message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4. | instruction | 0 | 60,155 | 22 | 120,310 |
Tags: binary search, math
Correct Solution:
```
a,b,c,d=map(int,input().split());l=1;r=10**10
while r-l>1:
m=(r+l)//2
both=m//(c*d)
f=[m//c,m//d]
other=m-f[0]-f[1]+both
f[0]-=both;f[1]-=both
if max(0,b-f[0])+max(0,a-f[1])<=other:
r=m
else:
l=m
print(r)
``` | output | 1 | 60,155 | 22 | 120,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4. | instruction | 0 | 60,156 | 22 | 120,312 |
Tags: binary search, math
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def check(mid,c1,c2,x,y):
a,b,c = mid//x,mid//y,mid//(x*y)
c1 = max(0,c1-b+c)
c2 = max(0,c2-a+c)
if c1+c2 <= mid-a-b+c:
return 1
return 0
def main():
c1,c2,x,y = map(int,input().split())
hi,lo,ans = 10**10,0,10**10
while hi >= lo:
mid = (hi+lo)//2
if check(mid,c1,c2,x,y):
ans = mid
hi = mid-1
else:
lo = mid+1
print(ans)
# Fast IO Region
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")
if __name__ == "__main__":
main()
``` | output | 1 | 60,156 | 22 | 120,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4. | instruction | 0 | 60,157 | 22 | 120,314 |
Tags: binary search, math
Correct Solution:
```
__author__ = 'neki'
import sys
#sys.stdin = open("friendsPresents.in", "r")
#sys.stdout = open("friendsPresents.out", "w")
words = input().split()
cnt1 = int(words[0])
cnt2 = int(words[1])
x = int(words[2])
y = int(words[3])
#print(cnt1, cnt2, x, y)
def determine(num, x, y, cnt1, cnt2):
like1 = num - num // x
like2 = num - num // y
like1and2 = like1 + like2 - num + num // (x*y)
#print(like1, like2, like1and2)
friend1 = like1 - like1and2
friend2 = like2 - like1and2
both = like1and2
#print(num, friend1, friend2, both)
cond1 = 0
cond2 = 0
if friend1 + both >= cnt1 and friend2 + both >= cnt2:
cond1 += 1
if friend1 + friend2 + both >= cnt1 + cnt2:
cond2 += 1
if cond1 > 0 and cond2 > 0:
return 1
return 0
lim = 0
lim1 = 1
lim2 = -1
while 1:
if lim2 < 0:
lim = lim1 * 2
else:
lim = (lim1 + lim2) // 2
if lim == lim1:
break
if determine(lim, x, y, cnt1, cnt2) == 1:
lim2 = lim
else:
lim1 = lim
#print("(", lim1, lim2, ")")
print(lim2)
``` | output | 1 | 60,157 | 22 | 120,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4. | instruction | 0 | 60,158 | 22 | 120,316 |
Tags: binary search, math
Correct Solution:
```
def check(v):
a = v // y - v // (x * y)
b = v // x - v // (x * y)
c = v - (a + b + v // (x * y))
fa = c + a
fb = min(c, fa - cnt1) + b
return (fa >= cnt1) and (fb >= cnt2)
cnt1, cnt2, x, y = map(int, input().split())
l, r = 0, (cnt1 + cnt2) * 2
while l <= r:
mid = (l + r) // 2
if check(mid):
ans = mid
r = mid - 1
else:
l = mid + 1
print(ans)
``` | output | 1 | 60,158 | 22 | 120,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
def main():
cnt1, cnt2, x, y = map(int, input().split())
lo = cnt3 = cnt1 + cnt2
hi, z = 10 ** 10, x * y
while lo < hi:
v = (lo + hi) // 2
if (v - v // x < cnt1) or (v - v // y < cnt2) or (v - v // z < cnt3):
lo = v + 1
else:
hi = v
print(lo)
if __name__ == '__main__':
main()
``` | instruction | 0 | 60,159 | 22 | 120,318 |
Yes | output | 1 | 60,159 | 22 | 120,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
c1, c2, x, y = map(int, input().split())
xy = x * y
n = 1
while n - n // x < c1 or n - n // y < c2 or n - n // xy < c1 + c2:
n = n << 1
l = 1
r = n
while r - l > 1:
m = (r + l) >> 1
if m - m // x < c1 or m - m // y < c2 or m - m // xy < c1 + c2:
l = m
else:
r = m
m = l
if m - m // x < c1 or m - m // y < c2 or m - m // xy < c1 + c2:
print(r)
else:
print(l)
``` | instruction | 0 | 60,160 | 22 | 120,320 |
Yes | output | 1 | 60,160 | 22 | 120,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
c1,c2,x,y=map(int,input().split())
def fn(val):
f=[val//x,val//y]
both=val//(x*y)
f=[i-both for i in f]
oth=val-f[0]-f[1]-both
cnt=[c1-f[1],c2-f[0]]
if cnt[0]<0:cnt[0]=0
if cnt[1] < 0: cnt[1] = 0
return (sum(cnt)<=oth)
l=0;r=int(1e18)
while r-l>1:
m=(r+l)//2
if fn(m):
r=m
else:l=m
print(r)
``` | instruction | 0 | 60,161 | 22 | 120,322 |
Yes | output | 1 | 60,161 | 22 | 120,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
cnt1, cnt2, x, y = list(map(int, input().split(' ')))
lo = 1
hi = 10**10
while lo < hi:
xx = cnt1
yy = cnt2
mid = (lo + hi) // 2
intx = mid // x
inty = mid // y
intxy = mid // (x*y)
notxy = mid - intx - inty + intxy
notx = mid - intx - notxy
noty = mid - inty - notxy
xx -= notx
yy -= noty
if max(xx,0) + max(yy,0) > notxy:
lo = mid + 1
else:
hi = mid
print(lo)
``` | instruction | 0 | 60,162 | 22 | 120,324 |
Yes | output | 1 | 60,162 | 22 | 120,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
inData = input().strip().split(" ")
c1,c2,x,y = int(inData[0]),int(inData[1]),int(inData[2]),int(inData[3])
arvud1 = []
count1 = c1
count2 = c2
limx = x
limy = y
if c2>c1:
count1 = c2
count2 = c1
limx = y
limy = x
n = 1
while (len(arvud1)<count1):
if (n%limx!=0):
arvud1.append(n)
n+=1
k = 1
print(arvud1[-1])
``` | instruction | 0 | 60,163 | 22 | 120,326 |
No | output | 1 | 60,163 | 22 | 120,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
a = input().split()
for i in range(len(a)):
a[i] = int(a[i])
c1=a[0]
c2=a[1]
x=a[2]
y=a[3]
i = a[0]+a[1]
while 1:
if c1 <= i - (c1+c2)//x and c2 <= i - (c1+c2)//y + min((c1+c2)//y, c1//y):
print(i)
exit()
i+=1
``` | instruction | 0 | 60,164 | 22 | 120,328 |
No | output | 1 | 60,164 | 22 | 120,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
cnt1, cnt2, x, y = map(int, input().split())
res = 1
a = 0
b = 0
res1 = 0
res2 = 0
if x-1 < cnt1:
if cnt1 % (x-1):
res1 = (cnt1 // (x-1)) * x + cnt1 % (x - 1)
else:
res1 = (cnt1 // (x-1)) * x - 1
else:
res1 = cnt1
if y-1 < cnt2:
if cnt2 % (y-1):
res2 = (cnt2 // (y-1)) * y + cnt2 % (y - 1)
else:
res2 = (cnt2 // (y-1)) * y - 1
else:
res2 = cnt2
if res1 > res2:
step = res1 // x
if res2 - step > 0:
res = res1 + res2 - step
if not res1 % y:
res -= 1
else:
res = res1
elif res1 < res2:
step = res2 // y
if res1 - step > 0:
res = res2 + res1 - step
if not res2 % x:
res -= 1
else:
res = res2
else:
step1 = res1 // x
step2 = res2 // y
if step1 > step2:
if res2 - step1 > 0:
res = res1 + res2 - step1
if not res1 % y:
res -= 1
else:
res = res1
else:
if res1 - step2 > 0:
res = res2 + res1 - step2
if not res2 % x:
res -= 1
else:
res = res2
print(res)
``` | instruction | 0 | 60,165 | 22 | 120,330 |
No | output | 1 | 60,165 | 22 | 120,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
Input
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 1 2 3
Output
5
Input
1 3 2 3
Output
4
Note
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
Submitted Solution:
```
cnt1, cnt2, x, y = map(int, input().split())
l = 0
r = (cnt1 // (x - 1) + 1) * x + (cnt2 // (y - 1) + 1) * y + 5
while r - l != 1:
m = l + (r - l) // 2
t_cnt1 = m // x * (x - 1) + m % x
t_cnt2 = m // y * (y - 1) + m % y
if t_cnt1 >= cnt1 and t_cnt2 >= cnt2 and t_cnt1 + t_cnt2 - m // (x * y) >= cnt1 + cnt2:
r = m
else:
l = m
print(r)
