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.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3. | instruction | 0 | 76,871 | 19 | 153,742 |
Tags: games, math, number theory
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int (input())
if n==2:
print ('Ashishgup')
continue
if n==1:
print ('FastestFinger')
continue
elif n&1:
print ('Ashishgup')
continue
elif n%2==0:
if (n//2)%2==1:
k=n//2
flag=0
for i in range(2,int(pow(k,0.5))+1):
if k%i==0:
flag=1
break
if flag==0:
print ('FastestFinger')
continue
else:
print ('Ashishgup')
continue
while not n&1:
n=n>>1
if n==1:
print ('FastestFinger')
continue
else:
print ('Ashishgup')
continue
``` | output | 1 | 76,871 | 19 | 153,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3. | instruction | 0 | 76,872 | 19 | 153,744 |
Tags: games, math, number theory
Correct Solution:
```
def primo(k):
if k == 1:
return False
i = 3
while i*i <= k:
if k % i == 0:
return False
i += 1
return True
t = int(input())
for test in range(t):
n = int(input())
if n == 1:
print("FastestFinger")
elif n % 2 or n == 2: # vince il primo
print("Ashishgup")
else:
if (n//2) % 2 and primo(n//2):
print("FastestFinger")
else:
while n % 2 == 0:
n = n // 2
if n == 1:
print("FastestFinger") # pow 2
else:
print("Ashishgup") # dispari>1 per pow2
``` | output | 1 | 76,872 | 19 | 153,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3. | instruction | 0 | 76,873 | 19 | 153,746 |
Tags: games, math, number theory
Correct Solution:
```
for s in[*open(0)][1:]:n=int(s);print(('FastestFinger','Ashishgup')[n==2or(n&-n<n)&(n&1|any(n//2%i<1for i in range(2,int(n**.5)+1)))])
``` | output | 1 | 76,873 | 19 | 153,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3. | instruction | 0 | 76,874 | 19 | 153,748 |
Tags: games, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import *
def factorize(n):
fct = []
for i in range(2, int(n**0.5)+1):
c = 0
while n%i==0:
n //= i
c += 1
if c>0:
fct.append((i, c))
if n>1:
fct.append((n, 1))
return fct
t = int(input())
for _ in range(t):
n = int(input())
fct = factorize(n)
cnt = defaultdict(int)
tot = 0
for f, c in fct:
cnt[f] = c
tot += c
if n==1:
print('FastestFinger')
elif n==2:
print('Ashishgup')
else:
if cnt[2]==0:
print('Ashishgup')
elif cnt[2]==tot:
print('FastestFinger')
elif cnt[2]>=2:
print('Ashishgup')
elif cnt[2]==1:
if tot==cnt[2]+1:
print('FastestFinger')
else:
print('Ashishgup')
``` | output | 1 | 76,874 | 19 | 153,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10000)
# default is 1000 in python
t = int(input())
# t = 1
for _ in range(t):
n = int(input())
if n ==1:
print("FastestFinger")
elif n ==2:
print("Ashishgup")
elif n % 2 != 0:
print("Ashishgup")
else:
# find odd divs
odd_divs = []
even_div = []
even_divs = 0
for i in range(2,int(n**0.5)+2):
if n % i == 0:
if i % 2 != 0:
odd_divs.append(i)
else:
even_div.append(i)
if (i // 2) % 2 == 0:
even_divs += 1
new_odd = odd_divs.copy()
for i in odd_divs + even_div:
if (n//i) % 2 != 0:
new_odd.append(n//i)
elif ((n//i) // 2) % 2 == 0:
even_divs += 1
new_odd = list(dict.fromkeys(new_odd))
if len(new_odd) > 0:
if len(new_odd) > 1:
print("Ashishgup")
else:
yes = 0
for i in new_odd:
if (n // i) % i == 0:
yes = 1
if yes == 1 or even_divs > 0:
print("Ashishgup")
else:
print("FastestFinger")
else:
print("FastestFinger")
