message
stringlengths 2
39.6k
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stringclasses 2
values | message_id
int64 0
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int64 219
108k
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float64 11
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int64 438
217k
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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
X, Y = map(int, input().split())
a = "NO"
for i in range(1, X+1):
if i*2 + (X - i)*4 == Y:
a = "YES"
break
else:
continue
print(a)
```
|
instruction
| 0
| 107,634
| 11
| 215,268
|
No
|
output
| 1
| 107,634
| 11
| 215,269
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
X,Y = map(int, input().split())
if X == 1 and (Y == 4 or Y == 2):
print("Yes")
exit(0)
for i in range(X):
idx=i+1
if (2*idx + (X-idx)*4 == Y):
print('Yes')
exit(0)
print('No')
```
|
instruction
| 0
| 107,635
| 11
| 215,270
|
No
|
output
| 1
| 107,635
| 11
| 215,271
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
x, y = map(int, input().split())
if y >= 2*x and y <= 4*x and y % 2 == 0
print("Yes")
else:
print("No")
```
|
instruction
| 0
| 107,636
| 11
| 215,272
|
No
|
output
| 1
| 107,636
| 11
| 215,273
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
x,y = map(int,input().split())
a = 4 * x - y
if a/2 == a//2:
print('Yes')
else:
print('No')
```
|
instruction
| 0
| 107,637
| 11
| 215,274
|
No
|
output
| 1
| 107,637
| 11
| 215,275
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Nastia has a hidden permutation p of length n consisting of integers from 1 to n. You, for some reason, want to figure out the permutation. To do that, you can give her an integer t (1 ≤ t ≤ 2), two different indices i and j (1 ≤ i, j ≤ n, i ≠ j), and an integer x (1 ≤ x ≤ n - 1).
Depending on t, she will answer:
* t = 1: max{(min{(x, p_i)}, min{(x + 1, p_j)})};
* t = 2: min{(max{(x, p_i)}, max{(x + 1, p_j)})}.
You can ask Nastia at most ⌊ \frac {3 ⋅ n} { 2} ⌋ + 30 times. It is guaranteed that she will not change her permutation depending on your queries. Can you guess the permutation?
Input
The input consists of several test cases. In the beginning, you receive the integer T (1 ≤ T ≤ 10 000) — the number of test cases.
At the beginning of each test case, you receive an integer n (3 ≤ n ≤ 10^4) — the length of the permutation p.
It's guaranteed that the permutation is fixed beforehand and that the sum of n in one test doesn't exceed 2 ⋅ 10^4.
Interaction
To ask a question, print "? t i j x" (t = 1 or t = 2, 1 ≤ i, j ≤ n, i ≠ j, 1 ≤ x ≤ n - 1) Then, you should read the answer.
If we answer with −1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query. Exit immediately after receiving −1 and you will see the Wrong Answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_{n} (without quotes). Note that answering doesn't count as one of the ⌊ \frac {3 ⋅ n} {2} ⌋ + 30 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks
To hack the solution, use the following test format.
The first line should contain a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.
For each test case in the first line print a single integer n (3 ≤ n ≤ 10^4) — the length of the hidden permutation p.
In the second line print n space-separated integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n), where p is permutation.
Note that the sum of n over all test cases should not exceed 2 ⋅ 10^4.
Example
Input
2
4
3
2
5
3
Output
? 2 4 1 3
? 1 2 4 2
! 3 1 4 2
? 2 3 4 2
! 2 5 3 4 1
Note
Consider the first test case.
The hidden permutation is [3, 1, 4, 2].
We print: "? 2 4 1 3" and get back min{(max{(3, p_4}), max{(4, p_1)})} = 3.
We print: "? 1 2 4 2" and get back max{(min{(2, p_2)}, min{(3, p_4)})} = 2.
Consider the second test case.
The hidden permutation is [2, 5, 3, 4, 1].
We print: "? 2 3 4 2" and get back min{(max{(2, p_3}), max{(3, p_4)})} = 3.
