text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Tags: games, math
Correct Solution:
```
def check_right(s):
found = False
partial_sum = 0
for i in range(len(s) - 1, 0, -1):
partial_sum += s[i]
if found:
if partial_sum % 2 == 1: return True
if s[i] % 2 == 0:
found = True
return False
def check_left(s):
found = False
partial_sum = 0
for i in range(len(s) - 1):
partial_sum += s[i]
if found:
if partial_sum % 2 == 1: return True
if s[i] % 2 == 0:
found = True
return False
def solve(l):
if sum(l) % 2 != 0: return "First"
if l[0] % 2 == 1 or l[len(l) - 1] % 2 == 1:
return "First"
if check_left(l) or check_right(l):
return "First"
return "Second"
n = int(input())
a = [int(i) for i in input().split()]
print(solve(a))
```
| 12,200 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Tags: games, math
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
for i in range(len(a)):
if a[i]%2!=0:
print('First')
exit()
print('Second')
```
| 12,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
def nik(rud,panda):
lfa=0
for x in rud:
if x%2!=0:
lfa=1
print("First") if panda%2!=0 or lfa==1 else print("Second")
n=int(input())
rud=list(map(int,input().split(" ")))
panda=sum(rud)
nik(rud,panda)
```
Yes
| 12,202 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
a = int(input())
b=input().split(' ')
x = 0
for i in b:
if int(i) % 2 == 1:
x += 1
c = 0
for z in b:
c += int(z)
if c % 2 == 1:
print('First')
elif c % 2 == 0 and x != 0:
print('First')
else:
print("Second")
```
Yes
| 12,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
n = int(input())
A = list(map(int, input().split()))
odd_counter = 0
for a in A:
if a % 2 == 0: # if a is even
# do nothing
pass
else:
odd_counter += 1
if odd_counter % 2 == 1:
print('First')
else:
if odd_counter >= 1:
print('First')
else:
print('Second')
```
Yes
| 12,204 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
import sys
# from collections import deque
# from collections import Counter
# from math import sqrt
# from math import log
# from math import ceil
# from bisect import bisect_left, bisect_right
# alpha=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
# mod=10**9+7
# mod=998244353
# def BinarySearch(a,x):
# i=bisect_left(a,x)
# if(i!=len(a) and a[i]==x):
# return i
# else:
# return -1
# def sieve(n):
# prime=[True for i in range(n+1)]
# p=2
# while(p*p<=n):
# if (prime[p]==True):
# for i in range(p*p,n+1,p):
# prime[i]=False
# p+=1
# prime[0]=False
# prime[1]=False
# s=set()
# for i in range(len(prime)):
# if(prime[i]):
# s.add(i)
# return s
def gcd(a, b):
if(a==0):
return b
return gcd(b%a,a)
fast_reader=sys.stdin.readline
fast_writer=sys.stdout.write
def input():
return fast_reader().strip()
def print(*argv):
fast_writer(' '.join((str(i)) for i in argv))
fast_writer('\n')
#____________________________________________________________________________________________________________________________________
n=int(input())
l=list(map(int, input().split()))
c=0
for i in range(n):
c+=int(l[i]%2)
if(c==0):
print('Second')
else:
print('First')
```
Yes
| 12,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
i=0
flag=0
while i<n:
su=0
if flag==0:
su+=l[i]
i+=1
while su%2!=1 and i<n:
su+=l[i]
i+=1
if su%2==1:
flag=1
elif flag==1:
su+=l[i]
i+=1
while su%2!=0 and i<n:
su+=l[i]
i+=1
if su%2==0:
flag=0
if flag==0:
print("Second")
else:
print("First")
```
No
| 12,206 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
import sys
def nik(rud):
for i in (rud):
if (i%2!=0):
print("First")
sys.exit(0)
else:
print("Second")
sys.exit(0)
n = int(input())
rud = list(map(int, input().split()))
nik(rud)
```
No
| 12,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
def kenti(a):
k = 0
for i in a:
if i%2==1:
k += 1
return k
n = int(input())
a = list(map(int,input().split()))
if sum(a)%2==1:
print('First')
elif sum(a)%2==0 and kenti(a)%2==0:
print('Second')
else:
print('First')
```
No
| 12,208 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?
Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).
Examples
Input
4
1 3 2 3
Output
First
Input
2
2 2
Output
Second
Note
In first sample first player remove whole array in one move and win.
In second sample first player can't make a move and lose.
Submitted Solution:
```
n = int(input())
data = list(map(int,input().split()))
ans = sum(data)
print('First' if ans%2==1 else 'Second')
```
No
| 12,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
n, x = map(int, input().split())
if n == 1:
print("YES")
print(x)
exit()
elif n == 2:
if x == 0:
print("NO")
else:
print("YES")
print(0, x)
exit()
l = [i for i in range(1, n-2)]
v = 0
for i in range(1,n-2):
v = v^i
r1 = x^v
r1 ^= 2**17 + 2**18
r2 = 2**17 + 2**19
r3 = 2**18 + 2**19
l.extend([r1,r2,r3])
print("YES")
print(*l)
```
| 12,210 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
n, x = map(int, input().split())
if n == 2 and x == 0:
print('NO')
exit()
if n == 1:
print('YES')
print(x)
exit()
a = [i for i in range(1, n-2)]
xr = 0
for i in a:
xr ^= i
if xr == x:
a += [2**17, 2**18, 2**17 ^ 2**18]
else:
if n >=3:
a += [0, 2**17, 2**17^xr^x]
else:
a += [2**17, 2**17^xr^x]
print('YES')
print(' '.join([str(i) for i in a]))
```
| 12,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
n, x = map(int, input().split())
if n == 2 and x == 0:
print("NO")
else:
print("YES")
if n > 1:
temp = x
for i in range(n-1):
temp ^= i
if temp:
for i in range(1, n-1): print(i, end = ' ')
print(2**17, 2**17 + temp)
else:
for i in range(n-2): print(i, end = ' ')
print(2**17, 2**17 + n-2)
else:
print(x)
```
| 12,212 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
def main():
n, x = map(int, input().split())
if n == 1:
print('YES')
print(x)
return
if n == 2:
if x == 0:
print('NO')
else:
print('YES')
print(0, x)
return
o = 0
for i in range(n - 1):
o ^= i
print('YES')
q = o ^ x
if q >= n - 1:
print(*([i for i in range(n - 1)]), q)
else:
if n - 2 != q:
print(*([i for i in range(n - 2)]), n - 2 + 262144, q + 262144)
else:
print(*([i for i in range(n - 3)]), n - 3 + 262144, n - 2, q + 262144)
main()
```