# try:
# raise Exception
# except:
# print("-1")
# thenos.sort(key=lambda x: x[2], reverse=True)
# int(math.log(max(numbers)+1,2))
# 2**3 (power)
# a,t = (list(x) for x in zip(*sorted(zip(a, t))))
# to copy lists use .copy()
# pow(p, si, 1000000007) for modular exponentiation
# my_dict.pop('key', None)
# This will return my_dict[key] if key exists in the dictionary, and None otherwise.
# bin(int('010101', 2))
``` | instruction | 0 | 76,875 | 19 | 153,750 |
Yes | output | 1 | 76,875 | 19 | 153,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
def isPrime(n):
if n==2 or n==3: return True
if n%2==0 or n<2: return False
for i in range(3, int(n**0.5)+1, 2): # only odd numbers
if n%i==0:
return False
return True
t=int(input())
pof2=set([2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824])
for _ in range(t):
n=int(input())
if n==1:
ans=("FastestFinger")
elif n==2:
ans=("Ashishgup")
elif n%2==1:
ans=("Ashishgup")
else:
if n in pof2:
ans=("FastestFinger")
else:
n=n//2
if n%2==0:
ans=("Ashishgup")
else:
while (n%2==0):
n=n//2
if isPrime(n):
ans=("FastestFinger")
else:
ans=("Ashishgup")
print(ans)
``` | instruction | 0 | 76,876 | 19 | 153,752 |
Yes | output | 1 | 76,876 | 19 | 153,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
def game(j):
if (j%2==1 and j!=1) or j==2:
return("Ashishgup")
elif j==1:
return("FastestFinger")
else:
for i in range(1,40):
if j%(2**i)==0:
ni=i
new=j//(2**ni)
if new==1:
return("FastestFinger")
else:
if (j//2)%2==1:
n1=j//2
for p in range(1,int(n1**(1/2))+1):
if n1%p==0:
n2=p
if n2==1:
return("FastestFinger")
return("Ashishgup")
t=int(input())
l=[]
for i in range(t):
n=int(input())
l.append(n)
for j in l:
print(game(j))
``` | instruction | 0 | 76,877 | 19 | 153,754 |
Yes | output | 1 | 76,877 | 19 | 153,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
def f(n):
i = 2
while(i<=n):
if i == n:
return(True)
else:
i = i*2
return(False)
def g(n):
l = int(n**(1/2))+1
c = 0
for i in range(2,l+1):
if n%i == 0:
if i%2 ==1 and (n//i)%2 == 1:
c = c+2
elif i%2 == 1 or (n//i)%2 == 1:
c = c+1
if c>1:
return(True)
return(False)
t = int(input())
while(t>0):
t = t-1
n = int(input())
if n == 1:
print("FastestFinger")
elif n == 2:
print("Ashishgup")
elif f(n) == True:
print("FastestFinger")
elif n%2 == 1:
print("Ashishgup")
elif (n//2)%2 == 0:
print("Ashishgup")
elif (n//2)%2 == 1:
if g(n//2) == True:
print("Ashishgup")
else:
print("FastestFinger")
``` | instruction | 0 | 76,878 | 19 | 153,756 |
Yes | output | 1 | 76,878 | 19 | 153,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
import math
def p(n):
j = []
i = 1
while i <= math.sqrt(n):
if (n % i == 0):
if (n / i == i):
if i%2 != 0 and i > 1:
j.append(i)
else:
if i%2 != 0 and i > 1:
j.append(i)
if (n/i)%2 != 0 and i > 1:
j.append(n / i)
i = i + 1
return j
a = int(input())
for x in range(a):
b = int(input())
c = "Ashishgup"
if b % 2 != 0:
print(c)
else:
j = p(b)
t = len(j)
if t == 0:
print("FastestFinger")
else:
if (b//j[t-1])%2 != 0 or (b//j[t-1]) == 2:
print("FastestFinger")
else:
print(c)
``` | instruction | 0 | 76,879 | 19 | 153,758 |
No | output | 1 | 76,879 | 19 | 153,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math, string, itertools, operator, fractions, heapq, collections, re, array, bisect, sys, functools
def solve(line):
n = int(line)
if n == 1: return "FastestFinger"
if n == 2: return "Ashishgup"
if n % 2 == 1: return "Ashishgup"
found = False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
found = True
break
return "FastestFinger" if not found else "Ashishgup"
T = int(sys.stdin.readline())
t = 0
while True:
line = sys.stdin.readline().rstrip()
if not line:
break
print(solve(line))
t += 1
``` | instruction | 0 | 76,880 | 19 | 153,760 |
No | output | 1 | 76,880 | 19 | 153,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
def notp(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return 1
return 0
k = int(input())
a,f = "Ashishgup","FastestFinger"
for i in range(k):
n = int(input())
if n ==2:
print(a)
elif (n==1):
print(f)
elif n%2==0 and notp(n//2)==0:
print(f)
else:
print(a)
``` | instruction | 0 | 76,881 | 19 | 153,762 |
No | output | 1 | 76,881 | 19 | 153,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ashishgup and FastestFinger play a game.
They start with a number n and play in turns. In each turn, a player can make any one of the following moves:
* Divide n by any of its odd divisors greater than 1.
* Subtract 1 from n if n is greater than 1.
Divisors of a number include the number itself.
The player who is unable to make a move loses the game.
Ashishgup moves first. Determine the winner of the game if both of them play optimally.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).
Output
For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).
Example
Input
7
1
2
3
4
5
6
12
Output
FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup
Note
In the first test case, n = 1, Ashishgup cannot make a move. He loses.
In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.
