text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Tags: binary search, brute force, dp, math
Correct Solution:
```
hashmap = [2]
counter = 5
for i in range(0, 10**5):
hashmap.append(hashmap[-1]+counter)
counter+=3
for _ in range(0, int(input())):
n = int(input())
i=0
res=0
while True:
if n < 2:
print(res)
break
if hashmap[i] == n:
res+=1
print(res)
break
if hashmap[i] > n:
res +=1
n-=hashmap[i-1]
i=0
else:
i+=1
```
| 1,000 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Tags: binary search, brute force, dp, math
Correct Solution:
```
import sys
input=sys.stdin.readline
import bisect as bi
I = lambda : list(map(int,input().split()))
a=[2];i=1;k=4;x=2
while x<10**9:
x+=i+k
a.append(x)
i+=1;k+=2
for _ in range(int(input())):
n,=I();ct=0
x=bi.bisect(a,n)
while n>1:
n-=a[x-1];ct+=1
x=bi.bisect(a,n)
#print(n,bi.bisect(a,n))
print(ct)
```
| 1,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Tags: binary search, brute force, dp, math
Correct Solution:
```
import bisect
start = [2]
for i in range(2,40000):
bot = (i*2) + i-1
start.append( start[-1] + bot )
def solve(num):
ans = 0
while num > 1:
pos = bisect.bisect_right( start , num )
num = num - start[ pos - 1 ]
ans += 1
return ans
# input = open('file.txt').readline
for _ in range( int( input() ) ):
print( solve( int( input() ) ) )
```
| 1,002 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Tags: binary search, brute force, dp, math
Correct Solution:
```
import bisect
pre=[]
dup=0
i=1
while(dup<1000000001):
temp=(i*((3*i)+1))//2
pre.append(temp)
dup=temp
i+=1
t=int(input())
for _ in range(t):
n=int(input())
cnt=0
while(True):
if(n<2):
break
x = bisect.bisect_left(pre,n,0,len(pre))
if(pre[x]==n):
cnt+=1
break
else:
x=x-1
n=n-pre[x]
cnt+=1
print(cnt)
```
| 1,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Tags: binary search, brute force, dp, math
Correct Solution:
```
from math import sqrt
def main():
t = int(input())
for __ in range(t):
n = int(input())
count = 0
while n >= 2:
count += 1
h = int((sqrt((24*n)+1)-1)/6)
#print(h)
n -= (h*((3*h)+1))//2
print(count)
if __name__ == '__main__':
main()
```
| 1,004 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
import sys
import math
input = lambda : sys.stdin.readline().strip()
t = int(input())
def cards(h):
return (3*h**2+h)//2
while t:
t-=1
n = int(input())
ans = 0
h = 1
while cards(h+1)<=n:
h+=1
while n>0 and h>0:
if cards(h)<=n:
n-=cards(h)
ans+=1
else:
h-=1
print(ans)
```
Yes
| 1,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
t = int(input())
for asd in range(t):
n = int(input())
out = 0
while(n > 1):
f=2
count = 1
num = f
while(f <= n):
count += 1
num = f
f = f+((3*count) -1)
n = n - num
# print(num)
out = out+1
# print(f)
print(out)
# print("a")
```
Yes
| 1,006 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
#importing bisect
import bisect
# defining input == t
t = int(input())
# Logical Execution
pyramid_heights = [2]
pyramid_height = 2
# sum
addition = 5
# comparison
while pyramid_height < 1000000000:
pyramid_height = pyramid_height + addition
# sum of += 3
addition += 3
pyramid_heights.append(pyramid_height)
# again while execution
while t !=0:
height = int(input())
#print answer
ans = 0
pos = (bisect.bisect_left(pyramid_heights, height))
# execution 2
while height >= 0:
# execution of if
if height == 0:
break
if height < pyramid_heights[0]:
break
else:
if height - pyramid_heights[pos] >= 0:
height = height - pyramid_heights[pos]
ans += 1
else:
pos -= 1
# printing of 2nd answer ts
print(ans)
t -= 1
```
Yes
| 1,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
import bisect
t=int(input())
const=[(3*(i+1)**2+i+1)//2 for i in range(33000)]
for i in range(t):
n=int(input())
cnt=0
while n>=2:
d=bisect.bisect_right(const,n)
n-=const[d-1]
cnt+=1
print(cnt)
```
Yes
| 1,008 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
n=int(input())
a=[]
for i in range(n):
a.append(int(input()))
k=5
p=2
d=0
for i in range(n):
if a[i]>=2:
while a[i]>=0:
while a[i]>=p:
p+=k
k+=3
p-=3
a[i]-=p
d+=1
k=5
p=2
a[i]=d
d=0
else:
a[i]=0
for i in range(n):
print(a[i])
```
No
| 1,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
dp = [0, 2]
i = 1
while dp[i] < n:
i+=1
dp.append(dp[i-1] +3*i - 1)
#i-=1
ans = 0
for j in range(i,-1,-1):
if dp[j] <=n:
n-=dp[j]
ans+=1
if n <= 1:
break
print(ans)
```
No
| 1,010 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
BUFSIZE = 8192
import os
import sys
import math
from io import BytesIO, IOBase
from bisect import bisect_left #c++ lowerbound bl(array,element)
from bisect import bisect_right #c++ upperbound br(array,element)
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")
a=[]
import bisect
for i in range(10**4):
s=(i*(i+1)//2)*3-i
a.append(s)
#print(a)
for _ in range(int(input())):
c=0
n=int(input())
while n>=2:
q=bisect.bisect_right(a,n)
n=n-a[q-1]
c+=1
print(c)
```
No
| 1,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A card pyramid of height 1 is constructed by resting two cards against each other. For h>1, a card pyramid of height h is constructed by placing a card pyramid of height h-1 onto a base. A base consists of h pyramids of height 1, and h-1 cards on top. For example, card pyramids of heights 1, 2, and 3 look as follows:
<image>
You start with n cards and build the tallest pyramid that you can. If there are some cards remaining, you build the tallest pyramid possible with the remaining cards. You repeat this process until it is impossible to build another pyramid. In the end, how many pyramids will you have constructed?
Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains a single integer n (1≤ n≤ 10^9) — the number of cards.
It is guaranteed that the sum of n over all test cases does not exceed 10^9.
Output
For each test case output a single integer — the number of pyramids you will have constructed in the end.
Example
Input
5
3
14
15
24
1
Output
1
2
1
3
0
Note
In the first test, you construct a pyramid of height 1 with 2 cards. There is 1 card remaining, which is not enough to build a pyramid.
In the second test, you build two pyramids, each of height 2, with no cards remaining.
In the third test, you build one pyramid of height 3, with no cards remaining.
In the fourth test, you build one pyramid of height 3 with 9 cards remaining. Then you build a pyramid of height 2 with 2 cards remaining. Then you build a final pyramid of height 1 with no cards remaining.
In the fifth test, one card is not enough to build any pyramids.
Submitted Solution:
```
a = [0, 2]
def triangle(h):
cards = 0
if len(a) > h:
cards = a[h]
else:
cards = triangle(h - 1) + 2 * h + h - 1
a.insert(h, cards)
return cards
k = int(input("Enter value "))
p = []
for i in range(0,k):
p.insert(i, int(input()))
def max(n):
m=0
while triangle(m)<=n:
m+=1
return n-triangle(m-1)
def number(b):
count=0
while b>=2:
b=max(b)
count+=1
return count
for s in range(0,k):
print(number(p[s]))
```
No
| 1,012 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter as count_values
import math
def input_data():
return list(map(int,input().split()))
def source():
string = input()
data = []
output = ''
j = 0
for i in range(2,len(string)+1,2):
data.append(string[j:i])
j = i
s = len(data)
i = 0
if s == 1:
return string
while True:
if i == s-1:
return output+data[i]
else:
output = output+data[i][0]
i +=1
return output
def main():
for _ in range(int(input())):
print(source())
# region fastio
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()
```
| 1,013 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
for z in range(int(input())):
b = list(input())
a = [b[0]]
for x in range(1, len(b), 2):
a.append(b[x])
print(''.join(a))
```
| 1,014 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
from sys import stdin
for _ in range (int(stdin.readline())):
b=stdin.readline().strip()
a=b[0]
for i in range (1,len(b)-1,2):
a+=b[i]
a+=b[len(b)-1]
print(a)
```
| 1,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
s=input()
we=len(s)
if we==2:
print(s)
else:
ans=s[0]
for i in range(1,we-1,2):
ans+=s[i]
ans+=s[-1]
print(ans)
```
| 1,016 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
n = int(input())
for i in range(n):
temp = input()
print(temp[0] + temp[1:len(temp) - 1:2] + temp[len(temp)-1])
```
| 1,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
t = int(input())
for z in range(t):
s = input()
if(len(s)<3):
print(s)
else:
r=s[0]
i=1
while(i<len(s)):
r+=s[i]
i+=2
print(r)
```
| 1,018 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
t=int(input())
for i in range(0,t):
b=input()
l=len(b)
a=b[0]
j=1
while j<(l-1) :
a=a+b[j]
j=j+2
a=a+b[l-1]
print(a)
```
| 1,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Tags: implementation, strings
Correct Solution:
```
def solve(b):
a = list()
l = len(b)
for i in range(0, l, 2):
a.append(b[i])
# append the last character if the length of b is even, it would be missed up the above loop
if l % 2 == 0:
a.append(b[-1])
return "".join(a)
if __name__ == "__main__":
t = int(input())
results = list()
for _ in range(0, t):
b = input()
results.append(solve(b))
for result in results:
print(result)
```
| 1,020 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
t=int(input())
ans=[]
for _ in range(t):
res=[]
s=list(input())
last=s[-1]
res.append(s[0])
s=s[1:]
s=s[:-1]
for i in s[::2]:
res.append(i)
res.append(last)
ans.append(''.join(res))
print('\n'.join(ans))
```
Yes
| 1,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
n = int(input())
for i in range(n):
string = input()
result = ""
for i in range(0, len(string), 2):
result += string[i]
result += string[-1]
print(result)
```
Yes
| 1,022 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
t = int(input())
for _ in range(t):
s = input()
# x = []
res= ''
x= [(s[i:i+2]) for i in range(0, len(s), 2)]
# print(x)
if len(x) ==1:
print(x[0])
else:
for i in range(len(x)-1):
if x[i][1]== x[i+1][0]:
res+=x[i][0]
else:
res+=x[i]
res+=x[-1]
print(res)
```
Yes
| 1,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
T = int(input())
for _ in range(T):
b = input()
print(b[0] + b[1::2])
```
Yes
| 1,024 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
def f(s,n):
res=""
res2=""
flag=1
for i in range(1,n):
if s[i]!=s[i-1]:
flag=0
break
if flag==0:
for i in range(1,n):
if s[i]!=s[i-1]:
res+=s[i-1]
else:
res2+=s[i]
else:
res+=s[n-1]
return res
else:
return s[:(n//2)+1]
for _ in range(int(input())):
s=input()
n=len(s)
print(f(s,n))
```
No
| 1,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
t=int(input())
while(t>0):
t-=1
n=input()
i=1
s=n[0]
while(i<len(n)-1):
s+=n[i]
i=i+2
s+=n[-1]
print(s)
```
No
| 1,026 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
n = input().split()
num = n[0]
cases = n[1:]
for case in cases:
new_case = ''
i = 0
while i < len(case):
if i == 0:
new_case += case[i]
i += 1
continue
letter = case[i]
try:
next_letter = case[i+1]
except IndexError:
if case[i-1] != letter:
new_case += letter
if letter == next_letter:
new_case += letter
i += 2
continue
i += 1
print(new_case)
```
No
| 1,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice guesses the strings that Bob made for her.
