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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input
The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4 3
1 2 3 4
Output
4
Input
4 2
-3 -2 -1 0
Output
2
Input
5 19
1 10 20 30 50
Output
1
Note
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | instruction | 0 | 273 | 14 | 546 |
Tags: binary search, combinatorics, two pointers
Correct Solution:
```
from collections import deque
from math import factorial
nd = input()
nd = nd.split()
n = int(nd[0])
d = int(nd[1])
xcorr = list(map(int, input().split()))
dq = deque()
g = 0
for x in xcorr:
dq.append(x)
while (x - dq[0]) > d:
if len(dq) >= 4:
g += (len(dq)-2)*(len(dq)-2-1)/2
#factorial(len(dq)-2)/(factorial(len(dq)-2-2)*2)
dq.popleft()
if len(dq) >= 3:
g += len(dq)*(len(dq)-1)*(len(dq)-2)/2/3
#factorial(len(dq))/(factorial(len(dq)-3)*2*3)
print(int(g))
``` | output | 1 | 273 | 14 | 547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input
The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4 3
1 2 3 4
Output
4
Input
4 2
-3 -2 -1 0
Output
2
Input
5 19
1 10 20 30 50
Output
1
Note
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | instruction | 0 | 274 | 14 | 548 |
Tags: binary search, combinatorics, two pointers
Correct Solution:
```
n, k = list(map(int, input().split()))
lis = list(map(int, input().split()))
def find(s, lis, a, k):
l = s
r = len(lis) - 1
while (r - l) > 1:
mid = (l + r) // 2
if lis[mid] < a + k:
l = mid
else:
r = mid
if lis[r] <= a + k:
return r
return l
nat = 0
def entekhab(y):
return int(y * (y - 1) / 2)
s = 0
for i in range(n):
now = lis[i]
loc = find(i, lis, now, k)
if now + k >= lis[loc]:
nat += entekhab(loc - i)
print(int(nat))
``` | output | 1 | 274 | 14 | 549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input
The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4 3
1 2 3 4
Output
4
Input
4 2
-3 -2 -1 0
Output
2
Input
5 19
1 10 20 30 50
Output
1
Note
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | instruction | 0 | 275 | 14 | 550 |
Tags: binary search, combinatorics, two pointers
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Mon May 16 17:14:14 2016
@author: Hangell
"""
from collections import deque
str1=input()
str2=input()
n=int(str1.split()[0])
d=int(str1.split()[1])
numlist=str2.split()
newdeque=deque()
Sum=0
cha=0
way=0
for i in range(n):
numlist[i]=int(numlist[i])
for i in range(n-1):
cha=numlist[i+1]-numlist[i]
newdeque.append(cha)
Sum+=cha
while Sum>d and len(newdeque)!=0:
Sum-=newdeque.popleft()
if len(newdeque)>1:
way+=(len(newdeque)-1)*len(newdeque)/2
print(int(way))
``` | output | 1 | 275 | 14 | 551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input
The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4 3
1 2 3 4
Output
4
Input
4 2
-3 -2 -1 0
Output
2
Input
5 19
1 10 20 30 50
Output
1
Note
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | instruction | 0 | 277 | 14 | 554 |
Tags: binary search, combinatorics, two pointers
Correct Solution:
```
from sys import stdin,stdout,stderr
def binary(lo,hi,value):
global a
ans=lo
while lo<=hi:
mid=(lo+hi)//2
if a[mid]<=value:
ans=mid
lo=mid+1
else:hi=mid-1
return ans
n,d=map(int,input().split())
a=[int(x)for x in input().split()]
ans=0
for i in range(n):
lo=binary(i+1,n-1,a[i]+d)
lo=(lo-i)
ans=ans+lo*(lo-1)/2
print(int(ans))
``` | output | 1 | 277 | 14 | 555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right.
Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy:
* Each piece should contain at least l numbers.
* The difference between the maximal and the minimal number on the piece should be at most s.
Please help Alexandra to find the minimal number of pieces meeting the condition above.
Input
The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105).
The second line contains n integers ai separated by spaces ( - 109 ≤ ai ≤ 109).
Output
Output the minimal number of strip pieces.
If there are no ways to split the strip, output -1.
Examples
Input
7 2 2
1 3 1 2 4 1 2
Output
3
Input
7 2 2
1 100 1 100 1 100 1
Output
-1
Note
For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2].
For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists. | instruction | 0 | 375 | 14 | 750 |
Tags: binary search, data structures, dp, two pointers
Correct Solution:
```
def split(a,n,s,l):
pieces = []
i = 1
tmpmin = a[0]
tmpmax = a[0]
tmppc = [a[0]]
while i<n:
if abs(a[i]-tmpmin)<=s and abs(a[i]-tmpmax)<=s:
tmppc.append(a[i])
if a[i]<tmpmin: tmpmin=a[i]
elif a[i]>tmpmax: tmpmax = a[i]
else:
pieces.append(tmppc)
tmppc = [a[i]]
tmpmin = a[i]
tmpmax = a[i]
i += 1
pieces.append(tmppc)
fail = False
for j in range(len(pieces)):
if len(pieces[j])<l:
if j>0:
prevpc = pieces[j-1]
minj = min(pieces[j])
maxj = max(pieces[j])
while len(pieces[j])<l:
tmp = prevpc.pop()
if abs(tmp-minj)<=s and abs(tmp-maxj)<=s:
pieces[j].insert(0,tmp)
if tmp<minj: minj=tmp
elif tmp>maxj: maxj=tmp
else:
return -1
if len(prevpc)<l:
return -1
else:
return -1
return len(pieces)
n,s,l = [int(s) for s in input().split()]
a = [int(s) for s in input().split()]
res = split(a,n,s,l)
if res<0:
a.reverse()
res = split(a,n,s,l)
print(res)
``` | output | 1 | 375 | 14 | 751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 471 | 14 | 942 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
l = list(map(int, input().split()))
l = sorted(l, reverse=True)
s = 0
for _ in range(n):
if l[0] <= 0:
break
if l[0] <= 2:
l[0] -= min(l[0],2)
s += 1
l[0] -= min(l[0],4)
l = sorted(l, reverse=True)
for _ in range(n):
if l[0] <= 0:
break
l[0] -= min(l[0],2)
l = sorted(l, reverse=True)
for _ in range(n):
if l[0] <= 0:
break
l[0] -= min(l[0],2)
l = sorted(l, reverse=True)