In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
Submitted Solution:
```
from sys import stdin, stdout
def II(): return int(stdin.readline())
oddprimes=[]
def ispowerof2(num):
if((num & (num - 1)) == 0):
return 1
return 0
def SieveOfEratosthenes(n):
global oddprimes
prime = [True for i in range(n + 1)]
prime[0]= False
prime[1]= False
p=2
while (p*p<= n):
if (prime[p]==True):
for i in range(p*2,n+1,p):
prime[i] = False
p += 1
oddprimes=[p for p in range(3,n+1) if prime[p]]
def TwoOrMoreOddDivisors(n):
global oddprimes
cnt=0
for p in oddprimes:
if cnt>1:
return True
while n%p==0:
n=n//p
cnt+=1
if cnt>1:
return True
if cnt>1:
return True
return False
SieveOfEratosthenes(31623)
t=II()
for _ in range(t):
n=II()
ff,ag=False,False
if n==1:
ff=True
elif n==2:
ag=True
elif n%2==1: #odd number>1, divide by number and make FF lose
ag=True
elif ispowerof2(n)==1: #even number, power of 2, have to subtract 1
ff=True
elif n%4==0:
#even number, not a power of 2, divided by product of all odd powers, and leave even number>=4
ag=True
elif TwoOrMoreOddDivisors(n):
ag=True
else: #divide by odd factor and leaves 2
ff=True
if ff:
stdout.write("FastestFinger\n")
else:
stdout.write("Ashishgup\n")
``` | instruction | 0 | 76,882 | 19 | 153,764 |
No | output | 1 | 76,882 | 19 | 153,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
zscoder has a deck of n+m custom-made cards, which consists of n cards labelled from 1 to n and m jokers. Since zscoder is lonely, he wants to play a game with himself using those cards.
Initially, the deck is shuffled uniformly randomly and placed on the table. zscoder has a set S which is initially empty.
Every second, zscoder draws the top card from the deck.
* If the card has a number x written on it, zscoder removes the card and adds x to the set S.
* If the card drawn is a joker, zscoder places all the cards back into the deck and reshuffles (uniformly randomly) the n+m cards to form a new deck (hence the new deck now contains all cards from 1 to n and the m jokers). Then, if S currently contains all the elements from 1 to n, the game ends. Shuffling the deck doesn't take time at all.
What is the expected number of seconds before the game ends? We can show that the answer can be written in the form P/Q where P, Q are relatively prime integers and Q ≠ 0 mod 998244353. Output the value of (P ⋅ Q^{-1}) modulo 998244353.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^{6}).
Output
Output a single integer, the value of (P ⋅ Q^{-1}) modulo 998244353.
Examples
Input
2 1
Output
5
Input
3 2
Output
332748127
Input
14 9
Output
969862773
Note
For the first sample, it can be proven that the expected time before the game ends is 5 seconds.
For the second sample, it can be proven that the expected time before the game ends is 28/3 seconds. | instruction | 0 | 76,883 | 19 | 153,766 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
import sys
import math
Q = 998244353
n, m = list(map(int, sys.stdin.readline().strip().split()))
z = [1, 1]
f = [1, 1]
for i in range (0, n):
f[0] = f[0] * (n - i) % Q
f[1] = f[1] * (n + m - i) % Q
z = [(z[0]*f[1]+z[1]*f[0]) % Q, (z[1]*f[1]) % Q]
ans = [z[0] * (m+1), z[1]]
for i in range (2, n + 1):
ans = [(ans[0] * i * z[1] + ans[1] * m * z[0]) % Q, (ans[1] * i * z[1]) % Q]
y = ans[1]
ans = ans[0]
q = Q - 2
while q > 0:
if q % 2 == 1:
ans = (ans * y) % Q
q = q // 2
y = (y * y) % Q
print(ans)
``` | output | 1 | 76,883 | 19 | 153,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
zscoder has a deck of n+m custom-made cards, which consists of n cards labelled from 1 to n and m jokers. Since zscoder is lonely, he wants to play a game with himself using those cards.
Initially, the deck is shuffled uniformly randomly and placed on the table. zscoder has a set S which is initially empty.
Every second, zscoder draws the top card from the deck.
* If the card has a number x written on it, zscoder removes the card and adds x to the set S.
* If the card drawn is a joker, zscoder places all the cards back into the deck and reshuffles (uniformly randomly) the n+m cards to form a new deck (hence the new deck now contains all cards from 1 to n and the m jokers). Then, if S currently contains all the elements from 1 to n, the game ends. Shuffling the deck doesn't take time at all.
What is the expected number of seconds before the game ends? We can show that the answer can be written in the form P/Q where P, Q are relatively prime integers and Q ≠ 0 mod 998244353. Output the value of (P ⋅ Q^{-1}) modulo 998244353.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^{6}).
Output
Output a single integer, the value of (P ⋅ Q^{-1}) modulo 998244353.
Examples
Input
2 1
Output
5
Input
3 2
Output
332748127
Input
14 9
Output
969862773
Note
For the first sample, it can be proven that the expected time before the game ends is 5 seconds.
For the second sample, it can be proven that the expected time before the game ends is 28/3 seconds. | instruction | 0 | 76,884 | 19 | 153,768 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
def modInverse(a, m) :
m0 = m; y = 0; x = 1
if (m == 1): return 0
while (a > 1):
q = a // m; t = m; m = a % m; a = t; t = y; y = x - q * y; x = t
if (x < 0): x = x + m0
return x
prime = 998244353
n, m = map(int, input().split())
z = 0
for i in range(1, n+1): z += modInverse(i, prime)
ans = (1+m*z)*(1+n*modInverse(m+1, prime))
print(ans % prime)
``` | output | 1 | 76,884 | 19 | 153,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
zscoder has a deck of n+m custom-made cards, which consists of n cards labelled from 1 to n and m jokers. Since zscoder is lonely, he wants to play a game with himself using those cards.