| 12,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Mon Jun 26 14:24:58 2017
"""
def xor(a,b):
a=bin(a)[2:]
b=bin(b)[2:]
while(len(a)>len(b)):
b='0' + b
while(len(b)>len(a)):
a='0' + a
c=['0']*len(a)
for i in range(len(a)):
if(a[i]!=b[i]):
c[i]='1'
a=''
for i in c:
a+=i
return(int(a,2))
n,x=input().split()
n=int(n)
x=int(x)
a=[]
ans=0
flag=False
if(n%4!=0):
for i in range(n-n%4):
a.append(100002+i)
if(n%4==1):
a.append(x)
elif(n%4==2):
if(x==0):
if(n==2):
flag=True
else:
for i in range(4):
del a[-1]
a.append(500001)
a.append(500002)
a.append(500007)
a.append(300046)
a.append(210218)
a.append(0)
else:
a.append(0)
a.append(x)
else:
if(x==0):
a.append(4)
a.append(7)
a.append(3)
elif(x==1):
a.append(8)
a.append(9)
a.append(0)
elif(x==2):
a.append(1)
a.append(4)
a.append(7)
else:
a.append(0)
a.append(1)
if(x%2==0):
a.append(x+1)
else:
a.append(x-1)
else:
for i in range(n-4):
a.append(100002+i)
a.append(500001)
a.append(200002)
a.append(306275)
a.append(x)
if(flag):
print("NO")
else:
print("YES")
temp=''
for i in a:
temp+=str(i)
temp+=' '
print(temp)
```
| 12,214 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
def comp(n) :
if n % 4 == 0 :
return n
if n % 4 == 1 :
return 1
if n % 4 == 2 :
return n + 1
return 0
n , x = map(int,input().split())
a=1<<17
if n==2:
if x==0:
print("NO")
else:
print("YES")
print(0,x)
elif n==1:
print("YES")
print(x)
else:
ans=[i for i in range(1,n-2)]
if comp(n-3)==x:
ans.append((a*2)^a)
ans.append(a)
ans.append(a*2)
else:
ans.append(0)
ans.append(a)
ans.append(a^x^comp(n-3))
print("YES")
print(*ans)
```
| 12,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
from sys import stdin, stdout
class Solve:
def __init__(self):
R = stdin.readline
W = stdout.write
n, x = map(int, R().split())
ans = []
if n == 2 and x == 0:
W('NO\n')
return
elif n == 1:
W('YES\n%d' % x)
return
elif n == 2:
W('YES\n%d %d' % (0,x))
return
ans = ['YES\n']
xor = 0
for i in range(1,n-2):
xor ^= i
ans.append(str(i) + ' ')
if xor == x:
ans += [str(2**17)+' ', str(2**18)+' ',str((2**17)^(2**18))]
else:
ans += ['0 ', str(2**17)+' ', str((2**17)^xor^x)]
W(''.join(ans))
def main():
s = Solve()
main()
```
| 12,216 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
def readints():
return [int(s) for s in input().strip().split()]
def to_s(int_objs):
return [str(obj) for obj in int_objs]
class Solver:
def main(self):
n, x = readints()
if n == 1:
print('Yes\n{}'.format(x))
elif x == 0 and n == 2:
print('No')
elif n == 2:
print('Yes\n{} {}'.format(0, x))
elif n == 3:
if x == 0:
print('Yes\n1 2 3\n')
else:
print('Yes\n{} {} {}'.format(
(1 << 17) + x,
(1 << 17) + (1 << 19),
1 << 19,
))
else:
results = [x]
searched = 0
current = 1
while len(results) < n - 3:
if current == x:
current += 1
results.append(current)
searched = searched ^ current
current += 1
results.append((1 << 17) + searched)
results.append((1 << 17) + (1 << 19))
results.append(1 << 19)
print('Yes\n' + ' '.join(to_s(results)))
Solver().main()
```
| 12,217 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Tags: constructive algorithms
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=1000000007
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return stdin.read().split()
range = xrange # not for python 3.0+
# main code
n,x=in_arr()
if n==2 and x==0:
pr('NO')
exit()
pr('YES\n')
ans=0
if n==1:
pr_num(x)
elif n==2:
pr_arr([x,0])
else:
pw=2**17
for i in range(1,n-2):
pr(str(i)+' ')
ans^=i
if ans==x:
pr_arr([pw,pw*2,pw^(pw*2)])
else:
pr_arr([0,pw,pw^x^ans])
```
| 12,218 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
n,x=list(map(int,input().split()))
xor = x
if(n==1):
print("YES")
print(x)
elif(n==2):
if(x==0):
print("NO")
else:
print("YES")
print(0,x)
else:
print("YES")
k=1
for u in range(n-2):
while(xor^k==0):
k+=1
xor=xor^k
print(k,end=" ")
k+=1
xor=xor^524287
print(524287,end=" ")
print(xor)
```
Yes
| 12,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
n, x = map(int, input().split())
if n == 1:
print("Yes")
print(x)
elif n == 2:
if x == 0:
print("No")
else:
print("Yes")
print(0, x)
else:
print("Yes")
for i in range(n-3):
print(i, end=' ')
x ^= i
print(1<<18, 1<<19, x^(1<<18)^(1<<19))
```
Yes
| 12,220 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
n,x=map(int,input().split())
if(n==1):
print("YES")
print(x)
elif(n==2):
if(x):
print("YES")
print(0,x)
else:print('NO')
else:
print("YES")
k=1
for _ in ' '*(n-2):
while(x^k==0):
k+=1
x=x^k
print(k,end=" ")
k+=1
x=x^524287
print(524287,x)
```
Yes
| 12,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
from functools import reduce
n, x = list(map(int, input().split(' ')))
if n == 1:
print('YES')
print(x)
elif n == 2:
if x == 0:
print('NO')
else:
print('YES')
print(0, x)
elif n == 3:
print('YES')
if x == 0:
print(1, 2, 3)
else:
print(0, (1 << 17), (1 << 17) ^ x)
else:
res = []
if x != 0:
res += [x]
n -= 1
i = 4
while n >= 3:
if n == 3 or n == 6:
res += [(0b11 << 17), (0b101 << 17), (0b110 << 17)]
n -= 3
if n == 3:
res += [(0b11 << 17) ^ 1, (0b101 << 17) ^ 2, (0b110 << 17) ^ 3]
n -= 3
elif n == 5:
#print('B', n)
res += [(0b100 << 17) ^ i, (0b101 << 17) ^ i, (0b110 << 17) ^ i, (0b111 << 17) ^ i, 0]
n -= 5
else:
#print('A', n)
res += [(0b100 << 17) ^ i, (0b101 << 17) ^ i, (0b110 << 17) ^ i, (0b111 << 17) ^ i]
n -= 4
#print('C', n)
i += 1
print('YES')
print(' '.join(map(str, res)))
#print(reduce(lambda a, b: a ^ b, res))
#print(len(set(res)))
```
Yes
| 12,222 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
n, x = tuple(map(int, input().split()))
# table n
# ---------
# | |
# 19 | |
# ---------
def get_4_n_numbrs(n):
a, b, c, d = 5, 10, 12, 3
res = []
for i in range(n):
pref = "1{0:014b}".format(i)
pref += '{0:04b}'
res.append((pref.format(a)))
res.append((pref.format(b)))
res.append((pref.format(c)))
res.append((pref.format(d)))
return res
def get_3_nums(x):
if x not in [1, 2]:
x = 1 ^ 2 ^ x
return [1, 2, x]
else:
x = 8 ^ 4 ^ x
return [8, 4, x]
def get_4_nums(x):
if x not in [1, 2, 4]:
x = 1 ^ 2 ^ 4 ^ x
return [1, 2, 4, x]
else:
x = 8 ^ 16 ^ 32 ^ x
return [8,16,32, x]
def get_2_nums(x):
if x == 0:
return [33, 3, 6, 12, 24, 48]
else:
return [0, x]