``` | instruction | 0 | 60,166 | 22 | 120,332 |
No | output | 1 | 60,166 | 22 | 120,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,199 | 22 | 120,398 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import math
def binarySearch(num,l,r,x):
while l<=r:
mid=(l+r)//2
if num[mid]==x:
return mid
elif num[mid]<x:
l=mid+1
else:
r=mid-1
def quadratic(b,c):
e=pow(b,2)
f=e-(4*c)
d=math.sqrt(f)
ans=(((-1)*b)+d)//2
return ans
n=int(input())
arr=list(map(int,input().split()))
matrix=[[0]*n for i in range(n)]
a=max(arr)
dict1={}
for i in range(n**2):
if arr[i] in dict1.keys():
dict1[arr[i]]+=1
else:
dict1[arr[i]]=1
lis=[]
for i in dict1.keys():
lis.append([i,dict1[i]])
lis.sort(key=lambda x:x[0],reverse=False)
num=[]
p=0
for i in lis:
num.append(i[0])
p+=1
a=math.sqrt(lis[-1][1])
ans=[[lis[-1][0],a]]
lis[-1][1]=0
i=p-2
k=1
while i>=0:
if lis[i][1]>0:
count=0
for j in range(k):
temp=math.gcd(ans[j][0],lis[i][0])
if temp==lis[i][0]:
count+=(ans[j][1]*2)
if count==0:
ans.append([lis[i][0],math.sqrt(lis[i][1])])
lis[i][1]=0
else:
ans.append([lis[i][0],quadratic(count,(-1)*lis[i][1])])
lis[i][1]=0
for j in range(k):
temp=math.gcd(ans[j][0],ans[-1][0])
if temp!=ans[-1][0]:
ind=binarySearch(num,0,p-1,temp)
lis[ind][1]-=(2*ans[j][1]*ans[-1][1])
k+=1
i-=1
answer=[]
for i in ans:
for j in range(int(i[1])):
answer.append(i[0])
print(" ".join(str(x) for x in answer))
``` | output | 1 | 60,199 | 22 | 120,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,200 | 22 | 120,400 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
from fractions import gcd
from random import randint, shuffle
from collections import Counter
def read_numbers():
return list(map(int, input().split()))
def get_original_array(n, numbers):
cnt = Counter(numbers)
array = []
for new_number in sorted(numbers, reverse=True):
if cnt[new_number]:
cnt[new_number] -= 1
for number in array:
table_entry = gcd(new_number, number)
cnt[table_entry] -= 2
array.append(new_number)
assert cnt.most_common()[0][1] == 0
return array
def test(n):
print(n)
array = [randint(0, 10**9) for _ in range(n)]
table = [gcd(a, b) for a in array for b in array]
shuffle(table)
print(sorted(array) == sorted(get_original_array(n, table)))
if __name__ == '__main__':
# n = 4
# numbers = [2, 1, 2, 3, 4, 3, 2, 6, 1, 1, 2, 2, 1, 2, 3, 2]
# print(get_original_array(n, numbers))
# test(10)
# test(100)
# test(200)
# test(300)
# test(400)
# test(500)
#else:
n = int(input())
numbers = read_numbers()
print(' '.join(map(str, get_original_array(n, numbers))))
``` | output | 1 | 60,200 | 22 | 120,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,201 | 22 | 120,402 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
import math
from collections import Counter
n = int(input())
a = Counter(map(int, input().rstrip().split()))
li = []
for i in range(n):
maxm = max(a)
a[maxm] -= 1
for x in li:
a[math.gcd(x, maxm)] -= 2
li += [maxm]
a += Counter()
print(*li)
``` | output | 1 | 60,201 | 22 | 120,403 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,202 | 22 | 120,404 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
import collections as cc
import math as mt
I=lambda:list(map(int,input().split()))
n,=I()
l=I()
f=cc.Counter(l)
ans=[]
for i in range(n):
now=max(f)
f[now]-=1
for j in ans:
f[mt.gcd(now,j)]-=2
ans.append(now)
f+=cc.Counter()
print(*ans)
``` | output | 1 | 60,202 | 22 | 120,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,203 | 22 | 120,406 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
from sys import stdin,stdout
from math import gcd
input = stdin.readline
def main():
#t = int(input())
t=1
for z in range(t):
n = int(input())
#n, m = map(int,input().split())
ai = list(map(int,input().split()))
if n == 1:
print(ai[0])
return
ai.sort()
ar2 = [ai[-1],ai[-2]]
d = {}
d[ai[-1]] = 0
d[ai[-2]] = 0
d[gcd(ai[-1],ai[-2])] = 2
for i in range(n**2-3,-1,-1):
if ai[i] in d:
if d[ai[i]] == 0:
for num in ar2:
temp = gcd(num,ai[i])
if temp in d:
d[temp] += 2
else:
d[temp] = 2
ar2 += [ai[i]]
else:
d[ai[i]] -= 1
else:
for num in ar2:
temp = gcd(num,ai[i])
if temp in d:
d[temp] += 2
else:
d[temp] = 2
ar2 += [ai[i]]
print(*ar2)
main()
``` | output | 1 | 60,203 | 22 | 120,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,204 | 22 | 120,408 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
import sys,math as mt
input = sys.stdin.readline
from collections import Counter as cc
I = lambda : list(map(int,input().split()))
n,=I()
l=I()
an=[];rq=cc(l)
for i in range(n):
mx=max(rq)
rq[mx]-=1
for x in an:
rq[mt.gcd(x,mx)]-=2
an.append(mx)
rq+=cc()
print(*an)
``` | output | 1 | 60,204 | 22 | 120,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,205 | 22 | 120,410 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
import math
from collections import Counter
n=int(input())
count_=Counter(map(int,input().split()))
ans=[]
for i in range(n):
big = max(count_)
count_[big]-=1
for other in ans:
a=math.gcd(other,big)#;b= math.gcd(other,big)
count_[a]-=2
ans.append(big)
count_+=Counter()
#print(count_)
print(*ans)#,count_)
``` | output | 1 | 60,205 | 22 | 120,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1 | instruction | 0 | 60,206 | 22 | 120,412 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
from sys import stdin,stdout
from math import gcd
from collections import Counter
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for _ in range(1):#nmbr()):
n=nmbr()
a=sorted(lst())
d=Counter(a)
f=1
ans=[]
while f:
mx=max(d)
# print(d,mx)
# ans+=[mx]
d[mx]-=1
if d[mx] == 0: del d[mx]
for v in ans:
g=gcd(v,mx)
d[g]-=2
if d[g]==0:del d[g]
ans+=[mx]
if len(ans)==n:f=0
print(*ans)
``` | output | 1 | 60,206 | 22 | 120,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
import os, 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")
from collections import Counter
from math import gcd
n=int(input())
a=[]
b=list(map(int,input().split()))
c=Counter(b)
for i in sorted(c,reverse=True):
while c[i]:
for j in a:
c[gcd(i,j)]-=2
a.append(i)
c[i]-=1
#print(c,i)
if len(a)==n:
break
print(*a)
``` | instruction | 0 | 60,207 | 22 | 120,414 |
Yes | output | 1 | 60,207 | 22 | 120,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
import math as ma
import sys
input=sys.stdin.readline
def fu(b):
for i in b:
if b[i]!=0:
return i
return -1
def gcd(a,b):
if a%b==0:
return b
else:
return gcd(b,a%b)
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
b={}
for i in range(n*n):
if a[i] in b.keys():
b[a[i]]+=1
else:
b[a[i]]=1
c=[]
for i in b:
c.append(i)
b[i]-=1
break
while 1>0:
if len(c)<n:
a=fu(b)
if a==-1:
break
else:
b[a]-=1
for i in range(len(c)):
b[gcd(a,c[i])]-=2
c.append(a)
else:
break
print(*c)
``` | instruction | 0 | 60,208 | 22 | 120,416 |
Yes | output | 1 | 60,208 | 22 | 120,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
def gcd(a,b):
if b==0: return a
return gcd(b,a%b)
n=int(input())
from collections import Counter