At first, Bob came up with the secret string a consisting of lowercase English letters. The string a has a length of 2 or more characters. Then, from string a he builds a new string b and offers Alice the string b so that she can guess the string a.
Bob builds b from a as follows: he writes all the substrings of length 2 of the string a in the order from left to right, and then joins them in the same order into the string b.
For example, if Bob came up with the string a="abac", then all the substrings of length 2 of the string a are: "ab", "ba", "ac". Therefore, the string b="abbaac".
You are given the string b. Help Alice to guess the string a that Bob came up with. It is guaranteed that b was built according to the algorithm given above. It can be proved that the answer to the problem is unique.
Input
The first line contains a single positive integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then t test cases follow.
Each test case consists of one line in which the string b is written, consisting of lowercase English letters (2 ≤ |b| ≤ 100) — the string Bob came up with, where |b| is the length of the string b. It is guaranteed that b was built according to the algorithm given above.
Output
Output t answers to test cases. Each answer is the secret string a, consisting of lowercase English letters, that Bob came up with.
Example
Input
4
abbaac
ac
bccddaaf
zzzzzzzzzz
Output
abac
ac
bcdaf
zzzzzz
Note
The first test case is explained in the statement.
In the second test case, Bob came up with the string a="ac", the string a has a length 2, so the string b is equal to the string a.
In the third test case, Bob came up with the string a="bcdaf", substrings of length 2 of string a are: "bc", "cd", "da", "af", so the string b="bccddaaf".
Submitted Solution:
```
t = int(input())
for i in range(t):
b = input()
for x in range(1,len(b)-1):
a = b[0] + b[1:len(b)-1:2] + b[len(b)-1]
print(a)
```
No
| 1,028 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
def nf(nn):
global fact
sim = True
for k in fact:
if nn % k == 0:
sim = False
break
if sim: fact.append(nn)
def rez(nn):
global df
k = [6, 10, 14]
j = 3
while nn - sum(k) > 0 and (nn - sum(k)) in k:
k[2] = df[j]
j += 1
k4 = nn - sum(k)
r = ""
if k4 > 0:
k.append(k4)
for n in k:
r += str(n) + " "
return r
fact = [2, 3, 5, 7]
for i in range(2, 7):
nf(i * 6 - 1)
nf(i * 6 + 1)
df = []
km = len(fact)
for i in range(km-1):
for j in range(i+1, km):
df.append(fact[i]*fact[j])
df.sort()
for i in range(int(input())):
x = int(input())
if x < 31: print('NO')
else:
print('YES')
print(rez(x))
```
| 1,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
if n<31:
print("NO")
elif n==44:
print("YES")
print(6,7,10,21)
elif n==40:
print("YES")
print(6,10,21,3)
elif n==36:
print("YES")
print(6,14,15,1)
else:
print("YES")
print(6,10,14,n-30)
```
| 1,030 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
import bisect
import os
import gc
import sys
from io import BytesIO, IOBase
from collections import Counter
from collections import deque
import heapq
import math
import statistics
def sin():
return input()
def ain():
return list(map(int, sin().split()))
def sain():
return input().split()
def iin():
return int(sin())
MAX = float('inf')
MIN = float('-inf')
MOD = 1000000007
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
s = set()
for p in range(2, n+1):
if prime[p]:
s.add(p)
return s
def readTree(n, m):
adj = [deque([]) for _ in range(n+1)]
for _ in range(m):
u,v = ain()
adj[u].append(v)
adj[v].append(u)
return adj
def main():
for _ in range(iin()):
n = iin()
if n<=30:
print("NO")
else:
print("YES")
if n-30 == 6:
print(6,10,15,n-31)
elif n-30 == 14:
print(6,10,15, n-31)
elif n-30 == 10:
print(6,14,15,n-35)
else:
print(6,10,14,n-30)
# Fast IO Template starts
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")
if os.getcwd() == 'D:\\code':
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# Fast IO Template ends
if __name__ == "__main__":
main()
```
| 1,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
if n<=30:
print('NO')
else:
print('YES')
if n==36:
print(5,6,10,15)
elif n==44:
print(6,7,10,21)
elif n==40:
print(6,10,15,9)
else:
print(6,10,14,n-30)
```
| 1,032 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
ans=[6,10,14,15,21,22,26]
check=False
for i in ans:
for j in ans:
for k in ans:
if n>i+j+k and n-i-j-k!=i and n-i-j-k!=j and n-i-j-k!=k and i!=j and j!=k and i!=k:
print("YES")
print(i,j,k,n-i-j-k)
check=True
break
if check:
break
if check:
break
if not check:
print("NO")
```
| 1,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
from sys import *
t = int(stdin.readline())
while t > 0:
n = int(stdin.readline())
if n >30:
print("YES")
if n-30 == 6:
print(6,10,15,n-31)
elif n-30 == 10:
print(6,10,15,n-31)
elif n-30 == 14:
print(6,10,21,n-37)
else:
print(6,10,14,n-30)
else:
print("NO")
t -= 1
```
| 1,034 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
t = int(input())
for aSRD in range(t):
n = int(input())
if n > (6+10+14):
k = n - 6 - 10 - 14
print("YES")
if k == 6 or k == 10 or k == 14:
print(6, 10, 15, k-1)
else:
print(6, 10, 14, k)
else:
print("NO")
```
| 1,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
t=int(input())
while(t>0):
t-=1
n=int(input())
if n<31:
print("NO")
else:
print("Yes")
if n==36:
print(15, 10, 6, 5)
elif n==40:
print(15, 10, 6, 9)
elif n==44:
print(15, 10, 6, 13)
else:
print(14, 10, 6, n-30)
```
| 1,036 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import combinations
pr=stdout.write
import heapq
raw_input = stdin.readline
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return (map(int,stdin.read().split()))
range = xrange # not for python 3.0+
arr=[6,10,14,15,21,22]
for t in range(ni()):
n=ni()
f=0
for com in combinations(arr,3):
tp=n-sum(com)
if tp>0 and tp not in com:
f=1
break
if f:
pr('YES\n')
pa(com+(tp,))
else:
pr('NO\n')
```
| 1,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
I = input
pr = print
def main():
for _ in range(int(I())):
n = int(I())
ar = [6,10,14]
if n<31:pr('NO');continue
else:
if (n-30) in ar: n-=5;ar[1]+=5
pr('YES');pr(*ar,n-30)
main()
```
Yes
| 1,038 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
def nearPrime(N):
if N <= 30:
print("NO")
else:
print("YES")
if N == 36 or N == 40 or N == 44:
print(6, 10, 15, N-31)
else:
print(6, 10, 14, N-30)
for i in range(int(input())):
N = int(input())
nearPrime(N)
```
Yes
| 1,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
tests = int(input())
for t in range(tests):
n = int(input())
if n < 31:
print('NO')
else:
print('YES')
if n == 40:
res = [6, 14, 15, 5]
elif n == 44:
res = [6,10,15,13]
elif n == 36:
res = [5, 6, 10, 15]
else:
res = [6,10,14, n-30]
print(' '.join(list(map(str,res))))
```
Yes
| 1,040 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
if n<=30:
print("NO")
elif n==36 or n==44 or n==40:
print("YES")
print(6, 10, 15, n-31)
else:
print("YES")
print(6, 10, 14, n-30)
```
Yes
| 1,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
import sys
input=sys.stdin.readline
T=int(input())
for _ in range(T):
#n,m=map(int,input().split())
n=int(input())
if (n<=30):
print("NO")
else:
print("YES")
print(6,10,14,n-30)
```
No
| 1,042 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
if n > 30:
print("YES")
print(6, 10, 14, n-30)
else:
print('NO')
```
No
| 1,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
import math
t = int(input())
ans = []
def solve(n):
if n < 30:
return ["NO"]
else:
if len({6,10,14,n-30})<4:
return [6,10,15,n-31]
if len({6,10,15,n-31})<4:
return [6,10,21,n-37]
return [6,10,14,n-30]
for i in range(t):
n = int(input())
ans.append(solve(n))
#print(ans)
for test in ans:
if len(test) == 1:
print(test[0])
else:
print("YES")
print(test[0],test[1],test[2],test[3])
```
No
| 1,044 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
import math
def simpl(x):
for i in range(2, int(math.sqrt(x))):
if x % i == 0:
return False
return True
if __name__ == "__main__":
t = int(input())
a = []
for i in range(10000):
if simpl(i):
a.append(i)
for i in range(t):
x = int(input())
if x >= 18 and x - 18 not in a:
print("YES")
print(6, 6, 6, x - 18)
else:
print("NO")
```
No
| 1,045 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.
Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.
Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.
Uncle Bogdan easily solved the task and joined the crew. Can you do the same?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.
Output
For each test case print:
* YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them);
* NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime.
You can print each character of YES or NO in any case.
Example
Input
7
7
23
31
36
44
100
258
Output
NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
Note
In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.
In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.
In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.
In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.
In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.
In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
pr=stdout.write
import heapq
raw_input = stdin.readline
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return (map(int,stdin.read().split()))
range = xrange # not for python 3.0+
for t in range(ni()):
n=ni()
if n<=30:
pr('NO\n')
continue
pr('YES\n')
pa([6,10,14,n-30])
```
No
| 1,046 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a_1, a_2, …, a_n of non-negative integers.
You need to find the largest number m of triples (i_1, j_1, k_1), (i_2, j_2, k_2), ..., (i_m, j_m, k_m) such that:
* 1 ≤ i_p < j_p < k_p ≤ n for each p in 1, 2, …, m;
* a_{i_p} = a_{k_p} = 0, a_{j_p} ≠ 0;
* all a_{j_1}, a_{j_2}, …, a_{j_m} are different;
* all i_1, j_1, k_1, i_2, j_2, k_2, …, i_m, j_m, k_m are different.
Input
The first line of input contains one integer t (1 ≤ t ≤ 500 000): the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 500 000).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n).
The total sum of n is at most 500 000.
Output
For each test case, print one integer m: the largest number of proper triples that you can find.
Example
Input
8
1
1
2
0 0
3
0 1 0
6
0 0 1 2 0 0
6
0 1 0 0 1 0
6
0 1 3 2 0 0
6
0 0 0 0 5 0
12
0 1 0 2 2 2 0 0 3 3 4 0
Output
0
0
1
2
1
1
1
2
Note
In the first two test cases, there are not enough elements even for a single triple, so the answer is 0.
In the third test case we can select one triple (1, 2, 3).
In the fourth test case we can select two triples (1, 3, 5) and (2, 4, 6).
In the fifth test case we can select one triple (1, 2, 3). We can't select two triples (1, 2, 3) and (4, 5, 6), because a_2 = a_5.
Submitted Solution:
```
from sys import stdout,stdin
from collections import defaultdict,deque
import math
t=int(stdin.readline())
for _ in range(t):
n=int(stdin.readline())
#n,e=map(int,stdin.readline().split())
l=list(map(int,stdin.readline().split()))
d=defaultdict(int)
if n<3:
print(0)
else:
for i in l:
d[i]+=1
nl=len(d)
if d[i]>0:
nl-=1
z=d[0]//2
if nl==0:
print(0)
else:
print(min(z,nl))
```
No
| 1,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a_1, a_2, …, a_n of non-negative integers.
You need to find the largest number m of triples (i_1, j_1, k_1), (i_2, j_2, k_2), ..., (i_m, j_m, k_m) such that:
* 1 ≤ i_p < j_p < k_p ≤ n for each p in 1, 2, …, m;
* a_{i_p} = a_{k_p} = 0, a_{j_p} ≠ 0;
* all a_{j_1}, a_{j_2}, …, a_{j_m} are different;
* all i_1, j_1, k_1, i_2, j_2, k_2, …, i_m, j_m, k_m are different.
Input
The first line of input contains one integer t (1 ≤ t ≤ 500 000): the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 500 000).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n).
The total sum of n is at most 500 000.
Output
For each test case, print one integer m: the largest number of proper triples that you can find.
Example
Input
8
1
1
2
0 0
3
0 1 0
6
0 0 1 2 0 0
6
0 1 0 0 1 0
6
0 1 3 2 0 0
6
0 0 0 0 5 0
12
0 1 0 2 2 2 0 0 3 3 4 0
Output
0
0
1
2
1
1
1
2
Note
In the first two test cases, there are not enough elements even for a single triple, so the answer is 0.
In the third test case we can select one triple (1, 2, 3).
In the fourth test case we can select two triples (1, 3, 5) and (2, 4, 6).
In the fifth test case we can select one triple (1, 2, 3). We can't select two triples (1, 2, 3) and (4, 5, 6), because a_2 = a_5.
Submitted Solution:
```
t = int(input())
def rainbow(n, arr):
zeros = []
for i in range(n):
if arr[i] == 0: zeros.append(i)
Count, p = 0, 0
start, end = 0, len(zeros) - 1
while start < end:
if zeros[end] - zeros[start] <= end - start + p: break
start, end = start + 1, end - 1
Count += 1
if zeros[start] == zeros[start - 1] + 1 and zeros[end] == zeros[end + 1] - 1:
p += 1
return Count
for i in range(t):
n = int(input())
arr = list(map(int, input().split()))
print(rainbow(n, arr))
```
No
| 1,048 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a_1, a_2, …, a_n of non-negative integers.
You need to find the largest number m of triples (i_1, j_1, k_1), (i_2, j_2, k_2), ..., (i_m, j_m, k_m) such that:
* 1 ≤ i_p < j_p < k_p ≤ n for each p in 1, 2, …, m;
* a_{i_p} = a_{k_p} = 0, a_{j_p} ≠ 0;
* all a_{j_1}, a_{j_2}, …, a_{j_m} are different;
* all i_1, j_1, k_1, i_2, j_2, k_2, …, i_m, j_m, k_m are different.
Input
The first line of input contains one integer t (1 ≤ t ≤ 500 000): the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 500 000).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n).
The total sum of n is at most 500 000.
Output
For each test case, print one integer m: the largest number of proper triples that you can find.
Example
Input
8
1
1
2
0 0
3
0 1 0
6
0 0 1 2 0 0
6
0 1 0 0 1 0
6
0 1 3 2 0 0
6
0 0 0 0 5 0
12
0 1 0 2 2 2 0 0 3 3 4 0
Output
0
0
1
2
1
1
1
2
Note
In the first two test cases, there are not enough elements even for a single triple, so the answer is 0.
In the third test case we can select one triple (1, 2, 3).
In the fourth test case we can select two triples (1, 3, 5) and (2, 4, 6).
In the fifth test case we can select one triple (1, 2, 3). We can't select two triples (1, 2, 3) and (4, 5, 6), because a_2 = a_5.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
a_reverse = a[::-1]
x = 0
y = 0
m = x
k = n - y - 1
differ_0_list = []
while(True):
if 0 in a[x:] and 0 in a_reverse[y:]:
x = a.index(0,x)
y = a_reverse.index(0,y)
else:
break
m = x
k = n - y - 1
if m >= k:
break
x += 1
y += 1
for differ_0 in a[m:k]:
if differ_0 != 0 and differ_0 not in differ_0_list:
differ_0_list.append(differ_0)
break
print(len(differ_0_list))
```
No
| 1,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a_1, a_2, …, a_n of non-negative integers.
You need to find the largest number m of triples (i_1, j_1, k_1), (i_2, j_2, k_2), ..., (i_m, j_m, k_m) such that:
* 1 ≤ i_p < j_p < k_p ≤ n for each p in 1, 2, …, m;
* a_{i_p} = a_{k_p} = 0, a_{j_p} ≠ 0;
* all a_{j_1}, a_{j_2}, …, a_{j_m} are different;
* all i_1, j_1, k_1, i_2, j_2, k_2, …, i_m, j_m, k_m are different.
Input
The first line of input contains one integer t (1 ≤ t ≤ 500 000): the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 500 000).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n).
The total sum of n is at most 500 000.
Output
For each test case, print one integer m: the largest number of proper triples that you can find.
Example
Input
8
1
1
2
0 0
3
0 1 0
6
0 0 1 2 0 0
6
0 1 0 0 1 0
6
0 1 3 2 0 0
6
0 0 0 0 5 0
12
0 1 0 2 2 2 0 0 3 3 4 0
Output
0
0
1
2
1
1
1
2
Note
In the first two test cases, there are not enough elements even for a single triple, so the answer is 0.
In the third test case we can select one triple (1, 2, 3).
In the fourth test case we can select two triples (1, 3, 5) and (2, 4, 6).
In the fifth test case we can select one triple (1, 2, 3). We can't select two triples (1, 2, 3) and (4, 5, 6), because a_2 = a_5.