for _ in range(s):
if l[0] <= 0:
break
l[0] -= min(l[0],1)
l = sorted(l, reverse=True)
if l[0] <= 0:
print("YES")
else:
print("NO")
``` | output | 1 | 471 | 14 | 943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 472 | 14 | 944 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
l=list(map(int,input().split()))
p=n
for i in range(m) :
b=l[i]//4
l[i]=l[i]-4*min(p,l[i]//4)
p=p-min(p,b)
p1=n*2
for i in range(m) :
b=l[i]//2
l[i]=l[i]-2*min(p1,l[i]//2)
p1=p1-min(p1,b)
p2=p+p1
p3=p
for i in range(m) :
b=l[i]//2
l[i]=l[i]-2*min(p2,l[i]//2)
p2=p2-min(p2,b)
k=0
p3+=p2
for i in range(m) :
k=k+l[i]
if k>p3 :
print('NO')
else :
print('YES')
``` | output | 1 | 472 | 14 | 945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 473 | 14 | 946 |
Tags: brute force, greedy, implementation
Correct Solution:
```
import sys
n, k = [ int(x) for x in input().split() ]
arr = list( map( int, input().split()) )
cnt_2 = 2 * n
cnt_4 = n
cnt_1 = 0
i = 0
while(cnt_4 > 0 and i < k):
tmp = int(arr[i] / 4)
if(tmp > 0 and cnt_4 > 0):
arr[i] = arr[i] - 4 * min(tmp, cnt_4)
cnt_4 = cnt_4 - min(tmp, cnt_4)
i = i + 1
if(cnt_4 > 0):
cnt_2 = cnt_2 + cnt_4
cnt_1 = cnt_1 + cnt_4
cnt_4 = 0
i = 0
while(cnt_2 > 0 and i < k):
tmp = int(arr[i] / 2)
if(tmp > 0 and cnt_2 > 0):
arr[i] = arr[i] - 2 * min(tmp, cnt_2)
cnt_2 = cnt_2 - min(tmp, cnt_2)
i = i + 1
if(cnt_2 > 0):
cnt_1 = cnt_1 + cnt_2
cnt_2 = 0
i = 0
while(i < k):
tmp = arr[i]
if(tmp > cnt_1):
print("NO")
sys.exit(0)
elif(tmp > 0 and tmp <= cnt_1):
cnt_1 = cnt_1 - tmp
arr[i] = arr[i] - tmp
i = i + 1
print("YES")
``` | output | 1 | 473 | 14 | 947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 474 | 14 | 948 |
Tags: brute force, greedy, implementation
Correct Solution:
```
from math import ceil
n,k = map(int, input().split())
a = list(map(int, input().split()))
def solve(n,k,a):
one,two,four = 0, n*2, n
for size in a:
while size >= 4:
if four > 0:
four -= 1
else:
two -= 2
size -= 4
while size > 0:
if four > 0:
four -= 1
if size == 2:
one += 1
elif size == 1:
two += 1
break
elif two > 0:
two -= 1
size -= 2
elif one > 0:
one -= 1
size -= 1
else:
return "NO"
return "YES"
print(solve(n,k,sorted(a, reverse=True)))
``` | output | 1 | 474 | 14 | 949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 475 | 14 | 950 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n,k = map(int,input().split(" "))
a = list(map(int,input().split(" ")))
count = 0
for i in range(0,len(a)):
if a[i]%2 == 0:
count = count + 1
if k-count <= n*8-sum(a):
if k-count < n and k==n*4:
print("NO")
else:
print('YES')
else:print('NO')
``` | output | 1 | 475 | 14 | 951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 476 | 14 | 952 |
Tags: brute force, greedy, implementation
Correct Solution:
```
def solve(a, n, k):
cnt =[0 for i in range(5)]
duplas, quads, sozinhos = 2*n, n, 0
for i in a:
for j in range(k):
while i >= 3:
if quads > 0:
i -= 4; quads -= 1
elif duplas > 0:
i -= 2; duplas -= 1
else: return "no"
if i > 0:
cnt[i] += 1
while cnt[2]:
if duplas > 0:
duplas -= 1; cnt[2] -= 1
elif quads > 0:
cnt[2] -= 1; quads -= 1; sozinhos += 1
else:
cnt[2] -= 1; cnt[1] += 2
if cnt[1] > sozinhos + duplas + 2*quads:
return "no"
return "yes"
n, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
print(solve(a, n, k))
# 1522848694321
``` | output | 1 | 476 | 14 | 953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 477 | 14 | 954 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
c = [0, 0, 0, 0]
for t in a:
c[0] += t//4
if t%4: c[t%4] += 1
c[0] += c[3]
c[3] = 0
if c[0] > n:
c[2] += 2*(c[0]-n)
c[0] = n
t = min(n-c[0], c[1], c[2])
c[0] += t
c[1] -= t
c[2] -= t
t = min(n-c[0], (c[1]+1)//2)
c[0] += t
c[1] -= min(c[1], t*2)
t = min(n-c[0], c[2])
c[0] += t
c[2] -= min(c[2], t+t//2)
c[2] += c[1]
c[1] = 0
print("YES" if c[2] <= 2*n else "NO")
``` | output | 1 | 477 | 14 | 955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6). | instruction | 0 | 478 | 14 | 956 |
Tags: brute force, greedy, implementation
Correct Solution:
```
# encoding:utf-8
def main():
n, k = list(map(int, input().split()))
nums = list(map(int, input().split()))
seat_two = n * 2
seat_four = n
seat_one = 0
n_four = sum([x // 4 for x in nums])
nums = [x % 4 for x in nums]
n_two = sum([x // 2 for x in nums])
n_one = sum([x % 2 for x in nums])
#print(n_one, n_two, n_four)
#print(seat_one, seat_two, seat_four)
if seat_four >= n_four:
# there is rest of 4 seat
seat_four -= n_four
n_four = 0
# break seat seat_one and seat_two
seat_two += seat_four
seat_one += seat_four
seat_four = 0
else:
# there is rest of 4 people
n_four -= seat_four
seat_four = 0
# break 4 people to 2, 2
n_two += n_four * 2
n_four = 0
#print(n_one, n_two, n_four)
#print(seat_one, seat_two, seat_four)
if seat_two >= n_two:
seat_two -= n_two
n_two = 0
if seat_two + seat_one >= n_one:
print("YES")
else:
print("NO")
else:
n_two -= seat_two
seat_two = 0
n_one += n_two * 2
if seat_one >= n_one:
print("YES")
else:
print("NO")
if __name__ == '__main__':
main()
1
``` | output | 1 | 478 | 14 | 957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
def solve(n, k, a):
fourSeatLeft = n
twoSeatLeft = n * 2
for i in range(k):
while a[i] >= 4:
if fourSeatLeft > 0:
fourSeatLeft -= 1
a[i] -= min(a[i], 4)
else:
break
for i in range(k):
while a[i] >= 3:
if fourSeatLeft > 0:
fourSeatLeft -= 1
a[i] -= min(a[i], 4)
else:
break
for i in range(k):
while a[i] >= 2:
if twoSeatLeft > 0:
twoSeatLeft -= 1
a[i] -= min(a[i], 2)
else:
break
for i in range(k):
while a[i] >= 1:
if twoSeatLeft > 0:
twoSeatLeft -= 1
a[i] -= min(a[i], 2)
else:
break
twoSeatLeft += fourSeatLeft
oneSeatLeft = fourSeatLeft
fourSeatLeft = 0
for i in range(k):
while a[i] >= 2:
if twoSeatLeft > 0:
twoSeatLeft -= 1
a[i] -= min(a[i], 2)
else:
break
oneSeatLeft += twoSeatLeft
twoSeatLeft = 0
for i in range(k):
while a[i] >= 1:
if oneSeatLeft > 0:
oneSeatLeft -= 1
a[i] -= min(a[i], 1)
else:
break
if sum(a) > 0:
return 'NO'
return 'YES'
##if __name__ == '__main__':
##
## import sys
##
## stdin = sys.stdin
## sys.stdin = open('cf839b.txt')
##
## testcaseNo = eval(input())
##
## for c in range(testcaseNo):
## [n, k] = [eval(v) for v in input().split()]
## a = [eval(v) for v in input().split()]
##
## result = solve(n, k, a)
##
## print('Case#' + str(c + 1) + ':', end=' ')
## print(result)
##
## sys.stdin = stdin
[n, k] = [eval(v) for v in input().split()]
a = [eval(v) for v in input().split()]
print(solve(n, k, a))
``` | instruction | 0 | 479 | 14 | 958 |
Yes | output | 1 | 479 | 14 | 959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
#coding=utf-8
def test():
n, k = map(int, input().split())
a = list(map(int, input().split()))
total = n * 8
place = 0
sum = 0
i = 0
temp = 0
flag = 0
while i < k:
sum += a[i]
if a[i] % 2 is 1:
place += 1
else:
flag += 1
if a[i] % 4 != 0:
temp += 1
i += 1
if sum == total:
if place!=0 or (flag%4==0 and n*4==k):
print("NO")
else:
print("YES")
else:
if (sum + place) > total or (n*4==k and (flag//4)==place ):
print("NO")
else:
print("YES")
test()
``` | instruction | 0 | 480 | 14 | 960 |
Yes | output | 1 | 480 | 14 | 961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
import sys
import os
import math
import re
n,k = map(int,input().split())
sol = list(map(int,input().split()))
totalSeats = 8*n
middle = n
leftRight = 2*n
single = 0
for i in range(k):
fours = min((sol[i]+1)//4,middle) #take the min of 4s needed + one empty and the middle
middle -= fours
sol[i] = max(0,sol[i]-4*fours) #soldiers remaining
leftRight += middle
single += middle
for i in range(k): #now deal with the twos
twos = min(sol[i]//2, leftRight)
leftRight -= twos
sol[i] = max(0, sol[i]-2*twos) #soldiers remaining
single += leftRight
for i in range(k): #all the singles that need a seat
single -= sol[i]
if single < 0:
print('No')
else:
print('Yes')
``` | instruction | 0 | 481 | 14 | 962 |
Yes | output | 1 | 481 | 14 | 963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
n,k=map(int,input().split())
wuya=[int(x) for x in input().split()]
n_4=sum([x//4 for x in wuya])
wuya=[x%4 for x in wuya]
n_2=sum([x//2 for x in wuya])
n_1=sum([x%2 for x in wuya])
p4=n
p4,n_4=p4-min(n_4,p4),n_4-min(n_4,p4)
p2=2*n+p4
n_2+=2*n_4
p2,n_2=p2-min(n_2,p2),n_2-min(n_2,p2)
p1=p4+p2
n_1+=n_2*2
if p1>=n_1:
print('YES')
else:
print('NO')
``` | instruction | 0 | 482 | 14 | 964 |
Yes | output | 1 | 482 | 14 | 965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
debug = 0
read = lambda: map(int, input().split())
k, n = read()
a = list(read())
cnt4 = k
cnt2 = 2 * k
for i in range(n):
cnt = min(cnt4, a[i] // 4)
a[i] -= cnt * 4
cnt4 -= cnt
if debug: print(*a)
for i in range(n):
cnt = min(cnt2, a[i] // 2)
a[i] -= cnt * 2
cnt2 -= cnt
if debug: print(' '.join(map(str, a)), ' ', cnt2, cnt4)
c = [0] * 20
for i in a:
c[min(i, 19)] += 1
d = min(cnt4, c[1], c[2])
c[1] -= d
c[2] -= d
cnt4 -= d
d = min(cnt2, c[2])
c[2] -= d
cnt2 -= d
d = min(cnt2, c[1])
c[1] -= d
cnt2 -= d
d = min(cnt4, c[2])
c[2] -= d
cnt4 -= d
if debug:
print('cnt4 = ', cnt4)
print('c[1] = ', c[1])
d = min(cnt4, (c[1] + 1) // 2)
c[1] -= d * 2
cnt4 -= d
print('YES' if sum(c[1:]) == 0 else 'NO')
``` | instruction | 0 | 483 | 14 | 966 |
No | output | 1 | 483 | 14 | 967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
n, k = map(int, input().split())
s = list(map(int, input().split()))
q = n
double = 2*n
for i in range(k):
if double>=s[i]//2+s[i]%2 :
s[i]=0
double-=s[i]//2+s[i]%2
else:
s[i]-=double*2
double=0
break
for i in range(k):
p = min(q, s[i]//4)
q -= p
s[i] -= (p*4)
if q != 0:
for i in range(k):
if s[i] == 0:
continue
if s[i] == 1 and q != 0:
q -= 0.5
s[i]=0
elif s[i] == 2:
if q % 1 == 0.5:
q -= 1.5
s[i]=0
elif q > 0:
q -= 1
s[i]=0
elif s[i] >= 3 and q >= 1:
q -= 1
s[i]=0
if double!=0:
for i in range(k):
if s[i]!=0:
if (double>=s[i]//2+s[i]%2):
double-=s[i]//2+s[i]%2
s[i]=0
else:
print('NO')
exit()
for i in s:
if i!=0:
print('NO')
exit()
print('YES')
``` | instruction | 0 | 484 | 14 | 968 |
No | output | 1 | 484 | 14 | 969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
def main():
n,_=[int(i) for i in input().split(" ")]
a=[int(i) for i in input().split(" ")]
sides=n*2
mid=n
single=0
for g in a:
while g>0:
if g>=4 and mid>0:
mid-=1
g-=4
elif g>=2 and sides>0:
sides-=1
g-=2
elif g>=2: #no sides g<4
mid-=1
single+=1
g-=2
elif g==1:
if single>0:
single-=1
g-=1
elif sides>0:
sides-=1
g-=1
elif mid>0:
mid-=1
g-=1
else:
print("NO")
return
else:
print("NO")
return
print("YES")
main()
``` | instruction | 0 | 485 | 14 | 970 |
No | output | 1 | 485 | 14 | 971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.
<image> A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.
Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.
Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.
The second line contains k integers a1, a2, a3, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.
It is guaranteed that a1 + a2 + ... + ak ≤ 8·n.
Output
If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).
You can choose the case (lower or upper) for each letter arbitrary.
Examples
Input
2 2
5 8
Output
YES
Input
1 2
7 1
Output
NO
Input
1 2
4 4
Output
YES
Input
1 4
2 2 1 2
Output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:
<image>
In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.
In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.
In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).