l=[int(i) for i in input().split()]
g=Counter(l)
ans=[]
while g:
m=max(g)
g[m]-=1
for i in ans:
g[gcd(m,i)]-=2
ans+=[m]
g+=Counter()
print(*ans)
``` | instruction | 0 | 60,209 | 22 | 120,418 |
Yes | output | 1 | 60,209 | 22 | 120,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
from collections import defaultdict
def gcd(a,b):
while b != 0:
b, a = a%b, b
return a
def main():
n = int(input())
numbers = list(map(int, input().split()))
numbers.sort()
answer = [numbers[-1]]
del numbers[-1]
how_many = defaultdict(int)
for x in numbers:
how_many[x] += 1
for _ in range(n-1):
max_value = 0
for number in how_many.keys():
max_value = max(max_value, number)
how_many[max_value] -= 1
if how_many[max_value] == 0:
del how_many[max_value]
for element in answer:
key = gcd(element,max_value)
how_many[key] -= 2
if how_many[key] == 0:
del how_many[key]
answer.insert(0, max_value)
for a in answer:
print(a, end = " ")
main()
``` | instruction | 0 | 60,210 | 22 | 120,420 |
Yes | output | 1 | 60,210 | 22 | 120,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
import sys
from math import gcd,sqrt,ceil,log2
from collections import defaultdict,Counter,deque
from bisect import bisect_left,bisect_right
import math
sys.setrecursionlimit(2*10**5+10)
import heapq
from itertools import permutations
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
aa='abcdefghijklmnopqrstuvwxyz'
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")
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_sum(bit,i):
s = 0
i+=1
while i>0:
s+=bit[i]
i-=i&(-i)
return s
def update(bit,n,i,v):
i+=1
while i<=n:
bit[i]+=v
i+=i&(-i)
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def primeFactors(n):
sa = []
# sa.add(n)
while n % 2 == 0:
sa.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
sa.append(i)
n = n // i
# sa.add(n)
if n > 2:
sa.append(n)
return sa
def seive(n):
pri = [True]*(n+1)
p = 2
while p*p<=n:
if pri[p] == True:
for i in range(p*p,n+1,p):
pri[i] = False
p+=1
return pri
def check_prim(n):
if n<0:
return False
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
return False
return True
def getZarr(string, z):
n = len(string)
# [L,R] make a window which matches
# with prefix of s
l, r, k = 0, 0, 0
for i in range(1, n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
def search(text, pattern):
# Create concatenated string "P$T"
concat = pattern + "$" + text
l = len(concat)
z = [0] * l
getZarr(concat, z)
ha = []
for i in range(l):
if z[i] == len(pattern):
ha.append(i - len(pattern) - 1)
return ha
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# n = int(input())
# l = list(map(int,input().split()))
#
# hash = defaultdict(list)
# la = []
#
# for i in range(n):
# la.append([l[i],i+1])
#
# la.sort(key = lambda x: (x[0],-x[1]))
# ans = []
# r = n
# flag = 0
# lo = []
# ha = [i for i in range(n,0,-1)]
# yo = []
# for a,b in la:
#
# if a == 1:
# ans.append([r,b])
# # hash[(1,1)].append([b,r])
# lo.append((r,b))
# ha.pop(0)
# yo.append([r,b])
# r-=1
#
# elif a == 2:
# # print(yo,lo)
# # print(hash[1,1])
# if lo == []:
# flag = 1
# break
# c,d = lo.pop(0)
# yo.pop(0)
# if b>=d:
# flag = 1
# break
# ans.append([c,b])
# yo.append([c,b])
#
#
#
# elif a == 3:
#
# if yo == []:
# flag = 1
# break
# c,d = yo.pop(0)
# if b>=d:
# flag = 1
# break
# if ha == []:
# flag = 1
# break
#
# ka = ha.pop(0)
#
# ans.append([ka,b])
# ans.append([ka,d])
# yo.append([ka,b])
#
# if flag:
# print(-1)
# else:
# print(len(ans))
# for a,b in ans:
# print(a,b)
def mergeIntervals(arr):
# Sorting based on the increasing order
# of the start intervals
arr.sort(key = lambda x: x[0])
# array to hold the merged intervals
m = []
s = -10000
max = -100000
for i in range(len(arr)):
a = arr[i]
if a[0] > max:
if i != 0:
m.append([s,max])
max = a[1]
s = a[0]
else:
if a[1] >= max:
max = a[1]
#'max' value gives the last point of
# that particular interval
# 's' gives the starting point of that interval
# 'm' array contains the list of all merged intervals
if max != -100000 and [s, max] not in m:
m.append([s, max])
return m
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def sol(n):
seti = set()
for i in range(1,int(sqrt(n))+1):
if n%i == 0:
seti.add(n//i)
seti.add(i)
return seti
def lcm(a,b):
return (a*b)//gcd(a,b)
#
# n,p = map(int,input().split())
#
# s = input()
#
# if n <=2:
# if n == 1:
# pass
# if n == 2:
# pass
# i = n-1
# idx = -1
# while i>=0:
# z = ord(s[i])-96
# k = chr(z+1+96)
# flag = 1
# if i-1>=0:
# if s[i-1]!=k:
# flag+=1
# else:
# flag+=1
# if i-2>=0:
# if s[i-2]!=k:
# flag+=1
# else:
# flag+=1
# if flag == 2:
# idx = i
# s[i] = k
# break
# if idx == -1:
# print('NO')
# exit()
# for i in range(idx+1,n):
# if
#
def moore_voting(l):
count1 = 0
count2 = 0
first = 10**18
second = 10**18
n = len(l)
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
elif count1 == 0:
count1+=1
first = l[i]
elif count2 == 0:
count2+=1
second = l[i]
else:
count1-=1
count2-=1
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
if count1>n//3:
return first
if count2>n//3:
return second
return -1
def find_parent(u,parent):
if u!=parent[u]:
parent[u] = find_parent(parent[u],parent)
return parent[u]
def dis_union(n,e):
par = [i for i in range(n+1)]
rank = [1]*(n+1)
for a,b in e:
z1,z2 = find_parent(a,par),find_parent(b,par)
if rank[z1]>rank[z2]:
z1,z2 = z2,z1
if z1!=z2:
par[z1] = z2
rank[z2]+=rank[z1]
else:
return a,b
def dijkstra(n,tot,hash):
hea = [[0,n]]
dis = [10**18]*(tot+1)
dis[n] = 0
boo = defaultdict(bool)
check = defaultdict(int)
while hea:
a,b = heapq.heappop(hea)
if boo[b]:
continue
boo[b] = True
for i,w in hash[b]:
if b == 1:
c = 0
if (1,i,w) in nodes:
c = nodes[(1,i,w)]
del nodes[(1,i,w)]
if dis[b]+w<dis[i]:
dis[i] = dis[b]+w
check[i] = c
elif dis[b]+w == dis[i] and c == 0:
dis[i] = dis[b]+w
check[i] = c
else:
if dis[b]+w<=dis[i]:
dis[i] = dis[b]+w
check[i] = check[b]
heapq.heappush(hea,[dis[i],i])
return check
def power(x,y,p):
res = 1
x = x%p
if x == 0:
return 0
while y>0:
if (y&1) == 1:
res*=x
x = x*x
y = y>>1
return res
import sys
from math import ceil,log2
INT_MAX = sys.maxsize
def minVal(x, y) :
return x if (x < y) else y
def getMid(s, e) :
return s + (e - s) // 2
def RMQUtil( st, ss, se, qs, qe, index) :
if (qs <= ss and qe >= se) :
return st[index]
if (se < qs or ss > qe) :
return INT_MAX
mid = getMid(ss, se)
return minVal(RMQUtil(st, ss, mid, qs,
qe, 2 * index + 1),
RMQUtil(st, mid + 1, se,
qs, qe, 2 * index + 2))
def RMQ( st, n, qs, qe) :
if (qs < 0 or qe > n - 1 or qs > qe) :
print("Invalid Input")
return -1
return RMQUtil(st, 0, n - 1, qs, qe, 0)
def constructSTUtil(arr, ss, se, st, si) :
if (ss == se) :
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = minVal(constructSTUtil(arr, ss, mid,
st, si * 2 + 1),
constructSTUtil(arr, mid + 1, se,
st, si * 2 + 2))
return st[si]
def constructST( arr, n) :
x = (int)(ceil(log2(n)))
max_size = 2 * (int)(2**x) - 1
st = [0] * (max_size)
constructSTUtil(arr, 0, n - 1, st, 0)