Initially, the deck is shuffled uniformly randomly and placed on the table. zscoder has a set S which is initially empty.
Every second, zscoder draws the top card from the deck.
* If the card has a number x written on it, zscoder removes the card and adds x to the set S.
* If the card drawn is a joker, zscoder places all the cards back into the deck and reshuffles (uniformly randomly) the n+m cards to form a new deck (hence the new deck now contains all cards from 1 to n and the m jokers). Then, if S currently contains all the elements from 1 to n, the game ends. Shuffling the deck doesn't take time at all.
What is the expected number of seconds before the game ends? We can show that the answer can be written in the form P/Q where P, Q are relatively prime integers and Q ≠ 0 mod 998244353. Output the value of (P ⋅ Q^{-1}) modulo 998244353.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^{6}).
Output
Output a single integer, the value of (P ⋅ Q^{-1}) modulo 998244353.
Examples
Input
2 1
Output
5
Input
3 2
Output
332748127
Input
14 9
Output
969862773
Note
For the first sample, it can be proven that the expected time before the game ends is 5 seconds.
For the second sample, it can be proven that the expected time before the game ends is 28/3 seconds. | instruction | 0 | 76,885 | 19 | 153,770 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
# modular inverse for positive a and b and nCk mod MOD depending on modinv
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a, m):
g, x, y = egcd(a, m)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return x % m
MOD = 998244353
n, m = map(int, input().split())
mul = n * modinv(m + 1,MOD) + 1
mul %= MOD
curN = 1
for i in range(1,n + 1):
curN += m * modinv(i,MOD)
curN %= MOD
curN *= mul
curN %= MOD
print(curN)
``` | output | 1 | 76,885 | 19 | 153,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
zscoder has a deck of n+m custom-made cards, which consists of n cards labelled from 1 to n and m jokers. Since zscoder is lonely, he wants to play a game with himself using those cards.
Initially, the deck is shuffled uniformly randomly and placed on the table. zscoder has a set S which is initially empty.
Every second, zscoder draws the top card from the deck.
* If the card has a number x written on it, zscoder removes the card and adds x to the set S.
* If the card drawn is a joker, zscoder places all the cards back into the deck and reshuffles (uniformly randomly) the n+m cards to form a new deck (hence the new deck now contains all cards from 1 to n and the m jokers). Then, if S currently contains all the elements from 1 to n, the game ends. Shuffling the deck doesn't take time at all.
What is the expected number of seconds before the game ends? We can show that the answer can be written in the form P/Q where P, Q are relatively prime integers and Q ≠ 0 mod 998244353. Output the value of (P ⋅ Q^{-1}) modulo 998244353.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^{6}).
Output
Output a single integer, the value of (P ⋅ Q^{-1}) modulo 998244353.
Examples
Input
2 1
Output
5
Input
3 2
Output
332748127
Input
14 9
Output
969862773
Note
For the first sample, it can be proven that the expected time before the game ends is 5 seconds.
For the second sample, it can be proven that the expected time before the game ends is 28/3 seconds. | instruction | 0 | 76,886 | 19 | 153,772 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
def b(a,m):
m0=m;y=0;x=1
if (m==1):return 0
while (a>1):
q=a//m;t=m;m=a%m;a=t;t=y;y=x-q*y;x=t
if (x<0):x=x+m0
return x
p=998244353
n,m=map(int,input().split())
z=0
for i in range(1,n+1): z+=b(i, p)
print(((1+m*z)*(1+n*b(m+1,p)))%p)
``` | output | 1 | 76,886 | 19 | 153,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
zscoder has a deck of n+m custom-made cards, which consists of n cards labelled from 1 to n and m jokers. Since zscoder is lonely, he wants to play a game with himself using those cards.
Initially, the deck is shuffled uniformly randomly and placed on the table. zscoder has a set S which is initially empty.
Every second, zscoder draws the top card from the deck.
* If the card has a number x written on it, zscoder removes the card and adds x to the set S.
* If the card drawn is a joker, zscoder places all the cards back into the deck and reshuffles (uniformly randomly) the n+m cards to form a new deck (hence the new deck now contains all cards from 1 to n and the m jokers). Then, if S currently contains all the elements from 1 to n, the game ends. Shuffling the deck doesn't take time at all.
What is the expected number of seconds before the game ends? We can show that the answer can be written in the form P/Q where P, Q are relatively prime integers and Q ≠ 0 mod 998244353. Output the value of (P ⋅ Q^{-1}) modulo 998244353.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^{6}).
Output
Output a single integer, the value of (P ⋅ Q^{-1}) modulo 998244353.
Examples
Input
2 1
Output
5
Input
3 2
Output
332748127
Input
14 9
Output
969862773
Note
For the first sample, it can be proven that the expected time before the game ends is 5 seconds.
For the second sample, it can be proven that the expected time before the game ends is 28/3 seconds.
