2, 2, 2) is turned off, it disrupts the control over CPU (2, 2, 3) by CPU (2, 2, 1).
In the second sample all processors except for the corner ones are critical.
In the third sample there is not a single processor controlling another processor, so the answer is 0. | instruction | 0 | 62,756 | 14 | 125,512 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
def put():
return map(int, input().split())
def safe(x,y,z):
return x>=0 and y>=0 and z>=0 and x<n and y<m and z<p
def check(x,y,z):
if mat[x][y][z]==0:
return 0
cur = [x,y,z]
for i in range(3):
src = cur.copy()
src[i]-=1
if (not safe(src[0], src[1], src[2])) or mat[src[0]][src[1]][src[2]]==0:
continue
for j in range(3):
des = cur.copy()
des[j]+=1
alt = src.copy()
alt[j]+=1
if safe(des[0], des[1], des[2]) and mat[des[0]][des[1]][des[2]]==1:
if j==i:
return 1
elif safe(alt[0], alt[1], alt[2]) and mat[alt[0]][alt[1]][alt[2]]==0:
return 1
return 0
n,m,p = put()
mat = []
for i in range(n):
tmp1 = []
for j in range(m):
s = input()
tmp2 = []
for k in range(p):
tmp2.append(int(s[k]))
tmp1.append(tmp2)
mat.append(tmp1)
if i!=n-1:
input()
#print(mat)
ans = 0
for i in range(n):
for j in range(m):
for k in range(p):
ans += check(i,j,k)
print(ans)
``` | output | 1 | 62,756 | 14 | 125,513 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,761 | 14 | 125,522 |
"Correct Solution:
```
n = int(input())
print((n + 2 - 1)//2)
``` | output | 1 | 62,761 | 14 | 125,523 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,762 | 14 | 125,524 |
"Correct Solution:
```
attendes = int(input())
print((attendes + 1)//2)
``` | output | 1 | 62,762 | 14 | 125,525 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,763 | 14 | 125,526 |
"Correct Solution:
```
from math import ceil
n = int(input())
# arr = [int(i) for i in input().split()]
print(ceil(n / 2))
``` | output | 1 | 62,763 | 14 | 125,527 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,764 | 14 | 125,528 |
"Correct Solution:
```
import sys, math
def rnd(x):
a = int(x)
b = x-a
if b>=0.5:
a+=1
return(a)
n = int(input())
print(rnd(n/2))
``` | output | 1 | 62,764 | 14 | 125,529 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,765 | 14 | 125,530 |
"Correct Solution:
```
n = int(input())
print((n - 1) // 2 + 1)
``` | output | 1 | 62,765 | 14 | 125,531 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,766 | 14 | 125,532 |
"Correct Solution:
```
n=int(input())
print((n+1)//2)
``` | output | 1 | 62,766 | 14 | 125,533 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,767 | 14 | 125,534 |
"Correct Solution:
```
import sys
import string
from collections import Counter, defaultdict
from math import fsum, sqrt, gcd, ceil, factorial
from itertools import combinations,permutations
# input = sys.stdin.readline
flush = lambda : sys.stdout.flush
comb = lambda x , y : (factorial(x) // factorial(y)) // factorial(x - y)
#inputs
# ip = lambda : input().rstrip()
ip = lambda : input()
ii = lambda : int(input())
r = lambda : map(int, input().split())
rr = lambda : list(r())
print(((ii()+1)//2))
``` | output | 1 | 62,767 | 14 | 125,535 |
Provide a correct Python 3 solution for this coding contest problem.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note | instruction | 0 | 62,768 | 14 | 125,536 |
"Correct Solution:
```
def zombi(n):
return (n + 1) // 2
print(zombi(int(input())))
``` | output | 1 | 62,768 | 14 | 125,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
import math
n = int(input())
print(math.ceil(n/2))
``` | instruction | 0 | 62,769 | 14 | 125,538 |
Yes | output | 1 | 62,769 | 14 | 125,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
import math
a = int(input())
print(math.ceil(a / 2))
``` | instruction | 0 | 62,770 | 14 | 125,540 |
Yes | output | 1 | 62,770 | 14 | 125,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
N = int(input())
print(N // 2 + N % 2)
``` | instruction | 0 | 62,771 | 14 | 125,542 |
Yes | output | 1 | 62,771 | 14 | 125,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
print((int(input())+1)//2)
``` | instruction | 0 | 62,772 | 14 | 125,544 |
Yes | output | 1 | 62,772 | 14 | 125,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
import math
N=int(input())
print(round(math.sqrt(N)))
``` | instruction | 0 | 62,773 | 14 | 125,546 |
No | output | 1 | 62,773 | 14 | 125,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
import math
N=int(input())
print(math.ceil(math.sqrt(N)))
``` | instruction | 0 | 62,774 | 14 | 125,548 |
No | output | 1 | 62,774 | 14 | 125,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
i = int(input())
if i <= 3: print(1);exit()
print((i+1)//2)
``` | instruction | 0 | 62,775 | 14 | 125,550 |
No | output | 1 | 62,775 | 14 | 125,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage.