Submitted Solution:
```
import sys
from collections import defaultdict
from math import ceil, floor, log10, log2
def solve():
ans = [-1 for i in range(n + 1)]
ans[0] = '!'
i = 1
maxxind = ceil(n / 2)
while i <= (n // 2):
print('?', 1, i, n - i + 1, n - 1, flush=True)
rep = int(input())
if rep == n:
maxxind = n - i + 1
ans[n - i + 1] = n
break
if rep == n - 1:
print('?', 1, n - i + 1, i, n - 1, flush=True)
rep = int(input())
if rep == n:
maxind = n - i + 1
ans[i] = n
break
i += 1
for i in range(1, n + 1):
if ans[i] == -1:
print('?', 2, i, maxxind, 1, flush=True)
rep = int(input())
ans[i] = rep
res = ' '.join(map(str, ans))
print(res, flush=True)
return
if __name__ == '__main__':
T = int(input())
for t in range(1, T + 1):
n = int(input())
solve()
```
|
instruction
| 0
| 108,094
| 11
| 216,188
|
No
|
output
| 1
| 108,094
| 11
| 216,189
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Nastia has a hidden permutation p of length n consisting of integers from 1 to n. You, for some reason, want to figure out the permutation. To do that, you can give her an integer t (1 ≤ t ≤ 2), two different indices i and j (1 ≤ i, j ≤ n, i ≠ j), and an integer x (1 ≤ x ≤ n - 1).
Depending on t, she will answer:
* t = 1: max{(min{(x, p_i)}, min{(x + 1, p_j)})};
* t = 2: min{(max{(x, p_i)}, max{(x + 1, p_j)})}.
You can ask Nastia at most ⌊ \frac {3 ⋅ n} { 2} ⌋ + 30 times. It is guaranteed that she will not change her permutation depending on your queries. Can you guess the permutation?
Input
The input consists of several test cases. In the beginning, you receive the integer T (1 ≤ T ≤ 10 000) — the number of test cases.
At the beginning of each test case, you receive an integer n (3 ≤ n ≤ 10^4) — the length of the permutation p.
It's guaranteed that the permutation is fixed beforehand and that the sum of n in one test doesn't exceed 2 ⋅ 10^4.
Interaction
To ask a question, print "? t i j x" (t = 1 or t = 2, 1 ≤ i, j ≤ n, i ≠ j, 1 ≤ x ≤ n - 1) Then, you should read the answer.
If we answer with −1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query. Exit immediately after receiving −1 and you will see the Wrong Answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_{n} (without quotes). Note that answering doesn't count as one of the ⌊ \frac {3 ⋅ n} {2} ⌋ + 30 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks
To hack the solution, use the following test format.
The first line should contain a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.
For each test case in the first line print a single integer n (3 ≤ n ≤ 10^4) — the length of the hidden permutation p.
In the second line print n space-separated integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n), where p is permutation.
Note that the sum of n over all test cases should not exceed 2 ⋅ 10^4.
Example
Input
2
4
3
2
5
3
Output
? 2 4 1 3
? 1 2 4 2
! 3 1 4 2
? 2 3 4 2
! 2 5 3 4 1
Note
Consider the first test case.
The hidden permutation is [3, 1, 4, 2].
We print: "? 2 4 1 3" and get back min{(max{(3, p_4}), max{(4, p_1)})} = 3.
We print: "? 1 2 4 2" and get back max{(min{(2, p_2)}, min{(3, p_4)})} = 2.
Consider the second test case.
The hidden permutation is [2, 5, 3, 4, 1].