Submitted Solution:
```
def main():
n, k = map(int, input().split())
cnt, a4 = [0] * 4, 0
for a in map(int, input().split()):
a4 += a // 4
cnt[a % 4] += 1
_, a1, a2, a3 = cnt
if (a3 + a4) * 2 < a2:
a4 += a3
a2 -= a4 * 2
n -= ((a2 + 2) // 3) if a2 > a1 * 3 else ((a2 + a1 + 3) // 4)
print(("YES", "NO")[n < a4])
else:
print(("YES", "NO")[n * 4 < (a4 + a3) * 2 + a2 + a1])
if __name__ == '__main__':
main()
``` | instruction | 0 | 486 | 14 | 972 |
No | output | 1 | 486 | 14 | 973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 109 + 7.
Input
Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.
N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.
Output
Print the number of possible assignments, modulo 1000000007 = 109 + 7.
Examples
Input
4
1 5
5 2
3 7
7 3
Output
6
Input
5
1 10
2 10
3 10
4 10
5 5
Output
5
Note
These are the possible assignments for the first example:
* 1 5 3 7
* 1 2 3 7
* 5 2 3 7
* 1 5 7 3
* 1 2 7 3
* 5 2 7 3 | instruction | 0 | 487 | 14 | 974 |
Tags: combinatorics, dfs and similar, dsu, graphs, trees
Correct Solution:
```
n = int(input())
m = 2 * n + 1
u = [[] for i in range(m)]
v = [0] * m
s = [0] * m
d = 10 ** 9 + 7
y = 1
for j in range(n):
a, b = map(int, input().split())
v[a] = b
if a != b:
s[b] += 1
u[b].append(a)
for b in range(m):
if not v[b]:
x = 0
p = [b]
while p:
x += 1
a = p.pop()
s[a] = -1
p += u[a]
y = (x * y) % d
for a in range(m):
if s[a] == 0:
b = v[a]
while s[b] == 1:
s[b] = -1
b = v[b]
s[b] -= 1
for a in range(m):
if s[a] == 1:
y = (2 * y) % d
while s[a]:
s[a] = 0
a = v[a]
print(y)
``` | output | 1 | 487 | 14 | 975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 109 + 7.
Input
Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.
N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.
Output
Print the number of possible assignments, modulo 1000000007 = 109 + 7.
Examples
Input
4
1 5
5 2
3 7
7 3
Output
6
Input
5
1 10
2 10
3 10
4 10
5 5
Output
5
Note
These are the possible assignments for the first example:
* 1 5 3 7
* 1 2 3 7
* 5 2 3 7
* 1 5 7 3
* 1 2 7 3
* 5 2 7 3
Submitted Solution:
```
import sys
input = sys.stdin.readline
M = 10**9 + 7
n = int(input())
G = [list() for _ in range(2*n+1)]
for _ in range(n):
u, v = map(int, input().split())
G[u].append(v)
G[v].append(u)
visited = [False]*(2*n+1)
sys.setrecursionlimit(10**6)
E, V, loop = None, None, False
def dfs(s):
global E, V, loop
V.add(s)
visited[s] = True
if len(G[s]) > 2:
loop = True
for v in G[s]:
E += 1
if not visited[v]:
dfs(v)
ans = 1
for v in range(2*n+1):
if not visited[v]:
E, V = 1, set()
loop = False
dfs(v)
cnt = 1
if E < 2*len(V):
cnt = len(V)
elif not loop:
cnt = 2
ans = (ans*cnt) % M
print(ans)
``` | instruction | 0 | 488 | 14 | 976 |
No | output | 1 | 488 | 14 | 977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 109 + 7.
Input
Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.
N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.
Output
Print the number of possible assignments, modulo 1000000007 = 109 + 7.
Examples
Input
4
1 5
5 2
3 7
7 3
Output
6
Input
5
1 10
2 10
3 10
4 10
5 5
Output
5
Note
These are the possible assignments for the first example:
* 1 5 3 7
* 1 2 3 7
* 5 2 3 7
* 1 5 7 3
* 1 2 7 3
* 5 2 7 3
Submitted Solution:
```
n = int(input())
m = 2 * n + 2
u = [[] for i in range(m)]
v = [0] * m
s = [0] * m
d = 10 ** 9 + 7
y = 1
for j in range(n):
a, b = map(int, input().split())
u[b].append(a)
v[a] = b
s[b] += 1
for b in range(m):
if not v[b]:
x = 0
p = [b]
while p:
x += 1
a = p.pop()
s[a] = -1
p += u[a]
y = (x * y) % d
for a in range(m):
if s[a] == 0:
b = v[a]
while s[b] == 1:
b = v[b]
s[b] -= 1
for a in range(m):
if s[a] == 1 and a != v[a]:
y = (2 * y) % d
while s[a]: a, s[a] = v[a], 0
print(y)
``` | instruction | 0 | 489 | 14 | 978 |
No | output | 1 | 489 | 14 | 979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 109 + 7.
Input
Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.
N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.
Output
Print the number of possible assignments, modulo 1000000007 = 109 + 7.
Examples
Input
4
1 5
5 2
3 7
7 3
Output
6
Input
5
1 10
2 10
3 10
4 10
5 5
Output
5
Note
These are the possible assignments for the first example:
* 1 5 3 7
* 1 2 3 7
* 5 2 3 7
* 1 5 7 3
* 1 2 7 3
* 5 2 7 3
Submitted Solution:
```
#!/bin/python3
# template
import sys
from collections import defaultdict
def read_ints(inp = sys.stdin):
return list(map(int,next(inp).strip().split()))
MOD = 10**9 + 7
def adjacency_lists_directed(e):
"""
form adjacency lists from sequences of (s,e) edges
:warning doesn't handle multi-edges
:param e: edge of non-directed graph
:return: dict of adjacency vertices for each vertex
"""
e_al = defaultdict(set)
for s,e in e:
e_al[s].add(e)
return e_al
def sol1(edges):
#revert
edges = [(v, u) for u, v in edges]
vs = set(u for u, _ in edges)
adj = adjacency_lists_directed(edges)
def count(u):
# return # of arrangement and visited vertices
# 1 if loop
# 2 if cycle
# nb nodes otherwise
visited = set()
stack = [u]
loop = False
cycle = False
while stack:
cur = stack.pop()
ch = adj[cur]
if cur in ch:
loop = True
if cur in visited:
cycle = True
continue
visited.add(cur)
stack.extend(ch)
return (1 if loop else (2 if cycle else len(visited))), visited
res = 1
while vs:
# each connected component
u = next(iter(vs))
c, vis = count(u)
res = (res * c) % MOD
vs -= vis
return res
def main(): # prevent variables at main scope
for i in range(1):
n, = read_ints()
E = []
for _ in range(n):
u,v = read_ints()
E.append((u,v))
d = sol1(E)
print(d)
if __name__=='__main__':
main()
``` | instruction | 0 | 490 | 14 | 980 |
No | output | 1 | 490 | 14 | 981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.
How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 109 + 7.
Input
Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.
N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.