They will be N guests tonight: N - 1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to N - 1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, N. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure:
The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.)
You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order:
1. survive the event (they experienced death already once and know it is no fun),
2. get as many brains as possible.
Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself.
What is the smallest number of brains that have to be in the chest for this to be possible?
Input
The only line of input contains one integer: N, the number of attendees (1 ≤ N ≤ 109).
Output
Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.
Examples
Input
1
Output
1
Input
4
Output
2
Note
Submitted Solution:
```
n = int(input())
f = lambda x: (1 + x) * x // 2
print(f((n - 1) // 2) + 1)
``` | instruction | 0 | 62,776 | 14 | 125,552 |
No | output | 1 | 62,776 | 14 | 125,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One department of some software company has n servers of different specifications. Servers are indexed with consecutive integers from 1 to n. Suppose that the specifications of the j-th server may be expressed with a single integer number c_j of artificial resource units.
In order for production to work, it is needed to deploy two services S_1 and S_2 to process incoming requests using the servers of the department. Processing of incoming requests of service S_i takes x_i resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service S_i is deployed using k_i servers, then the load is divided equally between these servers and each server requires only x_i / k_i (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
Input
The first line contains three integers n, x_1, x_2 (2 ≤ n ≤ 300 000, 1 ≤ x_1, x_2 ≤ 10^9) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains n space-separated integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^9) — the number of resource units provided by each of the servers.
Output
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers k_1 and k_2 (1 ≤ k_1, k_2 ≤ n) — the number of servers used for each of the services.
In the third line print k_1 integers, the indices of the servers that will be used for the first service.
In the fourth line print k_2 integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
Examples
Input
6 8 16
3 5 2 9 8 7
Output
Yes
3 2
1 2 6
5 4
Input
4 20 32
21 11 11 12
Output
Yes
1 3
1
2 3 4
Input
4 11 32
5 5 16 16
Output
No
Input
5 12 20
7 8 4 11 9
Output
No
Note
In the first sample test each of the servers 1, 2 and 6 will will provide 8 / 3 = 2.(6) resource units and each of the servers 5, 4 will provide 16 / 2 = 8 resource units.
In the second sample test the first server will provide 20 resource units and each of the remaining servers will provide 32 / 3 = 10.(6) resource units. | instruction | 0 | 62,876 | 14 | 125,752 |
Tags: binary search, implementation, sortings
Correct Solution:
```
n, x1, x2 = map(int, input().split())
c = list(map(int, input().split()))
c = [(ci, i) for i, ci in enumerate(c)]
c.sort(reverse=True)
def check(x1,x2,reverse=False):
sum1 = sum2 = 0
i = i1 = i2 = 0
while i < len(c):
sum1 += c[i][0]
i += 1
if sum1 >= x1 and 1.0*x1/i <= c[i-1][0]: break
i1 = i
if i1 == n: return False
while i < len(c):
sum2 += c[i][0]
i += 1
if sum2 >= x2 and 1.0*x2/(i-i1) <= c[i-1][0]:
print('Yes')
if reverse:
print(i-i1, i1)
print(' '.join(map(str, [ci[1]+1 for ci in c[i1:i]])))
print(' '.join(map(str, [ci[1]+1 for ci in c[:i1]])))
else:
print(i1, i-i1)
print(' '.join(map(str, [ci[1]+1 for ci in c[:i1]])))
print(' '.join(map(str, [ci[1]+1 for ci in c[i1:i]])))
return True
return False
if not (check(x1, x2) or check(x2, x1,reverse=True)):
print('No')
``` | output | 1 | 62,876 | 14 | 125,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One department of some software company has n servers of different specifications. Servers are indexed with consecutive integers from 1 to n. Suppose that the specifications of the j-th server may be expressed with a single integer number c_j of artificial resource units.
