message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,774 | 19 | 163,548 |
Tags: games
Correct Solution:
```
n=int(input())
piles = list(map(int,input().split()))
if(piles.count(min(piles))>n/2):
print("Bob")
else:
print("Alice")
``` | output | 1 | 81,774 | 19 | 163,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,775 | 19 | 163,550 |
Tags: games
Correct Solution:
```
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
n = int(input())
a = list(map(int,input().split()))
ct = a.count(min(a))
if(ct > n//2):
print("Bob")
else:
print("Alice")
``` | output | 1 | 81,775 | 19 | 163,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,776 | 19 | 163,552 |
Tags: games
Correct Solution:
```
x=int(input())
s=input().split()
a=[int(num)for num in s]
q=0
for i in a:
if i==min(a) and not max(a)==min(a):
q+=1
if max(a)==min(a):
print('Bob')
elif x/2<q:
print('Bob')
else :
print('Alice')
``` | output | 1 | 81,776 | 19 | 163,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,777 | 19 | 163,554 |
Tags: games
Correct Solution:
```
n = int(input())
a = [int(_) for _ in input().split()]
cnt = [0 for i in range(max(a) + 1)]
for i in a:
cnt[i] += 1
print("Bob" if cnt[min(a)] > (n >> 1) else "Alice")
``` | output | 1 | 81,777 | 19 | 163,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,778 | 19 | 163,556 |
Tags: games
Correct Solution:
```
n = int(input())
L = [int(i) for i in input().split()]
s = 0
m = L[0]
for i in L:
if i == 1:
s += 1
if i < m:
m = i
if s > n // 2:
print('Bob')
elif s <= n // 2 and s > 0:
print('Alice')
elif s == 0:
ss = 0
for j in L:
if j == m:
ss += 1
if ss <= n // 2:
print('Alice')
else:
print('Bob')
``` | output | 1 | 81,778 | 19 | 163,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,779 | 19 | 163,558 |
Tags: games
Correct Solution:
```
n=int(input())
s=list(map(int,input().split()))
print("Bob"if s.count(min(s))>n/2 else"Alice")
``` | output | 1 | 81,779 | 19 | 163,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,780 | 19 | 163,560 |
Tags: games
Correct Solution:
```
def getn():
return int(input())
def getns():
return [int(x)for x in input().split()]
# n=getn()
# ns=getns()
n=getn()
ns=getns()
mm=min(ns)
ans=0
for c in ns:
if c==mm:
ans+=1
if ans<=n//2:
print('Alice')
else:
print('Bob')
``` | output | 1 | 81,780 | 19 | 163,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 81,781 | 19 | 163,562 |
Tags: games
Correct Solution:
```
n = int(input())
ar = list(map(int, input().split()))
print("Alice" if ar.count(min(ar)) <= n//2 else "Bob")
``` | output | 1 | 81,781 | 19 | 163,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
def nim(lst):
count = 0
m1, m2 = min(lst), max(lst)
for elem in lst:
if elem == m1 and not m1 == m2:
count += 1
if m1 == m2:
return "Bob"
elif len(lst) // 2 < count:
return 'Bob'
return 'Alice'
n = int(input())
a = [int(i) for i in input().split()]
print(nim(a))
``` | instruction | 0 | 81,782 | 19 | 163,564 |
Yes | output | 1 | 81,782 | 19 | 163,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
n = int(input())
arr = [int(x) for x in input().split()]
arr.sort()
x = 1
while True:
if arr.count(x) > n//2:
print('Bob')
break
elif x in arr:
print('Alice')
break
else:
x += 1
``` | instruction | 0 | 81,783 | 19 | 163,566 |
Yes | output | 1 | 81,783 | 19 | 163,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
n = int(input())
s = input().split()
a = [int(num) for num in s]
a.sort()
if a.count(min(a)) > len(a)//2:
print("Bob")
else:
print("Alice")
``` | instruction | 0 | 81,784 | 19 | 163,568 |
Yes | output | 1 | 81,784 | 19 | 163,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
n = int(input())
l = input().split()
v = [int(x) for x in l]
v.sort()
if v[0] == v[n // 2]:
print("Bob")
else:
print("Alice")
``` | instruction | 0 | 81,785 | 19 | 163,570 |
Yes | output | 1 | 81,785 | 19 | 163,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
def test(n, a):
d = {}
for x in a:
if x not in d:
d[x] = 1
else:
d[x] += 1
hi = 0
most = None
for key, val in d.items():
if val > hi:
hi = val
most = key
if hi > n // 2:
print('Bob')
else:
print('Alice')
test(n, a)
``` | instruction | 0 | 81,786 | 19 | 163,572 |
No | output | 1 | 81,786 | 19 | 163,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
import math
import bisect
import heapq
from collections import defaultdict
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, x, y = egcd(b % a, a)
return (g, y - (b // a) * x, x)
def mulinv(b, n):
g, x, _ = egcd(b, n)
if g == 1:
return x % n
primes = []
def isprime(n):
for d in range(2, int(math.sqrt(n))+1):
if n % d == 0:
return False
return True
def argsort(ls):
return sorted(range(len(ls)), key=ls.__getitem__)
def f(p=0):
if p == 1:
return map(int, input().split())
elif p == 2:
return list(map(int, input().split()))
else:
return int(input())
n = f()
cl = f(2)
sm = 0