Submitted Solution:
```
t = int(input())
def rainbow(n, arr):
zeros = []
for i in range(n):
if arr[i] == 0: zeros.append(i)
Count = 0
start, end = 0, len(zeros) - 1
while start < end:
if zeros[end] - zeros[start] == end - start: break
start, end = start + 1, end - 1
Count += 1
return Count
for i in range(t):
n = int(input())
arr = list(map(int, input().split()))
print(rainbow(n, arr))
```
No
| 1,050 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
def voevoda(n, m):
if n == 1:
return m
if n == 2:
return (m - m % 4) + 2 * min(m % 4, 2)
return (n * m + 1) // 2
N, M = sorted([int(i) for i in input().split()])
print(voevoda(N, M))
```
| 1,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
if n>m:n,m=m,n
if n==1:print(m)
elif n==2:print((m//4)*4+min((m%4)*2,4))
else:print((n*m)-(n*m)//2)
```
| 1,052 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
def li():
return list(map(int, input().split(" ")))
n, m = li()
if n < m:
t =n
n = m
m = t
if m == 1:
print(n)
elif m == 2:
print(2*(2*(n//4) + min(n%4, 2)))
else:
print((n*m) - (n*m)//2)
```
| 1,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
s=input().split()
n=int(s[0])
m=int(s[1])
if (n>m):
a=n
n=m
m=a
if (n==1):
print(m)
elif (n==2):
if (m%4<=2):
print(m+m%4)
else:
print(m+1)
else:
print((n*m+1)//2)
```
| 1,054 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
if n > m:
n, m = m, n
if n > 2 and m > 2:
print(((n * m) + 1) // 2)
elif n == 1:
print(m)
else:
print(2 * (((m // 4) * 2) + min(m % 4, 2)))
# Made By Mostafa_Khaled
```
| 1,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,m=sorted(list(map(int,input().split())))
if n==1:
print(m)
elif n==2:
print(4*((n*m)//8)+min((n*m)%8,4))
else:
print(n*m//2+n*m%2)
```
| 1,056 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, m = sorted(map(int, input().split()))
k = 4 * (m >> 2)
print(m if n == 1 else k + 2 * min(2, m - k) if n == 2 else (m * n + 1 >> 1))
```
| 1,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
if n > m:
n, m = m, n
if n > 2 and m > 2:
print(((n * m) + 1) // 2)
elif n == 1:
print(m)
else:
print(2 * (((m // 4) * 2) + min(m % 4, 2)))
```
| 1,058 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Submitted Solution:
```
ab=input().split(' ')
a=int(ab[0])
b=int(ab[1])
c=a*b
print(c//2)
```
No
| 1,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Submitted Solution:
```
n,m=map(int,input().split())
if n%3==0:
c=n/3
else:
c=n//3+1
if m%3==0:
p=m/3
else:p=m//3+1
print(max(c*m,p*n))
```
No
| 1,060 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Submitted Solution:
```
ab=input().split(' ')
a=int(ab[0])
b=int(ab[1])
c=a*b
if c>1: print(c//2)
elif c==1: print(1)
```
No
| 1,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sir Lancelot, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exercises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one day Sir Lancelot had a major argument with the Fairy Godmother (there were rumors that the argument occurred after the general spoke badly of the Godmother's flying techniques. That seemed to hurt the Fairy Godmother very deeply).
As the result of the argument, the Godmother put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!"
The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x1, y1) and (x2, y2) equals exactly (x1 - x2)2 + (y1 - y2)2. Now not all nm squad soldiers can participate in the drill exercises as it was before the Fairy Godmother's curse. Unless, of course, the general wants the soldiers to fight with each other or even worse... For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2).
Your task is to help the general. You are given the size of the drill exercise field. You are asked to calculate the maximum number of soldiers that can be simultaneously positioned on this field, so that no two soldiers fall under the Fairy Godmother's curse.
Input
The single line contains space-separated integers n and m (1 ≤ n, m ≤ 1000) that represent the size of the drill exercise field.
Output
Print the desired maximum number of warriors.
Examples
Input
2 4
Output
4
Input
3 4
Output
6
Note
In the first sample test Sir Lancelot can place his 4 soldiers on the 2 × 4 court as follows (the soldiers' locations are marked with gray circles on the scheme):
<image>
In the second sample test he can place 6 soldiers on the 3 × 4 site in the following manner:
<image>
Submitted Solution:
```
ab=input().split(' ')
a=int(ab[0])
b=int(ab[1])
c=a*b
if c&2!=0:
if c>1: print(c//2+1)
elif c==1: print(1)
else:
if c>1: print(c//2)
elif c==1: print(1)
```
No
| 1,062 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset of powers of two. More precisely, for each i from 0 to n exclusive you have cnt_i elements equal to 2^i.
In one operation, you can choose any one element 2^l > 1 and divide it into two elements 2^{l - 1}.
You should perform q queries. Each query has one of two types:
* "1 pos val" — assign cnt_{pos} := val;
* "2 x k" — calculate the minimum number of operations you need to make at least k elements with value lower or equal to 2^x.
Note that all queries of the second type don't change the multiset; that is, you just calculate the minimum number of operations, you don't perform them.
Input
The first line contains two integers n and q (1 ≤ n ≤ 30; 1 ≤ q ≤ 2 ⋅ 10^5) — the size of array cnt and the number of queries.
The second line contains n integers cnt_0, cnt_1, ..., cnt_{n - 1} (0 ≤ cnt_i ≤ 10^6).
Next q lines contain queries: one per line. Each query has one of two types:
* "1 pos val" (0 ≤ pos < n; 0 ≤ val ≤ 10^6);
* "2 x k" (0 ≤ x < n; 1 ≤ k ≤ 10^{15}).
It's guaranteed that there is at least one query of the second type.
Output
For each query of the second type, print the minimum number of operations you need to make at least k elements with a value lower or equal to 2^x or -1 if there is no way to do it.
Example
Input
6 11
0 1 0 0 1 0
2 1 5
2 4 18
1 1 0
2 2 5
2 0 17
1 0 3
2 1 2
1 1 4
1 4 0
1 5 1
2 2 8
Output
4
16
4
-1
0
1
Submitted Solution:
```
n, q=list(map(int, input().split()))
cnt=list(map(int, input().split()))
for case in range(0, q):
o, x, y=list(map(int, input().split()))
if(o==1):
cnt[x]=y
else:
maxSplit=0
current=0
for i in range(0, x+1):
maxSplit+=(2**i)*cnt[i]
current+=cnt[i]
if(maxSplit>=y):
print(max(0, y-current))
else:
found=None
ans=0
for i in range(x+1, n):
maxSplit+=(2**i)*cnt[i]
if(maxSplit>=y):
found=i
maxSplit-=(2**i)*cnt[i]
break
else:
ans+=(2**(i-x)-1)*cnt[i]
current+=2**(i-x)*cnt[i]
if(found is None):
ans=-1
else:
for i in range(found, x, -1):
req=y-maxSplit
req=(req-1)//(2**i)+1
maxSplit+=(2**i)*(req-1)
ans+=(2**(i-x)-1)*(req-1)
current+=2**(i-x)*(req-1)
ans+=1
ans+=y-current
print(ans)
```
No
| 1,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset of powers of two. More precisely, for each i from 0 to n exclusive you have cnt_i elements equal to 2^i.
In one operation, you can choose any one element 2^l > 1 and divide it into two elements 2^{l - 1}.
You should perform q queries. Each query has one of two types:
* "1 pos val" — assign cnt_{pos} := val;
* "2 x k" — calculate the minimum number of operations you need to make at least k elements with value lower or equal to 2^x.
Note that all queries of the second type don't change the multiset; that is, you just calculate the minimum number of operations, you don't perform them.
Input
The first line contains two integers n and q (1 ≤ n ≤ 30; 1 ≤ q ≤ 2 ⋅ 10^5) — the size of array cnt and the number of queries.
The second line contains n integers cnt_0, cnt_1, ..., cnt_{n - 1} (0 ≤ cnt_i ≤ 10^6).
Next q lines contain queries: one per line. Each query has one of two types:
* "1 pos val" (0 ≤ pos < n; 0 ≤ val ≤ 10^6);
* "2 x k" (0 ≤ x < n; 1 ≤ k ≤ 10^{15}).
It's guaranteed that there is at least one query of the second type.
Output
For each query of the second type, print the minimum number of operations you need to make at least k elements with a value lower or equal to 2^x or -1 if there is no way to do it.
Example
Input
6 11
0 1 0 0 1 0
2 1 5
2 4 18
1 1 0
2 2 5
2 0 17
1 0 3
2 1 2
1 1 4
1 4 0
1 5 1
2 2 8
Output
4
16
4
-1
0
1
Submitted Solution:
```
from math import ceil
class cnt:
def __init__(self, n, a):
self.a = a
self.n = n-1
self.smm = [a[0]]
self.summ = [a[0]]
for i in range(1,n):
self.smm.append(self.smm[i-1] + 2**i * self.a[i])
self.summ.append(self.summ[i-1] + self.a[i])
# print('Created')
# print('c.a:', self.a, 'c.n:', self.n, 'c.smm:', self.smm, 'c.summ:', self.summ)
def f(self, pos, val):
ppos = pos
self.a[pos]=val
if pos == 0:
self.smm[0] = val
self.summ[0] = val
pos += 1
for ii in range(pos,n):
self.smm[ii]=(self.smm[ii-1] + 2**ii * self.a[ii])
self.summ[ii]=(self.summ[ii-1] + self.a[ii])
# print('F: pos:', ppos, 'val:', val, 'self.a:', self.a, 'self.smm:', self.smm, 'self.summ:', self.summ)
def ss(self, x, k):
if x == self.n:
# print('SS: x:', x, 'k:', k, 'res:', -1 if self.a[x] < k else 0)
return -1 if self.a[x] < k else 0
if self.a[x] >= k:
# print('SS: x:', x, 'k:', k, 'res:', 0)
return 0
kk = k
k = (k - self.a[x] + 1)//2
rs = self.ss(x+1, k)
if rs == -1:
# print('SS: x:', x, 'k:', kk, 'res:', -1)
return -1
# print('SS: x:', x, 'k:', kk, 'res:', k+rs)
return k + rs
def s(self, x, k):
if self.smm[self.n] < k:
# print('S: x:', x, 'k:', k, 'res:', -1)
return -1
if self.smm[x] >= k:
# print('S: x:', x, 'k:', k, 'res:', max(0, k-self.summ[x]))
return max(0, k-self.summ[x])
else:
kk = k
k = k - self.smm[x]
nd = ceil(k/(2**x))
rs = self.ss(x+1, nd)
if rs == -1:
# print('S: x:', x, 'k:', kk, 'res:', -1, 'nd:', nd)
return -1
# print('S: x:', x, 'k:', kk, 'res:', k+nd+rs, 'nd:',nd)
return max(0, k-nd*2-self.smm[x]) + nd + rs
n, q = map(int, input().split())
a = list(map(int, input().split()))
c = cnt(n, a)
acts = []
fpr = []
for _ in range(q):
qq, f, s = map(int, input().split())
if qq == 1:
c.f(f, s)
else:
fpr.append(c.s(f, s))
print(*fpr, sep='\n')
```
No
| 1,064 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset of powers of two. More precisely, for each i from 0 to n exclusive you have cnt_i elements equal to 2^i.