Output
Print the number of possible assignments, modulo 1000000007 = 109 + 7.
Examples
Input
4
1 5
5 2
3 7
7 3
Output
6
Input
5
1 10
2 10
3 10
4 10
5 5
Output
5
Note
These are the possible assignments for the first example:
* 1 5 3 7
* 1 2 3 7
* 5 2 3 7
* 1 5 7 3
* 1 2 7 3
* 5 2 7 3
Submitted Solution:
```
n = int(input())
m = 2 * n + 2
u = [[] for i in range(m)]
v = [0] * m
s = [0] * m
d = 10 ** 9 + 7
y = 1
for j in range(n):
a, b = map(int, input().split())
u[b].append(a)
v[a] = b
s[b] += 1
for b in range(m):
if not v[b]:
x = 0
p = [b]
while p:
x += 1
a = p.pop()
s[a] = -1
p += u[a]
y = (x * y) % d
for a in range(m):
if s[a] == 0:
b = v[a]
while s[b] == 1:
s[b] -= 1
b = v[b]
s[b] -= 1
for a in range(m):
if s[a] == 1 and a != v[a]:
y = (2 * y) % d
while s[a]:
s[a] = 0
a = v[a]
print(y)
``` | instruction | 0 | 491 | 14 | 982 |
No | output | 1 | 491 | 14 | 983 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 789 | 14 | 1,578 |
"Correct Solution:
```
from itertools import accumulate
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
N, T = map(int, input().split())
time = [0] * (T + 2)
for _ in range(N):
l, r = map(int, input().split())
time[l] += 1
time[r] -= 1
print(max(accumulate(time)))
``` | output | 1 | 789 | 14 | 1,579 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 791 | 14 | 1,582 |
"Correct Solution:
```
#!/usr/bin/env python3
# DSL_5_A: The Maximum Number of Customers
def run():
n, t = [int(i) for i in input().split()]
sch = [0] * (t+1)
for _ in range(n):
enter, leave = [int(i) for i in input().split()]
sch[enter] += 1
sch[leave] -= 1
c = 0
maxc = 0
for i in sch:
c += i
if c >= maxc:
maxc = c
print(maxc)
if __name__ == '__main__':
run()
``` | output | 1 | 791 | 14 | 1,583 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 792 | 14 | 1,584 |
"Correct Solution:
```
from itertools import accumulate
n,t=map(int,input().split())
timeline=[0 for i in range(t+1)]
for i in range(n):
a,b=map(int,input().split())
timeline[a]+=1
timeline[b]-=1
print(max(accumulate(timeline)))
``` | output | 1 | 792 | 14 | 1,585 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 795 | 14 | 1,590 |
"Correct Solution:
```
def main():
n,t = map(int,input().split())
t+=1
T = [0]*t
for _ in range(n):
l,r = map(int,input().split())
T[l]+=1
T[r]-=1
num = 0
res = 0
for i in range(t):
num += T[i]
res = max(num,res)
print (res)
if __name__ == '__main__':
main()
``` | output | 1 | 795 | 14 | 1,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 869 | 14 | 1,738 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
import math as mt
import sys,string
input=sys.stdin.readline
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
from collections import defaultdict
def dfs(n,dx):
vis[n]=1
dis[n]=dx
for i in adj[n]:
if(vis[i]==0):
dfs(i,dx+1)
n=I()
vis=[0]*(n+1)
dis=[0]*(n+1)
l=[]
adj=[]
for i in range(n+1):
adj.append([])
for i in range(n):
d=I()
if(d==-1):
l.append(i+1)
else:
adj[d].append(i+1)
for i in l:
dfs(i,1)
print(max(dis))
``` | output | 1 | 869 | 14 | 1,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 870 | 14 | 1,740 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
import sys
from collections import deque
class Graph(object):
def __init__(self, n):
self.root = []
self.adj = {}
self.make_adj(n)
def make_adj(self, n):
for i in range(1, n+1):
self.adj[i] = []
def adj_list(self, u):
return self.adj[u]
class BFSResult:
def __init__(self):
self.level = {}
self.parent = {}
self.max_level = None
def bfs(g, s, r):
r.parent[s] = None
r.level[s] = 0
queue = deque()
queue.append(s)
while queue:
u = queue.popleft()
for n in g.adj_list(u):
if n not in r.level:
r.parent[n] = u
r.level[n] = r.level[u] + 1
r.max_level = r.level[u] + 1
queue.append(n)
#return r
if __name__ == '__main__':
n = int(input().rstrip())
g = Graph(n)
i = 1
r = BFSResult()
for _ in range(n):
u = int(input().rstrip())
if u == -1:
g.root.append(i)
else:
g.adj[u].append(i)
i+=1
#print(g.root)
for s in g.root:
bfs(g, s,r)
max_l = 0
for i in r.level:
if r.level[i]>max_l:
max_l = r.level[i]
if max_l == 0:
print(1)
else:
print(max_l+1)
#print(r.level)
``` | output | 1 | 870 | 14 | 1,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 871 | 14 | 1,742 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
import sys
sys.setrecursionlimit(100000000)
n=int(input())
d={}
de=[-1 for i in range(n+1)]
for i in range(1,n+1):
d[i]=int(input())
def depth(v):
if de[v]!=-1:
return de[v]
if d[v]==-1:
return 0
else:
return 1+depth(d[v])
ma=-1
for i in d:
t=depth(i)
de[i]=t
if de[i]>ma:
ma=de[i]
print(ma+1)
``` | output | 1 | 871 | 14 | 1,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 872 | 14 | 1,744 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
def find(i) :
t = 0
while par[i] != 0 :
i = par[i]
t+=1
return (t)
n = int(input())
t = 0
par = [0 for i in range (n+1)]
d = {}
for _ in range (1,n+1) :
x = int(input())
if x != -1 :
par[_] = x
else :
par[_] = 0
#print(par[1:])
ans = 0
for i in range (1,n+1) :
r = find(i)
t = 0
ans = max(r , ans )
#print(r)
print(ans+1)
``` | output | 1 | 872 | 14 | 1,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 873 | 14 | 1,746 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
import sys
import threading
from collections import defaultdict
n=int(input())
adj=defaultdict(list)
for i in range(n):
adj[int(input())].append(i+1)
def fun(root,par,adj):
ans=0
for ch in adj.get(root,[]):
if ch != par:
ans=max(ans,fun(ch,root,adj))
return 1+ans
def main():
ans=-1
for i in adj[-1]:
ans=max(ans,fun(i,-1,adj))
print(ans)
if __name__=="__main__":
sys.setrecursionlimit(10**6)
threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
``` | output | 1 | 873 | 14 | 1,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 874 | 14 | 1,748 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
import math
import sys
p = [-1] * 5000
n = int(input())
f = -999999999
for i in range(0, n):
x = int(input())
p[i + 1] = x
for i in range(1, n + 1):
y = i
ans = 1
while(p[y] != -1):
ans = ans + 1
y = p[y]
f = max(f, ans)
print(f)
``` | output | 1 | 874 | 14 | 1,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 875 | 14 | 1,750 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
from collections import deque
n = int(input())
g = [[] for i in range(n)]
for i in range(n):
finish = int(input()) - 1
if finish >= 0:
g[i].append(finish)
visited2 = []