# Нужно уметь находить четное число чисел которые дают 0 в xor
# Начиная с четырех это риал.
# Тогда в зависимости от кратности четырем.
# Если остаток 1 то это то число которое хотели
# Если два то то что хотели и ноль.
# Нужно уметь исать 2 и 3 числа которые на хоре дают нужное
# Это брутфорсится
#
# #
def get_numbers(n, x):
if n == 1:
return [x]
fours = (n - 1) // 4
res = []
left = n - (fours * 4)
if left == 2 and x == 0:
if n == 2:
return -1
fours -= 1
res.extend([int(xx, 2) for xx in get_4_n_numbrs(fours)])
if left == 1:
res.append(x)
if left == 2:
res.extend(get_2_nums(x))
if left == 3:
res.extend(get_3_nums(x))
if left == 4:
res.extend(get_4_nums(x))
return res
res = get_numbers(n, x)
if res != -1:
print("YES")
print(' '.join(map(str, get_numbers(n, x))))
else:
print("NO")
```
No
| 12,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
n,x = map(int,input().split())
# 2,0 => NO
# 10**6 > 2**19
# 10**5 < 2**17
if n == 1:
print('YES')
print(x)
elif n == 2:
if x == 0:
print('NO')
else:
print('YES')
print(0,x)
else:
a = 1 << 19
c = a
if x+3 < n:
from itertools import chain
for i in chain(range(1,x+1),range(x+2,n-1)):
c ^= i
it = chain(range(1,x+1),range(x+2,n-1))
else:
for i in range(1,n-2):
c ^= i
it = range(1,n-2)
d = x+1
e = x ^ c ^ d
print('YES')
if n == 3:
print(a, d, e)
else:
print(a, ' '.join(map(str,it)), d, e)
```
No
| 12,224 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
def readints():
return [int(s) for s in input().strip().split()]
def to_s(int_objs):
return [str(obj) for obj in int_objs]
class Solver:
def main(self):
n, x = readints()
if n == 1:
print('Yes\n{}'.format(x))
return
if x == 0 and n == 2:
print('No')
return
if n == 2:
print('Yes\n{} {}'.format(0, x))
return
results = []
searched = 0
for i in range(n - 3):
results.append(i)
searched = searched ^ i
results.append((1 << 17) + searched)
results.append((1 << 17) + (1 << 19))
results.append(1 << 19)
print(' '.join(to_s(results)))
Solver().main()
```
No
| 12,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.
Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 106.
Input
The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.
Output
If there is no such set, print "NO" (without quotes).
Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.
Examples
Input
5 5
Output
YES
1 2 4 5 7
Input
3 6
Output
YES
1 2 5
Note
You can read more about the bitwise-xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>
For the first sample <image>.
For the second sample <image>.
Submitted Solution:
```
n,x=map(int,input().split())
a=x
A=[]
for i in range(1,n):
a^=i
A.append(i)
if a==0 or a>n-1:
A.append(a)
else:
a=0
if x>n-1:
A.pop(a-1)
A.append(0)
A.append(x)
else:
if x>a:
A.pop(x-1)
A.pop(a-1)
else:
A.pop(a-1)
A.pop(x-1)
A.append(500000)
A.append(x^500000)
A.append(0)
print("YES")
for i in A:
print(i,end=' ')
```
No
| 12,226 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
a = [int(s) for s in input().split(' ')]
e = 0
for i in range(1, n - 1):
if a[i] > max(a[i - 1], a[i + 1]) or a[i] < min(a[i - 1], a[i + 1]):
e += 1
print(e)
```
| 12,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
x = list(map(int,input().split()))
r = 0
if n<=2:
print(0)
else:
for i in range(1,n-1):
if x[i]>x[i-1] and x[i]>x[i+1]:
r+=1
if x[i]<x[i-1] and x[i]<x[i+1]:
r+=1
print(r)
```
| 12,228 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
from math import*
############ ---- Input Functions ---- ############
def inp():
return(int(input().strip()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
n=inp()
li=inlt()
cnt=0
for i in range(1,n-1):
if li[i]< li[i-1] and li[i]<li[i+1]:
cnt+=1
elif li[i]> li[i-1] and li[i]>li[i+1]:
cnt+=1
print(cnt)
```
| 12,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
number = int(input())
num = 0
lst = [int(i) for i in input().split()][:number]
for i in range(len(lst)):
if not(i == 0 or i == len(lst)-1) and (lst[i] > lst[i-1] and lst[i]>lst[i+1] or lst[i]<lst[i-1] and lst[i]<lst[i+1]):
num+=1
print(num)
```
| 12,230 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
def find_extrema(n, lst):
maxima = list()
minima = list()
if n >= 3:
for i in range(1, n - 1):
if lst[i] > lst[i - 1] and lst[i] > lst[i + 1]:
maxima.append(lst[i])
elif lst[i] < lst[i - 1] and lst[i] < lst[i + 1]:
minima.append(lst[i])
return len(minima) + len(maxima)
return 0
m = int(input())
a = [int(j) for j in input().split()]
print(find_extrema(m, a))
```
| 12,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
def main():
input()
nums = list(map(int, input().split()))
lastn = nums.pop(0)
mc = 0
while len(nums) > 1:
buff = nums.pop(0)
if (lastn < buff and buff > nums[0]) or (lastn > buff and buff < nums[0]):
mc += 1
lastn = buff
return mc
if __name__ == "__main__":
print(main())
```
| 12,232 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
x=input()
x=int(x)
y=input()
y=y.split()
min1=0
max1=0
for i in range(1,x-1):
if int(y[i]) < int(y[i-1]) and int(y[i])<int(y[i+1]):
min1=min1+1
if int(y[i])>int(y[i+1]) and int(y[i])>int(y[i-1]):
max1=max1+1
res=min1+max1
if x==2 or x==1:
print(0)
else:
print(res)
```
| 12,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Tags: brute force, implementation
Correct Solution:
```
# bsdk idhar kya dekhne ko aaya hai, khud kr!!!
# from math import *
# from itertools import *
# import random
n = int(input())
arr = list(map(int, input().split()))
count_ = 0
for i in range(1, n-1):
if (arr[i] < arr[i-1] and arr[i] < arr[i+1]) or (arr[i] > arr[i-1] and arr[i] > arr[i+1]):
count_ += 1
else:
continue
print(count_)
```
| 12,234 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
m=int(input())
a=list(map(int,input().split()))
c=0
for i in range(1,len(a)-1):
if (a[i]>a[i+1] and a[i]>a[i-1]) or (a[i]<a[i-1] and a[i]<a[i+1]) :
c+=1
else :
pass
print(c)
```
Yes
| 12,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n = int(input())
nums = list(map(int, input().split()))
res = 0
for i in range(1, n - 1):
f = -1
if(nums[i] > nums[i - 1]):f = 1
if(nums[i] < nums[i - 1]):f = 2
s = -2
if(nums[i] > nums[i + 1]):s = 1
if(nums[i] < nums[i + 1]):s = 2
if(s == f):
res += 1
print(res)
```
Yes
| 12,236 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n=int(input())
line=list(input().split(' '))
line=[int(i) for i in line]
s=0
for i in range(1,n-1):
if line[i-1]<line[i]>line[i+1]:
s+=1
elif line[i-1]>line[i]<line[i+1]:
s+=1
print(s)
```
Yes
| 12,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
count=0
for i in range(1,n-1):
if l[i]>l[i-1] and l[i]>l[i+1]:
count+=1
elif l[i]<l[i-1] and l[i]<l[i+1]:
count+=1
print(count)
```
Yes
| 12,238 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
n=int(input())
f=[int(i) for i in input().split()]
gh=0
b=0
t=0
for i in range(1,n-1):
b=f[i+1]
gh=f[i-1]
if i!=0 and i!=n-1 and (gh>f[i-1]<b or gh<f[i]>b):
t=t+1
print(t)
```
No
| 12,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
input()
data = [int(x) for x in input().split()]
al=0
for x in range(len(data)):
if x==0 or x==len(data)-1:
continue
if (data[x]<data[x+1] and data[x]<data[x-1]) or (data[x]>data[x+1] and data[x]>data[x-1]):
al+=x-1
print(al)
```
No
| 12,240 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
useless = input()
numbers = input().split()
def local_extremums(array):
numbers = list(map(int,array))
i = 1
result = 0
while i < len(numbers) - 1:
if (numbers[i] < numbers[i-1] and numbers[i] < numbers[i + 1]) or (numbers[i] > numbers[i-1] and numbers[i] > numbers[i + 1]):
result += 1
i += 1
return result
print(local_extremums)
```
No
| 12,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.