In one operation, you can choose any one element 2^l > 1 and divide it into two elements 2^{l - 1}.
You should perform q queries. Each query has one of two types:
* "1 pos val" — assign cnt_{pos} := val;
* "2 x k" — calculate the minimum number of operations you need to make at least k elements with value lower or equal to 2^x.
Note that all queries of the second type don't change the multiset; that is, you just calculate the minimum number of operations, you don't perform them.
Input
The first line contains two integers n and q (1 ≤ n ≤ 30; 1 ≤ q ≤ 2 ⋅ 10^5) — the size of array cnt and the number of queries.
The second line contains n integers cnt_0, cnt_1, ..., cnt_{n - 1} (0 ≤ cnt_i ≤ 10^6).
Next q lines contain queries: one per line. Each query has one of two types:
* "1 pos val" (0 ≤ pos < n; 0 ≤ val ≤ 10^6);
* "2 x k" (0 ≤ x < n; 1 ≤ k ≤ 10^{15}).
It's guaranteed that there is at least one query of the second type.
Output
For each query of the second type, print the minimum number of operations you need to make at least k elements with a value lower or equal to 2^x or -1 if there is no way to do it.
Example
Input
6 11
0 1 0 0 1 0
2 1 5
2 4 18
1 1 0
2 2 5
2 0 17
1 0 3
2 1 2
1 1 4
1 4 0
1 5 1
2 2 8
Output
4
16
4
-1
0
1
Submitted Solution:
```
import sys#,io,os;Z=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
Z=sys.stdin.readline
Y=lambda:map(int,Z().split())
n,q=Y();a=[*Y()];o=[]
b=a==[2,1,2,0,0,0,2,0,2,0]
c=0
for _ in range(q):
t,x,k=Y()
if t<2:a[x]=k;continue
c+=1
if b and c==2401:o.append('99'.join(map(str,a)))
p=sum(a[:x+1])
if p>=k:o.append('0');continue
k-=p;i=x+1;m=e=0
while i<n:
v=a[i]<<(i-x)
if v<=k:k-=v;m+=v-a[i];e+=v
else:break
i+=1
if i==n:
s=sum((a[i]<<i)-a[i] for i in range(x+1))+(e<<x)-e
if s>=k:o.append(str(m+k))
else:o.append('-1')
continue
u=k>>(i-x);v=u<<(i-x);k-=v;m+=v-u;e+=v
if not k:o.append(str(m));continue
s=sum((a[i]<<i)-a[i] for i in range(x+1))+(e<<x)-e;l=(1<<(i-x))-k
for j in range(i):
if l%2:
if s>=1<<j:s-=1<<j;l+=1
else:break
l//=2
ne=sum(map(int,bin(l)[2:]));m+=k-1+ne;o.append(str(m))
print('\n'.join(o))
```
No
| 1,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset of powers of two. More precisely, for each i from 0 to n exclusive you have cnt_i elements equal to 2^i.
In one operation, you can choose any one element 2^l > 1 and divide it into two elements 2^{l - 1}.
You should perform q queries. Each query has one of two types:
* "1 pos val" — assign cnt_{pos} := val;
* "2 x k" — calculate the minimum number of operations you need to make at least k elements with value lower or equal to 2^x.
Note that all queries of the second type don't change the multiset; that is, you just calculate the minimum number of operations, you don't perform them.
Input
The first line contains two integers n and q (1 ≤ n ≤ 30; 1 ≤ q ≤ 2 ⋅ 10^5) — the size of array cnt and the number of queries.
The second line contains n integers cnt_0, cnt_1, ..., cnt_{n - 1} (0 ≤ cnt_i ≤ 10^6).
Next q lines contain queries: one per line. Each query has one of two types:
* "1 pos val" (0 ≤ pos < n; 0 ≤ val ≤ 10^6);
* "2 x k" (0 ≤ x < n; 1 ≤ k ≤ 10^{15}).
It's guaranteed that there is at least one query of the second type.
Output
For each query of the second type, print the minimum number of operations you need to make at least k elements with a value lower or equal to 2^x or -1 if there is no way to do it.
Example
Input
6 11
0 1 0 0 1 0
2 1 5
2 4 18
1 1 0
2 2 5
2 0 17
1 0 3
2 1 2
1 1 4
1 4 0
1 5 1
2 2 8
Output
4
16
4
-1
0
1
Submitted Solution:
```
class cnt:
def __init__(self, n, a):
self.a = a
self.n = n-1
self.smm = [a[0]]
self.summ = [a[0]]
for i in range(1,n):
self.smm.append(self.smm[i-1] + 2**i * self.a[i])
self.summ.append(self.summ[i-1] + self.a[i])
# print('Created')
# print('c.a:', self.a, 'c.n:', self.n, 'c.smm:', self.smm, 'c.summ:', self.summ)
def f(self, pos, val):
ppos = pos
self.a[pos]=val
if pos == 0:
self.smm[0] = val
self.summ[0] = val
pos += 1
for ii in range(pos,n):
self.smm[ii]=(self.smm[ii-1] + 2**ii * self.a[ii])
self.summ[ii]=(self.summ[ii-1] + self.a[ii])
# print('F: pos:', ppos, 'val:', val, 'self.a:', self.a, 'self.smm:', self.smm, 'self.summ:', self.summ)
def ss(self, x, k):
if x == self.n:
# print('SS: x:', x, 'k:', k, 'res:', -1 if self.a[x] < k else 0)
return -1 if self.a[x] < k else 0
if self.a[x] >= k:
# print('SS: x:', x, 'k:', k, 'res:', 0)
return 0
kk = k
k = (k - self.a[x] + 1)//2
rs = self.ss(x+1, k)
if rs == -1:
# print('SS: x:', x, 'k:', kk, 'res:', -1)
return -1
# print('SS: x:', x, 'k:', kk, 'res:', k+rs)
return k + rs
def s(self, x, k):
if self.smm[self.n] < k:
# print('S: x:', x, 'k:', k, 'res:', -1)
return -1
if self.n==x:
# print('S (n=x): x:', x, 'k:', k, 'res:', max(0, k-self.summ[x]))
return max(0, k-self.summ[x])
if self.smm[x+1] >= k:
kk = k
k = k-self.summ[x]
if k <= 0:
# print('S (k<summ[x]): x:', x, 'k:', kk, 'res:', 0)
return 0
if k <= self.a[x+1]*2:
# print('S (k<=a[x+1]*2): x:', x, 'k:', kk, 'res:', (k+1)//2)
return (k+1)//2
else:
# print('S (k>a[x+1]*2): x:', x, 'k:', kk, 'res:', k-a[x+1])
return k-self.a[x+1]
kk = k
nd = (k-self.smm[x])//(2**x) + (1 if (k-self.smm[x])%(2**x) else 0)
if (nd-self.a[x+1])%2: nd += 1
rs = self.ss(x+1, nd)
if rs == -1:
# print('S: x:', x, 'k:', kk, 'res:', -1, 'nd:', nd)
return -1
# print('S: x:', x, 'k:', kk, 'res:', k+nd+rs, 'nd:',nd)
return max(0, k-self.smm[x]-nd*2-self.summ[x]) + nd + rs
n, q = map(int, input().split())
a = list(map(int, input().split()))
c = cnt(n, a)
acts = []
fpr = []
for _ in range(q):
qq, f, s = map(int, input().split())
if qq == 1:
c.f(f, s)
else:
fpr.append(c.s(f, s))
print(*fpr, sep='\n')
```
No
| 1,066 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
st="l"+input()+"r"
forw=[0 for i in range(n+2)]
back=[0 for i in range(n+2)]
i=1
while(i<=n):
if(st[i]=="R" and st[i+1]=="L"):
forw[i+1]=forw[i-1]+2
i+=2
elif(st[i]=="L"):
forw[i]=1
i+=1
else:
i+=1
i=n
while(i>=1):
if(st[i]=="L" and st[i-1]=="R"):
back[i-1]=back[i+1]+2
i-=2
elif(st[i]=="R"):
back[i]=1
i-=1
else:
i-=1
for i in range(n+1):
print(forw[i]+back[i+1]+1,end=' ')
print()
```
| 1,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
t= int(input().strip())
for i in range(t):
n=int(input().strip())
l1=[0]*(n+1)
l2=[0]*(n+1)
s=input()
if s[0]=="L":
l1[1]+=1
for j in range(1,n):
if s[j]=="L":
if l1[j]==0:
l1[j+1]=2+l1[j-1]
else:
l1[j+1]=1
if s[n-1]=="R":
l2[n-1]+=1
for k in range(n-2,-1,-1):
if s[k]=="R":
if l2[k+1]==0:
l2[k]=2 +l2[k+2]
else:
l2[k]=1
for m in range(n+1):
print(l1[m]+l2[m]+1,end=" ")
print()
```
| 1,068 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
N = int(input())
S = list(input().strip())
S_R = [""] * N
S_L = [""] * N
for i in range(N):
if i % 2 == 1:
S_L[i] = S[i]
if S[i] == "L":
S_R[i] = "R"
else:
S_R[i] = "L"
else:
S_R[i] = S[i]
if S[i] == "L":
S_L[i] = "R"
else:
S_L[i] = "L"
ans = [1] * (N + 1)
current_l = 0
current_r = 0
for i in range(N):
if S_L[i] == "L":
current_l += 1
else:
current_l = 0
if S_R[i] == "L":
current_r += 1
else:
current_r = 0
if i % 2 == 0:
ans[i + 1] += current_r
else:
ans[i + 1] += current_l
current_l = 0
current_r = 0
for i in range(N - 1, -1, -1):
if S_L[i] == "R":
current_l += 1
else:
current_l = 0
if S_R[i] == "R":
current_r += 1
else:
current_r = 0
if i % 2 == 0:
ans[i] += current_r
else:
ans[i] += current_l
print(*ans)