In order for production to work, it is needed to deploy two services S_1 and S_2 to process incoming requests using the servers of the department. Processing of incoming requests of service S_i takes x_i resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service S_i is deployed using k_i servers, then the load is divided equally between these servers and each server requires only x_i / k_i (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
Input
The first line contains three integers n, x_1, x_2 (2 ≤ n ≤ 300 000, 1 ≤ x_1, x_2 ≤ 10^9) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains n space-separated integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^9) — the number of resource units provided by each of the servers.
Output
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers k_1 and k_2 (1 ≤ k_1, k_2 ≤ n) — the number of servers used for each of the services.
In the third line print k_1 integers, the indices of the servers that will be used for the first service.
In the fourth line print k_2 integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
Examples
Input
6 8 16
3 5 2 9 8 7
Output
Yes
3 2
1 2 6
5 4
Input
4 20 32
21 11 11 12
Output
Yes
1 3
1
2 3 4
Input
4 11 32
5 5 16 16
Output
No
Input
5 12 20
7 8 4 11 9
Output
No
Note
In the first sample test each of the servers 1, 2 and 6 will will provide 8 / 3 = 2.(6) resource units and each of the servers 5, 4 will provide 16 / 2 = 8 resource units.
In the second sample test the first server will provide 20 resource units and each of the remaining servers will provide 32 / 3 = 10.(6) resource units. | instruction | 0 | 62,877 | 14 | 125,754 |
Tags: binary search, implementation, sortings
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=300006, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n,x,y=map(int,input().split())
l=list(map(int,input().split()))
l=[(l[i],i+1) for i in range(n)]
l.sort()
t=1
f=-1
for i in range(n-1,0,-1):
if l[i][0]*t>=x:
f=i
break
t+=1
t=1
f1 = -1
if f!=-1:
for i in range(f-1,-1,-1):
if l[i][0] * t >= y:
f1=i
break
t += 1
if f1!=-1:
q=[]
q1=[]
for i in range(f1,f):
q.append(l[i][1])
for i in range(f,n):
q1.append(l[i][1])
print("Yes")
print(len(q1),len(q))
print(*q1)
print(*q)
sys.exit(0)
t=1
f=-1
for i in range(n-1,0,-1):
if l[i][0]*t>=y:
f=i
break
t+=1
t=1
f1=-1
if f!=-1:
for i in range(f-1,-1,-1):
if l[i][0] * t >= x:
f1=i
break
t += 1
if f1!=-1:
q=[]
q1=[]
for i in range(f1,f):
q.append(l[i][1])
for i in range(f,n):
q1.append(l[i][1])
print("Yes")
print(len(q),len(q1))
print(*q)
print(*q1)
sys.exit(0)
print("No")
``` | output | 1 | 62,877 | 14 | 125,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One department of some software company has n servers of different specifications. Servers are indexed with consecutive integers from 1 to n. Suppose that the specifications of the j-th server may be expressed with a single integer number c_j of artificial resource units.
In order for production to work, it is needed to deploy two services S_1 and S_2 to process incoming requests using the servers of the department. Processing of incoming requests of service S_i takes x_i resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service S_i is deployed using k_i servers, then the load is divided equally between these servers and each server requires only x_i / k_i (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
Input
The first line contains three integers n, x_1, x_2 (2 ≤ n ≤ 300 000, 1 ≤ x_1, x_2 ≤ 10^9) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains n space-separated integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^9) — the number of resource units provided by each of the servers.