Submitted Solution:
```
MOD = 998244353
def inv(x):
return pow(x, MOD - 2, MOD)
n, m = map(int, input().split())
mult = n * inv(m + 1) + 1
mult %= MOD
out = m + 1
for currN in range(1, n):
out += m * inv(currN)
out %= MOD
out *= mult
print(out % MOD)
``` | instruction | 0 | 76,887 | 19 | 153,774 |
No | output | 1 | 76,887 | 19 | 153,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
def indivisibility(n):
return n - (n//2 + n//3 + n//5 + n//7) + (n//6 + n//10 + n//14 + n//15 + n//21 + n//35) - (n//30+n//42+n//105+n//70) + (n//210)
n = int(input())
print(indivisibility(n))
``` | instruction | 0 | 77,159 | 19 | 154,318 |
Yes | output | 1 | 77,159 | 19 | 154,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
n=int(input())
t=[False]*2521
sum=0
for a in range(2521):
t[a]=(a%2==0 or a%3==0 or a%4==0 or a%5==0 or a%6==0 or a%7==0 or a%8==0 or a%9==0 or a%10==0)
if not t[a]:
sum+=1
ret=sum*(n//2520)
for a in range(n%2520+1):
ret+=(not t[a])
print(ret)
``` | instruction | 0 | 77,160 | 19 | 154,320 |
Yes | output | 1 | 77,160 | 19 | 154,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
n = int(input())
a = [2, 3 ,5, 7]
ans = 0
import operator
import itertools
import functools
for i in range(1, 5):
for p in itertools.combinations(a, i):
x = functools.reduce(operator.mul, p)
ans += (-1) ** (i + 1) * (n // x)
print(n - ans)
``` | instruction | 0 | 77,161 | 19 | 154,322 |
Yes | output | 1 | 77,161 | 19 | 154,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
def main():
n = int(input())
one = n // 2 + n // 3 + n // 5 + n // 7
two = n // (2*3) + n // (2*5) + n // (2*7) + n // (3*5) + n // (3*7) + n // (5*7)
three = n // (2*3*5) + n // (2*3*7) + n // (2*5*7) + n // (3*5*7)
four = n // (2*3*5*7)
print(n - one + two - three + four)
if __name__ == '__main__':
main()
``` | instruction | 0 | 77,162 | 19 | 154,324 |
Yes | output | 1 | 77,162 | 19 | 154,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
n = int(input())
p = [2 , 3 , 5 , 7]
np = [6 , 10 , 14 , 15 , 21 , 35]
ans = 0
for i in p:
ans += (n // i)
for i in np:
ans -= (n // i)
print(n - ans)
``` | instruction | 0 | 77,163 | 19 | 154,326 |
No | output | 1 | 77,163 | 19 | 154,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
n = int(input())
print(n - (n - 2) // 2 - (n - 3) // 6 - (n - 5) // 30 - (n - 7) // 210 - 4)
``` | instruction | 0 | 77,164 | 19 | 154,328 |
No | output | 1 | 77,164 | 19 | 154,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
n = int(input())
def Solution(n):
return n-n//2-n//3-n//5-n//7+n//6+n//10+n//14+n//15+n//21+n//35-n//30-n//42-n//70-n//105+n//210
Solution(n)
``` | instruction | 0 | 77,165 | 19 | 154,330 |
No | output | 1 | 77,165 | 19 | 154,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.
Output
Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.
Examples
Input
12
Output
2
Submitted Solution:
```
LCM=2520
sieve=[1]*LCM
for i in range(2,10):
for j in range(i,LCM,i):
sieve[j-1]=0
n=int(input())
print(sum(sieve)*(n//LCM)+sum(sieve[:n%LCM]))
``` | instruction | 0 | 77,166 | 19 | 154,332 |
No | output | 1 | 77,166 | 19 | 154,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,237 | 19 | 154,474 |
Tags: implementation
Correct Solution:
```
s = list(map(int,input().split()))
cups = s[0]
swapping = s[2]
l = []
for i in range(cups):
l.append(False)
holes = list(map(int,input().split()))
p = True
for i in holes:
if i == 1:
p = False
break
l[i-1] = True
aux = 1
for i in range(swapping):
if p:
swapp = list(map(int,input().split()))
if swapp[0] == aux:
if l[swapp[1]-1]:
p = False
aux = swapp[1]
elif swapp[1] == aux:
if l[swapp[0]-1]:
p = False
aux = swapp[0]
else:
input()
print(aux)
``` | output | 1 | 77,237 | 19 | 154,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,238 | 19 | 154,476 |
Tags: implementation
Correct Solution:
```
import math
import random
def main(n,arr,holes):
if 1 in holes:
return 1
bone_pos=1
for i in range(len(arr)):
j,k=arr[i]
if j==bone_pos:
if k in holes:
return k
else:
bone_pos=k
elif k==bone_pos:
if j in holes:
return j
else:
bone_pos=j
return bone_pos
n,m,k=list(map(int,input().split()))
holes=set(map(int,input().split()))
arr=[]
for i in range(k):
arr.append(list(map(int,input().split())))
print(main(n,arr,holes))
``` | output | 1 | 77,238 | 19 | 154,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,239 | 19 | 154,478 |
Tags: implementation
Correct Solution:
```
from sys import stdin
import re
def main():
l = list(map(int, filter(None, re.split(r'\W', stdin.read()))))
n, m, k = l[:3]
mm = [False] * (n + 1)
for x in l[3:3 + m]:
mm[x] = True
x = 1
for u, v in zip(l[3 + m::2], l[4 + m::2]):