if __name__ == '__main__':
main()
```
| 1,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
def main():
T = eval(input())
for _ in range(T):
N=eval(input())
str=input()
dp1=[1]*N
dp2=[1]*N
ans=[1]*(N+1)
for i in range(1,N):
if str[i]!=str[i-1]:
dp1[i]+=dp1[i-1]
for i in range(N-2,-1,-1):
if str[i]!=str[i+1]:
dp2[i]+=dp2[i+1]
for i in range(N+1):
if i!=N and str[i]=='R':
ans[i]+=dp2[i]
if i!=0 and str[i-1]=='L':
ans[i]+=dp1[i-1]
print(*ans)
if __name__=='__main__':
main()
```
| 1,070 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
import collections
from collections import deque
import heapq
import itertools
from collections import defaultdict
from collections import Counter
for _ in range(int(input())):
n = int(input())
s = input().rstrip()
ans = [0]*(n+1)
for i in range(n+1):
r = 0
l = 0
if ans[i]==0:
right = False
while i-l-1>=0 and right==(s[i-l-1]=='R'):
l+=1
right = not right
right = True
while i+r<n and right ==(s[i+r]=='R'):
r+=1
right = not right
val = l+r+1
for j in range(i,i+r+1,2):
ans[j] = val
for num in ans:
print(num, end = " ")
print()
```
| 1,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
import sys
for tc in range(int(input())):
n = int(sys.stdin.readline().strip())
p = sys.stdin.readline().strip()
L = [0]
Lt = ""
Lc = 0
for i in range(n):
if p[i] == "R":
if Lt == "L":
Lc += 1
else: Lc = 1
Lt = "R"
L.append(0)
else:
if Lt == "R":
Lc += 1
else: Lc = 1
Lt = "L"
L.append(Lc)
R = [0]
Rt = ""
Rc = 0
p = p[::-1]
for i in range(n):
if p[i] == "L":
if Rt == "R":
Rc += 1
else: Rc = 1
Rt = "L"
R.append(0)
else:
if Rt == "L":
Rc += 1
else: Rc = 1
Rt = "R"
R.append(Rc)
R = R[::-1]
ans = []
for i in range(n+1):
ans.append(L[i] + R[i] + 1)
print(" ".join(map(str, ans)))
```
| 1,072 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
for i in range(int(input())):
n = int(input())
s = input()
t1 = [0]
if s[0] == 'L':
t1.append(1)
else:t1.append(0)
cnt = 1
for i in range(1,n):
if s[i] != s[i-1]:
cnt += 1
else:cnt = 1
if s[i] == 'L':t1.append(cnt)
else:t1.append(0)
t2 = [0]
if s[-1] == 'R':t2.append(1)
else:t2.append(0)
cnt = 1
for i in range(n-2,-1,-1):
if s[i] != s[i+1]:
cnt += 1
else:cnt = 1
if s[i] == 'R':
t2.append(cnt)
else:t2.append(0)
for i in range(n+1):
print(t1[i] + t2[n-i] + 1,end = ' ')
print()
```
| 1,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Tags: dfs and similar, dp, dsu, implementation
Correct Solution:
```
from collections import defaultdict
from itertools import accumulate
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
sys.stdout.write(" ".join(map(str,ANS))+"\n")
'''
inf = 100000000000000000 # 1e17
mod = 998244353
for CASES in range(int(input())):
n = int(input())
S = input().strip()
dp1=[1]*n
for i in range(n-2,-1,-1):
if S[i]!=S[i+1]:
dp1[i]=dp1[i]+dp1[i+1]
dp2=[1]*n
for i in range(1,n):
if S[i]!=S[i-1]:
dp2[i]=dp2[i]+dp2[i-1]
print(dp1[0]+1 if S[0]=='R' else 1,end=" ")
for i in range(1,n):
tmp1=dp2[i-1] if S[i-1]=='L' else 0
tmp2=dp1[i] if S[i]=='R' else 0
print(tmp1+1+tmp2,end=" ")
print(dp2[n-1]+1 if S[n-1]=='L' else 1)
#for now in range(n):
```
| 1,074 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
t = int(input())
for _ in range(t):
def main():
n = int(input())
s = list(input())
ans = [0]*(n+1)
cur = 1
start = 0
for i in range(n):
if (i % 2 and s[i] == 'L') or (not i % 2 and s[i] == 'R'):
cur += 1
else:
if i % 2:
end = i-1
ns = i+1
else:
end = i
ns = i+2
for j in range(start+2, end+2, 2):
ans[j] = cur
ans[start] = max(ans[start], cur)
start = ns
cur = 1
if (n-1) % 2:
end = n
else:
end = n-1
for j in range(start + 2, end + 2, 2):
ans[j] = cur
if start <= n:
ans[start] = max(ans[start], cur)
# print(ans)
cur = 1
start = 1
for i in range(n):
if (i % 2 and s[i] == 'R') or (not i % 2 and s[i] == 'L'):
cur += 1
else:
if i % 2:
end = i
ns = i+2
else:
end = i-1
ns = i+1
for j in range(start+2, end+2, 2):
ans[j] = cur
ans[start] = max(ans[start], cur)
start = ns
cur = 1
if (n-1) % 2:
end = n-1
else:
end = n
for j in range(start+2, end+2, 2):
ans[j] = cur
if start <= n:
ans[start] = max(ans[start], cur)
return print(' '.join(map(str, ans)))
main()
```
Yes
| 1,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
import sys
for _ in range(int(input())):
n = int(input())
s = input()
A = [[1,1] for _ in range(n+1)]
ans = [0]*(n+1)
for i in range(1,n+1):
if s[i-1] == "L":
A[i][0] = 1+A[i-1][1]
else:
A[i][1] = 1+A[i-1][0]
ans[i] += A[i][0]
A = [[1,1] for _ in range(n+1)]
for i in range(n-1,-1,-1):
if s[i] == "R":
A[i][0] = 1+A[i+1][1]
else:
A[i][1] = 1+A[i+1][0]
ans[i] += A[i][0]-1
ans[0] += 1
print(*ans)
```
Yes
| 1,076 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
from collections import defaultdict,OrderedDict,Counter
from sys import stdin,stdout
from bisect import bisect_left,bisect_right
# import numpy as np
from queue import Queue,PriorityQueue
from heapq import heapify,heappop,heappush
from statistics import median
from math import gcd,sqrt,floor,factorial,ceil,log2,log10
import copy
from copy import deepcopy
import sys
sys.setrecursionlimit(10**7)
import math
import os
import bisect
import collections
mod=pow(10,9)+7
def ncr(n, r, p=mod):
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
# input=stdin.readline
# print=stdout.write
inf=float("inf")
adj=defaultdict(set)
visited=defaultdict(int)
def addedge(a,b):
adj[a].add(b)
adj[b].add(a)
def bfs(v):
q=Queue()
q.put(v)
visited[v]=1
while q.qsize()>0:
s=q.get_nowait()
print(s)
for i in adj[s]:
if visited[i]==0:
q.put(i)
visited[i]=1
def dfs(v,visited):
if visited[v]==1:
return;
visited[v]=1
print(v)
for i in adj[v]:
dfs(i,visited)
# a9=pow(10,6)+10
# prime = [True for i in range(a9 + 1)]
# def SieveOfEratosthenes(n):
# p = 2
# while (p * p <= n):
# if (prime[p] == True):
# for i in range(p * p, n + 1, p):
# prime[i] = False
# p += 1
def get_list():
return list(map(int,input().split()))
def get_str_list():
return list(input())
def get_map():
return map(int,input().split())
def get_int():
return int(input())
def fun1(l):
for i in range(n+1):
if i==0:
left[i]=1;
elif i==1:
if l[i-1]=='L':
left[1]=2;
else:
left[1]=1;
else:
if l[i-1]=='L':
if l[i-2]=='R':
left[i]=2+left[i-2]
else:
left[i]=2;
else:
left[i]=1;
def fun2(l):
i=n;
while i>=0:
if i==n:
right[i]=1;
elif i==n-1:
if l[i]=='R':
right[i]=2;
else:
right[i]=1;
else:
if l[i]=='R':
if l[i+1]=='L':
right[i]=right[i+2]+2
else:
right[i]=2;
else:
right[i]=1;
i-=1
t=int(input())
for i in range(t):
n=get_int();
l=get_str_list()
m=[]
for i in l:
if i=='L':
m.append('R')
else:
m.append('L')
left=defaultdict(int)
right=defaultdict(int)
fun1(m)
fun2(m);
for i in range(n+1):
count=1
if i==0:
if l[0]=='R':
count+=right[1]
elif i==n:
if l[i-1]=='L':
count+=left[i-1]
else:
if l[i]=='R':
count+=right[i+1]
if l[i-1]=='L':
count+=left[i-1]
print(count,end=" ")
print("")
```
Yes
| 1,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
import collections
import sys
import functools
import heapq
import bisect
import collections
import math
input = sys.stdin.readline
mod = 10**9 + 7
t = int(input())
for _ in range(t):
n = int(input().strip())
s = input().strip()
right = [[0, 0] for i in range(n)]
left = [[0, 0] for i in range(n)]
if s[-1] == 'L':
right[-1][1] = 1
else:
right[-1][0] = 1
for i in range(n-2, -1, -1):
if s[i] == 'L':
right[i][1] = 1 + right[i+1][0]
else:
right[i][0] = 1 + right[i+1][1]
if s[0] == 'L':
left[0][1] = 1