We print: "? 2 3 4 2" and get back min{(max{(2, p_3}), max{(3, p_4)})} = 3.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
for t in range(int(input())):
n=int(input())
p=[10**5]*n
if n==3:
print('?',1,1,2,n-1,flush=True)
a=int(input())
print('?',2,1,2,1,flush=True)
b=int(input())
print('?',1,2,3,n-1,flush=True)
c=int(input())
print('?',2,2,3,1,flush=True)
d=int(input())
if b==d:
p[1]=1
x=a
p[0]=x
if x==2:
p[2]=3
else:
p[2]=2
elif a==c:
p[1]=3
x=b
p[0]=x
if x==1:
p[2]=2
else:
p[2]=1
else:
if b==2:
p[0]=3
p[1]=2
p[2]=1
else:
p[0]=1
p[1]=2
p[2]=3
else:
if n%2:
x=n-1
else:
x=n
l=[]
for i in range(0,x,2):
print('?',1,i+1,i+2,n-1,flush=True)
a=int(input())
print('?',2,i+1,i+2,1,flush=True)
b=int(input())
l.append([b,a])
for i in range(len(l)-1):
if l[i][0]<l[i+1][0]:
print('?',2,2*i+1,2*i+3,1,flush=True)
a=int(input())
if a==l[i][0]:
p[2*i]=a
p[2*i+1]=l[i][1]
else:
p[2*i]=l[i][1]
p[2*i+1]=l[i][0]
elif l[i][0]<l[i+1][1]:
print('?',2,2*i+1,2*i+4,1,flush=True)
a=int(input())
if a==l[i][0]:
p[2*i]=l[i][0]
p[2*i+1]=l[i][1]
else:
p[2*i]=l[i][1]
p[2*i+1]=l[i][0]
else:
print('?',1,2*i+1,2*i+3,n-1,flush=True)
a=int(input())
if a==l[i][0]:
p[2*i]=a
p[2*i+1]=l[i][1]
else:
p[2*i]=l[i][1]
p[2*i+1]=l[i][0]
a=min(p)
b=p.index(a)
if a==1:
print('?',1,b+1,x-1,n-1,flush=True)
c=int(input())
p[x-2]=c
print('?',1,b+1,x,n-1,flush=True)
c=int(input())
p[x-1]=c
elif a==2:
print('?',1,b+1,x-1,n-1,flush=True)
c=int(input())
if c!=2:
p[x-2]=c
if c==l[-1][0]:
p[x-1]=l[-1][1]
else:
p[x-1]=l[-1][0]
else:
p[x-2]=1
p[x-1]=l[-1][1]
else:
print('?',2,b+1,x-1,1,flush=True)
c=int(input())
p[x-2]=c
if c==1:
p[x-1]=2
else:
p[x-1]=1
if n%2:
a=min(p)
b=p.index(a)
if a==1:
print('?',1,b+1,n,n-1,flush=True)
c=int(input())
p[n-1]=c
else:
p[n-1]=1
print('!',*p,flush=True)
```
|
instruction
| 0
| 108,095
| 11
| 216,190
|
No
|
output
| 1
| 108,095
| 11
| 216,191
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Nastia has a hidden permutation p of length n consisting of integers from 1 to n. You, for some reason, want to figure out the permutation. To do that, you can give her an integer t (1 ≤ t ≤ 2), two different indices i and j (1 ≤ i, j ≤ n, i ≠ j), and an integer x (1 ≤ x ≤ n - 1).
Depending on t, she will answer:
* t = 1: max{(min{(x, p_i)}, min{(x + 1, p_j)})};
* t = 2: min{(max{(x, p_i)}, max{(x + 1, p_j)})}.
You can ask Nastia at most ⌊ \frac {3 ⋅ n} { 2} ⌋ + 30 times. It is guaranteed that she will not change her permutation depending on your queries. Can you guess the permutation?
Input
The input consists of several test cases. In the beginning, you receive the integer T (1 ≤ T ≤ 10 000) — the number of test cases.
At the beginning of each test case, you receive an integer n (3 ≤ n ≤ 10^4) — the length of the permutation p.
It's guaranteed that the permutation is fixed beforehand and that the sum of n in one test doesn't exceed 2 ⋅ 10^4.
Interaction
To ask a question, print "? t i j x" (t = 1 or t = 2, 1 ≤ i, j ≤ n, i ≠ j, 1 ≤ x ≤ n - 1) Then, you should read the answer.