for i in range(n):
if cl[i]==1:
print('Alice')
exit()
else:
sm+=cl[i]-2
t = sm//(n//2)
if t%2==0:
print('Bob')
else:
print('Alice')
``` | instruction | 0 | 81,787 | 19 | 163,574 |
No | output | 1 | 81,787 | 19 | 163,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
d = [0 for i in range(52)]
for i in range(n):
d[a[i]] += 1
flag = False
for i in range(52):
if d[i] % 2 != 0:
flag = True
if flag:
print("Alice")
else:
print("Bob")
``` | instruction | 0 | 81,788 | 19 | 163,576 |
No | output | 1 | 81,788 | 19 | 163,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 ≤ n ≤ 50) — the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
n2 = n//2
a.sort(reverse=True)
Alice = True
while True:
if a[n2] > 1:
a[:n2] = [x-1 for x in a[:n2]]
a.sort(reverse=True)
Alice=(Alice==False)
else:
break
if Alice:
print('Alice')
else:
print('Bob')
``` | instruction | 0 | 81,789 | 19 | 163,578 |
No | output | 1 | 81,789 | 19 | 163,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1. | instruction | 0 | 82,243 | 19 | 164,486 |
Tags: binary search, data structures, number theory, two pointers
Correct Solution:
```
import sys
def get_primes(n: int):
from itertools import chain
from array import array
primes = {2, 3}
is_prime = (array('b', (0, 0, 1, 1, 0, 1, 0)) +
array('b', (1, 0, 0, 0, 1, 0))*((n-1)//6))
for i in chain.from_iterable((range(5, n+1, 6), range(7, n+1, 6))):
if is_prime[i]:
primes.add(i)
for j in range(i*3, n+1, i*2):
is_prime[j] = 0
return is_prime, primes
n, k = map(int, input().split())
cards = list(map(int, input().split()))
_, primes = get_primes(32000)
div, div_cnt = [], []
for p in primes:
if k % p == 0:
div.append(p)
div_cnt.append(0)
while k % p == 0:
div_cnt[-1] += 1
k //= p
if k > 1:
div.append(k)
div_cnt.append(1)
m = len(div)
acc = [[0]*m for _ in range(n+1)]
for i, x in enumerate(cards, start=1):
for j in range(m):
acc[i][j] += acc[i-1][j]
while x % div[j] == 0:
acc[i][j] += 1
x //= div[j]
ans = 0
j = 0
for i in range(n):
j = max(j, i+1)
while j <= n and any(acc[j][k]-acc[i][k] < div_cnt[k] for k in range(m)):
j += 1
if j > n:
break
ans += n - j + 1
print(ans)
``` | output | 1 | 82,243 | 19 | 164,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1. | instruction | 0 | 82,244 | 19 | 164,488 |
Tags: binary search, data structures, number theory, two pointers
Correct Solution:
```
n,k=map(int,input().split())
l=list(map(int,input().split()))
pf=[]
needed=[]
for i in range(2,40000):
if k%i==0:
pf.append(i)
c=0
while k%i==0:
k//=i
c+=1
needed.append(c)
if k>1:
pf.append(k)
needed.append(1)
pfl=len(pf)
cnt=[[0]*n for i in range(pfl)]
for i in range(n):
for j in range(len(pf)):
c=0
while l[i]%pf[j]==0:
c+=1
l[i]//=pf[j]
cnt[j][i]=c
have=[sum(i) for i in cnt]
pos=n
def ok():
for i in range(len(pf)):
if have[i]<needed[i]:
return False
return True
if not ok():
print(0)
quit()
for i in range(n-1,0,-1):
for j in range(len(pf)):
have[j]-=cnt[j][i]
if not ok():
for j in range(len(pf)):
have[j]+=cnt[j][i]
break
pos=i
ans=n-pos+1
for x in range(n-1):
for j in range(len(pf)):
have[j]-=cnt[j][x]
if pos==(x+1):
for j in range(len(pf)):
have[j]+=cnt[j][pos]
pos+=1
while pos<n:
if ok():
break
else:
for i in range(len(pf)):
have[i]+=cnt[i][pos]
pos+=1
if ok():
ans+=n-pos+1
else:
break
print(ans)
``` | output | 1 | 82,244 | 19 | 164,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1. | instruction | 0 | 82,245 | 19 | 164,490 |
Tags: binary search, data structures, number theory, two pointers
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,k=map(int,input().split())
l=list(map(int,input().split()))
div=defaultdict(int)
su=defaultdict(list)
for i in range(2,int(math.sqrt(k))+1):
while(k%i==0):
k//=i
div[i]+=1
if k>1:
div[k]+=1
for j in div:
su[j].append(0)
def find(ind):
st=0
end=ind
ans=st
while(st<=end):
mid=(st+end)//2
f=0
for j in div:
if su[j][ind+1]-su[j][mid]>=div[j]:
continue
else:
f=1
break
if f==0:
ans=mid
st=mid+1
else:
end=mid-1
return ans
ans=0
for i in range(n):
f=0
for j in div:
c=0
while(l[i]%j==0):
l[i]//=j
c+=1
su[j].append(su[j][-1]+c)
if su[j][i+1]<div[j]:
f=1
if f==1:
continue
u=find(i)
ans+=u+1
print(ans)
``` | output | 1 | 82,245 | 19 | 164,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1. | instruction | 0 | 82,246 | 19 | 164,492 |
Tags: binary search, data structures, number theory, two pointers
Correct Solution:
```
def gcd(a,b):
if a == 0:
return b
return gcd(b%a,a)
n,k = [int(x) for x in input().split()]
a = [gcd(int(x),k) for x in input().split()]
if k == 1:
print(((n+1)*(n+2))//2-n-1)
else:
s = 0
e = 0
total = ((n+1)*(n+2))//2-1-n
#print(total)
#extra = {}
c = 1
while e < n:
flag = False
while c%k != 0 and e < n:
total -= e-s
c *= a[e]
e += 1
while c%k == 0 and s < e:
c //= a[s]
s += 1
total -= e-s
print(total)
``` | output | 1 | 82,246 | 19 | 164,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1.