Output
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers k_1 and k_2 (1 ≤ k_1, k_2 ≤ n) — the number of servers used for each of the services.
In the third line print k_1 integers, the indices of the servers that will be used for the first service.
In the fourth line print k_2 integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
Examples
Input
6 8 16
3 5 2 9 8 7
Output
Yes
3 2
1 2 6
5 4
Input
4 20 32
21 11 11 12
Output
Yes
1 3
1
2 3 4
Input
4 11 32
5 5 16 16
Output
No
Input
5 12 20
7 8 4 11 9
Output
No
Note
In the first sample test each of the servers 1, 2 and 6 will will provide 8 / 3 = 2.(6) resource units and each of the servers 5, 4 will provide 16 / 2 = 8 resource units.
In the second sample test the first server will provide 20 resource units and each of the remaining servers will provide 32 / 3 = 10.(6) resource units. | instruction | 0 | 62,878 | 14 | 125,756 |
Tags: binary search, implementation, sortings
Correct Solution:
```
def fin(c, x):
return (x + c - 1) // c
def ck(x, b):
r = (n, n)
for i in range(b, n):
r = min(r, (i + fin(c[i][0], x), i))
return r
def sol(r, l):
if r[0] <= n and l[0] <= n and r[1] < n and l[1] < n :
print("Yes")
print(r[0] - r[1], l[0]- l[1])
print(' '.join([str(x[1]) for x in c[r[1]:r[0]]]))
print(' '.join([str(x[1]) for x in c[l[1]:l[0]]]))
return True
else:
return False
n, x1, x2 = [int(x) for x in input().split()]
c = sorted([(int(x), i + 1) for i, x in enumerate(input().split())])
r1 = ck(x1, 0)
l1 = ck(x2, r1[0])
r2 = ck(x2, 0)
l2 = ck(x1, r2[0])
if not sol(r1, l1) and not sol(l2, r2):
print("No")
# 6 8 16
# 3 5 2 9 8 7
``` | output | 1 | 62,878 | 14 | 125,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One department of some software company has n servers of different specifications. Servers are indexed with consecutive integers from 1 to n. Suppose that the specifications of the j-th server may be expressed with a single integer number c_j of artificial resource units.
In order for production to work, it is needed to deploy two services S_1 and S_2 to process incoming requests using the servers of the department. Processing of incoming requests of service S_i takes x_i resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service S_i is deployed using k_i servers, then the load is divided equally between these servers and each server requires only x_i / k_i (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
Input
The first line contains three integers n, x_1, x_2 (2 ≤ n ≤ 300 000, 1 ≤ x_1, x_2 ≤ 10^9) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains n space-separated integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^9) — the number of resource units provided by each of the servers.
Output
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers k_1 and k_2 (1 ≤ k_1, k_2 ≤ n) — the number of servers used for each of the services.
In the third line print k_1 integers, the indices of the servers that will be used for the first service.
In the fourth line print k_2 integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
Examples
Input
6 8 16
3 5 2 9 8 7
Output
Yes
3 2
1 2 6
5 4
Input
4 20 32
21 11 11 12
Output
Yes
1 3
1
2 3 4
Input
4 11 32
5 5 16 16
Output
No
Input
5 12 20
7 8 4 11 9
Output
No
Note
In the first sample test each of the servers 1, 2 and 6 will will provide 8 / 3 = 2.(6) resource units and each of the servers 5, 4 will provide 16 / 2 = 8 resource units.
In the second sample test the first server will provide 20 resource units and each of the remaining servers will provide 32 / 3 = 10.(6) resource units.