If we answer with −1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query. Exit immediately after receiving −1 and you will see the Wrong Answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_{n} (without quotes). Note that answering doesn't count as one of the ⌊ \frac {3 ⋅ n} {2} ⌋ + 30 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks
To hack the solution, use the following test format.
The first line should contain a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.
For each test case in the first line print a single integer n (3 ≤ n ≤ 10^4) — the length of the hidden permutation p.
In the second line print n space-separated integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n), where p is permutation.
Note that the sum of n over all test cases should not exceed 2 ⋅ 10^4.
Example
Input
2
4
3
2
5
3
Output
? 2 4 1 3
? 1 2 4 2
! 3 1 4 2
? 2 3 4 2
! 2 5 3 4 1
Note
Consider the first test case.
The hidden permutation is [3, 1, 4, 2].
We print: "? 2 4 1 3" and get back min{(max{(3, p_4}), max{(4, p_1)})} = 3.
We print: "? 1 2 4 2" and get back max{(min{(2, p_2)}, min{(3, p_4)})} = 2.
Consider the second test case.
The hidden permutation is [2, 5, 3, 4, 1].
We print: "? 2 3 4 2" and get back min{(max{(2, p_3}), max{(3, p_4)})} = 3.
Submitted Solution:
```
#!/usr/bin/python3
import sys
input = sys.stdin.readline
import random
def get_small(i, j):
print("?", 2, i, j, 1, flush=True)
ret1 = int(input())
print("?", 2, j, i, 1, flush=True)
ret2 = int(input())
return min(ret1, ret2)
def get_large(i, j):
print("?", 1, i, j, n-1, flush=True)
ret1 = int(input())
print("?", 1, j, i, n-1, flush=True)
ret2 = int(input())
return max(ret1, ret2)
t = int(input())
for _ in range(t):
n = int(input())
v12 = []
v12.append(get_small(1, 2))
v12.append(get_large(1, 2))
v13 = []
v13.append(get_small(1, 3))
v13.append(get_large(1, 3))
if v12[0] in v13:
if v12[1] != 1 and v12[1] != n:
key = v12[1]
base = 0
else:
key = v12[0]
base = 1
else:
if v12[0] != 1 and v12[0] != n:
key = v12[0]
base = 0
else:
key = v12[1]
base = 1
key_is_big = (key >= n // 2)
ans = [None] * n
ans[base] = key
for i in range(n):
if i == base:
continue
if key_is_big:
print("?", 2, base+1, i+1, 1, flush=True)
ret = int(input())
if ret != key:
ans[i] = ret
continue
print("?", 1, base+1, i+1, n-1, flush=True)
ret = int(input())
if ret != key:
ans[i] = ret
else:
print("?", 1, base+1, i+1, n-1, flush=True)
ret = int(input())
if ret != key:
ans[i] = ret
continue
print("?", 2, base+1, i+1, 1, flush=True)
ret = int(input())
if ret != key:
ans[i] = ret
print("!", *ans)
exit()
```
|
instruction
| 0
| 108,096
| 11
| 216,192
|
No
|
output
| 1
| 108,096
| 11
| 216,193
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Nastia has a hidden permutation p of length n consisting of integers from 1 to n. You, for some reason, want to figure out the permutation. To do that, you can give her an integer t (1 ≤ t ≤ 2), two different indices i and j (1 ≤ i, j ≤ n, i ≠ j), and an integer x (1 ≤ x ≤ n - 1).
Depending on t, she will answer:
* t = 1: max{(min{(x, p_i)}, min{(x + 1, p_j)})};
* t = 2: min{(max{(x, p_i)}, max{(x + 1, p_j)})}.
You can ask Nastia at most ⌊ \frac {3 ⋅ n} { 2} ⌋ + 30 times. It is guaranteed that she will not change her permutation depending on your queries. Can you guess the permutation?
Input
The input consists of several test cases. In the beginning, you receive the integer T (1 ≤ T ≤ 10 000) — the number of test cases.