if mm[x]:
break
if x == u:
x = v
elif x == v:
x = u
print(x)
if __name__ == '__main__':
main()
``` | output | 1 | 77,239 | 19 | 154,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,240 | 19 | 154,480 |
Tags: implementation
Correct Solution:
```
if __name__ == "__main__":
n,m,k = map(int,input().split())
holes = {int(i):0 for i in input().split()}
poss = [False]*(n+1)
poss[1] = True
ans = 1
if 1 in holes:
ans = 1
else:
for i in range(k):
x,y = map(int,input().split())
if poss[x] and y in holes:
ans = y
for j in range(i+1,k):
input()
break
if poss[y] and x in holes:
ans = x
for j in range(i+1,k):
input()
break
if poss[y]:
poss[y] = False
poss[x] = True
ans = x
elif poss[x]:
poss[x] = False
poss[y] = True
ans = y
print(ans)
``` | output | 1 | 77,240 | 19 | 154,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,241 | 19 | 154,482 |
Tags: implementation
Correct Solution:
```
n, m, k = [int(x) for x in input().split()]
holes = set([int(x) for x in input().split()])
now = 1
if(now in holes):
print(1)
exit(0)
for t in range(k):
i, j = [int (x) for x in input().split()]
if(i == now): now = j
elif(j == now): now = i
if now in holes:
print(now)
exit(0)
print(now)
``` | output | 1 | 77,241 | 19 | 154,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,242 | 19 | 154,484 |
Tags: implementation
Correct Solution:
```
n,m,k=map(int,input().split())
l=set(map(int,input().split()))
i=1
for _ in range(k):
u,v=map(int,input().split())
if i in l: break
if i==u: i=v
elif i==v: i=u
print(i)
# Made By Mostafa_Khaled
``` | output | 1 | 77,242 | 19 | 154,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,243 | 19 | 154,486 |
Tags: implementation
Correct Solution:
```
def solve():
res = 1
n,m,k = map(int,input().split())
holes = [False]*(n+1)
tmp = list(map(int,input().split()))
for i in tmp:
holes[i] = True
if holes[1] == True:
return 1
else:
while k != 0:
k = k-1
a,b = map(int,input().split())
if holes[res] == True:
continue
if a == res:
res = b
elif b == res:
res = a
return res
print(solve())
``` | output | 1 | 77,243 | 19 | 154,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground. | instruction | 0 | 77,244 | 19 | 154,488 |
Tags: implementation
Correct Solution:
```
def inp(n):
if n == 1:
return map(int, stdin.readline().split())
elif n == 2:
return map(float, stdin.readline().split())
else:
return map(str, stdin.readline().split())
def arr_inp():
return [int(x) for x in stdin.readline().split()]
def arr_2d(n):
return [[int(x) for x in stdin.readline().split()] for i in range(n)]
from sys import stdin
from collections import defaultdict
n, m, k = inp(1)
h, swap, bone = arr_inp(), arr_2d(k), 1
h = defaultdict(int, {h[i]: 1 for i in range(m)})
if h[bone]:
exit(print(bone))
for i in swap:
if i[0] == bone:
bone = i[1]
elif i[1] == bone:
bone = i[0]
if h[bone]:
exit(print(bone))
print(bone)
``` | output | 1 | 77,244 | 19 | 154,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
n,m,k=map(int,input().split())
h=set(int(x) for x in input().split())
b = 1
In=[]
for i in range(k):
u,v = map(int,input().split())
p= u
u = v
v = p
if b in h:
In.append(b)
break
if u == b :
b = v
elif v == b:
b = u
if In==[]:
print(b)
else:
print(In[0])
``` | instruction | 0 | 77,245 | 19 | 154,490 |
Yes | output | 1 | 77,245 | 19 | 154,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
n, m, k = map(int, input().split())
otv = set(input().split())
x = '1'
for _ in range(k):
if x in otv:
break
o, p = input().split()
if x == o:
x = p
elif x == p:
x = o
print(x)
``` | instruction | 0 | 77,246 | 19 | 154,492 |
Yes | output | 1 | 77,246 | 19 | 154,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
n,m,k=map(int,input().split())
h={int(x) for x in input().split()}
p=1
for t in range(k):
u,v=map(int,input().split())
if p in h: break
elif p==u: p=v
elif p==v: p=u
print(p)
``` | instruction | 0 | 77,247 | 19 | 154,494 |
Yes | output | 1 | 77,247 | 19 | 154,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
[n,m,k] = [int(i) for i in input().split()]
arr = [0 for i in range(n)]
for i in input().split():
arr[int(i) - 1] = 1
current = int(1)
hole = False
if(arr[0] == 0):
for i in range(k):
[a,b] = [int(p) for p in input().split()]
if(a == current or b == current):
if(a == current and not hole):
current = b
if(arr[b - 1] != 0):
hole = True
elif (not hole):
current = a
if(arr[a - 1] != 0):
hole = True
print(current)
``` | instruction | 0 | 77,248 | 19 | 154,496 |
Yes | output | 1 | 77,248 | 19 | 154,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
x = input().split(' ')
y = input().split(' ')
x = list(map(int, x))