Submitted Solution:
```
n, x1, x2 = map(int, input().split())
c = list(map(int, input().split()))
c = [(ci, i) for i, ci in enumerate(c)]
c.sort(reverse=True)
def check(x1,x2,reverse=False):
sum1 = sum2 = 0
i = i1 = i2 = 0
while i < len(c):
sum1 += c[i][0]
i += 1
if sum1 >= x1: break
i1 = i
while i < len(c):
sum2 += c[i][0]
i += 1
if sum2 >= x2: break
i2 = i
# print(c,i1,i2)
if 1.0*x1/i1 <= c[i1-1][0] and 1.0*x2/(i2-i1) <= c[i2-1][0] :
print('Yes')
if reverse:
print(i2-i1, i1)
print(' '.join(map(str, [ci[1]+1 for ci in c[i1:i2]])))
print(' '.join(map(str, [ci[1]+1 for ci in c[:i1]])))
else:
print(i1, i2-i1)
print(' '.join(map(str, [ci[1]+1 for ci in c[:i1]])))
print(' '.join(map(str, [ci[1]+1 for ci in c[i1:i2]])))
return True
return False
if not (check(x1, x2) or check(x2, x1,reverse=True)):
print('No')
``` | instruction | 0 | 62,879 | 14 | 125,758 |
No | output | 1 | 62,879 | 14 | 125,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One department of some software company has n servers of different specifications. Servers are indexed with consecutive integers from 1 to n. Suppose that the specifications of the j-th server may be expressed with a single integer number c_j of artificial resource units.
In order for production to work, it is needed to deploy two services S_1 and S_2 to process incoming requests using the servers of the department. Processing of incoming requests of service S_i takes x_i resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service S_i is deployed using k_i servers, then the load is divided equally between these servers and each server requires only x_i / k_i (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
Input
The first line contains three integers n, x_1, x_2 (2 ≤ n ≤ 300 000, 1 ≤ x_1, x_2 ≤ 10^9) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains n space-separated integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^9) — the number of resource units provided by each of the servers.
Output
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers k_1 and k_2 (1 ≤ k_1, k_2 ≤ n) — the number of servers used for each of the services.
In the third line print k_1 integers, the indices of the servers that will be used for the first service.
In the fourth line print k_2 integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
Examples
Input
6 8 16
3 5 2 9 8 7
Output
Yes
3 2
1 2 6
5 4
Input
4 20 32
21 11 11 12
Output
Yes
1 3
1
2 3 4
Input
4 11 32
5 5 16 16
Output
No
Input
5 12 20
7 8 4 11 9
Output
No
Note
In the first sample test each of the servers 1, 2 and 6 will will provide 8 / 3 = 2.(6) resource units and each of the servers 5, 4 will provide 16 / 2 = 8 resource units.
In the second sample test the first server will provide 20 resource units and each of the remaining servers will provide 32 / 3 = 10.(6) resource units.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=300006, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n,x,y=map(int,input().split())
l=list(map(int,input().split()))
l=[(l[i],i+1) for i in range(n)]
l.sort()
t=1
f=-1
for i in range(n-1,0,-1):
if l[i][0]*t>=x:
f=i
break
t+=1
t=1
f1 = -1
if f!=-1:
for i in range(f-1,-1,-1):
if l[i][0] * t >= y:
f1=i
break
t += 1
if f1!=-1:
q=[]
q1=[]
for i in range(f1,f):
q.append(l[i][1])
for i in range(f,n):
q1.append(l[i][1])
print("YES")
print(len(q1),len(q))
print(*q1)
print(*q)
sys.exit(0)
t=1
f=-1
for i in range(n-1,0,-1):
if l[i][0]*t>=y:
f=i
break
t+=1
t=1
f1=-1
if f!=-1:
for i in range(f-1,-1,-1):
if l[i][0] * t >= x:
f1=i
break
t += 1
if f1!=-1:
q=[]
q1=[]
for i in range(f1,f):
q.append(l[i][1])
for i in range(f,n):
q1.append(l[i][1])
print("YES")
print(len(q),len(q1))
print(*q)
print(*q1)
sys.exit(0)
print("NO")
``` | instruction | 0 | 62,880 | 14 | 125,760 |
No | output | 1 | 62,880 | 14 | 125,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One department of some software company has n servers of different specifications. Servers are indexed with consecutive integers from 1 to n. Suppose that the specifications of the j-th server may be expressed with a single integer number c_j of artificial resource units.