return st
# t = int(input())
# for _ in range(t):
#
# n = int(input())
# l = list(map(int,input().split()))
# # x,y = 0,10
# st = constructST(l, n)
#
# pre = [0]
# suf = [0]
# for i in range(n):
# pre.append(max(pre[-1],l[i]))
# for i in range(n-1,-1,-1):
# suf.append(max(suf[-1],l[i]))
#
#
# i = 1
# # print(pre,suf)
# flag = 0
# x,y,z = -1,-1,-1
# # suf.reverse()
# print(suf)
# while i<len(pre):
#
# z = pre[i]
# j = bisect_left(suf,z)
# if suf[j] == z:
# while i<n and l[i]<=z:
# i+=1
# if pre[i]>z:
# break
# while j<n and l[n-j]<=z:
# j+=1
# if suf[j]>z:
# break
# # j-=1
# print(i,n-j)
# # break/
# if RMQ(st,n,i,j) == z:
# c = i+j-i+1
# x,y,z = i,j-i+1,n-c
# break
# else:
# i+=1
#
# else:
# i+=1
#
#
#
# if x!=-1:
# print('Yes')
# print(x,y,z)
# else:
# print('No')
# t = int(input())
#
# for _ in range(t):
#
# def debug(n):
# ans = []
# for i in range(1,n+1):
# for j in range(i+1,n+1):
# if (i*(j+1))%(j-i) == 0 :
# ans.append([i,j])
# return ans
#
#
# n = int(input())
# print(debug(n))
# import sys
# input = sys.stdin.readline
# import bisect
#
# t=int(input())
# for tests in range(t):
# n=int(input())
# A=list(map(int,input().split()))
#
# LEN = len(A)
# Sparse_table = [A]
#
# for i in range(LEN.bit_length()-1):
# j = 1<<i
# B = []
# for k in range(len(Sparse_table[-1])-j):
# B.append(min(Sparse_table[-1][k], Sparse_table[-1][k+j]))
# Sparse_table.append(B)
#
# def query(l,r): # [l,r)におけるminを求める.
# i=(r-l).bit_length()-1 # 何番目のSparse_tableを見るか.
#
# return min(Sparse_table[i][l],Sparse_table[i][r-(1<<i)]) # (1<<i)個あれば[l, r)が埋まるので, それを使ってminを求める.
#
# LMAX=[A[0]]
# for i in range(1,n):
# LMAX.append(max(LMAX[-1],A[i]))
#
# RMAX=A[-1]
#
# for i in range(n-1,-1,-1):
# RMAX=max(RMAX,A[i])
#
# x=bisect.bisect(LMAX,RMAX)
# #print(RMAX,x)
# print(RMAX,x,i)
# if x==0:
# continue
#
# v=min(x,i-1)
# if v<=0:
# continue
#
# if LMAX[v-1]==query(v,i)==RMAX:
# print("YES")
# print(v,i-v,n-i)
# break
#
# v-=1
# if v<=0:
# continue
# if LMAX[v-1]==query(v,i)==RMAX:
# print("YES")
# print(v,i-v,n-i)
# break
# else:
# print("NO")
#
#
#
#
#
#
#
#
#
# t = int(input())
#
# for _ in range(t):
#
# x = int(input())
# mini = 10**18
# n = ceil((-1 + sqrt(1+8*x))/2)
# for i in range(-100,1):
# z = x+-1*i
# z1 = (abs(i)*(abs(i)+1))//2
# z+=z1
# # print(z)
# n = ceil((-1 + sqrt(1+8*z))/2)
#
# y = (n*(n+1))//2
# # print(n,y,z,i)
# mini = min(n+y-z,mini)
# print(n+y-z,i)
#
#
# print(mini)
#
#
#
# n,m = map(int,input().split())
# l = []
# hash = defaultdict(int)
# for i in range(n):
# la = list(map(int,input().split()))[1:]
# l.append(set(la))
# # for i in la:
# # hash[i]+=1
#
# for i in range(n):
#
# for j in range(n):
# if i!=j:
#
# if len(l[i].intersection(l[j])) == 0:
# for k in range(n):
#
#
# else:
# break
#
#
#
#
#
#
# practicing segment_trees
# t = int(input())
#
# for _ in range(t):
# n = int(input())
# l = []
# for i in range(n):
# a,b = map(int,input().split())
# l.append([a,b])
#
# l.sort()
# n,m = map(int,input().split())
# l = list(map(int,input().split()))
#
# hash = defaultdict(int)
# for i in range(1,2**n,2):
# count = 0
# z1 = bin(l[i]|l[i-1])[2:]
# z1+='0'*(17-len(z1)) + z1
# for k in range(len(z1)):
# if z1[k] == '1':
# hash[k]+=1
# for i in range(m):
# a,b = map(int,input().split())
# a-=1
# init = a
# if a%2 == 0:
# a+=1
# z1 = bin(l[a]|l[a-1])[2:]
# z1+='0'*(17-len(z1)) + z1
# for k in range(len(z1)):
# if z1[k] == '1':
# hash[k]-=1
# l[init] = b
# a = init
# if a%2 == 0:
# a+=1
# z1 = bin(l[a]|l[a-1])[2:]
# z1+='0'*(17-len(z1)) + z1
# for k in range(len(z1)):
# if z1[k] == '1':
# hash[k]+=1
# ans = ''
# for k in range(17):
# if n%2 == 0:
# if hash[k]%2 == 0:
# ans+='0'
# else:
# ans+='1'
# else:
# if hash[k]%2 == 0:
# if hash[k]
#
#
#
def bfs1(p):
level = [10**18]*(n+1)
boo = [False]*(n+1)
par = [i for i in range(n+1)]
queue = [p]
boo[p] = True
level[p] = 0
while queue:
z = queue.pop(0)
for i in hash[z]:
if not boo[i]:
boo[i] = True
queue.append(i)
level[i] = level[z]+1
par[i] = z
return par,level
n = int(input())
l = list(map(int,input().split()))
hash = defaultdict(int)
for i in l:
hash[i]+=1
k = list(set(l))
l.sort()
k.sort()
ans = []
poin = n-1
yo = defaultdict(int)
for i in k[::-1]:
if len(ans) == n:
break
z1 = 0
if hash[i] == 0:
continue
for j in ans:
if gcd(j,i) == i:
z1+=1
hash[gcd(j,i)]-=2
# if i == 3:
# print(ans,z1)
z = -2*z1+sqrt((2*z1)**2+4*hash[i])
if z<=0:
continue
z//=2
for j in range(int(z)):
ans.append(i)
yo[i]+=1
print(*ans)
``` | instruction | 0 | 60,211 | 22 | 120,422 |
No | output | 1 | 60,211 | 22 | 120,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
n=int(input())
l=[int(i) for i in input().split()]
if (len(set(l)))==1:
print(*([l[0]]*n))
exit()
s=set(l)
a=[]
for i in range(n):
a.append(max(s))
s.remove(max(s))
print(*a)
``` | instruction | 0 | 60,212 | 22 | 120,424 |
No | output | 1 | 60,212 | 22 | 120,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
import sys
from math import gcd,sqrt,ceil,log2
from collections import defaultdict,Counter,deque
from bisect import bisect_left,bisect_right
import math
sys.setrecursionlimit(2*10**5+10)
import heapq
from itertools import permutations
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
aa='abcdefghijklmnopqrstuvwxyz'
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")
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_sum(bit,i):
s = 0
i+=1
while i>0:
s+=bit[i]
i-=i&(-i)