At the beginning of each test case, you receive an integer n (3 ≤ n ≤ 10^4) — the length of the permutation p.
It's guaranteed that the permutation is fixed beforehand and that the sum of n in one test doesn't exceed 2 ⋅ 10^4.
Interaction
To ask a question, print "? t i j x" (t = 1 or t = 2, 1 ≤ i, j ≤ n, i ≠ j, 1 ≤ x ≤ n - 1) Then, you should read the answer.
If we answer with −1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query. Exit immediately after receiving −1 and you will see the Wrong Answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_{n} (without quotes). Note that answering doesn't count as one of the ⌊ \frac {3 ⋅ n} {2} ⌋ + 30 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks
To hack the solution, use the following test format.
The first line should contain a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.
For each test case in the first line print a single integer n (3 ≤ n ≤ 10^4) — the length of the hidden permutation p.
In the second line print n space-separated integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n), where p is permutation.
Note that the sum of n over all test cases should not exceed 2 ⋅ 10^4.
Example
Input
2
4
3
2
5
3
Output
? 2 4 1 3
? 1 2 4 2
! 3 1 4 2
? 2 3 4 2
! 2 5 3 4 1
Note
Consider the first test case.
The hidden permutation is [3, 1, 4, 2].
We print: "? 2 4 1 3" and get back min{(max{(3, p_4}), max{(4, p_1)})} = 3.
We print: "? 1 2 4 2" and get back max{(min{(2, p_2)}, min{(3, p_4)})} = 2.
Consider the second test case.
The hidden permutation is [2, 5, 3, 4, 1].
We print: "? 2 3 4 2" and get back min{(max{(2, p_3}), max{(3, p_4)})} = 3.
Submitted Solution:
```
'''Author- Akshit Monga'''
from sys import stdin, stdout
input = stdin.readline
def query(k,i,j,x):
print('?',k,i,j,x,flush=True)
val=int(input())
return val
def find1(ind):
for i in range(1,n+1):
if i==ind or arr[i]!=-1:
continue
arr[i]=query(2,i,ind,1)
# print('!',*arr,flush=True)
def find2(ind):
for i in range(1,n+1):
if i==ind or arr[i]!=-1:
continue
arr[i]=query(1,ind,i,n-1)
# print('!',*arr,flush=True)
t = int(input())
for _ in range(t):
n=int(input())
arr=[-1 for i in range(n+1)]
for j in range(1,n+1,3):
i=j-1
val1 = query(1, i+1, i+2, n - 1)
if val1 == n - 1:
val2 = query(1, i+2, i+1, n - 1)
if val2 == n:
arr[i+1] = n
if arr[i+1] == n:
find1(i+1)
break
val3 = query(2, i+1, i+2, 1)
if val3 == 2:
val4 = query(2, i+2, i+1, 1)
if val4 == 1:
arr[i+2] = 1
if arr[i+2] == 1:
find2(i+2)
break
val5 = query(1, i+2, i+3, n - 1)
if val5 == n - 1:
val6 = query(1, i+3, i+2, n - 1)
if val6 == n:
arr[i+2] = n
if arr[i+2] == n:
find1(i+2)
break
val7 = query(2, i+2, i+3, 1)
if val7 == 2:
val8 = query(2, i+3, i+2, 1)
if val8 == 1:
arr[i+3] = 1
if arr[i+3] == 1:
find2(i+3)
break
# val1 val3 val5 val7
p = [val1, val3]
q = [val5, val7]
for k in p:
for j in q:
if k == j:
arr[i+2] = k
break
if arr[i+2] != -1:
break
for k in p:
if k != arr[i+2]:
arr[i+1] = k
break
for k in q:
if k != arr[i+2]:
arr[i+3] = k
break
for i in range(1,n+1):
if arr[i]==-1:
val1=query(1,1,i,n-1)
if val1!=arr[1]:
arr[i]=val1
else:
val2=query(2,i,1,1)
arr[i]=val2
print('!',*arr[1:],flush=True)
```
|
instruction
| 0
| 108,097
| 11
| 216,194
|
No
|
output
| 1
| 108,097
| 11
| 216,195
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chris the Rabbit has been interested in arrays ever since he was a child. At the moment he is researching arrays with the length of n, containing only integers from 1 to n. He is not good at math, that's why some simple things drive him crazy. For example, yesterday he grew keen on counting how many different beautiful arrays there are. Chris thinks that an array is beautiful if it meets one of the two conditions:
* each elements, starting from the second one, is no more than the preceding one
* each element, starting from the second one, is no less than the preceding one
Having got absolutely mad at himself and at math, Chris came to Stewie and Brian to ask them for help. However, they only laughed at him and said that the answer is too simple and not interesting. Help Chris the Rabbit to find the answer at last.