In order for production to work, it is needed to deploy two services S_1 and S_2 to process incoming requests using the servers of the department. Processing of incoming requests of service S_i takes x_i resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service S_i is deployed using k_i servers, then the load is divided equally between these servers and each server requires only x_i / k_i (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
Input
The first line contains three integers n, x_1, x_2 (2 ≤ n ≤ 300 000, 1 ≤ x_1, x_2 ≤ 10^9) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains n space-separated integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^9) — the number of resource units provided by each of the servers.
Output
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers k_1 and k_2 (1 ≤ k_1, k_2 ≤ n) — the number of servers used for each of the services.
In the third line print k_1 integers, the indices of the servers that will be used for the first service.
In the fourth line print k_2 integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
Examples
Input
6 8 16
3 5 2 9 8 7
Output
Yes
3 2
1 2 6
5 4
Input
4 20 32
21 11 11 12
Output
Yes
1 3
1
2 3 4
Input
4 11 32
5 5 16 16
Output
No
Input
5 12 20
7 8 4 11 9
Output
No
Note
In the first sample test each of the servers 1, 2 and 6 will will provide 8 / 3 = 2.(6) resource units and each of the servers 5, 4 will provide 16 / 2 = 8 resource units.
In the second sample test the first server will provide 20 resource units and each of the remaining servers will provide 32 / 3 = 10.(6) resource units.
Submitted Solution:
```
def fin(c, x):
return x // c if x // c * c != x else x // c - 1
def ck(x, b):
r = (n, n)
for i in range(b, n):
r = min(r, (i + fin(c[i][0], x), i))
return r
def sol(r, l):
if r[0] <= n and l[0] <= n and r[1] < n and l[1] < n :
print("Yes")
print(r[0] - r[1], l[0]- l[1])
solr = ""
for i in range(r[1], r[0]):
solr += str(c[i][1]) + " "
print(solr)
soll = ""
for i in range(l[1], l[0]):
soll += str(c[i][1]) + " "
print(soll)
return True
else:
return False
n, x1, x2 = [int(x) for x in input().split()]
c = sorted([(int(x), i + 1) for i, x in enumerate(input().split())])
r1 = ck(x1, 0)
l1 = ck(x2, r1[0])
r2 = ck(x2, 0)
l2 = ck(x1, r2[0])
if not sol(r1, l1) and not sol(l2, r2):
print("No")
``` | instruction | 0 | 62,881 | 14 | 125,762 |
No | output | 1 | 62,881 | 14 | 125,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One department of some software company has n servers of different specifications. Servers are indexed with consecutive integers from 1 to n. Suppose that the specifications of the j-th server may be expressed with a single integer number c_j of artificial resource units.
In order for production to work, it is needed to deploy two services S_1 and S_2 to process incoming requests using the servers of the department. Processing of incoming requests of service S_i takes x_i resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service S_i is deployed using k_i servers, then the load is divided equally between these servers and each server requires only x_i / k_i (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
Input
The first line contains three integers n, x_1, x_2 (2 ≤ n ≤ 300 000, 1 ≤ x_1, x_2 ≤ 10^9) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains n space-separated integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^9) — the number of resource units provided by each of the servers.
Output
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers k_1 and k_2 (1 ≤ k_1, k_2 ≤ n) — the number of servers used for each of the services.
In the third line print k_1 integers, the indices of the servers that will be used for the first service.
In the fourth line print k_2 integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
Examples
Input
6 8 16
3 5 2 9 8 7
Output
Yes
3 2
1 2 6
5 4
Input
4 20 32
21 11 11 12
Output
Yes
1 3
1
2 3 4
Input
4 11 32
5 5 16 16
Output
No
Input
5 12 20
7 8 4 11 9
Output
No
Note
In the first sample test each of the servers 1, 2 and 6 will will provide 8 / 3 = 2.(6) resource units and each of the servers 5, 4 will provide 16 / 2 = 8 resource units.
In the second sample test the first server will provide 20 resource units and each of the remaining servers will provide 32 / 3 = 10.(6) resource units.