return s
def update(bit,n,i,v):
i+=1
while i<=n:
bit[i]+=v
i+=i&(-i)
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def primeFactors(n):
sa = []
# sa.add(n)
while n % 2 == 0:
sa.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
sa.append(i)
n = n // i
# sa.add(n)
if n > 2:
sa.append(n)
return sa
def seive(n):
pri = [True]*(n+1)
p = 2
while p*p<=n:
if pri[p] == True:
for i in range(p*p,n+1,p):
pri[i] = False
p+=1
return pri
def check_prim(n):
if n<0:
return False
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
return False
return True
def getZarr(string, z):
n = len(string)
# [L,R] make a window which matches
# with prefix of s
l, r, k = 0, 0, 0
for i in range(1, n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
def search(text, pattern):
# Create concatenated string "P$T"
concat = pattern + "$" + text
l = len(concat)
z = [0] * l
getZarr(concat, z)
ha = []
for i in range(l):
if z[i] == len(pattern):
ha.append(i - len(pattern) - 1)
return ha
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# n = int(input())
# l = list(map(int,input().split()))
#
# hash = defaultdict(list)
# la = []
#
# for i in range(n):
# la.append([l[i],i+1])
#
# la.sort(key = lambda x: (x[0],-x[1]))
# ans = []
# r = n
# flag = 0
# lo = []
# ha = [i for i in range(n,0,-1)]
# yo = []
# for a,b in la:
#
# if a == 1:
# ans.append([r,b])
# # hash[(1,1)].append([b,r])
# lo.append((r,b))
# ha.pop(0)
# yo.append([r,b])
# r-=1
#
# elif a == 2:
# # print(yo,lo)
# # print(hash[1,1])
# if lo == []:
# flag = 1
# break
# c,d = lo.pop(0)
# yo.pop(0)
# if b>=d:
# flag = 1
# break
# ans.append([c,b])
# yo.append([c,b])
#
#
#
# elif a == 3:
#
# if yo == []:
# flag = 1
# break
# c,d = yo.pop(0)
# if b>=d:
# flag = 1
# break
# if ha == []:
# flag = 1
# break
#
# ka = ha.pop(0)
#
# ans.append([ka,b])
# ans.append([ka,d])
# yo.append([ka,b])
#
# if flag:
# print(-1)
# else:
# print(len(ans))
# for a,b in ans:
# print(a,b)
def mergeIntervals(arr):
# Sorting based on the increasing order
# of the start intervals
arr.sort(key = lambda x: x[0])
# array to hold the merged intervals
m = []
s = -10000
max = -100000
for i in range(len(arr)):
a = arr[i]
if a[0] > max:
if i != 0:
m.append([s,max])
max = a[1]
s = a[0]
else:
if a[1] >= max:
max = a[1]
#'max' value gives the last point of
# that particular interval
# 's' gives the starting point of that interval
# 'm' array contains the list of all merged intervals
if max != -100000 and [s, max] not in m:
m.append([s, max])
return m
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def sol(n):
seti = set()
for i in range(1,int(sqrt(n))+1):
if n%i == 0:
seti.add(n//i)
seti.add(i)
return seti
def lcm(a,b):
return (a*b)//gcd(a,b)
#
# n,p = map(int,input().split())
#
# s = input()
#
# if n <=2:
# if n == 1:
# pass
# if n == 2:
# pass
# i = n-1
# idx = -1
# while i>=0:
# z = ord(s[i])-96
# k = chr(z+1+96)
# flag = 1
# if i-1>=0:
# if s[i-1]!=k:
# flag+=1
# else:
# flag+=1
# if i-2>=0:
# if s[i-2]!=k:
# flag+=1
# else:
# flag+=1
# if flag == 2:
# idx = i
# s[i] = k
# break
# if idx == -1:
# print('NO')
# exit()
# for i in range(idx+1,n):
# if
#
def moore_voting(l):
count1 = 0
count2 = 0
first = 10**18
second = 10**18
n = len(l)
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
elif count1 == 0:
count1+=1
first = l[i]
elif count2 == 0:
count2+=1
second = l[i]
else:
count1-=1
count2-=1
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
if count1>n//3:
return first
if count2>n//3:
return second
return -1
def find_parent(u,parent):
if u!=parent[u]:
parent[u] = find_parent(parent[u],parent)
return parent[u]
def dis_union(n,e):
par = [i for i in range(n+1)]
rank = [1]*(n+1)
for a,b in e:
z1,z2 = find_parent(a,par),find_parent(b,par)
if rank[z1]>rank[z2]:
z1,z2 = z2,z1
if z1!=z2:
par[z1] = z2
rank[z2]+=rank[z1]
else:
return a,b
def dijkstra(n,tot,hash):
hea = [[0,n]]
dis = [10**18]*(tot+1)
dis[n] = 0
boo = defaultdict(bool)
check = defaultdict(int)
while hea:
a,b = heapq.heappop(hea)
if boo[b]:
continue
boo[b] = True
for i,w in hash[b]:
if b == 1:
c = 0
if (1,i,w) in nodes:
c = nodes[(1,i,w)]
del nodes[(1,i,w)]
if dis[b]+w<dis[i]:
dis[i] = dis[b]+w
check[i] = c
elif dis[b]+w == dis[i] and c == 0:
dis[i] = dis[b]+w
check[i] = c
else:
if dis[b]+w<=dis[i]:
dis[i] = dis[b]+w
check[i] = check[b]
heapq.heappush(hea,[dis[i],i])
return check
def power(x,y,p):
res = 1
x = x%p
if x == 0:
return 0
while y>0:
if (y&1) == 1:
res*=x
x = x*x
y = y>>1
return res
import sys
from math import ceil,log2
INT_MAX = sys.maxsize
def minVal(x, y) :
return x if (x < y) else y
def getMid(s, e) :
return s + (e - s) // 2
def RMQUtil( st, ss, se, qs, qe, index) :
if (qs <= ss and qe >= se) :
return st[index]
if (se < qs or ss > qe) :
return INT_MAX
mid = getMid(ss, se)
return minVal(RMQUtil(st, ss, mid, qs,
qe, 2 * index + 1),
RMQUtil(st, mid + 1, se,
qs, qe, 2 * index + 2))
def RMQ( st, n, qs, qe) :
if (qs < 0 or qe > n - 1 or qs > qe) :
print("Invalid Input")
return -1
return RMQUtil(st, 0, n - 1, qs, qe, 0)
def constructSTUtil(arr, ss, se, st, si) :
if (ss == se) :
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = minVal(constructSTUtil(arr, ss, mid,
st, si * 2 + 1),
constructSTUtil(arr, mid + 1, se,
st, si * 2 + 2))
return st[si]
def constructST( arr, n) :
x = (int)(ceil(log2(n)))
max_size = 2 * (int)(2**x) - 1
st = [0] * (max_size)
constructSTUtil(arr, 0, n - 1, st, 0)