def dfs(start):
visited.append(start)
for k in g[start]:
if k not in visited:
dfs(k)
visited2.append(len(visited))
for i in range(n):
visited = []
dfs(i)
print(max(visited2))
``` | output | 1 | 875 | 14 | 1,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 876 | 14 | 1,752 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
n = int(input())
p=[int(input()) for i in range(n)]
g=0
c=0
for i in range(n):
c=0
while i>=0:
i=p[i]-1
c+=1
g=max(g,c)
print(g)
``` | output | 1 | 876 | 14 | 1,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
import sys
import math
def dfs(graph, start):
stack = [start]
c = 0
while stack:
vertex = stack.pop()
c += 1
stack.extend(graph[vertex])
return c
n = int(input())
d = []
for i in range(n + 1):
d.append(set())
vmax = 0
for i in range(1, n + 1):
v = int(input())
if(v != -1):
d[i].add(v)
for i in range(1, n + 1):
k = dfs(d, i)
if(k > vmax):
vmax = k
print(vmax)
``` | instruction | 0 | 877 | 14 | 1,754 |
Yes | output | 1 | 877 | 14 | 1,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
def BFS_level(vertex):
s=[(vertex,1)]
while(s):
n=s.pop(0)
for i in range(len(d[n[0]])):
s.append((d[n[0]][i],n[1]+1))
if len(s)==0 and len(d[n[0]])==0:
return n[1]
n=int(input())
d={i+1:[] for i in range(n)}
roots=[]
for i in range(1,n+1):
a=int(input())
if a==-1:
roots.append(i)
else:
d[a].append(i)
max_level=0
for i in roots:
w=BFS_level(i)
if w>max_level:
max_level=w
print(max_level)
``` | instruction | 0 | 878 | 14 | 1,756 |
Yes | output | 1 | 878 | 14 | 1,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
n=int(input());r=[];m=0
for _ in [0]*n:
r.append(int(input()))
for i in range(n):
a=i;l=0
while r[a]!=-1:
a=r[a]-1;l+=1
m=max(m,l)
print(m+1)
``` | instruction | 0 | 879 | 14 | 1,758 |
Yes | output | 1 | 879 | 14 | 1,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
n = int(input())
freedom = []
# myBoss = [[] for x in range(n+1)]
Imboss = [[] for x in range(n+1)]
for i in range(n):
inn = int(input())
if inn == -1:
# myBoss[i+1] = -1
freedom.append(i+1)
else:
# myBoss[i+1].append(inn)
Imboss[inn].append(i+1)
temp = []
group =0
while(len(freedom)!=0):
group+=1
for f in freedom:
temp = temp + Imboss[f]
freedom = temp[::]
temp=[]
print(group)
``` | instruction | 0 | 880 | 14 | 1,760 |
Yes | output | 1 | 880 | 14 | 1,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
def inc_level_sons(emp):
for e in emps[emp]:
emps_level[e]+=1
#inc_level_sons(e)
n = int(input())
emps = [[]] * 2005
emps_level = [1] * 2005
for i in range(n):
e = int(input())
if e != -1:
emps[e].append(i+1)
emps_level[i+1] = emps_level[e] + 1
#inc_level_sons(i+1)
print(str(max(emps_level)))
``` | instruction | 0 | 881 | 14 | 1,762 |
No | output | 1 | 881 | 14 | 1,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
n = int(input())
m = [[0,0]]*n
for i in range(n):
sup = int(input())
if sup == -1:
m[i] = [i+1, 1]
else:
m[i] = [i+1, m[sup-1][1]+1]
max = 0
for i in range(n):
if m[i][1]>max:
max = m[i][1]
print(max)
``` | instruction | 0 | 882 | 14 | 1,764 |
No | output | 1 | 882 | 14 | 1,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
def dfs(graph, node, visited, height, parent):
visited[node] = True
if parent != -1:
height[node] = height[parent] + 1
for i in graph[node]:
if not visited[i]:
dfs(graph, i, visited, height, node)
v = int(input())
graph = []
for i in range(v+1):
graph.append([])
edges = []
for i in range(v):
edges.append(int(input()))
for i in range(len(edges)):
if edges[i] != -1:
graph[edges[i]].append(i+1)
height = [1]*(v+1)
visited = [False]*(v+1)
for i in range(len(graph)):
if not visited[i]:
dfs(graph, i, visited, height, -1)
print(max(height))
``` | instruction | 0 | 883 | 14 | 1,766 |
No | output | 1 | 883 | 14 | 1,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 17 13:44:48 2020
@author: pctab
"""
from collections import defaultdict
graph=defaultdict(set)
n=int(input())
for i in range(n):
m_i=int(input())
if m_i!=-1:
graph[m_i].add(i+1)
else:
graph[i+1]
max_depth=0
def dfs(visited,u,l):
global max_depth
visited[u]=True
max_depth=max(max_depth,l)
for v in graph[u]:
if not visited[v]:
dfs(visited,v,l+1)
visited=[False]*(n+1)
for i in range(1,n+1):
if not visited[i]:
dfs(visited,i,1)
print(max_depth)
``` | instruction | 0 | 884 | 14 | 1,768 |
No | output | 1 | 884 | 14 | 1,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The capital of Berland has the only movie theater in the country. Besides, it consists of only one room. The room is divided into n rows, each row consists of m seats.
There are k people lined up to the box office, each person wants to buy exactly one ticket for his own entertainment. Before the box office started selling tickets, each person found the seat that seemed best for him and remembered it as a pair of coordinates (xi, yi), where xi is the row number, and yi is the seat number in this row.
It is possible that some people have chosen the same place, then when some people see their favorite seat taken in the plan of empty seats in the theater, they choose and buy a ticket to another place. Each of them has the following logic: let's assume that he originally wanted to buy a ticket to seat (x1, y1), then when he comes to the box office, he chooses such empty seat (x2, y2), which satisfies the following conditions:
* the value of |x1 - x2| + |y1 - y2| is minimum
* if the choice is not unique, then among the seats that satisfy the first condition, this person selects the one for which the value of x2 is minimum
* if the choice is still not unique, among the seats that satisfy the first and second conditions, this person selects the one for which the value of y2 is minimum
Your task is to find the coordinates of a seat for each person.
Input
The first input line contains three integers n, m, k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ min(n·m, 105) — the number of rows in the room, the number of seats in each row and the number of people in the line, correspondingly. Each of the next k lines contains two integers xi, yi (1 ≤ xi ≤ n, 1 ≤ yi ≤ m) — the coordinates of the seat each person has chosen. Numbers on the same line are separated by a space. The pairs of coordinates are located in the order, in which people stand in the line, starting from the head (the first person in the line who stands in front of the box office) to the tail (the last person in the line).