Submitted Solution:
```
n, x1, x2 = map(int, input().split())
c = list(map(int, input().split()))
c = [(ci, i) for i, ci in enumerate(c)]
c.sort(reverse=True)
def check(x1,x2):
sum1 = sum2 = 0
i = i1 = i2 = 0
while i < len(c):
sum1 += c[i][0]
i += 1
if sum1 >= x1: break
i1 = i
while i < len(c):
sum2 += c[i][0]
i += 1
if sum2 >= x2: break
i2 = i
if 1.0*x1/i1 <= c[i1-1][0] and 1.0*x2/(i2-i1) <= c[i2-1][0] :
print('Yes')
print(i1, i2-i1)
print(' '.join(map(str, [ci[1]+1 for ci in c[:i1]])))
print(' '.join(map(str, [ci[1]+1 for ci in c[i1:i2]])))
return True
return False
if not (check(x1, x2) or check(x2, x1)):
print('No')
``` | instruction | 0 | 62,882 | 14 | 125,764 |
No | output | 1 | 62,882 | 14 | 125,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7 | instruction | 0 | 63,165 | 14 | 126,330 |
Tags: *special, dfs and similar, trees
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
b = [n]
i = n - 2
while a[i] != 1:
b.append(a[i])
i = a[i] - 2
b.append(1)
b.reverse()
print(*b)
``` | output | 1 | 63,165 | 14 | 126,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7 | instruction | 0 | 63,166 | 14 | 126,332 |
Tags: *special, dfs and similar, trees
Correct Solution:
```
n = int(input())
p = [int(i) for i in input().split()]
ans = []
t = n
while t != 1:
ans.append(t)
t = p[t-2]
ans.append(1)
for i in range(len(ans)):
print(ans[len(ans) - 1 - i], end=' ')
``` | output | 1 | 63,166 | 14 | 126,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7 | instruction | 0 | 63,167 | 14 | 126,334 |
Tags: *special, dfs and similar, trees
Correct Solution:
```
import math
x = int(input())
p = list(map(int,input().split()))
ans = []
sz = 0
while x > 1:
ans.append(x)
x = p[x - 2]
sz += 1
ans.append(1)
sz += 1
for i in range(1,sz + 1):
print (ans[sz - i])
``` | output | 1 | 63,167 | 14 | 126,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7 | instruction | 0 | 63,168 | 14 | 126,336 |
Tags: *special, dfs and similar, trees
Correct Solution:
```
n = int(input())
parents = list(map(int, input().strip().split()))
cur = parents[-1]-2
path = [n]
while 1:
path.append(cur+2)
if cur == -1:
break
cur = parents[cur]-2
path.reverse()
for i in path:
print(i, end = " ")
``` | output | 1 | 63,168 | 14 | 126,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7 | instruction | 0 | 63,169 | 14 | 126,338 |
Tags: *special, dfs and similar, trees
Correct Solution:
```
from collections import deque
class Graph():
def __init__(self, n):
self.size = n
self.g = [[] for _ in range(n)]
def addEdge(self, u, v):
self.g[u].append(v)
self.g[v].append(u)
def findPath(self, _from, _to):
queue = deque()
visited = [False] * self.size
parent = [i for i in range(self.size)]
queue.append(_from)
while queue:
u = queue.popleft()
visited[u] = True
if u == _to:
break
for v in self.g[u]:
if not visited[v]:
parent[v] = u
queue.append(v)
path = []
curr = _to
while curr != parent[curr]:
path.append(curr)
curr = parent[curr]
path.append(_from)
path.reverse()
return path
def main():
n = int(input())
g = Graph(n)
for u, v in enumerate(map(int, input().split())):
g.addEdge(u+1, v-1)
path = g.findPath(0, n-1)
print(*(i+1 for i in path))
if __name__ == '__main__':
main()
``` | output | 1 | 63,169 | 14 | 126,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7 | instruction | 0 | 63,171 | 14 | 126,342 |
Tags: *special, dfs and similar, trees
Correct Solution:
```
# ��������� ������� ������
N = int(input())
input_str = str(input())
input_str = '0 ' + input_str
L = input_str.split()
for i in range(len(L)):
L[i] = int(L[i])