return st
# t = int(input())
# for _ in range(t):
#
# n = int(input())
# l = list(map(int,input().split()))
# # x,y = 0,10
# st = constructST(l, n)
#
# pre = [0]
# suf = [0]
# for i in range(n):
# pre.append(max(pre[-1],l[i]))
# for i in range(n-1,-1,-1):
# suf.append(max(suf[-1],l[i]))
#
#
# i = 1
# # print(pre,suf)
# flag = 0
# x,y,z = -1,-1,-1
# # suf.reverse()
# print(suf)
# while i<len(pre):
#
# z = pre[i]
# j = bisect_left(suf,z)
# if suf[j] == z:
# while i<n and l[i]<=z:
# i+=1
# if pre[i]>z:
# break
# while j<n and l[n-j]<=z:
# j+=1
# if suf[j]>z:
# break
# # j-=1
# print(i,n-j)
# # break/
# if RMQ(st,n,i,j) == z:
# c = i+j-i+1
# x,y,z = i,j-i+1,n-c
# break
# else:
# i+=1
#
# else:
# i+=1
#
#
#
# if x!=-1:
# print('Yes')
# print(x,y,z)
# else:
# print('No')
# t = int(input())
#
# for _ in range(t):
#
# def debug(n):
# ans = []
# for i in range(1,n+1):
# for j in range(i+1,n+1):
# if (i*(j+1))%(j-i) == 0 :
# ans.append([i,j])
# return ans
#
#
# n = int(input())
# print(debug(n))
# import sys
# input = sys.stdin.readline
# import bisect
#
# t=int(input())
# for tests in range(t):
# n=int(input())
# A=list(map(int,input().split()))
#
# LEN = len(A)
# Sparse_table = [A]
#
# for i in range(LEN.bit_length()-1):
# j = 1<<i
# B = []
# for k in range(len(Sparse_table[-1])-j):
# B.append(min(Sparse_table[-1][k], Sparse_table[-1][k+j]))
# Sparse_table.append(B)
#
# def query(l,r): # [l,r)におけるminを求める.
# i=(r-l).bit_length()-1 # 何番目のSparse_tableを見るか.
#
# return min(Sparse_table[i][l],Sparse_table[i][r-(1<<i)]) # (1<<i)個あれば[l, r)が埋まるので, それを使ってminを求める.
#
# LMAX=[A[0]]
# for i in range(1,n):
# LMAX.append(max(LMAX[-1],A[i]))
#
# RMAX=A[-1]
#
# for i in range(n-1,-1,-1):
# RMAX=max(RMAX,A[i])
#
# x=bisect.bisect(LMAX,RMAX)
# #print(RMAX,x)
# print(RMAX,x,i)
# if x==0:
# continue
#
# v=min(x,i-1)
# if v<=0:
# continue
#
# if LMAX[v-1]==query(v,i)==RMAX:
# print("YES")
# print(v,i-v,n-i)
# break
#
# v-=1
# if v<=0:
# continue
# if LMAX[v-1]==query(v,i)==RMAX:
# print("YES")
# print(v,i-v,n-i)
# break
# else:
# print("NO")
#
#
#
#
#
#
#
#
#
# t = int(input())
#
# for _ in range(t):
#
# x = int(input())
# mini = 10**18
# n = ceil((-1 + sqrt(1+8*x))/2)
# for i in range(-100,1):
# z = x+-1*i
# z1 = (abs(i)*(abs(i)+1))//2
# z+=z1
# # print(z)
# n = ceil((-1 + sqrt(1+8*z))/2)
#
# y = (n*(n+1))//2
# # print(n,y,z,i)
# mini = min(n+y-z,mini)
# print(n+y-z,i)
#
#
# print(mini)
#
#
#
# n,m = map(int,input().split())
# l = []
# hash = defaultdict(int)
# for i in range(n):
# la = list(map(int,input().split()))[1:]
# l.append(set(la))
# # for i in la:
# # hash[i]+=1
#
# for i in range(n):
#
# for j in range(n):
# if i!=j:
#
# if len(l[i].intersection(l[j])) == 0:
# for k in range(n):
#
#
# else:
# break
#
#
#
#
#
#
# practicing segment_trees
# t = int(input())
#
# for _ in range(t):
# n = int(input())
# l = []
# for i in range(n):
# a,b = map(int,input().split())
# l.append([a,b])
#
# l.sort()
# n,m = map(int,input().split())
# l = list(map(int,input().split()))
#
# hash = defaultdict(int)
# for i in range(1,2**n,2):
# count = 0
# z1 = bin(l[i]|l[i-1])[2:]
# z1+='0'*(17-len(z1)) + z1
# for k in range(len(z1)):
# if z1[k] == '1':
# hash[k]+=1
# for i in range(m):
# a,b = map(int,input().split())
# a-=1
# init = a
# if a%2 == 0:
# a+=1
# z1 = bin(l[a]|l[a-1])[2:]
# z1+='0'*(17-len(z1)) + z1
# for k in range(len(z1)):
# if z1[k] == '1':
# hash[k]-=1
# l[init] = b
# a = init
# if a%2 == 0:
# a+=1
# z1 = bin(l[a]|l[a-1])[2:]
# z1+='0'*(17-len(z1)) + z1
# for k in range(len(z1)):
# if z1[k] == '1':
# hash[k]+=1
# ans = ''
# for k in range(17):
# if n%2 == 0:
# if hash[k]%2 == 0:
# ans+='0'
# else:
# ans+='1'
# else:
# if hash[k]%2 == 0:
# if hash[k]
#
#
#
def bfs1(p):
level = [10**18]*(n+1)
boo = [False]*(n+1)
par = [i for i in range(n+1)]
queue = [p]
boo[p] = True
level[p] = 0
while queue:
z = queue.pop(0)
for i in hash[z]:
if not boo[i]:
boo[i] = True
queue.append(i)
level[i] = level[z]+1
par[i] = z
return par,level
n = int(input())
l = list(map(int,input().split()))
hash = defaultdict(int)
for i in l:
hash[i]+=1
k = list(set(l))
l.sort()
k.sort()
ans = []
poin = n-1
yo = defaultdict(int)
for i in k[::-1]:
if len(ans) == n:
break
z1 = 0
if hash[i] == 0:
continue
for j in ans:
if gcd(j,i) == i:
z1+=1
else:
hash[gcd(j,i)]-=2
# if i == 3:
# print(ans,z1)
z = -2*z1+sqrt((2*z1)**2+4*hash[i])
if z<=0:
continue
z//=2
for j in range(int(z)):
ans.append(i)
yo[i]+=1
print(*ans)
``` | instruction | 0 | 60,213 | 22 | 120,426 |
No | output | 1 | 60,213 | 22 | 120,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The GCD table G of size n × n for an array of positive integers a of length n is defined by formula
<image>
Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both x and y, it is denoted as <image>. For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:
<image>
Given all the numbers of the GCD table G, restore array a.
Input
The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.
All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.
Output
In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.
Examples
Input
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
Output
4 3 6 2
Input
1
42
Output
42
Input
2
1 1 1 1
Output
1 1
Submitted Solution:
```
def main():
n = int(input())
numbers = list(map(int, input().split()))
numbers.sort()
i = 0
limit = n**2 - n
anserw = []
while i < limit and i<n**2-1:
if numbers[i] != numbers[i+1]:
anserw.append(numbers[i])
limit += 1
i -= 1
elif numbers[i] != numbers[i-1] and i !=0 and numbers[i] != numbers[i+2]:
anserw.append(numbers[i])
anserw.append(numbers[i+1])
limit += 2
i += 2
anserw += numbers[limit:]
for x in anserw:
print(x, end=" ")
main()
``` | instruction | 0 | 60,214 | 22 | 120,428 |
No | output | 1 | 60,214 | 22 | 120,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.
Let [n] denote the set {1, ..., n}. We will also write f: [x] → [y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.
Now then, you are given a function f: [n] → [n]. Your task is to find a positive integer m, and two functions g: [n] → [m], h: [m] → [n], such that g(h(x)) = x for all <image>, and h(g(x)) = f(x) for all <image>, or determine that finding these is impossible.
Input
The first line contains an integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers — values f(1), ..., f(n) (1 ≤ f(i) ≤ n).
Output
If there is no answer, print one integer -1.
Otherwise, on the first line print the number m (1 ≤ m ≤ 106). On the second line print n numbers g(1), ..., g(n). On the third line print m numbers h(1), ..., h(m).
If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.
Examples
Input
3
1 2 3
Output
3
1 2 3
1 2 3
Input
3
2 2 2
Output
1
1 1 1
2
Input
2
2 1
Output
-1 | instruction | 0 | 60,273 | 22 | 120,546 |
Tags: constructive algorithms, dsu, math
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return list(map(int, input().split()))
def II(): return int(input())
def LS(): return input().split()
def S(): return input()
def main():
n = II()
a = list(map(lambda x: x-1, LI()))
g = [-1] * n
h = list(set(a))
l = len(h)
d = {}
for i in range(l):
hi = h[i]
g[hi] = i
for i in range(n):
if g[i] == -1:
g[i] = g[a[i]]
elif g[i] != g[a[i]]:
return -1
print(l)
print(' '.join(map(lambda x: str(x+1), g)))
return ' '.join(map(lambda x: str(x+1), h))
print(main())
``` | output | 1 | 60,273 | 22 | 120,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.
Let [n] denote the set {1, ..., n}. We will also write f: [x] → [y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.
Now then, you are given a function f: [n] → [n]. Your task is to find a positive integer m, and two functions g: [n] → [m], h: [m] → [n], such that g(h(x)) = x for all <image>, and h(g(x)) = f(x) for all <image>, or determine that finding these is impossible.
Input
The first line contains an integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers — values f(1), ..., f(n) (1 ≤ f(i) ≤ n).
Output
If there is no answer, print one integer -1.
Otherwise, on the first line print the number m (1 ≤ m ≤ 106). On the second line print n numbers g(1), ..., g(n). On the third line print m numbers h(1), ..., h(m).
If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.
Examples
Input
3
1 2 3
Output
3
1 2 3
1 2 3
Input
3
2 2 2
Output
1
1 1 1
2
Input
2
2 1
Output
-1 | instruction | 0 | 60,274 | 22 | 120,548 |
Tags: constructive algorithms, dsu, math
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
i = 0
h = dict()
g = []
for x in a:
cur = x
if a[cur - 1] != cur:
print("-1")
quit()
if not h.__contains__(cur):
h[cur] = len(g)
g.append(cur)
print(len(g))
for x in a:
print(h[x] + 1, end=" ")
print()
for x in g:
print(x, end=" ")
``` | output | 1 | 60,274 | 22 | 120,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. How many tuples (A,B,C) of positive integers satisfy A \times B + C = N?
Constraints
* 2 \leq N \leq 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer.