Submitted Solution:
```
n,k=map(int,input().split())
l=list(map(int,input().split()))
pf=[]
needed=[]
for i in range(2,40000):
if k%i==0:
pf.append(i)
c=0
while k%i==0:
k//=i
c+=1
needed.append(c)
pfl=len(pf)
cnt=[[0]*n for i in range(pfl)]
for i in range(n):
for j in range(len(pf)):
c=0
while l[i]%pf[j]==0:
c+=1
l[i]//=pf[j]
cnt[j][i]=c
have=[sum(i) for i in cnt]
pos=n
def ok():
for i in range(len(pf)):
if have[i]<needed[i]:
return False
return True
if not ok():
print(0)
quit()
for i in range(n-1,0,-1):
for j in range(len(pf)):
have[j]-=cnt[j][i]
if not ok():
for j in range(len(pf)):
have[j]+=cnt[j][i]
break
pos=i
ans=n-pos+1
for x in range(n):
if pos==(x+1):
pos+=1
for j in range(len(pf)):
have[j]-=cnt[j][x]
if pos==(x+1):
for j in range(len(pf)):
have[j]+=cnt[j][pos]
pos+=1
while pos<n:
if ok():
break
else:
for i in range(len(pf)):
have[i]+=cnt[i][pos]
pos+=1
if ok():
ans+=n-pos+1
else:
break
print(ans)
``` | instruction | 0 | 82,247 | 19 | 164,494 |
No | output | 1 | 82,247 | 19 | 164,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1.
Submitted Solution:
```
# cook your dish here
inp= [int(x) for x in input().split()]
n= inp[0]
k= inp[1]
a= [int(x) for x in input().split()]
count= 0
productPrefix= []
productPrefix.insert(len(productPrefix),a[0])
for i in range(1,n):
productPrefix.insert(len(productPrefix),a[i]*productPrefix[i-1])
#finding the last postion of the first subarray which is divisible by k.
end=-1
for i in range(0,n):
if productPrefix[i]%k==0:
end= i
product= productPrefix[i]
break
if end==-1:
print(0)
else:
count+=(n-end)
for i in range(1,n):
product/= productPrefix[i-1]
if product%k==0:
count+=(n-end)
else:
if end<n:
end= end+1
while end<n:
product*=(a[end])
if product%k==0:
break
count+=(n-end)
print(count)
``` | instruction | 0 | 82,248 | 19 | 164,496 |
No | output | 1 | 82,248 | 19 | 164,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1.
Submitted Solution:
```
read = lambda: map(int, input().split())
n, k = read()
a = list(read())
p = 1
ans = 0
r = 0
for i in range(n):
r = max(r, i)
if r == i: p = a[i]
while r + 1 < n and p % k:
p *= a[r + 1]
r += 1
if p % k == 0: ans += n - r
print(ans)
``` | instruction | 0 | 82,249 | 19 | 164,498 |
No | output | 1 | 82,249 | 19 | 164,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving the deck and the magic number, Vova removes x (possibly x = 0) cards from the top of the deck, y (possibly y = 0) cards from the bottom of the deck, and the rest of the deck is his new deck (Vova has to leave at least one card in the deck after removing cards). So Vova's new deck actually contains cards x + 1, x + 2, ... n - y - 1, n - y from the original deck.
Vova's new deck is considered valid iff the product of all numbers written on the cards in his new deck is divisible by k. So Vova received a deck (possibly not a valid one) and a number k, and now he wonders, how many ways are there to choose x and y so the deck he will get after removing x cards from the top and y cards from the bottom is valid?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards.
Output
Print the number of ways to choose x and y so the resulting deck is valid.
Examples
Input
3 4
6 2 8
Output
4
Input
3 6
9 1 14
Output
1
Note
In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1.
Submitted Solution:
```
# cook your dish here
inp= [int(x) for x in input().split()]
n= inp[0]
k= inp[1]
a= [int(x) for x in input().split()]
count= 0
productPrefix= []
productPrefix.insert(len(productPrefix),a[0])
for i in range(1,n):
productPrefix.insert(len(productPrefix),a[i]*productPrefix[i-1])
#finding the last postion of the first subarray which is divisible by k.
end=-1
for i in range(0,n):
if productPrefix[i]%k==0:
end= i
product= productPrefix[i]
prevKatbo= a[0]
break
if end==-1:
print(0)
else:
count+=(n-end)
for i in range(1,n):
if i>end:
product=1
for q in range(i,n):
product*=a[q]
if product%k==0:
end= q
count+=(n-end)
prevKatbo=a[i]
break
# print(i,count,end)
else:
product/= prevKatbo
prevKatbo=a[i]
if product%k==0:
count+=(n-end)
else:
while end<n:
end= end+1
if end>=n:
break
product*=(a[end])
if product%k==0:
break
count+=(n-end)
print(count)
``` | instruction | 0 | 82,250 | 19 | 164,500 |
No | output | 1 | 82,250 | 19 | 164,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Slime and Orac are holding a turn-based game. In a big room, there are n players sitting on the chairs, looking forward to a column and each of them is given a number: player 1 sits in the front of the column, player 2 sits directly behind him; player 3 sits directly behind player 2, and so on; player n sits directly behind player n-1. Each player wears a hat that is either black or white. As each player faces forward, player i knows the color of player j's hat if and only if i is larger than j.
At the start of each turn, Orac will tell whether there exists a player wearing a black hat in the room.
After Orac speaks, if the player can uniquely identify the color of his hat, he will put his hat on the chair, stand up and leave the room. All players are smart, so if it is possible to understand the color of their hat using the obtained information during this and previous rounds, they will understand it.
In each turn, all players who know the color of their hats will leave at the same time in this turn, which means a player can only leave in the next turn if he gets to know the color of his hat only after someone left the room at this turn.