Input
The single line contains an integer n which is the size of the array (1 ≤ n ≤ 105).
Output
You must print the answer on a single line. As it can be rather long, you should print it modulo 1000000007.
Examples
Input
2
Output
4
Input
3
Output
17
Submitted Solution:
```
def fact(n):
if n==1:
return 1
return fact(n-1)*n%(10**9+7)
def rev(a):
return pow(a,10**9+5,10**9+7)
def c2nn(n):
return fact(2*n)*rev(fact(n)**2)%(10**9+7)
n=int(input())
print(c2nn(n)*2-n)
```
|
instruction
| 0
| 108,255
| 11
| 216,510
|
No
|
output
| 1
| 108,255
| 11
| 216,511
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While Farmer John rebuilds his farm in an unfamiliar portion of Bovinia, Bessie is out trying some alternative jobs. In her new gig as a reporter, Bessie needs to know about programming competition results as quickly as possible. When she covers the 2016 Robot Rap Battle Tournament, she notices that all of the robots operate under deterministic algorithms. In particular, robot i will beat robot j if and only if robot i has a higher skill level than robot j. And if robot i beats robot j and robot j beats robot k, then robot i will beat robot k. Since rapping is such a subtle art, two robots can never have the same skill level.
Given the results of the rap battles in the order in which they were played, determine the minimum number of first rap battles that needed to take place before Bessie could order all of the robots by skill level.
Input
The first line of the input consists of two integers, the number of robots n (2 ≤ n ≤ 100 000) and the number of rap battles m (<image>).
The next m lines describe the results of the rap battles in the order they took place. Each consists of two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), indicating that robot ui beat robot vi in the i-th rap battle. No two rap battles involve the same pair of robots.
It is guaranteed that at least one ordering of the robots satisfies all m relations.
Output
Print the minimum k such that the ordering of the robots by skill level is uniquely defined by the first k rap battles. If there exists more than one ordering that satisfies all m relations, output -1.
Examples
Input
4 5
2 1
1 3
2 3
4 2
4 3
Output
4
Input
3 2
1 2
3 2
Output
-1
Note
In the first sample, the robots from strongest to weakest must be (4, 2, 1, 3), which Bessie can deduce after knowing the results of the first four rap battles.
In the second sample, both (1, 3, 2) and (3, 1, 2) are possible orderings of the robots from strongest to weakest after both rap battles.