y = list(map(int, y))
a = [True] * x[0]
pair = []
for i in range(len(y)):
a[y[i]-1] = False
for i in range(x[2]):
z = input().split(' ')
z = list(map(int, z))
pair.append(z)
print(pair)
if a[0] == False:
#if initial position is a hole, the bone falls, don't need check the swap
print(1)
else:
# the position of the bone
cur = 1
for i in range(len(pair)):
# swap cup has bone
if pair[i][0] == cur:
#print(pair[i][1])
cur = pair[i][1]
#print(cur)
if a[cur-1] == False:
# there is a hole, don't need to check next
break
elif pair[i][1] == cur:
#print(pair[i][0])
cur = pair[i][0]
#print(cur)
if a[cur-1] == False:
# there is a hole, don't need to check next
break
print(cur)
``` | instruction | 0 | 77,249 | 19 | 154,498 |
No | output | 1 | 77,249 | 19 | 154,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
n, m, k = [int(x) for x in input().strip().split(' ')]
holes = set([int(x) for x in input().strip().split(' ')])
bone = 1
fall = False
for _ in range(k):
a, b = [int(x) for x in input().strip().split(' ')]
if a == bone and not fall:
bone = b
if not fall and bone in holes:
fall = True
print(bone)
``` | instruction | 0 | 77,250 | 19 | 154,500 |
No | output | 1 | 77,250 | 19 | 154,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
n, m, k = map(int, input().split())
A = set(map(int, input().split()))
for i in range(k):
x, y = map(int, input().split())
if x in A:
print(x)
break
x = y
else:
print(x)
``` | instruction | 0 | 77,251 | 19 | 154,502 |
No | output | 1 | 77,251 | 19 | 154,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Zane the wizard is going to perform a magic show shuffling the cups.
There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.
The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.
Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.
<image>
Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.
The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.
Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.
Output
Print one integer — the final position along the x-axis of the bone.
Examples
Input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
Output
1
Input
5 1 2
2
1 2
2 4
Output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.
In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
Submitted Solution:
```
numberOfHoles, numberOfCups, numberOfSwaps = [int(x) for x in input().split(' ')]
positionsOfHoles = set([int(x) for x in input().split(' ')])
positionOfBone = 1
isInHole = False
if (positionOfBone in positionsOfHoles):
print(positionOfBone)
for x in range(numberOfSwaps):
positionOne, positionTwo = [int(x) for x in input().split(' ')]
if positionOfBone == positionOne:
positionOfBone = positionTwo
if positionOfBone in positionsOfHoles:
break
if positionOfBone == positionTwo:
positionOfBone = positionOne
if positionOfBone in positionsOfHoles:
break
print(positionOfBone)
``` | instruction | 0 | 77,252 | 19 | 154,504 |
No | output | 1 | 77,252 | 19 | 154,505 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,421 | 19 | 154,842 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10000000)
from math import ceil
def grundy(a, k):
if a < k:
return 0
d, m = divmod(a, k)
if m == 0:
return d
d1 = d + 1
if d > k:
return grundy(a - d1, k)
e = ceil(m / d1) * d1
return grundy(a - e, k)
n = int(input())
nim = 0
for _ in range(n):
a, k = map(int, input().split())
nim ^= grundy(a, k)
print('Takahashi' if nim > 0 else 'Aoki')
``` | output | 1 | 77,421 | 19 | 154,843 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,422 | 19 | 154,844 |
"Correct Solution:
```
def main():
import sys
input = sys.stdin.buffer.readline
N = int(input())
win = 0
for _ in range(N):
a, k = map(int, input().split())
while True:
if a < k:
g = 0
break
elif a % k == 0:
g = a // k
break
else:
d = (a//k) + 1
a -= d * ((a - k * (a//k))//d)
if a < k:
g = 0
break
elif a % k == 0:
g = a // k
break
else:
a -= d
win ^= g
if win:
print('Takahashi')
else:
print('Aoki')
if __name__ == '__main__':
main()
``` | output | 1 | 77,422 | 19 | 154,845 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,423 | 19 | 154,846 |
"Correct Solution:
```
def main():
n = int(input())
res = 0
for (a, k) in [map(int, input().split()) for _ in range(n)]:
while a > k and a % k != 0:
m = a // k
x = (a - k * m - 1) // (m + 1)
a -= max(1, x) * (m + 1)
res ^= a // k
print('Takahashi' if res != 0 else 'Aoki')
if __name__ == '__main__':
main()
``` | output | 1 | 77,423 | 19 | 154,847 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,424 | 19 | 154,848 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7
def solve(n, k):