Examples
Input
3
Output
3
Input
100
Output
473
Input
1000000
Output
13969985
Submitted Solution:
```
import math
N=int(input())
sum=0
def make_prime_list_2(num):
if num < 2:
return []
# 0のものは素数じゃないとする
prime_list = [i for i in range(num + 1)]
prime_list[1] = 0 # 1は素数ではない
num_sqrt = math.sqrt(num)
for prime in prime_list:
if prime == 0:
continue
if prime > num_sqrt:
break
for non_prime in range(2 * prime, num, prime):
prime_list[non_prime] = 0
return [prime for prime in prime_list if prime != 0]
def search_divisor_num_1(num):
if num < 0:
return None
elif num == 1:
return 1
else:
num_sqrt = math.floor(math.sqrt(num))
prime_list = make_prime_list_2(num_sqrt)
divisor_num = 1
for prime in prime_list:
count = 1
while num % prime == 0:
num //= prime
count += 1
divisor_num *= count
if num != 1:
divisor_num *= 2
return divisor_num
for c in range(1,N):
sum+=search_divisor_num_1(c)
print(sum)
``` | instruction | 0 | 60,395 | 22 | 120,790 |
No | output | 1 | 60,395 | 22 | 120,791 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,527 | 22 | 121,054 |
"Correct Solution:
```
n=int(input())
d=dict()
def get(n):
if(n==1):return 1
if(n==0):return 0
if(n in d.keys()):return d[n]
if(n%2==0):
d[n]=2*get(n//2)+get(n//2-1)
else:
d[n]=2*get(n//2)+get(n//2+1)
d[n]%=(10**9+7)
return d[n]
def check(u,v,n):
for a in range(n+1):
b=u ^ a
if(a+b==v):
return 1
return 0
#for e in range(1,n):
# ans=0
# for i in range(e+1):
## for j in range(e+1):
# ans+=check(i,j,e)
# print(ans,get(e+1))
print(get(n+1))
``` | output | 1 | 60,527 | 22 | 121,055 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,528 | 22 | 121,056 |
"Correct Solution:
```
#社協
N = int(input())
binN = format(N, "b").zfill(60)
mod = 7 + 10 ** 9
DP = [[0, 0, 0] for _ in range(60)]
if binN[0] == "0":
DP[0] = [1, 0, 0]
else:
DP[0] = [1, 1, 0]
for i in range(1, 60):
if binN[i] == "0":
DP[i][0] = (DP[i-1][0] + DP[i-1][1]) % mod
DP[i][1] = DP[i-1][1]
DP[i][2] = (DP[i-1][2] * 3 + DP[i-1][1]) % mod
else:
DP[i][0] = DP[i-1][0]
DP[i][1] = (DP[i-1][0] + DP[i-1][1]) % mod
DP[i][2] = (DP[i-1][2] * 3 + DP[i-1][1] * 2) % mod
print(sum(DP[-1]) % mod)
``` | output | 1 | 60,528 | 22 | 121,057 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,529 | 22 | 121,058 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
from bisect import bisect_left, bisect_right
import random
from itertools import permutations, accumulate, combinations
import sys
import string
INF = float('inf')
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return sys.stdin.readline().strip()
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
n = bin(I())[2:]
dp0, dp1, dp2 = 1, 0, 0
for d in n:
if d == '1':
dp0, dp1, dp2 = dp0 % mod, (dp0 + dp1) % mod, (dp1 * 2 + dp2 * 3) % mod
else:
dp0, dp1, dp2 = (dp0 + dp1) % mod, dp1 % mod, (dp1 + dp2 * 3) % mod
print((dp0 + dp1 + dp2) % mod)
``` | output | 1 | 60,529 | 22 | 121,059 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,530 | 22 | 121,060 |
"Correct Solution:
```
MOD = 10**9 + 7
n = int(input())
dp = [[0 for _ in range(3)] for _ in range(61)]
dp[60][0] = 1
for i in range(59, -1, -1):
if (n>>i)&1 == 1:
dp[i][0] += dp[i+1][0]
dp[i][1] += dp[i+1][0] + dp[i+1][1]
dp[i][2] += 2*dp[i+1][1] + 3*dp[i+1][2]
else:
dp[i][0] += dp[i+1][0] + dp[i+1][1]
dp[i][1] += dp[i+1][1]
dp[i][2] += dp[i+1][1] + 3*dp[i+1][2]
for j in range(3):
dp[i][j] %= MOD
print(sum(dp[0])%MOD)
``` | output | 1 | 60,530 | 22 | 121,061 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,531 | 22 | 121,062 |
"Correct Solution:
```
MOD = 10 ** 9 + 7
N = int(input())
binN = list(map(int, bin(N)[2:]))
dp = [[0] * 3 for _ in range(len(binN) + 1)]
dp[0][0] = 1
for i, Ni in enumerate(binN):
dp[i + 1][0] += dp[i][0] * 1
dp[i + 1][1] += dp[i][0] * Ni
dp[i + 1][0] += dp[i][1] * (1 - Ni)
dp[i + 1][1] += dp[i][1] * 1
dp[i + 1][2] += dp[i][1] * (1 + Ni)
dp[i + 1][2] += dp[i][2] * 3
dp[i + 1][0] %= MOD
dp[i + 1][1] %= MOD
dp[i + 1][2] %= MOD
print(sum(dp[-1]) % MOD)
``` | output | 1 | 60,531 | 22 | 121,063 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,532 | 22 | 121,064 |
"Correct Solution:
```
N = int(input())
binN = format(N, "b").zfill(60)
mod = 7 + 10 ** 9
DP = [[0, 0, 0] for _ in range(60)]
if binN[0] == "0":
DP[0] = [1, 0, 0]
else:
DP[0] = [1, 1, 0]
for i in range(1, 60):
if binN[i] == "0":
DP[i][0] = (DP[i-1][0] + DP[i-1][1]) % mod
DP[i][1] = DP[i-1][1]
DP[i][2] = (DP[i-1][2] * 3 + DP[i-1][1]) % mod
else:
DP[i][0] = DP[i-1][0]
DP[i][1] = (DP[i-1][0] + DP[i-1][1]) % mod
DP[i][2] = (DP[i-1][2] * 3 + DP[i-1][1] * 2) % mod
print(sum(DP[-1]) % mod)
``` | output | 1 | 60,532 | 22 | 121,065 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,533 | 22 | 121,066 |
"Correct Solution:
```
N= int(input())
bit = bin(N)[2:]
k = len(bit)
dp = [[0] * 2 for i in range(len(bit) + 1)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(k):
j = int(bit[k-i-1])
a,b,c = [(1,0,0),(1,1,0)][j]
d,e,f = [(1, 1, 1),(0, 1, 2)][j]
dp[i+1][0] = (a * dp[i][0] + b * dp[i][1] + c * 3**i) % 1000000007
dp[i+1][1] = d * dp[i][0] + e * dp[i][1] + f * 3**i % 1000000007
print(dp[k][0])
``` | output | 1 | 60,533 | 22 | 121,067 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179 | instruction | 0 | 60,534 | 22 | 121,068 |
"Correct Solution:
```
n = int(input())
b = bin(n)[2:]
MOD = 10 ** 9 + 7
dp1, dp2, dp3 = 1, 0, 0
for d in b:
if d == '1':
dp1, dp2, dp3 = dp1 % MOD, (dp1 + dp2) % MOD, (dp2 * 2 + dp3 * 3) % MOD
else:
dp1, dp2, dp3 = (dp1 + dp2) % MOD, dp2, (dp2 + dp3 * 3) % MOD
print((dp1 + dp2 + dp3) % MOD)
``` | output | 1 | 60,534 | 22 | 121,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179
Submitted Solution:
```
n=int(input())
mod=10**9+7
bt=bin(n)[2:]
k=len(bt)
dp=[[0]*2 for i in range(k+1)]
dp[0][0]=1
dp[0][1]=1
for i in range(k):
j=int(bt[k-i-1])
a,b,c=[(1,0,0),(1,1,0)][j]
d,e,f=[(1,1,1),(0,1,2)][j]
dp[i+1][0]=(a*dp[i][0]+b*dp[i][1]+c*(3**i))%mod
dp[i+1][1]=(d*dp[i][0]+e*dp[i][1]+f*(3**i))%mod
print(dp[k][0])
``` | instruction | 0 | 60,535 | 22 | 121,070 |
Yes | output | 1 | 60,535 | 22 | 121,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
* 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Examples
Input
3
Output
5
Input
1422
Output
52277
Input
1000000000000000000
Output
787014179
Submitted Solution:
```
tg = 0
ls = 1
md = 2
MOD = 10**9+7
n = int(input())
l = n.bit_length()
# ^tight/loose, +tight/loose/middle, bitnum
dp = [[[0] * l for _ in range(3)] for i in range(2)]
for i in range(l):
if i == 0:
# 10
dp[tg][tg][0] = 1
# 00
dp[ls][md][0] = 1
continue
# i-th bit is 1
if n & (1 << l - i - 1):
# 10
dp[tg][tg][i] += dp[tg][tg][i-1]
dp[ls][ls][i] += dp[ls][md][i-1]
dp[ls][tg][i] += dp[ls][tg][i-1]
dp[ls][ls][i] += dp[ls][ls][i-1]
# 00
dp[ls][md][i] += dp[tg][tg][i-1]
dp[ls][ls][i] += dp[ls][md][i-1]
dp[ls][md][i] += dp[ls][tg][i-1]
dp[ls][ls][i] += dp[ls][ls][i-1]
# 11
dp[ls][md][i] += dp[ls][md][i-1]
dp[ls][ls][i] += dp[ls][ls][i-1]
# i-th bit is 0
else:
# 10
dp[ls][ls][i] += dp[ls][ls][i-1]
dp[ls][md][i] += dp[ls][md][i-1]
# 00
dp[tg][tg][i] += dp[tg][tg][i-1]
dp[ls][ls][i] += dp[ls][md][i-1]
dp[ls][ls][i] += dp[ls][ls][i-1]
dp[ls][tg][i] += dp[ls][tg][i-1]
# 11
dp[ls][tg][i] += dp[ls][md][i-1]
dp[ls][ls][i] += dp[ls][ls][i-1]
for xp in dp:
for pp in xp:
pp[i] %= MOD
ans = 0
for xp in dp:
for pp in xp:
ans += pp[-1]
ans %= MOD
print(ans)
``` | instruction | 0 | 60,537 | 22 | 121,074 |
Yes | output | 1 | 60,537 | 22 | 121,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob received a gift from his friend recently — a recursive sequence! He loves this sequence very much and wants to play with it.