Output
Print the number of local extrema in the given array.
Examples
Input
3
1 2 3
Output
0
Input
4
1 5 2 5
Output
2
Submitted Solution:
```
input()
a = b = s = 0
for c in map(int, input().split()):
if a and a > b < c or a < b > c: s += 1
a, b = b, c
print(s)
```
No
| 12,242 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = int(input())
b = int(input())
ans = 6
cnt = 0
cur = 2
cnt += 2 * ((n - b) // a)
while cnt < 4:
cur += 1
cnt += (n // a)
ans = min(ans, cur)
if b * 2 <= n:
cur, cnt = 0, 0
cur = 1
cnt += ((n - 2 * b) // a)
while cnt < 4:
cur += 1
cnt += (n // a)
ans = min(ans, cur)
print(ans)
```
| 12,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Thu Nov 30 12:11:39 2017
@author: vishal
"""
n=int(input())
a=int(input())
b=int(input())
if(4*a+2*b<=n):
print(1)
elif(2*a+b<=n or a+2*b<=n and 3*a<=n):
print(2)
elif(a+b<=n and 2*a<=n or 2*b<=n and 2*a<=n or 4*a<=n):
print(3)
elif(2*a<=n or a+b<=n):
print(4)
elif(2*b<=n):
print(5)
else:
print(6)
```
| 12,244 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
a=int(input())
b=int(input())
if a>=b:
l=a
s=b
if n-a>=3*l+2*b:
print(1)
if n-a<3*l+2*b and n-a>=a+b:
print(2)
if n-a<a+b and n-a>=a:
print(3)
if n-a<a and n-a>=b:
print(4)
if n-a<b and n-b>=b:
print(5)
if n-a<b and n-b<b:
print(6)
else:
l=b
s=a
if n-l>=1*l+4*s:
print(1)
if n-l<l+4*s and n-l>=2*s:
print(2)
if n-l<2*s and n-l>=s:
print(3)
if n-l<s and n>=4*s:
print(3)
if n-l<s and n>=2*s and n<4*s:
print(4)
if n-l<s and n-s<s:
print(6)
```
| 12,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
x=int(input())
a=int(input())
b=int(input())
ans=6
if a+b>x and 2*b<=x and 2*a>x:
ans=5
if (a+b<=x and 2*a>=x) or (2*b>x and 2*a<=x):
ans=4
if (2*a<=x and 2*b<=x) or (4*a<=x and 2*b>x) or (2*a<=x and a+b<=x):
ans=3
if 2*a+b<=x:
ans=2
if 4*a +2*b<=x:
ans=1
print(ans)
```
| 12,246 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = int(input())
b = int(input())
na = 4
nb = 2
cnt=0
while True:
len = n
cnt+=1
while len>0:
resa = len-min(int(len/a),na)*a
resb = len-min(int(len/b),nb)*b
if resa<resb and na>0 and len>=a:
len-=a
na-=1
elif nb>0 and len>=b:
len-=b
nb-=1
else:
break
if na==nb==0:
break
print(cnt)
```
| 12,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = int(input())
b = int(input())
if n >= 4 * a + 2 * b:
ans = 1
#print(1)
elif n >= 4 * a + b:
ans = 2
#print(2)
elif n >= 4 * a and 2 * b <= n:
ans = 2
#print(3)
elif n >= 3 * a + 2 * b:
ans = 2
#print(-7)
elif n >= 3 * a and n >= a + 2 * b:
ans = 2
#print(-6)
elif n >= 2 * a + b or n >= 2 * a + 2 * b:
ans = 2
#print(5)
elif n >= 2 * a and (n >= 2 * b or n >= a + b):
ans = 3
#else:####
# ans = 4
#print(6)
elif n >= a + 2 * b:######
ans = 4
#print(7)
elif n >= a + b:
ans = 4
#print(8)
elif n >= 2 * b:
if 3 * a <= n:
ans = 3
else:
ans = 5
#print(9)
else:
if 4 * a <= n:
ans = 3
elif 3 * a <= n:
ans = 4
elif 2 * a <= n:
ans = 4
else:
ans = 6
#print(10)
print(ans)
```
| 12,248 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
N,a,b=int(input()),int(input()),int(input())
queue=[(N,1,4,2)]
res=10000
while queue:
n,amount,x,y=queue.pop()
if x==0 and y==0:res=min(res,amount);continue
if x>0:
if n>=a:queue.append((n-a,amount,x-1,y))
else:queue.append((N-a,amount+1,x-1,y))
if y>0:
if n>=b:queue.append((n-b,amount,x,y-1))
else:queue.append((N-b,amount+1,x,y-1))
print(res)
```
| 12,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Tags: greedy, implementation
Correct Solution:
```
import sys
#needed: 4 a, 2b
n = int(sys.stdin.readline().strip())
am = int(sys.stdin.readline().strip())
bm = int(sys.stdin.readline().strip())
def recurse(a, b):
global n
#print("foo", a, b)
pairs = set()
if(a <= 0 and b <= 0):
return 0;
for i in range(0, a + 1):
for j in range(0, b + 1):
pairs.add((i, j))
woods = 1
answer = float("inf")
for (i, j) in pairs:
if(n >= (am * i + bm * j) and (am * i + bm * j) is not 0):
rec = recurse(a - i, b - j)
#print(n, (am * i + bm * j), rec)
answer = min(answer, rec)
return woods + answer
print(recurse(4, 2))
```
| 12,250 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
import math
[n,a,b],r,i,j=[int(input())for x in range(3)],6,4,5
while i>=0:
l,c,o=[b if x in[i,j]else a for x in range(6)],0,n
for k in l:
if o<k:
o,c=n-k,c+1
else:o-=k
r=min(r,c if o==n else c+1)
j-=1
if i==j:i,j=i-1,5
print(r)
# Made By Mostafa_Khaled
```
Yes
| 12,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
ax, bx = 4, 2
x = 0
z = False
if a*2+b < n//2:
print(1)
elif a*2+b == n:
print(2)
elif a >= b:
while ax >= 0 and bx >= 0:
if ax == bx == 0:
print(x)
exit()
for i in range(ax, -1, -1):
for j in range(bx, -1, -1):
# print(i ,j)
if (a*i)+(b*j) <= n:
# print('yes')
ax -= i
bx -= j
x += 1
z = True
break
if z:
z = not z
break
else:
while ax >= 0 and bx >= 0:
if ax == bx == 0:
print(x)
exit()
for i in range(bx, -1, -1):
for j in range(ax, -1, -1):
# print(i ,j)
if (a*j)+(b*i) <= n:
# print('yes')
ax -= j
bx -= i
x += 1
z = True
break
if z:
z = not z
break
```
Yes
| 12,252 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
'''input
6
4
2
'''
def list_input():
return list(map(int,input().split()))
def map_input():
return map(int,input().split())
def map_string():
return input().split()
def f(n,a,b,left,cnta = 4,cntb = 2):