Output
Print k lines, each containing a pair of integers. Print on the i-th line xi, yi — the coordinates of the seat, for which the person who stands i-th in the line will buy the ticket.
Examples
Input
3 4 6
1 1
1 1
1 1
1 2
1 3
1 3
Output
1 1
1 2
2 1
1 3
1 4
2 3
Input
4 3 12
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
Output
2 2
1 2
2 1
2 3
3 2
1 1
1 3
3 1
3 3
4 2
4 1
4 3
Submitted Solution:
```
from datetime import *
time1 = datetime.now()
import sys
try: file = open('test.in');fin=file.readline
except: fin = sys.stdin.readline
n,m,k=map(int,fin().split())
dis=[0]*(m+2)*(n+2)
mulm=[(m+2)*i for i in range(n+2)]
f1=dict()
f2=dict()
ansx,ansy=0,0
rangeIJ=((i,j) for j in range(-1,2) for i in range(-1,2))
def renew(x,y,d,i):
global ansx,ansy
yy = i%(m+2); xx = i//(m+2)
if (xx<=n)and(yy>0)and(yy<=m)and(abs(xx-x)+abs(yy-y)<=d):
if (xx<ansx)or(xx==ansx)and(yy<ansy):
ansx, ansy = xx, yy
def findset1(x):
if x in f1:f1[x]=findset1(f1[x])
return f1[x] if (x in f1) else x
def findset2(x):
if x in f2:f2[x]=findset2(f2[x])
return f2[x] if (x in f2) else x
def check(x,y,d):
global ansx,ansy
ansx,ansy=3000,3000
x0 = max(1,x-d);y0 = x-x0+y-d
if(y0>0):renew(x,y,d,findset1(mulm[x0]+y0))
y0 = d-x+x0+y
if(y0<=m):renew(x,y,d,findset2(mulm[x0]+y0))
if ansx<3000:
#print(ansx,ansy)
ind=mulm[ansx]+ansy
ind1=mulm[ansx+1]+ansy
f1[ind] = ind1-1
f2[ind] = ind1+1
return 1
y0 = max(y-d,1); x0 = d+x-y+y0
if(x0<=n):renew(x,y,d,findset2(mulm[x0]+y0))
y0 = min(y+d,m); x0 = d+x-y0+y
if(x0<=n):renew(x,y,d,findset1(mulm[x0]+y0))
if ansx<3000:
#print(ansx,ansy)
ind=mulm[ansx]+ansy
ind1=mulm[ansx+1]+ansy
f1[ind] = ind1-1
f2[ind] = ind1+1
return 1
return 0
for _ in range(k):
x,y=map(int,fin().split())
ind0=mulm[x]+y
for i,j in rangeIJ:
dis[ind0]=max(dis[ind0],dis[mulm[x+i]+y+j]-abs(i)-abs(j))
while 1:
if(check(x,y,dis[ind0])):break
dis[ind0]+=1
print(datetime.now()-time1)
try:file.close()
except:pass
'''
import sys
import random
fout=open('test.in','w')
fout.write('2000 2000 100000\n')
for i in range(100000):
fout.write(str(random.randint(417,419))+' '+str(random.randint(1306,1308))+'\n')
'''
``` | instruction | 0 | 1,127 | 14 | 2,254 |
No | output | 1 | 1,127 | 14 | 2,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The capital of Berland has the only movie theater in the country. Besides, it consists of only one room. The room is divided into n rows, each row consists of m seats.
There are k people lined up to the box office, each person wants to buy exactly one ticket for his own entertainment. Before the box office started selling tickets, each person found the seat that seemed best for him and remembered it as a pair of coordinates (xi, yi), where xi is the row number, and yi is the seat number in this row.
It is possible that some people have chosen the same place, then when some people see their favorite seat taken in the plan of empty seats in the theater, they choose and buy a ticket to another place. Each of them has the following logic: let's assume that he originally wanted to buy a ticket to seat (x1, y1), then when he comes to the box office, he chooses such empty seat (x2, y2), which satisfies the following conditions:
* the value of |x1 - x2| + |y1 - y2| is minimum
* if the choice is not unique, then among the seats that satisfy the first condition, this person selects the one for which the value of x2 is minimum
* if the choice is still not unique, among the seats that satisfy the first and second conditions, this person selects the one for which the value of y2 is minimum
Your task is to find the coordinates of a seat for each person.
Input
The first input line contains three integers n, m, k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ min(n·m, 105) — the number of rows in the room, the number of seats in each row and the number of people in the line, correspondingly. Each of the next k lines contains two integers xi, yi (1 ≤ xi ≤ n, 1 ≤ yi ≤ m) — the coordinates of the seat each person has chosen. Numbers on the same line are separated by a space. The pairs of coordinates are located in the order, in which people stand in the line, starting from the head (the first person in the line who stands in front of the box office) to the tail (the last person in the line).
Output
Print k lines, each containing a pair of integers. Print on the i-th line xi, yi — the coordinates of the seat, for which the person who stands i-th in the line will buy the ticket.
Examples
Input
3 4 6
1 1
1 1
1 1
1 2
1 3
1 3
Output
1 1
1 2
2 1
1 3
1 4
2 3
Input
4 3 12
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
Output
2 2
1 2
2 1
2 3
3 2
1 1
1 3
3 1
3 3
4 2
4 1
4 3
Submitted Solution:
```
n, m, k = map(int, input().split())
step = {}
seki = set()
an = []
def walk(x, y):
i, j, k = step.get((x, y), (0, 0, 1))
if k==1:
k += 1
else:
k -= 1
j += 1
if j>2*i:
i += 1
j = 0
while True:
while j<=2*i:
n_x = j-i
m_x = n_x+x
if n>=m_x>0:
while k<=2:
if n_x<0:
if k==1:
m_y = -n_x-i+y
else:
m_y = n_x+i+y
else:
if k==1:
m_y = n_x-i+y
else:
m_y = i-n_x+y
if m>=m_y>0 and (m_x, m_y) not in seki:
step[(x, y)] = (i, j, k)
seki.add((m_x, m_y))
return m_x, m_y
k += 1
k = 1
j += 1
j = 0
i += 1
'''
x, y = map(int, input().split())
inps = [[x, y, 1]]
for i in range(k-1):
x, y = map(int, input().split())
if x!=inps[-1][0] or y!=inps[-1][1]:
inps.append([x, y, 1])
else:
inps[-1][2]+=1
for i in inps:
for j in range(i[2]):
an.append(walk(i[0], i[1]))
'''
x, y = map(int, input().split())
if x==552 and y==1028:
for i in range(60000):
an.append(walk(552, 1028))
for i in range(k-60000):
x, y = map(int, input().split())
an.append(walk(x, y))
else:
an.append(walk(x, y))
for i in range(k-1):
x, y = map(int, input().split())
an.append(walk(x, y))
for a in an:
print(a[0], a[1])
``` | instruction | 0 | 1,128 | 14 | 2,256 |
No | output | 1 | 1,128 | 14 | 2,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The capital of Berland has the only movie theater in the country. Besides, it consists of only one room. The room is divided into n rows, each row consists of m seats.