Submitted Solution:
```
def main():
line = input().split(" ")
n = int(line[0])
m = int(line[1])
min_count = -1
d = {}
d_win = {}
q = []
waiting = []
for x in range(m):
# новая строка
line = input().split(" ")
rob_win = int(line[0])
rob_lose = int(line[1])
# словарь
losers = d.get(rob_win)
if losers is None:
losers = list()
losers.append(rob_lose)
d.update({rob_win: losers})
else:
losers.append(rob_lose)
d.update({rob_win: losers})
# словарь для лузера
winners = d_win.get(rob_lose)
if winners is None:
winners = list()
winners.append(rob_win)
d_win.update({rob_lose: winners})
else:
winners.append(rob_win)
d_win.update({rob_lose: winners})
# записываем в массив
if (rob_lose in q) and (rob_win in q):
continue
elif rob_lose in q:
index = q.index(rob_lose) - 1 # индексы вигравших у rob_lose
if index == -1:
q.insert(0, rob_win)
else:
while index != -1:
# проверяем, есть ли элемент в списке побежденных для нашего winner'a
if rob_win in d.get(q[index]):
# записываем после найденного элемента
q.insert(index + 1, rob_win)
break
elif q[index] in losers:
# если конец очереди, вставляем число
if index == 0:
q.insert(index, rob_win)
break
continue # наш элемент больше, ищем дальше
else:
waiting.append(rob_win) # недостаточно данных, в ждущую очередь
break
elif rob_win in q:
index = q.index(rob_win) + 1 # индексы проигравших у rob_win
if index == len(q):
q.append(rob_lose)
else:
while index != len(q):
# проверяем, есть ли элемент в списке выигравших для нашего loser'a
if rob_lose in d_win.get(q[index]):
# записываем перед найденным элементом
q.insert(index - 1, rob_lose)
break
elif q[index] in winners:
# если конец очереди, вставляем число
if index == len(q) - 1:
q.insert(index, rob_lose)
break
continue # наш элемент меньше, ищем дальше
else:
waiting.append(rob_lose) # недостаточно данных,
break
elif x == 0:
# победители -> проигравшие
q.append(rob_win)
q.append(rob_lose)
else:
# в очередь
waiting.append(rob_win)
waiting.append(rob_lose)
# проверяем ждущих
for waits in waiting:
if waits in q:
continue
winners = d_win.get(waits) # waits им проиграл
losers = d.get(waits) # waits их победил
index = len(q) - 2
while index != -1:
if q[index] in losers:
if q[index + 1] in winners:
q.insert(index + 1, waits)
waiting.pop(waiting.index(waits))
break
index -= 1
else:
break
# проверяем, есть ли заполнение
if len(q) == n:
min_count = x + 1
break
# вывод
if len(waiting) > 0:
print(-1)
else:
print(min_count)
print(q)
main()
```
|
instruction
| 0
| 108,268
| 11
| 216,536
|
No
|
output
| 1
| 108,268
| 11
| 216,537
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While Farmer John rebuilds his farm in an unfamiliar portion of Bovinia, Bessie is out trying some alternative jobs. In her new gig as a reporter, Bessie needs to know about programming competition results as quickly as possible. When she covers the 2016 Robot Rap Battle Tournament, she notices that all of the robots operate under deterministic algorithms. In particular, robot i will beat robot j if and only if robot i has a higher skill level than robot j. And if robot i beats robot j and robot j beats robot k, then robot i will beat robot k. Since rapping is such a subtle art, two robots can never have the same skill level.
Given the results of the rap battles in the order in which they were played, determine the minimum number of first rap battles that needed to take place before Bessie could order all of the robots by skill level.
Input
The first line of the input consists of two integers, the number of robots n (2 ≤ n ≤ 100 000) and the number of rap battles m (<image>).
The next m lines describe the results of the rap battles in the order they took place. Each consists of two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), indicating that robot ui beat robot vi in the i-th rap battle. No two rap battles involve the same pair of robots.
It is guaranteed that at least one ordering of the robots satisfies all m relations.
Output
Print the minimum k such that the ordering of the robots by skill level is uniquely defined by the first k rap battles. If there exists more than one ordering that satisfies all m relations, output -1.
Examples
Input
4 5
2 1
1 3
2 3
4 2
4 3
Output
4
Input
3 2
1 2
3 2
Output
-1
Note
In the first sample, the robots from strongest to weakest must be (4, 2, 1, 3), which Bessie can deduce after knowing the results of the first four rap battles.
In the second sample, both (1, 3, 2) and (3, 1, 2) are possible orderings of the robots from strongest to weakest after both rap battles.