if(cnta == 0 and cntb == 0): return 0
if(cnta < 0 or cntb < 0): return 100000000000000000000
if a <= left and cnta and b <= left and cntb:
return min(f(n,a,b,left-a,cnta-1,cntb),f(n,a,b,left-b,cnta,cntb-1))
if a <= left and cnta:
return f(n,a,b,left-a,cnta-1,cntb)
if b <= left and cntb:
return f(n,a,b,left-b,cnta,cntb-1)
return 1+min(f(n,a,b,n-a,cnta-1,cntb),f(n,a,b,n-b,cnta,cntb-1))
n = int(input())
a = int(input())
b = int(input())
print(f(n,a,b,0))
```
Yes
| 12,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
#from dust i have come dust i will be
n=int(input())
a=int(input())
b=int(input())
cnt=1
r=n
qa,qb=0,0
while 1:
if r>=a and qa<4:
r-=a
qa+=1
if r>=b and qb<2:
r-=b
qb+=1
if qa==4 and qb==2:
print(cnt)
exit(0)
if (r<a or qa==4) and (r<b or qb==2):
r=n
cnt+=1
```
Yes
| 12,254 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def main():
n = int(input())
a = int(input())
b = int(input())
for i in range(1, 7):
left = [a, a, b, a, a, b]
left.sort(reverse=True)
for _ in range(i):
if not left:
break
sz = n
while sz >= left[0]:
sz -= left[0]
left.pop(0)
if not left:
break
if not left:
break
while sz >= left[-1]:
sz -= left[-1]
left.pop()
if not left:
print(i)
exit()
if __name__ == '__main__':
main()
```
No
| 12,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
cnt = 0
lenth = 0
cnt = 0
max_num = max(a,b)
min_num = min(a,b)
if max_num == a:
max_cnt = 4
min_cnt = 2
else:
max_cnt = 2
min_cnt = 4
rand_num = min_num
for i in range(6):
if max_cnt == 0 and min_cnt == 0:
break
if lenth < rand_num:
lenth = n
if lenth % (2*a + b) == 0:
cnt = 2
break
cnt += 1
if lenth >= max_num and max_cnt > 0:
lenth -= max_num
max_cnt -= 1
elif lenth >= min_num and min_cnt > 0:
lenth -= min_num
min_cnt -= 1
if min_cnt == 0:
rand_num = max_num
print(cnt)
```
No
| 12,256 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
from math import floor
n = int(input())
a = int(input())
b = int(input())
if 4 * a + 2 * b <= n:
print(1)
exit(0)
if 2 * a + b <= n:
print(2)
exit(0)
if 4 * a <= n:
print(3)
exit(0)
if 2 * b <= n:
if a + 2 * b <= n:
if n // a >= 3:
print(2)
exit(0)
elif n // a == 2:
print(3)
exit(0)
else:
print(4)
exit(0)
else:
if n // a >= 4:
print(2)
exit(0)
elif n // a == 3:
print(3)
exit(0)
elif n // a == 2:
print(4)
exit(0)
if a + b <= n:
print(4)
exit(0)
print(6)
```
No
| 12,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input
The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.
The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.
Output
Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.
Examples
Input
8
1
2
Output
1
Input
5
3
4
Output
6
Input
6
4
2
Output
4
Input
20
5
6
Output
2
Note
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
Submitted Solution:
```
n=int(input())
a=int(input())
b=int(input())
ma=max(a,b)
mi=min(a,b)
L=[]
if mi==a:
L.append(mi)
L.append(mi)
L.append(mi)
L.append(mi)
elif mi==b:
L.append(mi)
L.append(mi)
if ma==a:
L.append(ma)
L.append(ma)
L.append(ma)
L.append(ma)
elif ma==b:
L.append(ma)
L.append(ma)
ini=n
count=1
start=0
for i in L:
if i<=n:
n-=i
if start==1:
count+=1
else:
count+=1
n=ini
start=1
print(count)
```
No
| 12,258 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a node of the tree with index 1 and with weight 0. Let cnt be the number of nodes in the tree at any instant (initially, cnt is set to 1). Support Q queries of following two types:
* <image> Add a new node (index cnt + 1) with weight W and add edge between node R and this node.
* <image> Output the maximum length of sequence of nodes which
1. starts with R.
2. Every node in the sequence is an ancestor of its predecessor.
3. Sum of weight of nodes in sequence does not exceed X.
4. For some nodes i, j that are consecutive in the sequence if i is an ancestor of j then w[i] ≥ w[j] and there should not exist a node k on simple path from i to j such that w[k] ≥ w[j]
The tree is rooted at node 1 at any instant.
Note that the queries are given in a modified way.
Input
First line containing the number of queries Q (1 ≤ Q ≤ 400000).
Let last be the answer for previous query of type 2 (initially last equals 0).
Each of the next Q lines contains a query of following form:
* 1 p q (1 ≤ p, q ≤ 1018): This is query of first type where <image> and <image>. It is guaranteed that 1 ≤ R ≤ cnt and 0 ≤ W ≤ 109.
* 2 p q (1 ≤ p, q ≤ 1018): This is query of second type where <image> and <image>. It is guaranteed that 1 ≤ R ≤ cnt and 0 ≤ X ≤ 1015.
<image> denotes bitwise XOR of a and b.
It is guaranteed that at least one query of type 2 exists.
Output
Output the answer to each query of second type in separate line.
Examples
Input
6
1 1 1
2 2 0
2 2 1
1 3 0
2 2 0
2 2 2
Output
0
1
1
2
Input
6
1 1 0
2 2 0
2 0 3
1 0 2
2 1 3
2 1 6
Output
2
2
3
2
Input
7
1 1 2
1 2 3
2 3 3
1 0 0
1 5 1
2 5 0
2 4 0
Output
1
1
2
Input
7
1 1 3
1 2 3
2 3 4
1 2 0
1 5 3
2 5 5
2 7 22
Output
1
2
3
Note
In the first example,
last = 0
- Query 1: 1 1 1, Node 2 with weight 1 is added to node 1.
- Query 2: 2 2 0, No sequence of nodes starting at 2 has weight less than or equal to 0. last = 0
- Query 3: 2 2 1, Answer is 1 as sequence will be {2}. last = 1
- Query 4: 1 2 1, Node 3 with weight 1 is added to node 2.