else:
left[0][0] = 1
for i in range(1, n):
if s[i] == 'L':
left[i][1] = 1 + left[i-1][0]
else:
left[i][0] = 1 + left[i-1][1]
#print(left,right)
for i in range(n+1):
if i == 0:
print(1+right[i][0],end=' ')
elif i == n:
print(1+left[i-1][1],end=' ')
else:
print(1+left[i-1][1]+right[i][0],end=' ')
print()
```
Yes
| 1,078 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
from math import ceil
def main():
T = int(input())
for ___ in range(T):
n = input().strip()
n = int(n)
d = input().strip()
L = [-1 for i in range(n)]
R = [-1 for i in range(n)]
L[0] = 0
R[n - 1] = n - 1
for i in range(1, n):
if d[i] == d[i - 1]:
L[i] = i
else:
L[i] = L[i - 1]
for i in range(n - 2, -1, -1):
if d[i] == d[i + 1]:
R[i] = i
else:
R[i] = R[i + 1]
ans = [0 for i in range(n + 1)]
if d[0] == 'R':
ans[0] = R[0]
else:
ans[0] = 1
if d[n - 1] == 'L':
ans[n] = (n - L[n - 1]) + 1
else:
ans[n] = 1
for i in range(1, n):
ans[i] = 1
if d[i - 1] == 'L':
ans[i] += i - L[i - 1]
if d[i] == 'R':
ans[i] += R[i] - i + 1
# print(L)
# print(R)
for item in ans:
print(item, end=' ')
print()
# region fastio
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
No
| 1,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
st=input()
dp_left=[i for i in range(n+1)]
for i in range (n+1):
if i==0 or st[i-1]=='R':
dp_left[i]=i
elif st[i-1]=='L' and st[i-2]=='R':
dp_left[i]=dp_left[i-2]
else:
dp_left[i]=i-1
#for i in dp_left:
# print(i,end=" ")
dp_right = [i for i in range(n + 1)]
for i in range(n,-1,-1):
if i==n or st[i]=='L':
dp_right[i]=i
elif i==n-1 or st[i+1]=='R':
dp_right[i]=i+1
else:
dp_right[i]=dp_right[i+2]
#for i in dp_right:
#print(i,end=" ")
print(end="\n")
for i in range(n + 1):
print((dp_right[i] - dp_left[i]) + 1, end=" ")
```
No
| 1,080 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
a=int(input())
for i in range(a):
s=int(input())
s=input()
ans=[]
for i in range(len(s)):
if(s[i]=='L'):
ans.append(1)
else:
ans.append(0)
bac=[]
for i in range(len(ans)):
if(ans[i]==0):
bac.append(0)
continue;
if(i==0 and ans[i]==1):
bac.append(1)
continue;
if(ans[i]==1):
if(i>=2):
if(bac[-1]==0):
bac.append(bac[-2]+2)
else:
bac.append(1)
else:
if(ans[0]=='R'):
bac.append(2)
else:
bac.append(1)
forw=[0 for i in range(len(ans))]
for i in range(len(ans)-2,-1,-1):
if(ans[i+1]==1):
forw[i]=0
continue;
else:
if(i+2<=len(ans)-1 and ans[i+2]==1):
forw[i]=forw[i+2]+2
elif(i+2<=len(ans)-1 and ans[i+2]==0):
forw[i]=1
fin=[0 for i in range(len(forw)+1)]
for i in range(len(bac)):
fin[i+1]=bac[i]+forw[i]+1
cnt=0
for i in range(len(s)):
if(i%2==0):
if(s[i]=='R'):
cnt+=1
else:
break;
else:
if(s[i]=='L'):
cnt+=1
else:
break;
fin[0]=cnt+1
print(*fin)
```
No
| 1,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).
Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.
A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.
The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.
Example
Input
2
6
LRRRLL
3
LRL
Output
1 3 2 3 1 3 2
1 4 1 4
Submitted Solution:
```
answers = []
def solve(string):
length = len(string)
distances = [0 for i in range(length+1)]
store = ['C']
for ele in string:
store.append(ele)
store.append('C')
# print(store)
length = len(store)
final = []
for i, ele in enumerate(store):
if ele == 'C':
switch = 0
current = 1
left = None
right = None
left = i - 1
right = i + 1
while left >= 0:
if switch == 0:
if store[left] == 'L':
current += 1
switch = 1
left -= 2
else:
break
else:
if store[left] == 'R':
current += 1
switch = 0
left -= 2
else:
break
switch = 0
while right <= length-1:
if switch == 0:
if store[right] == 'R':
current += 1
switch = 1
right += 2
else:
break
else:
if store[right] == 'L':
current += 1
switch = 0
right += 2
else:
break
final.append(current)
# print(final)
answers.append(final)
T = int(input())
while T:
N = int(input())
S = input()
solve(S)
T -= 1
for ans in answers:
print(ans)
```
No
| 1,082 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
from sys import stdin, stdout
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
s = stdin.readline(n)
stdin.readline(1)
if s[0] == '0' or s[n-1] == '0':
stdout.write("NO\n")
continue
a = []
b = []
va, vb = 0, 0
for i in range(n):
if s[i] == '1':
if va + vb < 4:
va += 1
vb += 1
a.append('(')
b.append('(')
else:
va -= 1
vb -= 1
a.append(')')
b.append(')')
elif s[i] == '0':
if va <= 1:
va += 1
vb -= 1
a.append('(')
b.append(')')
else:
va -= 1
vb += 1
a.append(')')
b.append('(')
if not (va == vb == 2):
stdout.write("NO\n")
else:
a[-1] = ')'
b[-1] = ')'
stdout.write("YES\n")
stdout.write(''.join(a))
stdout.write("\n")
stdout.write(''.join(b))
stdout.write("\n")
```
| 1,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys,functools,collections,bisect,math,heapq
input = sys.stdin.readline
#print = sys.stdout.write
sys.setrecursionlimit(200000)
mod = 10**9 + 7
t = int(input())
for _ in range(t):
n = int(input())
s = input().strip()
if s[0] == '0' or s[-1] == '0':
print('NO')
continue
d = collections.Counter(s)
if d['0']%2 or d['1']%2:
print('NO')
continue
res1 = [1]*n
res2 = [1]*n
x = d['1']//2
i = 0
while x:
if s[i] == '1':
res1[i] = '('
res2[i] = '('
x -= 1
i += 1
if x == 0:
while i < n:
if s[i] == '1':
res1[i] = ')'
res2[i] = ')'
i += 1
c = 0
for i in range(n):
if s[i] == '0':
if c%2:
res1[i] = ')'
res2[i] = '('
else:
res1[i] = '('
res2[i] = ')'
c += 1
print('YES')
print(''.join(res1))
print(''.join(res2))
```
| 1,084 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
n = int(input())
S = input().strip()
l = S.count("1")
o = S.count("0")
if l % 2 == 1:
print("NO")
return
if S[0] == "0" or S[-1] == "0":
print("NO")
return
print("YES")
cnt_o = 0
cnt_l = 0
T1 = []
T2 = []
for s in S:
if s == "1":
if cnt_l < l // 2:
T1.append("(")
T2.append("(")
else:
T1.append(")")
T2.append(")")
cnt_l += 1
else:
cnt_o += 1
if cnt_o % 2 == 0:
T1.append("(")
T2.append(")")
else:
T1.append(")")
T2.append("(")
print("".join(T1))
print("".join(T2))
for _ in range(int(input())):
main()
```
| 1,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
s=input()
a=[]
b=[]
count1=0
for i in range(n):
if s[i]=='1':
count1+=1
count0=n-count1
if count0%2:
print('NO')
continue
leftbr0=count0//2
rightbr0=count0//2
leftbr1=n//2-leftbr0
score=0
stack=[]
for i in range(n):
if s[i]=='1':
if leftbr1:
a.append('(')
score+=1
leftbr1-=1
else:
a.append(')')
score-=1
if score<0 and stack:
rightbr0+=1
score+=2
k=stack.pop()
a[k]='('
else:
if rightbr0 and score>0:
a.append(')')
score-=1
rightbr0-=1
stack.append(i)
else:
a.append('(')
score+=1
b=[]
for i in range(n):
if s[i]=='1':
b.append(a[i])
else:
if a[i]=='(':
b.append(')')
else:
b.append('(')
score1=0
score2=0
flag=0
for i in range(n):
if a[i]=='(':
score1+=1
else:
score1-=1
if b[i]=='(':
score2+=1
else:
score2-=1
if score1<0 or score2<0:
flag=1
break
if flag or score1!=0 or score2!=0:
print('NO')
else:
print('YES')
print(''.join(a))
print(''.join(b))
```
| 1,086 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
import time
#start_time = time.time()
#def TIME_(): print(time.time()-start_time)
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string, heapq as h
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def getBin(): return list(map(int,list(input())))
def isInt(s): return '0' <= s[0] <= '9'
def ceil_(a,b): return a//b + (a%b > 0)
MOD = 10**9 + 7
"""
The ones must be split in half
Let's make the first half of the ones (, the second half )