Submitted Solution:
```
n, m = list(map(int, input().split()))
forces = [[set(), set()] for i in range(n)]
goal = 0
for i in range(m):
a, b = list(map(int, input().split()))
if a not in forces[b-1][0]:
goal += 1
forces[b-1][0].add(a)
if b not in forces[a-1][1]:
forces[a-1][1].add(b)
goal += 1
for j in forces[a-1][0]:
if b not in forces[j-1][1]:
forces[j-1][1].add(b)
goal += 1
if j not in forces[b-1][0]:
goal += 1
forces[b-1][0].add(j)
for j in forces[b-1][1]:
if a not in forces[j-1][0]:
forces[j-1][0].add(a)
goal += 1
if j not in forces[a-1][1]:
forces[a-1][1].add(j)
goal += 1
if goal >= n * (n - 1):
print(i + 1)
break
if goal < n * (n - 1):
print(-1)
```
|
instruction
| 0
| 108,269
| 11
| 216,538
|
No
|
output
| 1
| 108,269
| 11
| 216,539
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While Farmer John rebuilds his farm in an unfamiliar portion of Bovinia, Bessie is out trying some alternative jobs. In her new gig as a reporter, Bessie needs to know about programming competition results as quickly as possible. When she covers the 2016 Robot Rap Battle Tournament, she notices that all of the robots operate under deterministic algorithms. In particular, robot i will beat robot j if and only if robot i has a higher skill level than robot j. And if robot i beats robot j and robot j beats robot k, then robot i will beat robot k. Since rapping is such a subtle art, two robots can never have the same skill level.
Given the results of the rap battles in the order in which they were played, determine the minimum number of first rap battles that needed to take place before Bessie could order all of the robots by skill level.
Input
The first line of the input consists of two integers, the number of robots n (2 ≤ n ≤ 100 000) and the number of rap battles m (<image>).
The next m lines describe the results of the rap battles in the order they took place. Each consists of two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), indicating that robot ui beat robot vi in the i-th rap battle. No two rap battles involve the same pair of robots.
It is guaranteed that at least one ordering of the robots satisfies all m relations.
Output
Print the minimum k such that the ordering of the robots by skill level is uniquely defined by the first k rap battles. If there exists more than one ordering that satisfies all m relations, output -1.
Examples
Input
4 5
2 1
1 3
2 3
4 2
4 3
Output
4
Input
3 2
1 2
3 2
Output
-1
Note
In the first sample, the robots from strongest to weakest must be (4, 2, 1, 3), which Bessie can deduce after knowing the results of the first four rap battles.
In the second sample, both (1, 3, 2) and (3, 1, 2) are possible orderings of the robots from strongest to weakest after both rap battles.
Submitted Solution:
```
n, m = list(map(int, input().split()))
forces = [[set(), set()] for i in range(n)]
goal = 0
for i in range(m):
a, b = list(map(int, input().split()))
if a not in forces[b-1][0]:
goal += 1
forces[b-1][0].add(a)
if b not in forces[a-1][1]:
forces[a-1][1].add(b)
goal += 1
for j in forces[a-1][0]:
for k in forces[a - 1][1]:
if k not in forces[j-1][1]:
forces[j-1][1].add(k)
goal += 1
if j not in forces[k-1][0]:
goal += 1
forces[k-1][0].add(j)
if b not in forces[j-1][1]:
forces[j-1][1].add(b)
goal += 1
if j not in forces[b-1][0]:
goal += 1
forces[b-1][0].add(j)
for j in forces[b-1][1]:
for k in forces[b-1][0]:
if k not in forces[j-1][0]:
forces[j-1][0].add(k)
goal += 1
if k not in forces[a-1][1]:
forces[a-1][1].add(k)
goal += 1
if a not in forces[j-1][0]:
forces[j-1][0].add(a)
goal += 1
if j not in forces[a-1][1]:
forces[a-1][1].add(j)
goal += 1
print(a, b, goal, forces)
if goal >= n * (n - 1):
print(i + 1)
break
if goal < n * (n - 1):
print(-1)
```
|
instruction
| 0
| 108,270
| 11
| 216,540
|
No
|
output
| 1
| 108,270
| 11
| 216,541
|
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