- Query 5: 2 3 1, Answer is 1 as sequence will be {3}. Node 2 cannot be added as sum of weights cannot be greater than 1. last = 1
- Query 6: 2 3 3, Answer is 2 as sequence will be {3, 2}. last = 2
Submitted Solution:
```
from bisect import bisect_right
def sol(Q):
last = 0
weight = [0]
parent = [0]
# id -> cumulated weight for each possible length
# (computed lazyly)
lss = {1 : [0]}
def get_ls(idx):
todo = []
i = idx
while i not in lss:
todo.append(i)
i=parent[i-1]
break
for i in reversed(todo):
cur = weight[i-1]
ls = [cur]
p = parent[i-1]
while p>0:
w = weight[p-1]
if w >= cur:
ls.append(cur+w)
cur = w
p = parent[p-1]
lss[i] = ls
return lss[idx]
for q, r, w in Q:
r ^= last
w ^= last
if q == 1:
parent.append(r)
weight.append(w)
else:
ls = get_ls(r)
last = bisect_right(ls, w)
print(last)
if 0:
n = int(input())
Q = []
for _ in range(n):
Q.append(map(int, input().strip().split()))
else:
Q = [(1,1,1),(2,2,0),(2,2,1),(1,3,0),(2,2,0),(2,2,2)]
#Q = [(1,1,0),(2,2,0),(2,0,3),(1,0,2),(2,1,3),(2,1,6)]
sol(Q)
```
No
| 12,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
f = lambda x: f(x // 2) * 2 + (x + 1) // 2 if x else 0
print(f(int(input()) - 1))
```
| 12,260 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
def slow_cal(x):
if x == 0:
return 0
else:
return slow_cal(x - 1) + min(x ^ v for v in range(x))
def med_cal(x):
if x == 0:
return 0
return sum(min(i ^ j for j in range(i)) for i in range(1, x+1))
def fast_cal(x):
res = 0
for bit in range(50):
res += (x//(1 << (bit)) - x//(1 << (bit + 1)))*(1 << bit)
return res
"""
for i in range(700):
assert fast_cal(i) == slow_cal(i) == med_cal(i)
print("ALL_CORRECT")
"""
n = int(input()) - 1
print(fast_cal(n))
```
| 12,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
n = int(input())
ans = 0
_pow = 2
while 2 * n > _pow:
ans += (_pow //2 )* (n // _pow + (1 if n % _pow > _pow // 2 else 0))
_pow *= 2
print(ans)
```
| 12,262 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
from math import log
##n = int(input())
##
####print(n-1 + int(log(n-1)/log(2)))
##
##def ord2(n): ## max power of 2 that divides n
## ret = 1
##
## while n%ret == 0:
## ret*=2
##
## return ret//2
##
##total = 0
##
##for i in range(1, n):
## total += ord2(i)
# fast way?
n = int(input()) - 1
divider = 2**int(log(n)/log(2))
total = 0
doubled = 0
while divider > 0:
total += (n//divider - doubled)*divider
##print('A total of', n//divider, 'work, and the number of doubled', doubled)
doubled += n//divider - doubled
divider //=2
## total = 0
## power = 0
##
## while 2**power <= n:
##
## total += n//(2**power)
##
## power += 1
print(total)
```
| 12,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
n=int(input())-1;a=0
for i in range(0,40):
k=1<<i;a+=k*((n+(k<<1)-k)//(k<<1))
print(a)
```
| 12,264 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
def func(n):
if n==0:
return 0
if n==1:
return 1
if(n%2):
return func(n//2)*2+(n//2)+1
else:
return func(n//2)*2+(n//2)
n=int(input())
a=[1,2,1,4,1,2,1,8,1,2,1,4,1,2,1]
print(func(n-1))
```
| 12,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
import math
n = int(input())
ans = 0
cur = 1
while cur < n:
cnt = math.ceil((n-cur)/(cur << 1))
ans += cnt*cur
cur <<= 1
print(ans)
```
| 12,266 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Tags: bitmasks, dp, graphs, implementation, math
Correct Solution:
```
n=int(input())-1;a=0
for i in range(40):
k=1<<i;a+=(n+k)//(k<<1)*k
print(a)
```
| 12,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
L=input().split();n=int(L[0])-1;k=0
while (1<<k)<n:
k += 1
ans = 0
for i in range(0, k+1):
ans+=(1<<i)*((n+(1<<(i+1))-(1<<i))//(1<<(i+1)))
print(ans)
```
Yes
| 12,268 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
def f(n):
if n == 2:
return 1
if n == 3:
return 3
if n % 2 == 0:
return 2 * f(n // 2) + n // 2
else:
return 2 * f(n // 2 + 1) + n // 2
n = int(input())
print(f(n))
```
Yes
| 12,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
"""
ATSTNG's ejudge Python3 solution template
(actual solution is below)
"""
import sys, queue
try:
import dev_act_ffc429465ab634 # empty file in directory
DEV = True
except:
DEV = False
def log(*s):
if DEV: print('LOG', *s)
class EJudge:
def __init__(self, problem="default", reclim=1<<30):
self.problem = problem
sys.setrecursionlimit(reclim)
def use_files(self, infile='', outfile=''):
if infile!='':
self.infile = open(infile)
sys.stdin = self.infile
if outfile!='':
self.outfile = open(outfile, 'w')
sys.stdout = self.outfile
def use_bacs_files(self):
self.use_files(self.problem+'.in', self.problem+'.out')
def get_tl(self):
while True: pass
def get_ml(self):
tmp = [[[5]*100000 for _ in range(1000)]]
while True: tmp.append([[5]*100000 for _ in range(1000)])
def get_re(self):
s = (0,)[8]
def get_wa(self, wstr='blablalblah'):
for _ in range(3): print(wstr)
exit()
class IntReader:
def __init__(self):
self.ost = queue.Queue()
def get(self):
return int(self.sget())
def sget(self):
if self.ost.empty():
for el in input().split():
self.ost.put(el)
return self.ost.get()
def release(self):
res = []
while not self.ost.empty():
res.append(self.ost.get())
return res
def tokenized(s):
""" Parses given string into tokens with default rules """
word = []
for ch in s.strip():
if ch == ' ':
if word: yield ''.join(word); word = []
elif 'a' <= ch <= 'z' or 'A' <= ch <= 'Z' or '0' <= ch <= '9':
word.append(ch)
else:
if word: yield ''.join(word); word = []
yield ch
if word: yield ''.join(word); word = []
###############################################################################
ej = EJudge( )
int_reader = IntReader()
fmap = lambda f,*l: list(map(f,*l))
parse_int = lambda: fmap(int, input().split())
# input
n = bin(int(input())-1)[2:][::-1]
log(n)
bit_full_cost = [(1 << i) * (i+1) for i in range(100)]
bit_cost = [1<<i for i in range(100)]
log(bit_cost)
ans = 0
for i in range(len(n)):
if n[i] == '1':
if i > 0: ans += bit_full_cost[i-1]
ans += bit_cost[i]
print(ans)
'''
1 2 3 4 5 6 7 8
8 7
0 5
0
'''
```
Yes
| 12,270 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
import sys
import math
import collections
dy = [1, 0, -1, 0]
dx = [0, 1, 0, -1]
r = sys.stdin.readline
N = int(r())
N -= 1
i = 1
ans = N
while 2**i <= N:
ans += N//(2**i)*(2**(i-1))
i += 1
print(ans)
```
Yes
| 12,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
n = int(input())
ans = 0
cur = 1
ecnt = 0
while cur < n:
ans += cur
cur *= 2
ecnt += 1
ans += (n - 1 - ecnt)
print(ans)
```
No
| 12,272 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
n = int(input())
k = (n - 1) // 2
if (n - 1) % 2 == 1:
print(k * (k + 1) + k + 1)
else:
print(k * (k + 1))
```
No
| 12,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
n=int(input())-1;a=0
for i in range(0,35):
k=1<<i;a+=k*((n+(k<<1)-k)//(k<<1))
print(a)
```
No
| 12,274 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight <image> (where <image> is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)). Can you find the weight of the minimum spanning tree of that graph?
You can read about complete graphs in <https://en.wikipedia.org/wiki/Complete_graph>
You can read about the minimum spanning tree in <https://en.wikipedia.org/wiki/Minimum_spanning_tree>
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: <image> The weight of the minimum spanning tree is 1+2+1=4.
Submitted Solution:
```
li = []
size, xor = 2, 1
while size <= 10**12:
li.append((size, xor))
size *= 2
xor *= 2
n = int(input())
if n == 1:
print(0)
exit(0)
ans = 0
for i in range(len(li)):
size, xor = li[i]
if size >= n:
ans += xor
break;
if not i: ans += (n//2)
else:
prev_size, nothing = li[i-1]
ans += ((n+prev_size-1)//size) * xor
print(ans)
```
No
| 12,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a set of size m with integer elements between 0 and 2^{n}-1 inclusive. Let's build an undirected graph on these integers in the following way: connect two integers x and y with an edge if and only if x \& y = 0. Here \& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Count the number of connected components in that graph.