Note that when the player needs to leave, he will put the hat on the chair before leaving, so the players ahead of him still cannot see his hat.
The i-th player will know who exactly left the room among players 1,2,…,i-1, and how many players among i+1,i+2,…,n have left the room.
Slime stands outdoor. He watches the players walking out and records the numbers of the players and the time they get out. Unfortunately, Slime is so careless that he has only recorded some of the data, and this given data is in the format "player x leaves in the y-th round".
Slime asked you to tell him the color of each player's hat. If there are multiple solutions, you can find any of them.
Input
The first line contains a integer n\ (1≤ n≤ 200 000).
The second line contains n integers t_1,t_2,...,t_n\ (0≤ t_i≤ 10^{15}). If t_i=0, then there are no data about player i; otherwise it means player i leaves in the t_i-th round.
At least one solution exists for the given input.
Output
Print one binary string of n characters. The i-th character of the string should be '1' if player i wears a black hat and should be '0', otherwise. If there are multiple solutions, you can print any of them.
Examples
Input
5
0 1 1 0 0
Output
00000
Input
5
0 2 2 0 0
Output
00001
Input
5
0 0 0 0 0
Output
00000
Input
5
4 4 0 4 4
Output
00100
Note
In the first example, for the given solution, all the players wear white hats. In the first turn, Orac tells all the players that there are no players wearing a black hat, so each player knows that he is wearing a white hat, and he will leave in the first turn.
In the second example, for the given solution, the player 5 wears a black hat, other players wear white hats. Orac tells all the players that there exists a player wearing a black hat, and player 5 know that the other players are all wearing white hats, so he can infer that he is wearing a black hat; therefore he leaves in the first turn, other players leave in the second turn. Note that other players can infer that they are wearing white hats immediately after player 5 leaves, but they have to wait for the next turn to leave according to the rule.
In the third example, there is no information about the game, so any output is correct.
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
d={}
o=[0]*n
s=''
t=0
te=0
o1=o[:]
x=0
for i in range(n):
try:
d[l[i]]+=1
except:
d[l[i]]=1
if(l[i]>t):
t=l[i]
ind1=i
def fun(ind,tem):
global t,l,n,ind1,o,o1,x
#print(ind,tem,*o,len(l))
#if(tem>t or ind>ind1):
#print("returning tem>t or ")
#return
if(ind==(len(l)-1)):
#print("01")
o1=o[:]
x=1
return
#elif(l[ind]!=0 and l[ind]!=tem):
#return
else:
for i in range(ind+1,len(l)):
if(l[i]>-1):
#print(*o)
tem+=(n-i)
o[i]=1
if(l[i]==0 or l[i]==tem):
fun(i,tem)
else:
return
if(x==1):
#print("break")
break
#else:
#fun(i,tem)
o[i]=0
tem-=(n-i)
'''else:
tem+=l[i]
o[i]=1
fun(i,tem)
if(x==1):
print("break")
break
o[i]=0
tem-=l[i]'''
return
try:
if(d[0]==n or d[1]>1):
for j in range(n):
s+=str(o[j])
print(s)
pass
else:
fun(-1,0)
for j in range(n):
s+=str(o1[j])
print(s)
except:
if(d[t]>1):
try:
if(d[t-1]!=0):
for q in range(n-1,-1,-1):
if(l[q]==t):
l[q]=0
print(*l)
fun(-1,0)
except:
for q in range(n-1,-1,-1):
if(l[q]==t):
l[q]=0
if(l.count(0)==n):
l[-1]=t-1
#print(*l)
fun(-1,0)
else:
for q in range(n-1,-1,-1):
if(q == (n-1) and l[q]!=0):
l=l[:q+1]
break
if(l[q]==0 and l[q-1]!=0):
l=l[:q]
break
#print(*l)
fun(-1,0)
for j in range(n):
s+=str(o1[j])
print(s)
'''10
0 0 0 24 30 0 0 0 0 0
0111100000
10
0 0 8 0 0 12 0 0 16 15
Participant's output
1000001010
Jury's answer
0001001101
'''
``` | instruction | 0 | 82,741 | 19 | 165,482 |
No | output | 1 | 82,741 | 19 | 165,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Slime and Orac are holding a turn-based game. In a big room, there are n players sitting on the chairs, looking forward to a column and each of them is given a number: player 1 sits in the front of the column, player 2 sits directly behind him; player 3 sits directly behind player 2, and so on; player n sits directly behind player n-1. Each player wears a hat that is either black or white. As each player faces forward, player i knows the color of player j's hat if and only if i is larger than j.
At the start of each turn, Orac will tell whether there exists a player wearing a black hat in the room.
After Orac speaks, if the player can uniquely identify the color of his hat, he will put his hat on the chair, stand up and leave the room. All players are smart, so if it is possible to understand the color of their hat using the obtained information during this and previous rounds, they will understand it.
In each turn, all players who know the color of their hats will leave at the same time in this turn, which means a player can only leave in the next turn if he gets to know the color of his hat only after someone left the room at this turn.
Note that when the player needs to leave, he will put the hat on the chair before leaving, so the players ahead of him still cannot see his hat.
The i-th player will know who exactly left the room among players 1,2,…,i-1, and how many players among i+1,i+2,…,n have left the room.
Slime stands outdoor. He watches the players walking out and records the numbers of the players and the time they get out. Unfortunately, Slime is so careless that he has only recorded some of the data, and this given data is in the format "player x leaves in the y-th round".
Slime asked you to tell him the color of each player's hat. If there are multiple solutions, you can find any of them.
Input
The first line contains a integer n\ (1≤ n≤ 200 000).