Let f_1, f_2, …, f_i, … be an infinite sequence of positive integers. Bob knows that for i > k, f_i can be obtained by the following recursive equation:
$$$f_i = \left(f_{i - 1} ^ {b_1} ⋅ f_{i - 2} ^ {b_2} ⋅ ⋅⋅⋅ ⋅ f_{i - k} ^ {b_k}\right) mod p,$$$
which in short is
$$$f_i = \left(∏_{j = 1}^{k} f_{i - j}^{b_j}\right) mod p,$$$
where p = 998 244 353 (a widely-used prime), b_1, b_2, …, b_k are known integer constants, and x mod y denotes the remainder of x divided by y.
Bob lost the values of f_1, f_2, …, f_k, which is extremely troublesome – these are the basis of the sequence! Luckily, Bob remembers the first k - 1 elements of the sequence: f_1 = f_2 = … = f_{k - 1} = 1 and the n-th element: f_n = m. Please find any possible value of f_k. If no solution exists, just tell Bob that it is impossible to recover his favorite sequence, regardless of Bob's sadness.
Input
The first line contains a positive integer k (1 ≤ k ≤ 100), denoting the length of the sequence b_1, b_2, …, b_k.
The second line contains k positive integers b_1, b_2, …, b_k (1 ≤ b_i < p).
The third line contains two positive integers n and m (k < n ≤ 10^9, 1 ≤ m < p), which implies f_n = m.
Output
Output a possible value of f_k, where f_k is a positive integer satisfying 1 ≤ f_k < p. If there are multiple answers, print any of them. If no such f_k makes f_n = m, output -1 instead.
It is easy to show that if there are some possible values of f_k, there must be at least one satisfying 1 ≤ f_k < p.
Examples
Input
3
2 3 5
4 16
Output
4
Input
5
4 7 1 5 6
7 14187219
Output
6
Input
8
2 3 5 6 1 7 9 10
23333 1
Output
1
Input
1
2
88888 66666
Output
-1
Input
3
998244352 998244352 998244352
4 2
Output
-1
Input
10
283 463 213 777 346 201 463 283 102 999
2333333 6263423
Output
382480067
Note
In the first sample, we have f_4 = f_3^2 ⋅ f_2^3 ⋅ f_1^5. Therefore, applying f_3 = 4, we have f_4 = 16. Note that there can be multiple answers.
In the third sample, applying f_7 = 1 makes f_{23333} = 1.
In the fourth sample, no such f_1 makes f_{88888} = 66666. Therefore, we output -1 instead. | instruction | 0 | 60,672 | 22 | 121,344 |
Tags: math, matrices, number theory
Correct Solution:
```
from math import ceil, sqrt
p = 998244353
def bsgs(g, h):
'''
Solve for x in h = g^x mod p given a prime p.
'''
N = ceil(sqrt(p - 1)) # phi(p) is p-1 if p is prime
# Store hashmap of g^{1...m} (mod p). Baby step.
tbl = {pow(g, i, p): i for i in range(N)}
# Precompute via Fermat's Little Theorem
c = pow(g, N * (p - 2), p)
# Search for an equivalence in the table. Giant step.
for j in range(N):
y = (h * pow(c, j, p)) % p
if y in tbl:
return j * N + tbl[y]
def gcd(a, b):
return b if a % b == 0 else gcd(b, a % b)
def xgcd(a, b):
"""return (g, x, y) such that a*x + b*y = g = gcd(x, y)"""
x0, x1, y0, y1 = 0, 1, 1, 0
while a != 0:
q, b, a = b // a, a, b % a
y0, y1 = y1, y0 - q * y1
x0, x1 = x1, x0 - q * x1
return b, x0, y0
def inv(a, m):
g, x, y = xgcd(a, m)
if g != 1:
return None
return (x + m) % m
# solve a = bx (mod m)
def div(a, b, m):
k = gcd(b, m)
if a % k != 0:
return None
ak = a // k
bk = b // k
mk = m // k
inv_bk = inv(bk, mk)
return (ak * inv_bk) % mk
def matmul(A, B):
m = len(A)
C = [[0 for _ in range(m)] for _ in range(m)]
for i in range(m):
for k in range(m):
for j in range(m):
C[i][j] += A[i][k]*B[k][j] % (p-1)
for i in range(m):
for j in range(m):
C[i][j] %= (p-1)
return C
def Id(n):
return [[int(i == j) for i in range(n)] for j in range(n)]
def matexp(A, n):
if n == 0:
return Id(len(A))
h = matexp(A, n//2)
R = matmul(h, h)
if n % 2 == 1:
R = matmul(A, R)
return R
def solve():
k = int(input())
A = [[0 for _ in range(k)] for _ in range(k)]
A[0] = [int(x) for x in input().split()]
for i in range(k-1):
A[i+1][i] = 1
n, v = [int(x) for x in input().split()]
e = matexp(A, n-k)[0][0]
g = 3
u = bsgs(g, v)
x = div(u, e, p-1)
if x is not None:
print(pow(g, x, p))
else:
print(-1)
if __name__ == '__main__':
solve()
``` | output | 1 | 60,672 | 22 | 121,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3]. | instruction | 0 | 60,705 | 22 | 121,410 |
Tags: math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
p=dict()
count1=0
count2=0
count0=0
ans=0
for i in a:
if i%3==1:
count1+=1
elif i%3==2:
count2+=1
elif i%3==0:
count0+=1
if count2<count1:
ans=count0+count2+(count1-count2)//3
else:
ans=count0+count1+(count2-count1)//3
print(ans)
``` | output | 1 | 60,705 | 22 | 121,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3]. | instruction | 0 | 60,706 | 22 | 121,412 |
Tags: math
Correct Solution:
```
def main():
answer = ""
for i in range(int(input())):
n = int(input())
nums = list(map(int,input().split(' ')))
new_nums = []
for i in nums:
new_nums.append(i%3)
counts = [0,0,0]
for i in new_nums:
counts[i] += 1
ans = counts[0] + min(counts[1],counts[2]) + (max(counts[1],counts[2])-min(counts[1],counts[2]))//3
answer += str(ans) + "\n"
print(answer)
main()
``` | output | 1 | 60,706 | 22 | 121,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3]. | instruction | 0 | 60,707 | 22 | 121,414 |
Tags: math
Correct Solution:
```
# B. Merge it!
t=int(input())
for i in range(t):
n=int(input())
k = [int(i) % 3 for i in input().split()]
a=k.count(0)
b=k.count(1)
c=k.count(2)
m=a+min(b,c)+abs(b-c)//3
print(m)
``` | output | 1 | 60,707 | 22 | 121,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3]. | instruction | 0 | 60,708 | 22 | 121,416 |
Tags: math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
c=[0,0,0]
for i in a:
if i%3==0:
c[0]+=1
elif i%3==1:
c[1]+=1
elif i%3==2:
c[2]+=1
print(c[0]+min(c[1],c[2])+(max(c[1],c[2])-min(c[1],c[2]))//3)
``` | output | 1 | 60,708 | 22 | 121,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3]. | instruction | 0 | 60,709 | 22 | 121,418 |
Tags: math
Correct Solution:
```
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
nd3=0
nd2=0
nd1=0
for x in a:
if(x%3==0):
nd3+=1
elif(x%3==1):
nd1+=1
else:
nd2+=1
res=0
res+=nd3
if(nd1>nd2):
res+=nd2
res+=(nd1-nd2)//3
elif(nd1==nd2):
res+=nd2
else:
res+=nd1
res+=(nd2-nd1)//3
print(res)
``` | output | 1 | 60,709 | 22 | 121,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3]. | instruction | 0 | 60,711 | 22 | 121,422 |
Tags: math
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = [int(x) for x in input().split()]
mod0 = mod1= mod2 = 0
for i in a:
if i%3==0:
mod0+=1
elif i%3==1:
mod1+=1
else:
mod2+=1
if mod1==mod2:
mod0+=mod1
elif mod1>mod2:
mod1-=mod2
mod0+=mod2
mod0+=mod1//3
else:
mod2-=mod1
mod0+=mod1
mod0+=mod2//3
print(mod0)
``` | output | 1 | 60,711 | 22 | 121,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3].
Submitted Solution:
```
#code
for i in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
c=0
y1=0
y2=0
for x in l:
if x%3==0:
c+=1
elif x%3==1:
y1+=1
else:
y2+=1
m=min(y1,y2)
y1-=m
y2-=m
x=y1//3+y2//3
print(x+c+m)
``` | instruction | 0 | 60,713 | 22 | 121,426 |
Yes | output | 1 | 60,713 | 22 | 121,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3].
Submitted Solution:
```
qcount=int(input())
for qnumb in range(qcount):
n= int(input())
k=list(map(int,input().split()))
w = [0,0,0]
for i in k:
w[i%3]+=1
z=min([w[1],w[2]])
w[0]+=z
w[0]+=(w[1]-z)//3
w[0]+=(w[2]-z)//3
print(w[0])
``` | instruction | 0 | 60,715 | 22 | 121,430 |
Yes | output | 1 | 60,715 | 22 | 121,431 |
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