Input
In the first line of input there are two integers n and m (0 ≤ n ≤ 22, 1 ≤ m ≤ 2^{n}).
In the second line there are m integers a_1, a_2, …, a_m (0 ≤ a_{i} < 2^{n}) — the elements of the set. All a_{i} are distinct.
Output
Print the number of connected components.
Examples
Input
2 3
1 2 3
Output
2
Input
5 5
5 19 10 20 12
Output
2
Note
Graph from first sample:
<image>
Graph from second sample:
<image>
Tags: bitmasks, dfs and similar, dsu, graphs
Correct Solution:
```
n, m = map(int, input().split())
a = set(map(int, input().split()))
y = 2 ** n
mk = [0] * (2 * y)
cur = 0
for x in a:
if mk[x]: continue
mk[x] = 1
st = [x]
while st:
u = st.pop()
if u < y:
if not mk[y + u]:
mk[y + u] = 1
st.append(y + u)
else:
for b in range(n):
v = u | 1 << b
if u < v and not mk[v]:
mk[v] = 1
st.append(v)
v = y - 1 - (u - y)
if v in a and not mk[v]:
mk[v] = 1
st.append(v)
cur += 1
print(cur)
```
| 12,276 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a set of size m with integer elements between 0 and 2^{n}-1 inclusive. Let's build an undirected graph on these integers in the following way: connect two integers x and y with an edge if and only if x \& y = 0. Here \& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Count the number of connected components in that graph.
Input
In the first line of input there are two integers n and m (0 ≤ n ≤ 22, 1 ≤ m ≤ 2^{n}).
In the second line there are m integers a_1, a_2, …, a_m (0 ≤ a_{i} < 2^{n}) — the elements of the set. All a_{i} are distinct.
Output
Print the number of connected components.
Examples
Input
2 3
1 2 3
Output
2
Input
5 5
5 19 10 20 12
Output
2
Note
Graph from first sample:
<image>
Graph from second sample:
<image>
Tags: bitmasks, dfs and similar, dsu, graphs
Correct Solution:
```
n, m = map(int, input().split())
a = set(map(int, input().split()))
y = 2 ** n
mk = [0] * (2 * y)
cur = 0
for x in a:
if mk[x]: continue
mk[x] = 1
st = [x]
def push(v):
if not mk[v]: mk[v] = 1; st.append(v)
while st:
u = st.pop()
if u < y:
push(y + u)
else:
for b in range(n):
v = u | 1 << b
push(v)
v = y - 1 - (u - y)
if v in a: push(v)
cur += 1
print(cur)
```
| 12,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a set of size m with integer elements between 0 and 2^{n}-1 inclusive. Let's build an undirected graph on these integers in the following way: connect two integers x and y with an edge if and only if x \& y = 0. Here \& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Count the number of connected components in that graph.
Input
In the first line of input there are two integers n and m (0 ≤ n ≤ 22, 1 ≤ m ≤ 2^{n}).
In the second line there are m integers a_1, a_2, …, a_m (0 ≤ a_{i} < 2^{n}) — the elements of the set. All a_{i} are distinct.
Output
Print the number of connected components.
Examples
Input
2 3
1 2 3
Output
2
Input
5 5
5 19 10 20 12
Output
2
Note
Graph from first sample:
<image>
Graph from second sample:
<image>
Tags: bitmasks, dfs and similar, dsu, graphs
Correct Solution:
```
import sys
inp = list(map(int, sys.stdin.buffer.read().split()))
n, m = inp[:2]
a = set(inp[2:])
y = 2 ** n
mk = [0] * (2 * y)
cur = 0
for x in a:
if mk[x]: continue
mk[x] = 1
st = [x]
def push(v):
if not mk[v]: mk[v] = 1; st.append(v)
while st:
u = st.pop()
if u < y:
push(y + u)
else:
for b in range(n):
v = u | 1 << b
push(v)
v = y - 1 - (u - y)
if v in a: push(v)
cur += 1
print(cur)
```
| 12,278 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of size m with integer elements between 0 and 2^{n}-1 inclusive. Let's build an undirected graph on these integers in the following way: connect two integers x and y with an edge if and only if x \& y = 0. Here \& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Count the number of connected components in that graph.
Input
In the first line of input there are two integers n and m (0 ≤ n ≤ 22, 1 ≤ m ≤ 2^{n}).
In the second line there are m integers a_1, a_2, …, a_m (0 ≤ a_{i} < 2^{n}) — the elements of the set. All a_{i} are distinct.
Output
Print the number of connected components.
Examples
Input
2 3
1 2 3
Output
2
Input
5 5
5 19 10 20 12
Output
2
Note
Graph from first sample:
<image>
Graph from second sample:
<image>
Submitted Solution:
```
import math
n,m=input().strip().split(' ')
n,m=[int(n),int(m)]
arr=list(map(int,input().strip().split(' ')))
alist=[]
for i in arr:
alist.append(i)
for i in range(m):
for j in range(i+1,m):
if(arr[i]&arr[j]==0):
alist[j]=alist[i]
print(len(list(set(alist))))
```
No
| 12,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of size m with integer elements between 0 and 2^{n}-1 inclusive. Let's build an undirected graph on these integers in the following way: connect two integers x and y with an edge if and only if x \& y = 0. Here \& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Count the number of connected components in that graph.
Input
In the first line of input there are two integers n and m (0 ≤ n ≤ 22, 1 ≤ m ≤ 2^{n}).
In the second line there are m integers a_1, a_2, …, a_m (0 ≤ a_{i} < 2^{n}) — the elements of the set. All a_{i} are distinct.
Output
Print the number of connected components.
Examples
Input
2 3
1 2 3
Output
2
Input
5 5
5 19 10 20 12
Output
2
Note
Graph from first sample:
<image>
Graph from second sample:
<image>
Submitted Solution:
```
import bisect
import collections
import copy
import functools
import heapq
import itertools
import math
import random
import re
import sys
import time
import string
from typing import List
sys.setrecursionlimit(99999)
n,m = map(int,input().split())
# st = "3 8 13 16 18 19 21 22 24 25 26 28 29 31 33 42 44 46 49 50 51 53 54 57 58 59 60 61 62 63"
s = set(map(int,input().split()))
l = (1<<n)-1
ans = 0
vis = set()
def dfs(mask):
if mask in vis:
return
vis.add(mask)
if mask in s:
s.remove(mask)
dfs(l^mask)
t = mask
while t:
dfs(t)
t=mask&(t-1)
for i in range(1,l+1):
if i in s:
ans+=1
dfs(i)
print(ans)
```
No
| 12,280 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of size m with integer elements between 0 and 2^{n}-1 inclusive. Let's build an undirected graph on these integers in the following way: connect two integers x and y with an edge if and only if x \& y = 0. Here \& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Count the number of connected components in that graph.
Input
In the first line of input there are two integers n and m (0 ≤ n ≤ 22, 1 ≤ m ≤ 2^{n}).
In the second line there are m integers a_1, a_2, …, a_m (0 ≤ a_{i} < 2^{n}) — the elements of the set. All a_{i} are distinct.
Output
Print the number of connected components.