The second line contains n integers t_1,t_2,...,t_n\ (0≤ t_i≤ 10^{15}). If t_i=0, then there are no data about player i; otherwise it means player i leaves in the t_i-th round.
At least one solution exists for the given input.
Output
Print one binary string of n characters. The i-th character of the string should be '1' if player i wears a black hat and should be '0', otherwise. If there are multiple solutions, you can print any of them.
Examples
Input
5
0 1 1 0 0
Output
00000
Input
5
0 2 2 0 0
Output
00001
Input
5
0 0 0 0 0
Output
00000
Input
5
4 4 0 4 4
Output
00100
Note
In the first example, for the given solution, all the players wear white hats. In the first turn, Orac tells all the players that there are no players wearing a black hat, so each player knows that he is wearing a white hat, and he will leave in the first turn.
In the second example, for the given solution, the player 5 wears a black hat, other players wear white hats. Orac tells all the players that there exists a player wearing a black hat, and player 5 know that the other players are all wearing white hats, so he can infer that he is wearing a black hat; therefore he leaves in the first turn, other players leave in the second turn. Note that other players can infer that they are wearing white hats immediately after player 5 leaves, but they have to wait for the next turn to leave according to the rule.
In the third example, there is no information about the game, so any output is correct.
Submitted Solution:
```
n=int(input())
a=[int(i) for i in input().split()][ :n]
if a[0]==0:
print("{0:b}".format(a[1]-1))
else:
print("{0:b}".format(a[0]))
``` | instruction | 0 | 82,742 | 19 | 165,484 |
No | output | 1 | 82,742 | 19 | 165,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Slime and Orac are holding a turn-based game. In a big room, there are n players sitting on the chairs, looking forward to a column and each of them is given a number: player 1 sits in the front of the column, player 2 sits directly behind him; player 3 sits directly behind player 2, and so on; player n sits directly behind player n-1. Each player wears a hat that is either black or white. As each player faces forward, player i knows the color of player j's hat if and only if i is larger than j.
At the start of each turn, Orac will tell whether there exists a player wearing a black hat in the room.
After Orac speaks, if the player can uniquely identify the color of his hat, he will put his hat on the chair, stand up and leave the room. All players are smart, so if it is possible to understand the color of their hat using the obtained information during this and previous rounds, they will understand it.
In each turn, all players who know the color of their hats will leave at the same time in this turn, which means a player can only leave in the next turn if he gets to know the color of his hat only after someone left the room at this turn.
Note that when the player needs to leave, he will put the hat on the chair before leaving, so the players ahead of him still cannot see his hat.
The i-th player will know who exactly left the room among players 1,2,…,i-1, and how many players among i+1,i+2,…,n have left the room.
Slime stands outdoor. He watches the players walking out and records the numbers of the players and the time they get out. Unfortunately, Slime is so careless that he has only recorded some of the data, and this given data is in the format "player x leaves in the y-th round".
Slime asked you to tell him the color of each player's hat. If there are multiple solutions, you can find any of them.
Input
The first line contains a integer n\ (1≤ n≤ 200 000).
The second line contains n integers t_1,t_2,...,t_n\ (0≤ t_i≤ 10^{15}). If t_i=0, then there are no data about player i; otherwise it means player i leaves in the t_i-th round.
At least one solution exists for the given input.
Output
Print one binary string of n characters. The i-th character of the string should be '1' if player i wears a black hat and should be '0', otherwise. If there are multiple solutions, you can print any of them.
Examples
Input
5
0 1 1 0 0
Output
00000
Input
5
0 2 2 0 0
Output
00001
Input
5
0 0 0 0 0
Output
00000
Input
5
4 4 0 4 4
Output
00100
Note
In the first example, for the given solution, all the players wear white hats. In the first turn, Orac tells all the players that there are no players wearing a black hat, so each player knows that he is wearing a white hat, and he will leave in the first turn.
In the second example, for the given solution, the player 5 wears a black hat, other players wear white hats. Orac tells all the players that there exists a player wearing a black hat, and player 5 know that the other players are all wearing white hats, so he can infer that he is wearing a black hat; therefore he leaves in the first turn, other players leave in the second turn. Note that other players can infer that they are wearing white hats immediately after player 5 leaves, but they have to wait for the next turn to leave according to the rule.
In the third example, there is no information about the game, so any output is correct.
Submitted Solution:
```
def makeSum(a,b,sum):
#print(a,b,sum)
initial=sum
mylist=[]
if a==0:
for i in range(b,a,-1):
if i<=sum:
sum-=i
mylist.append(i)
if sum==0:
break
else:
for i in range(a,b+1):
if i<=sum:
sum-=i
mylist.append(i)
if sum==0:
break
if sum:
mylist=[initial-1]
return mylist
n=int(input())
array=input().split()
ch=["0"]*n
c=0
flag=False
foo=-1
prev=0
prevNon=0
boo=0
mydict={}
for i in array:
if((int)(i)<prev):
foo=i
if array.count(i)>1 and (int)(i) != 0:
flag=True
c=i
break
#print(i,array.index(i),prevNon,(int)(i)-prevNon<(n-array.index(i)))
if (int)(i) and array.index(i) and ((int)(i)-prevNon)<(n-array.index(i)):
boo=array.index((str)(prevNon))+(n-2*(int)(i)+prevNon)
mydict[str(prevNon)]=boo
if ((int)(i)):
prevNon = (int)(i)
prev=(int)(i)
if(flag):
for i in makeSum(0,n,(int)(c)-1):
ch[n-i]='1'
else:
c=0
for i in array:
if i in mydict.keys():
ch[mydict[i]]="1"
i='0'
continue
if foo!= -1 and i==foo:
continue
if(not (int)(i)):
continue
for j in makeSum(n-array.index(i),n,(int)(i)-c):
ch[n-j]='1'
c=(int)(i)
i='0'
print(''.join(ch))
``` | instruction | 0 | 82,743 | 19 | 165,486 |
No | output | 1 | 82,743 | 19 | 165,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Slime and Orac are holding a turn-based game. In a big room, there are n players sitting on the chairs, looking forward to a column and each of them is given a number: player 1 sits in the front of the column, player 2 sits directly behind him; player 3 sits directly behind player 2, and so on; player n sits directly behind player n-1. Each player wears a hat that is either black or white. As each player faces forward, player i knows the color of player j's hat if and only if i is larger than j.