There are k people lined up to the box office, each person wants to buy exactly one ticket for his own entertainment. Before the box office started selling tickets, each person found the seat that seemed best for him and remembered it as a pair of coordinates (xi, yi), where xi is the row number, and yi is the seat number in this row.
It is possible that some people have chosen the same place, then when some people see their favorite seat taken in the plan of empty seats in the theater, they choose and buy a ticket to another place. Each of them has the following logic: let's assume that he originally wanted to buy a ticket to seat (x1, y1), then when he comes to the box office, he chooses such empty seat (x2, y2), which satisfies the following conditions:
* the value of |x1 - x2| + |y1 - y2| is minimum
* if the choice is not unique, then among the seats that satisfy the first condition, this person selects the one for which the value of x2 is minimum
* if the choice is still not unique, among the seats that satisfy the first and second conditions, this person selects the one for which the value of y2 is minimum
Your task is to find the coordinates of a seat for each person.
Input
The first input line contains three integers n, m, k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ min(n·m, 105) — the number of rows in the room, the number of seats in each row and the number of people in the line, correspondingly. Each of the next k lines contains two integers xi, yi (1 ≤ xi ≤ n, 1 ≤ yi ≤ m) — the coordinates of the seat each person has chosen. Numbers on the same line are separated by a space. The pairs of coordinates are located in the order, in which people stand in the line, starting from the head (the first person in the line who stands in front of the box office) to the tail (the last person in the line).
Output
Print k lines, each containing a pair of integers. Print on the i-th line xi, yi — the coordinates of the seat, for which the person who stands i-th in the line will buy the ticket.
Examples
Input
3 4 6
1 1
1 1
1 1
1 2
1 3
1 3
Output
1 1
1 2
2 1
1 3
1 4
2 3
Input
4 3 12
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
Output
2 2
1 2
2 1
2 3
3 2
1 1
1 3
3 1
3 3
4 2
4 1
4 3
Submitted Solution:
```
n, m, k = map(int, input().split())
step = {}
seki = set()
an = []
def walk(x, y):
i, j, k = step.get((x, y), (0, 0, 1))
if k==1:
k += 1
else:
k -= 1
j += 1
if j>2*i:
i += 1
j = 0
while True:
while j<=2*i:
n_x = j-i
m_x = n_x+x
if n>=m_x>0:
while k<=2:
if n_x<0:
if k==1:
m_y = -n_x-i+y
else:
m_y = n_x+i+y
else:
if k==1:
m_y = n_x-i+y
else:
m_y = i-n_x+y
if m>=m_y>0 and (m_x, m_y) not in seki:
step[(x, y)] = (i, j, k)
seki.add((m_x, m_y))
return m_x, m_y
k += 1
k = 1
j += 1
j = 0
i += 1
for i in range(k):
x, y = map(int, input().split())
if x==552 and y==1028:
for i in range(k):
an.append(walk(552, 1028))
break
an.append(walk(x, y))
for a in an:
print(a[0], a[1])
``` | instruction | 0 | 1,129 | 14 | 2,258 |
No | output | 1 | 1,129 | 14 | 2,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The capital of Berland has the only movie theater in the country. Besides, it consists of only one room. The room is divided into n rows, each row consists of m seats.
There are k people lined up to the box office, each person wants to buy exactly one ticket for his own entertainment. Before the box office started selling tickets, each person found the seat that seemed best for him and remembered it as a pair of coordinates (xi, yi), where xi is the row number, and yi is the seat number in this row.
It is possible that some people have chosen the same place, then when some people see their favorite seat taken in the plan of empty seats in the theater, they choose and buy a ticket to another place. Each of them has the following logic: let's assume that he originally wanted to buy a ticket to seat (x1, y1), then when he comes to the box office, he chooses such empty seat (x2, y2), which satisfies the following conditions:
* the value of |x1 - x2| + |y1 - y2| is minimum
* if the choice is not unique, then among the seats that satisfy the first condition, this person selects the one for which the value of x2 is minimum
* if the choice is still not unique, among the seats that satisfy the first and second conditions, this person selects the one for which the value of y2 is minimum
Your task is to find the coordinates of a seat for each person.
Input
The first input line contains three integers n, m, k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ min(n·m, 105) — the number of rows in the room, the number of seats in each row and the number of people in the line, correspondingly. Each of the next k lines contains two integers xi, yi (1 ≤ xi ≤ n, 1 ≤ yi ≤ m) — the coordinates of the seat each person has chosen. Numbers on the same line are separated by a space. The pairs of coordinates are located in the order, in which people stand in the line, starting from the head (the first person in the line who stands in front of the box office) to the tail (the last person in the line).
Output
Print k lines, each containing a pair of integers. Print on the i-th line xi, yi — the coordinates of the seat, for which the person who stands i-th in the line will buy the ticket.
Examples
Input
3 4 6
1 1
1 1
1 1
1 2
1 3
1 3
Output
1 1
1 2
2 1
1 3
1 4
2 3
Input
4 3 12
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
Output
2 2
1 2
2 1
2 3
3 2
1 1
1 3
3 1
3 3
4 2
4 1
4 3
Submitted Solution:
```
n,m,k = list(map(int,input().split()))
booked_seats = []
for i in range(k):
x,y = list(map(int,input().split()))
booked_seat = []
if [x,y] not in booked_seats:
booked_seat = [x,y]
else:
delta = 0
while(booked_seat == []):
delta += 1
delta_x = delta+1
delta_y = -1
available_seats = []
while((delta_y <= delta) and (delta_x >= 0)):
delta_x -= 1
delta_y += 1
x_new = x-delta_x
y_new = y-delta_y
if ([x_new,y_new] not in booked_seats) and (x_new * y_new > 0) and (x_new <= n) and (y_new <= m):
if [x_new,y_new] not in available_seats:
available_seats.append([x_new,y_new])
continue
x_new = x-delta_x
y_new = y+delta_y
if ([x_new,y_new] not in booked_seats) and (x_new * y_new > 0) and (x_new <= n) and (y_new <= m):
if [x_new,y_new] not in available_seats:
available_seats.append([x_new,y_new])
continue
x_new = x+delta_x
y_new = y-delta_y
if ([x_new,y_new] not in booked_seats) and (x_new * y_new > 0) and (x_new <= n) and (y_new <= m):
if [x_new,y_new] not in available_seats:
available_seats.append([x_new,y_new])
continue
x_new = x+delta_x
y_new = y+delta_y
if ([x_new,y_new] not in booked_seats) and (x_new * y_new > 0) and (x_new <= n) and (y_new <= m):
if [x_new,y_new] not in available_seats:
available_seats.append([x_new,y_new])
continue
if(available_seats != []):
booked_seat = min(available_seats)
booked_seats.append(booked_seat)
print(booked_seat[0],booked_seat[1])
``` | instruction | 0 | 1,130 | 14 | 2,260 |
No | output | 1 | 1,130 | 14 | 2,261 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.