Examples
Input
2 3
1 2 3
Output
2
Input
5 5
5 19 10 20 12
Output
2
Note
Graph from first sample:
<image>
Graph from second sample:
<image>
Submitted Solution:
```
import math
n,m=input().strip().split(' ')
n,m=[int(n),int(m)]
arr=list(map(int,input().strip().split(' ')))
alist=[]
for i in arr:
alist.append(i)
for i in range(m):
for j in range(i+1,m):
if(arr[i]&arr[j]==0):
if(alist[j]==arr[j]):
alist[j]=alist[i]
else:
alist[i]=alist[j]
print(len(list(set(alist))))
```
No
| 12,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of size m with integer elements between 0 and 2^{n}-1 inclusive. Let's build an undirected graph on these integers in the following way: connect two integers x and y with an edge if and only if x \& y = 0. Here \& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Count the number of connected components in that graph.
Input
In the first line of input there are two integers n and m (0 ≤ n ≤ 22, 1 ≤ m ≤ 2^{n}).
In the second line there are m integers a_1, a_2, …, a_m (0 ≤ a_{i} < 2^{n}) — the elements of the set. All a_{i} are distinct.
Output
Print the number of connected components.
Examples
Input
2 3
1 2 3
Output
2
Input
5 5
5 19 10 20 12
Output
2
Note
Graph from first sample:
<image>
Graph from second sample:
<image>
Submitted Solution:
```
n,m = map(int , input().split(" "))
l=list(map(int ,input().split(" ")))
nombre=0
for i in range(1,m):
if l[0] & l[1]==0:
nombre+=2
l.pop(0)
print(nombre)
```
No
| 12,282 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
from itertools import combinations
n = int(input())
l = list(map(int,input().split()))
c = combinations(l,3)
ans=0
for (a,b,c) in c:
if a+b>c and b+c>a and c+a>b and a!=b and b!=c and a!=c:
ans+=1
print(ans)
```
| 12,283 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
n = int(input())
L = sorted(list(map(int,input().split())),reverse = True)
ans = 0
for i in L:
for j in L:
for k in L:
if i<j<k:
if i + j>k:
ans += 1
print(ans)
```
| 12,284 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
from itertools import *
n=int(input())
a=list(map(int,input().split()))
a.sort()
ans=0
for i,j,k in combinations(a,3):
if i<j<k and i+j>k:
ans+=1
print(ans)
```
| 12,285 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
import itertools
n=int(input())
l=list(map(int,input().split()))
cnt=0
for a,b,c in itertools.combinations(l,3):
if a!=b and b!=c and c!=a and abs(b-c)<a and a<b+c :
cnt+=1
print(cnt)
```
| 12,286 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
n = int(input())
listL = list(map(int, input().split()))
count = 0
for l1 in listL:
for l2 in listL:
for l3 in listL:
if l2 > l1 and l3 > l2 and l1+l2 > l3:
count+=1
print(count)
```
| 12,287 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
n=int(input())
l=list(map(int, input().split()))
l.sort()
ans=0
for i in range(n):
for j in range(i+1,n):
for k in range(j+1,n):
if l[i]+l[j]>l[k] and l[i]!=l[j] and l[j]!=l[k]:
ans+=1
print(ans)
```
| 12,288 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
from itertools import combinations
n = int(input())
l = map(int, input().split())
count = 0
for i in combinations(l, 3):
c = sorted(set(i))
if len(c) == 3:
if c[0] + c[1] > c[2]:
count += 1
print(count)
```
| 12,289 |
Provide a correct Python 3 solution for this coding contest problem.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
"Correct Solution:
```
n=int(input())
Ns=list(map(int, input().split() ) )
ans=0
for i in range(n):
for j in range(i,n):
for k in range(j,n):
a , b , c = sorted([Ns[i] , Ns[j] , Ns[k]])
if a+b>c and a!=b and b!=c:
ans+=1
print(ans)
```
| 12,290 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
from itertools import combinations as ic
N = int(input())
L = map(int, input().split())
A = list(ic(L, 3))
ans = 0
for i in A:
if 2 * max(i) - sum(i) < 0 and len(set(i)) == 3:
ans += 1
else:
print(ans)
```
Yes
| 12,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
N = int(input())
A = list(map(int,input().split()))
A.sort()
cnt = 0
for i in range(0,N):
for j in range(i+1,N):
for k in range(j+1,N):
if (A[i]!=A[j]!=A[k]):
if (A[i]+A[j])>A[k]:
cnt+=1
print(cnt)
```
Yes
| 12,292 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
import itertools
n = int(input())
a = list(map(int, input().split()))
a.sort()
cnt = 0
for v in itertools.combinations(a, 3):
x, y, z = v[0], v[1], v[2]
if x + y > z and x != y and y != z and z != x:
cnt += 1
print(cnt)
```
Yes
| 12,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
from itertools import combinations
n = int(input())
l = list(map(int,input().split()))
c = combinations(l,3)
k=0
for (a,b,c) in c:
if a+b>c and b+c>a and c+a>b and a!=b and b!=c and a!=c:
k+=1
print(k)
```
Yes
| 12,294 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split()))
l.sort(reverse=True)
ans = 0
for k in range(0, n-2):
for j in range(k+1, n-1):
if l[k]/2 > l[j]:
break
for i in range(j+1, n):
if l[k] > l[j] + l[i]:
continue
else:
ans += 1
print(ans)
```
No
| 12,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
N=int(input())
L=list(map(int,input().split()))
count=0
for i in range(N):
for j in range(i+1,N):
for k in range(j+1,N):
if L[i]!=L[j]!=L[k]:
sum_=L[i]+L[j]+L[k]
if max(L[i],L[j],L[k])*2<sum_:
count+=1
print(count)
```
No
| 12,296 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
N=int(input())
L=list(map(int, input().split()))
count=0
for i in range(N):
for j in range(i+1,N):
for k in range(j+1,N):
L2=[L[i],L[j],L[k]]
L2.sort()
print(L2)
if L2[0]+L2[1]>L2[2] and L2[0] != L2[1] and L2[1] != L2[2] and L2[2] != L2[0]:
count+=1
print(count)
```
No
| 12,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.
In how many ways can we choose three of the sticks with different lengths that can form a triangle?
That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* L_i, L_j, and L_k are all different.
* There exists a triangle whose sides have lengths L_i, L_j, and L_k.
Constraints
* 1 \leq N \leq 100
* 1 \leq L_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 \cdots L_N
Output
Print the number of ways to choose three of the sticks with different lengths that can form a triangle.
Examples
Input
5
4 4 9 7 5
Output
5
Input
6
4 5 4 3 3 5
Output
8
Input
10
9 4 6 1 9 6 10 6 6 8
Output
39
Input
2
1 1
Output
0
Submitted Solution:
```
N = int(input())
L = list(map(int, input().split()))
L.sort()
cnt = 0
for i in range(N):
Li = L[i]
for j in range(N):
Lj = L[j]
if Li != Lj:
for k in range(N):
Lk = L[k]
if Lk != Lj and Lk != Li:
if Lk < Li + Lj:
cnt += 1
print(cnt)
```
No
| 12,298 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string S of length N consisting of `R`, `G`, and `B`.
Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.
* j - i \neq k - j.
Constraints
* 1 \leq N \leq 4000
* S is a string of length N consisting of `R`, `G`, and `B`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of triplets in question.
Examples
Input
4
RRGB
Output
1
Input
39
RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB
Output
1800
"Correct Solution:
```
N = int(input())
S = input()
A = S.count("R") * S.count("G") * S.count("B")
for i in range(N):
for j in range(i+1, N):
k = j-i+j
if k<N and S[k] != S[j] and S[k] != S[i] and S[i] !=S[j]:
A -= 1
print(A)
```
| 12,299 |
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