At the start of each turn, Orac will tell whether there exists a player wearing a black hat in the room.
After Orac speaks, if the player can uniquely identify the color of his hat, he will put his hat on the chair, stand up and leave the room. All players are smart, so if it is possible to understand the color of their hat using the obtained information during this and previous rounds, they will understand it.
In each turn, all players who know the color of their hats will leave at the same time in this turn, which means a player can only leave in the next turn if he gets to know the color of his hat only after someone left the room at this turn.
Note that when the player needs to leave, he will put the hat on the chair before leaving, so the players ahead of him still cannot see his hat.
The i-th player will know who exactly left the room among players 1,2,…,i-1, and how many players among i+1,i+2,…,n have left the room.
Slime stands outdoor. He watches the players walking out and records the numbers of the players and the time they get out. Unfortunately, Slime is so careless that he has only recorded some of the data, and this given data is in the format "player x leaves in the y-th round".
Slime asked you to tell him the color of each player's hat. If there are multiple solutions, you can find any of them.
Input
The first line contains a integer n\ (1≤ n≤ 200 000).
The second line contains n integers t_1,t_2,...,t_n\ (0≤ t_i≤ 10^{15}). If t_i=0, then there are no data about player i; otherwise it means player i leaves in the t_i-th round.
At least one solution exists for the given input.
Output
Print one binary string of n characters. The i-th character of the string should be '1' if player i wears a black hat and should be '0', otherwise. If there are multiple solutions, you can print any of them.
Examples
Input
5
0 1 1 0 0
Output
00000
Input
5
0 2 2 0 0
Output
00001
Input
5
0 0 0 0 0
Output
00000
Input
5
4 4 0 4 4
Output
00100
Note
In the first example, for the given solution, all the players wear white hats. In the first turn, Orac tells all the players that there are no players wearing a black hat, so each player knows that he is wearing a white hat, and he will leave in the first turn.
In the second example, for the given solution, the player 5 wears a black hat, other players wear white hats. Orac tells all the players that there exists a player wearing a black hat, and player 5 know that the other players are all wearing white hats, so he can infer that he is wearing a black hat; therefore he leaves in the first turn, other players leave in the second turn. Note that other players can infer that they are wearing white hats immediately after player 5 leaves, but they have to wait for the next turn to leave according to the rule.
In the third example, there is no information about the game, so any output is correct.
Submitted Solution:
```
w = float(input())
if (w%4)==0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 82,744 | 19 | 165,488 |
No | output | 1 | 82,744 | 19 | 165,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors. | instruction | 0 | 82,916 | 19 | 165,832 |
Tags: bitmasks, brute force, implementation
Correct Solution:
```
from itertools import combinations
from copy import copy
def variants(a):
for i in range(6):
for j in combinations(a, i):
yield j
def main():
input()
cards = set((i[0], int(i[1])) for i in input().split())
values = [0 for i in range(6)]
masks = {i: 0 for i in "RGBYW"}
for i in cards:
values[i[1]] += 1
masks[i[0]] += 1
result = float("inf")
for i in variants("RGBYW"):
for j in variants(list(range(1, 6))):
c = copy(cards)
v, m = values[:], copy(masks)
for k in copy(c):
if k[0] in i and k[1] in j:
v[k[1]] -= 1
m[k[0]] -= 1
c.remove(k)
for l in range(11):
for k in copy(c):
if (v[k[1]] == 1 and k[1] in j or
m[k[0]] == 1 and k[0] in i):
v[k[1]] -= 1
m[k[0]] -= 1
c.remove(k)
if len(c) <= 1:
result = min(result, len(i) + len(j))
print(result)
main()
``` | output | 1 | 82,916 | 19 | 165,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors. | instruction | 0 | 82,917 | 19 | 165,834 |
Tags: bitmasks, brute force, implementation
Correct Solution:
```
def main():
p = [1]
for i in range(1,11):
p.append(p[i-1]*2)
hr = '12345RGBYW'
gt = {'1': 0,'2': 1,'3': 2,'4': 3,'5': 4,'R': 5,'G': 6,'B': 7,'Y': 8,'W': 9}
def check(c,mask):
tmp = gt[c]
if (mask // p[tmp]) % 2 == 1:
return True
return False
def count(mask):
tmp = 0
while mask>0:
tmp+=mask % 2
mask//=2
return tmp
n = int(input())
a = list(set(input().split()))
n = len(a)
ans = 10
for mask in range(1023):
f = True
for i in range(n):
for j in range(i+1,n):
if a[i] == a[j]:
pass
elif (a[i][1] != a[j][1]) and (check(a[i][1],mask) or check(a[j][1],mask)):
pass
elif (a[i][0] != a[j][0]) and (check(a[i][0],mask) or check(a[j][0],mask)):
pass
else:
f = False
break
if not f:
break
if f:
ans = min(ans,count(mask))
print(ans)
main()
``` | output | 1 | 82,917 | 19 | 165,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors. | instruction | 0 | 82,918 | 19 | 165,836 |