def find_path(l):
# � ��� ���������� ������ ����� ��������������, ������ �������� �� ����
k = len(l)
output_list = [k]
while not k == 1:
k = l[k-1]
output_list.append(k)
# ������� ������
output_str = str(output_list[len(output_list)-1])
for i in range(len(output_list)-2, -1, -1):
output_str += ' '
output_str += str(output_list[i])
return output_str
# main
print(find_path(L))
``` | output | 1 | 63,171 | 14 | 126,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7 | instruction | 0 | 63,172 | 14 | 126,344 |
Tags: *special, dfs and similar, trees
Correct Solution:
```
'''input
6
1 2 3 4 5
'''
n=int(input())
l=list(map(int,input().strip().split(' ')))
v=n
route=[v]
while v!=1:
v=l[v-2]
route.append(v)
print(*route[::-1])
``` | output | 1 | 63,172 | 14 | 126,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7
Submitted Solution:
```
n = int(input())
p = list(map(int, input().split()))
ls = [n]
while n > 1:
n = p[n - 2]
ls.append(n)
print(*ls[::-1])
``` | instruction | 0 | 63,173 | 14 | 126,346 |
Yes | output | 1 | 63,173 | 14 | 126,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7
Submitted Solution:
```
n = int(input()) #2e5
pl = list(map(int,input().split())) #p2开始
ll = [n]
while ll[-1]>1:
ll.append(pl[ll[-1]-2])
ll.reverse()
print(*ll)
``` | instruction | 0 | 63,176 | 14 | 126,352 |
Yes | output | 1 | 63,176 | 14 | 126,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
k=[n]
i=n-2
temp=0
while(temp!=1):
k.append(l[i])
i=l[i]-2
temp=l[i]
print(*set(sorted([1]+k)))
``` | instruction | 0 | 63,178 | 14 | 126,356 |
No | output | 1 | 63,178 | 14 | 126,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.
Input
The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7
Submitted Solution:
```
N = int(input())-1
arr = list(map(int, input().split()))
arr.append(N+1)
arr1 = []
for i in range(N):
if arr[i] < arr[i+1] and arr[i] not in arr1:
arr1.append(arr[i])
arr1.append(N+1)
arr1.sort()
print(*arr1)
``` | instruction | 0 | 63,180 | 14 | 126,360 |
No | output | 1 | 63,180 | 14 | 126,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 1000) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
It is guaranteed that ∑ n ≤ 2 ⋅ 10^5 (sum of n over all queries does not exceed 2 ⋅ 10^5).
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4 | instruction | 0 | 63,237 | 14 | 126,474 |
Tags: dfs and similar, dsu, math
Correct Solution:
```
q = int(input())
for _ in range(q):
n = int(input())
perm = list(map(int, input().split()))
remain = set(perm)
result = {i+1: 0 for i in range(n)}
while len(remain) > 0:
m = remain.pop()
cycle = [m]
while True:
m = perm[m-1]
if m not in remain:
break
remain.remove(m)
cycle.append(m)
k = len(cycle)
for j in cycle:
result[j] = str(k)
print(' '.join(result.values()))
``` | output | 1 | 63,237 | 14 | 126,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 1000) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
It is guaranteed that ∑ n ≤ 2 ⋅ 10^5 (sum of n over all queries does not exceed 2 ⋅ 10^5).
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4 | instruction | 0 | 63,238 | 14 | 126,476 |
Tags: dfs and similar, dsu, math
Correct Solution:
```
t = int(input())
while t>0:
t -= 1
n = int(input())
a = list(map(int,input().split()))
for i in range(n):
a[i] -= 1
res = [0 for _ in range(n)]
for i in range(n):
if res[i]!=0:
continue
x = [i]
j = a[i]
ans = 1
while j!=i:
j = a[j]
ans += 1
x.append(j)
for k in x:
res[k] = ans
for i in res:
print(i,end=' ')
print()
``` | output | 1 | 63,238 | 14 | 126,477 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.