if n % k == 0:
return n // k
diff = (n // k) + 1
target = n - n % k
if (n - target) % diff == 0:
return solve(target, k)
n -= -((-(n - target)) // diff) * diff
return solve(n, k)
n = I()
ret = 0
for i in range(n):
a, k = LI()
ret ^= solve(a, k)
print("Takahashi" if ret else "Aoki")
``` | output | 1 | 77,424 | 19 | 154,849 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,425 | 19 | 154,850 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
AK = [list(map(int, input().split())) for _ in range(N)]
# N = 40
# AK = [(i+1, 4) for i in range(N)]
def grundy(A, K):
if K == 1:
return A
X = (A+K-1)//K
num = X*K - A
while X*K-num > K*K:
#print(X, num)
if num%K == 0:
return X - num//K
if num//K == X//K:
num = num%K
else:
num = X%K+1 + (K-1)*(num//K) + (num%K-1)
X = X - X//K
S = X*K - num
while S >= K:
if S%K == 0:
return S//K
delta = S//K + 1
S -= max(((S%K)//delta)*delta, delta)
return 0
g = 0
for A, K in AK:
# print(A, A//K, grundy(A, K))
# print()
g ^= grundy(A, K)
print("Takahashi" if g else "Aoki")
``` | output | 1 | 77,425 | 19 | 154,851 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,426 | 19 | 154,852 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7
def f(ai, ki):
if ai < ki: return 0
elif ai % ki == 0 :return ai // ki
d = ai // ki + 1
if ai % ki % d:
return f(ai - (ai % ki // d + 1) * d, ki)
return f(ai - ai % k, k)
n = I()
ret = 0
for i in range(n):
a, k = LI()
ret ^= f(a, k)
print("Takahashi" if ret else "Aoki")
``` | output | 1 | 77,426 | 19 | 154,853 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,427 | 19 | 154,854 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 8)
n = int(input())
info = [list(map(int, input().split())) for i in range(n)]
# 愚直解
def mex(array):
if not array:
return 0
set_ = set(array)
for i in range(10000):
if i not in set_:
return i
"""
res = 0
for a, k in info:
grundy = [0] * (a + 1)
for num in range(1, a + 1):
length = num // k
grundy[num] = mex(grundy[num - length:num])
res ^= grundy[-1]
if res == 0:
print("Aoki")
else:
print("Takahashi")
"""
# a(3n)=n, otherwise a(n)=a(floor(2n/3))
"""
def solve(n, k):
if n % k == 0:
return n // k
return solve(n * (k - 1) // k, k)
"""
def solve(n, k):
if n % k == 0:
return n // k
diff = (n // k) + 1
target = n - n % k
if (n - target) % diff == 0:
return solve(target, k)
n -= -((-(n - target)) // diff) * diff
return solve(n, k)
res = 0
for a, k in info:
res ^= solve(a, k)
if res == 0:
print("Aoki")
else:
print("Takahashi")
``` | output | 1 | 77,427 | 19 | 154,855 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi | instruction | 0 | 77,428 | 19 | 154,856 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**6)
def grun(A, K):
if A < K:
return 0
a, b = divmod(A, K)
if b == 0:
return a
a += 1
r = A % a
return grun((((a-1)*K-r)//a)*a+r, K)
def solve():
N = int(input())
X = 0
for _ in range(N):
A, K = map(int, input().split())
X ^= grun(A, K)
print('Takahashi' if X else 'Aoki')
if __name__ == '__main__':
solve()
``` | output | 1 | 77,428 | 19 | 154,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi and Aoki are playing a stone-taking game. Initially, there are N piles of stones, and the i-th pile contains A_i stones and has an associated integer K_i.
Starting from Takahashi, Takahashi and Aoki take alternate turns to perform the following operation:
* Choose a pile. If the i-th pile is selected and there are X stones left in the pile, remove some number of stones between 1 and floor(X/K_i) (inclusive) from the pile.
The player who first becomes unable to perform the operation loses the game. Assuming that both players play optimally, determine the winner of the game. Here, floor(x) represents the largest integer not greater than x.
Constraints
* 1 \leq N \leq 200
* 1 \leq A_i,K_i \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 K_1
:
A_N K_N
Output
If Takahashi will win, print `Takahashi`; if Aoki will win, print `Aoki`.
Examples
Input
2
5 2
3 3
Output
Aoki
Input
3
3 2
4 3
5 1
Output
Takahashi
Input
3
28 3
16 4
19 2
Output
Aoki
Input
4
3141 59
26535 897
93 23
8462 64
Output
Takahashi
Submitted Solution:
```
"""
各山のgrundy数がわかればよい
基本的にループする
"""
N = int(input())
ans = 0
for i in range(N):
A,K = map(int,input().split())
while True:
if A % K == 0:
ans ^= A // K
break
elif A == K-1:
ans ^= 0
break
else:
if A // K != (A - A//K -1) // K:
A = A - A // K - 1
else:
A -= (A - (A // K) * K) // (A // K+1) * (A//K+1)
if ans == 0:
print ("Aoki")
else:
print ("Takahashi")
``` | instruction | 0 | 77,429 | 19 | 154,858 |
Yes | output | 1 | 77,429 | 19 | 154,859 |
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