Tags: bitmasks, brute force, implementation
Correct Solution:
```
from collections import Counter
n = int(input())
pcards = set(input().split())
qs = 'RGBYW12345'
best = 10
for mask in range(1024):
cnt = 0
tmp = mask
while tmp:
tmp &= tmp - 1
cnt += 1
knows = [qs[i] for i in range(10) if ((mask >> i) & 1)]
cards = [c for c in pcards if ((not c[0] in knows) or (not c[1] in knows))]
if not cards:
best = min(best, cnt)
continue
for q in knows:
cou = sum(1 for c in cards if q in c)
if cou > 1:
cnt = 100500
break
coun = sum(1 for c in cards if not c[0] in knows and not c[1] in knows)
if coun > 1:
cnt = 100500
best = min(best, cnt)
print(best)
``` | output | 1 | 82,918 | 19 | 165,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors. | instruction | 0 | 82,919 | 19 | 165,838 |
Tags: bitmasks, brute force, implementation
Correct Solution:
```
A = {'R':0, 'G':1, 'B':2, 'Y':3, 'W':4}
def bit_m(s):
num = 1 << (int(s[1]) - 1)
symb = 1 << (A[s[0]] + 5)
return num | symb
n = int(input())
cards = list(set(input().split()))
res = [0 for i in range(len(cards))]
mask = [bit_m(cards[i]) for i in range(len(cards))]
def make_res(Q, cards):
for i in range(len(cards)):
if mask[i] & Q:
res[i] |= Q
def check():
for i in range(len(res)):
for j in range(i + 1, len(res)):
if res[i] == res[j]:
return 0
return 1
quest = [1 << i for i in range(10)]
cnt = 11
for i in range(0, 1023):
cnt_now = 0
res = [0 for i in range(len(cards))]
for j in range(10):
if (1 << j) & i != 0:
cnt_now += 1
make_res(quest[j], cards)
if check():
cnt = min(cnt, cnt_now)
print(cnt)
``` | output | 1 | 82,919 | 19 | 165,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors. | instruction | 0 | 82,920 | 19 | 165,840 |
Tags: bitmasks, brute force, implementation
Correct Solution:
```
n = int(input())
data = set(input().split())
cards = []
lets = 'RGBYW'
ans = 10
for i in data:
p1 = lets.find(i[0])
p2 = int(i[1]) - 1
cards.append(2 ** p1 + 2 ** (p2 + 5))
for i in range(1024):
good = True
for j in range(len(cards) - 1):
for k in range(j + 1, len(cards)):
if (cards[j] & i) == (cards[k] & i):
good = False
if good:
ones = 0
now = i
while now != 0:
ones += now % 2
now //= 2
ans = min(ans, ones)
print(ans)
``` | output | 1 | 82,920 | 19 | 165,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors. | instruction | 0 | 82,921 | 19 | 165,842 |
Tags: bitmasks, brute force, implementation
Correct Solution:
```
input()
p = {(1 << 'RGBYW'.index(c)) + (1 << int(k) + 4) for c, k in input().split()}
print(min(bin(t).count('1') for t in range(1024) if len({t & q for q in p}) == len(p)))
# Made By Mostafa_Khaled
``` | output | 1 | 82,921 | 19 | 165,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors. | instruction | 0 | 82,922 | 19 | 165,844 |
Tags: bitmasks, brute force, implementation
Correct Solution:
```
n = int(input())
data = set(input().split())
temp = 'RGBYW'
c = []
res = 10
for i in data:
p1 = temp.find(i[0])
p2 = int(i[1]) - 1
c.append(2 ** p1 + 2 ** (p2 + 5))
for i in range(1024):
ok = True
for j in range(len(c) - 1):
for k in range(j + 1, len(c)):
if (c[j] & i) == (c[k] & i):
ok = False
if ok:
tt = 0
inow = i
while inow != 0:
tt += inow % 2
inow //= 2
res = min(res, tt)
print(res)
``` | output | 1 | 82,922 | 19 | 165,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors.
Submitted Solution:
```
A = {'R':0, 'G':1, 'B':2, 'Y':3, 'W':4}
def bit_m(s):
num = 1 << int(s[1])
symb = 1 << (A[s[0]] + 6)
return num | symb
n = int(input())
cards = list(set(input().split()))
res = [0 for i in range(len(cards))]
mask = [bit_m(cards[i]) for i in range(len(cards))]
def make_res(Q, cards):
for i in range(len(cards)):
if mask[i] & Q:
res[i] |= Q
def check():
for i in range(len(res)):
for j in range(i + 1, len(res)):
if res[i] == res[j]:
return 0
return 1
quest = [1 << i for i in range(10)]
cnt = 11
for i in range(0, 1023):
cnt_now = 0
res = [0 for i in range(len(cards))]
for j in range(10):
if (1 << j) & i != 0:
cnt_now += 1
make_res(quest[j], cards)
if check():
cnt = min(cnt, cnt_now)
print(cnt)
``` | instruction | 0 | 82,923 | 19 | 165,846 |
No | output | 1 | 82,923 | 19 | 165,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors.
Submitted Solution:
```
p = [1]
for i in range(1,11):
p.append(p[i-1]*2)
hr = '12345RGBYW'
gt = {'1': 0,'2': 1,'3': 2,'4': 3,'5': 4,'R': 5,'G': 6,'B': 7,'Y': 8,'W': 9}
def check(c,mask):
tmp = gt[c]
if (mask // p[tmp]) % 2 == 1:
return True
return False
def count(mask):
tmp = 0
while mask>0:
tmp+=mask % 2
mask//=2
return tmp
n = int(input())
a = list(input().split())
ans = 10
for mask in range(1024):
f = True
for i in range(n):
for j in range(n):
if a[i] == a[j]:
pass
elif (a[i][0] == a[j][0]) and (check(a[i][1],mask) or check(a[j][1],mask)):
pass
elif (a[i][1] == a[j][1]) and (check(a[i][0],mask) or check(a[j][0],mask)):
pass
elif (check(a[i][1],mask) or check(a[j][1],mask)) or (check(a[i][0],mask) or check(a[j][0],mask)):
pass
else:
f = False
if f:
ans = min(ans,count(mask))
print(ans)
``` | instruction | 0 | 82,924 | 19 | 165,848 |
No | output | 1 | 82,924 | 19 | 165,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.
Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).
The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.
A color hint goes like that: a player names some color and points at all the cards of this color.
Similarly goes the value hint. A player names some value and points at all the cards that contain the value.
Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.
Output
Print a single integer — the minimum number of hints that the other players should make.
Examples
Input
2
G3 G3
Output
0
Input
4
G4 R4 R3 B3
Output
2
Input
5
B1 Y1 W1 G1 R1
Output
4
Note
In the first sample Borya already knows for each card that it is a green three.
In the second sample we can show all fours and all red cards.
In the third sample you need to make hints about any four colors.
Submitted Solution:
```
from itertools import combinations
from copy import copy
def variants(a):
for i in range(6):
for j in combinations(a, i):
yield j
def main():
input()
cards = set((i[0], int(i[1])) for i in input().split())
values = [0 for i in range(6)]
masks = {i: 0 for i in "RGBYW"}
for i in cards:
values[i[1]] += 1
masks[i[0]] += 1
result = float("inf")
for i in variants("RGBYW"):
for j in variants(list(range(1, 6))):
c = copy(cards)
v, m = values[:], copy(masks)
for k in copy(c):
if k[0] in i and k[1] in j:
v[k[1]] -= 1
m[k[0]] -= 1
c.remove(k)
for l in range(11):
for k in copy(c):
if (v[k[1]] == 1 and m[k[0]] == 1 or
v[k[1]] == 1 and k[1] in j or
m[k[0]] == 1 and k[0] in i):
v[k[1]] -= 1
m[k[0]] -= 1
c.remove(k)
if not c:
result = min(result, len(i) + len(j))
print(result)
main()
``` | instruction | 0 | 82,925 | 19 | 165,850 |
No | output | 1 | 82,925 | 19 | 165,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,116 | 19 | 166,232 |
Tags: games, math
Correct Solution:
```
money = int(input())
if money %2 == 1:
print("Ehab")
else:
print("Mahmoud")
``` | output | 1 | 83,116 | 19 | 166,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,117 | 19 | 166,234 |
Tags: games, math
Correct Solution:
```
a=int(input())
if a&1:
print("Ehab")
else:
print("Mahmoud")
``` | output | 1 | 83,117 | 19 | 166,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,118 | 19 | 166,236 |
Tags: games, math
Correct Solution:
```
start = int(input())
if start % 2 == 0: # even
print("Mahmoud")
else:
print("Ehab")
``` | output | 1 | 83,118 | 19 | 166,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,119 | 19 | 166,238 |
Tags: games, math
Correct Solution:
```
n = int(input())
print("MEahhambo u d"[n%2::2])
``` | output | 1 | 83,119 | 19 | 166,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,120 | 19 | 166,240 |
Tags: games, math
Correct Solution:
```
num = int(input(""))
if num % 2 == 0:
print("Mahmoud")
else:
print("Ehab")
``` | output | 1 | 83,120 | 19 | 166,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,121 | 19 | 166,242 |
Tags: games, math
Correct Solution:
```
n = int(input())
if n == 1:
print("Ehab")
elif n == 2:
print("Mahmoud")
elif n % 2 == 0:
print("Mahmoud")
elif n % 2 != 0:
print("Ehab")
``` | output | 1 | 83,121 | 19 | 166,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,122 | 19 | 166,244 |
Tags: games, math
Correct Solution:
```
n = int(input())
if n % 2 == 0:
print('Mahmoud')
elif n % 2 != 0:
print('Ehab')
``` | output | 1 | 83,122 | 19 | 166,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | instruction | 0 | 83,123 | 19 | 166,246 |
Tags: games, math
Correct Solution:
```
n = int(input())
print ("Ehab") if n & 1 else print("Mahmoud")
``` | output | 1 | 83,123 | 19 | 166,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
Submitted Solution:
```
print(['Ehab','Mahmoud'][int(input())%2 == 0])
``` | instruction | 0 | 83,124 | 19 | 166,248 |
Yes | output | 1 | 83,124 | 19 | 166,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
Submitted Solution:
```
n=int(input())
if n==1:
print('Ehab')
elif n%2==0:
print('Mahmoud')
elif n%2!=0 and n!=1:
print('Ehab')
``` | instruction | 0 | 83,125 | 19 | 166,250 |
Yes | output | 1 | 83,125 | 19 | 166,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Jul 26 07:04:59 2020
@author: Tanmay
"""
n=int(input())
chance=0
ans=0
while(n!=0):
if(chance%2==0):
chance+=1
ans+=2*(n//2)
n=n-ans
else:
chance+=1
if(n%2==0):
ans=n-1
else:
ans=n
n=n-ans
if(chance%2==0):
print("Ehab")
else:
print("Mahmoud")
``` | instruction | 0 | 83,126 | 19 | 166,252 |
Yes | output | 1 | 83,126 | 19 | 166,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:
* 1 ≤ a ≤ n.
* If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input
The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.
Output
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Examples
Input
1
Output
Ehab
Input
2
Output
Mahmoud
Note
In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
Submitted Solution:
```
N = int(input())
if (N%10)%2==0:
print("Mahmoud")
else: print("Ehab")
``` | instruction | 0 | 83,127 | 19 | 166,254 |
Yes | output | 1 | 83,127 | 19 | 166,255 |
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