"""
def solve():
N = getInt()
S = getBin()
ones = S.count(1)
if ones % 2:
print("NO")
return
ones //= 2
curr1 = curr2 = 0
A = []
B = []
one_count = 0
for i in range(N):
if S[i] == 1:
if one_count < ones:
A.append('(')
B.append('(')
curr1 += 1
curr2 += 1
one_count += 1
else:
A.append(')')
B.append(')')
curr1 -= 1
curr2 -= 1
else:
if curr1 < curr2:
A.append('(')
B.append(')')
curr1 += 1
curr2 -= 1
else:
A.append(')')
B.append('(')
curr1 -= 1
curr2 += 1
if min(curr1,curr2) < 0:
print("NO")
return
if curr1 != 0 or curr2 != 0:
print("NO")
return
print("YES")
print(''.join(A))
print(''.join(B))
return
for _ in range(getInt()):
#print(solve())
solve()
#TIME_()
```
| 1,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
#------------------Important Modules------------------#
from sys import stdin,stdout
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import *
from random import *
input=stdin.readline
prin=stdout.write
from random import sample
from collections import Counter,deque
from fractions import *
from math import sqrt,ceil,log2,gcd
#dist=[0]*(n+1)
mod=10**9+7
"""
class DisjSet:
def __init__(self, n):
self.rank = [1] * n
self.parent = [i for i in range(n)]
def find(self, x):
if (self.parent[x] != x):
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def union(self, x, y):
xset = self.find(x)
yset = self.find(y)
if xset == yset:
return
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
else:
self.parent[yset] = xset
self.rank[xset] = self.rank[xset] + 1
"""
def f(arr,i,j,d,dist):
if i==j:
return
nn=max(arr[i:j])
for tl in range(i,j):
if arr[tl]==nn:
dist[tl]=d
#print(tl,dist[tl])
f(arr,i,tl,d+1,dist)
f(arr,tl+1,j,d+1,dist)
#return dist
def ps(n):
cp=0;lk=0;arr={}
#print(n)
#cc=0;
while n%2==0:
n=n//2
#arr.append(n);arr.append(2**(lk+1))
lk+=1
for ps in range(3,ceil(sqrt(n))+1,2):
lk=0
cc=0
while n%ps==0:
n=n//ps
#cc=1
#arr.append(n);arr.append(ps**(lk+1))
lk+=1
if n!=1:
#lk+=1
#arr[n]=1
#ans*=n;
lk+=1
return False
#return arr
#print(arr)
return True
#count=0
#dp=[[0 for i in range(m)] for j in range(n)]
#[int(x) for x in input().strip().split()]
def gcd(x, y):
while(y):
x, y = y, x % y
return x
# Driver Code
def factorials(n,r):
#This calculates ncr mod 10**9+7
slr=n;dpr=r
qlr=1;qs=1
mod=10**9+7
for ip in range(n-r+1,n+1):
qlr=(qlr*ip)%mod
for ij in range(1,r+1):
qs=(qs*ij)%mod
#print(qlr,qs)
ans=(qlr*modInverse(qs))%mod
return ans
def modInverse(b):
qr=10**9+7
return pow(b, qr - 2,qr)
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
def power(arr):
listrep = arr
subsets = []
for i in range(2**len(listrep)):
subset = []
for k in range(len(listrep)):
if i & 1<<k:
subset.append(listrep[k])
subsets.append(subset)
return subsets
def pda(n) :
list = []
for i in range(1, int(sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
list.append(i)
else :
list.append(n//i);list.append(i)
# The list will be printed in reverse
return list
def dis(xa,ya,xb,yb):
return sqrt((xa-xb)**2+(ya-yb)**2)
#### END ITERATE RECURSION ####
#===============================================================================================
#----------Input functions--------------------#
def ii():
return int(input())
def ilist():
return [int(x) for x in input().strip().split()]
def islist():
return list(map(str,input().split().rstrip()))
def inp():
return input().strip()
def google(test,ans):
return "Case #"+str(test)+": "+str(ans);
def overlap(x1,y1,x2,y2):
if x2>y1:
return y1-x2
if y1>y2:
return y2-x2
return y1-x2;
###-------------------------CODE STARTS HERE--------------------------------###########
def pal(s):
k=len(s)
n=len(s)//2
for i in range(n):
if s[i]==s[k-1-i]:
continue
else:
return 0
return 1
#########################################################################################
t=int(input())
#t=1
for p in range(t):
n=ii()
a=inp();b=a
xy=a.count('0');#yz=b.count('0')
if xy%2==1 or a[0]=='0' or a[-1]=='0':
print("No")
continue
ac=''
bc=''
az=a.count('1')//2
one=0
bit=0
for i in range(n):
if a[i]=='0':
if bit==0:
ac+='('
bc+=')'
else:
ac+=')'
bc+='('
bit^=1
else:
if one<az:
ac+='('
bc+='('
else:
ac+=')'
bc+=')'
one+=1
#if debug(ac) and debug(bc):
prin("YES"+'\n');prin(ac+'\n');prin(bc+'\n')
```
| 1,088 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for t1 in range(0,t):
n=int(input())
s=input()
if s[0]=='0' or s[-1]=='0':
print("NO")
else:
same=[]
diff=[]
for i in range(0,n):
if s[i]=='1':
same.append(i)
else:
diff.append(i)
ans1=list(s)
ans2=list(s)
if len(diff)%2==1:
print("NO")
else:
flag=1
for i in diff:
if flag==1:
ans1[i]=')'
ans2[i]='('
else:
ans1[i]='('
ans2[i]=')'
flag=(flag+1)%2
start=0
end=len(same)-1
while(start<end):
xx=same[start]
yy=same[end]
ans1[xx]='('
ans2[xx]='('
ans1[yy]=')'
ans2[yy]=')'
start+=1
end-=1
print("YES")
print(''.join(ans1))
print(''.join(ans2))
```
| 1,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
lines = sys.stdin.readlines()
lineidx = 0
def readLine():
global lineidx
s = lines[lineidx].strip()
lineidx += 1
return s
t = int(readLine())
for testcase in range(t):
n = int(readLine())
s = readLine()
cntone = s.count('1')
cntzero = s.count('0')
if (cntone % 2 != 0) or (cntzero % 2 != 0):
print("NO")
continue
cntone /= 2
a = ['x'] * n
b = ['x'] * n
bala = balb = 0
for i in range(n):
if s[i] == '1':
if cntone > 0:
cntone -= 1
a[i] = b[i] = '('
else:
a[i] = b[i] = ')'
else:
if cntzero % 2 == 0:
a[i] = '('
b[i] = ')'
else:
a[i] = ')'
b[i] = '('
cntzero -= 1
if a[i] == '(':
bala += 1
else:
bala -= 1
if b[i] == '(':
balb += 1
else:
balb -= 1
if bala < 0 or balb < 0:
break
if bala != 0 or balb != 0:
print("NO")
else:
print("YES")
print("".join(a))
print("".join(b))
```
| 1,090 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n = int(input())
a = input()
ans1 = ["*"] * n
ans2 = ["*"] * n
cnt1 = 0
for elem in a:
if elem == "1":
cnt1 += 1
if cnt1 % 2:
print("NO")
continue
cntPlus = cnt1 // 2
tmp = 0
for i in range(n):
if a[i] == "1":
if tmp < cntPlus:
ans1[i] = "("
ans2[i] = "("
tmp += 1
else:
ans1[i] = ")"
ans2[i] = ")"
count1 = 0
count2 = 0
flag = True
for i in range(n):
if a[i] == "1":
if ans1[i] == "(":
count1 += 1
count2 += 1
else:
count1 -= 1
count2 -= 1
if count1 < 0 or count2 < 0:
flag = False
break
else:
if count1 >= count2:
ans1[i] = ")"
ans2[i] = "("
count1 -= 1
count2 += 1
if count1 < 0:
flag = False
break
else:
ans1[i] = "("
ans2[i] = ")"
count1 += 1
count2 -= 1
if count2 < 0:
flag = False
break
if flag:
print("YES")
print("".join(ans1))
print("".join(ans2))
else:
print("NO")
# region fastio
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")
# endregion
if __name__ == "__main__":
main()
```
Yes
| 1,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
# IMPORTS GO HERE
# *************************************************************************
# CLASSES YOU MAKE GO HERE
# *************************************************************************
# ALREADY MADE FUNCTIONS
# To initialize an array of n values with 0
def init_array(n):
arr1d = [0] * n
return arr1d
# To initialize a 2D array of rows * columns 0
def init_array2d(arr_rows, arr_columns):
arr2d = [[0 for a in range(arr_columns)] for b in range(arr_rows)]
return arr2d
# To print a 2d array => helpful in debugging
def print_array2d(arr2d):
# len(arr) = no of rows, len(arr[0]) = no of columns
for f in range(len(arr2d)):
for q in range(len(arr2d[0])):
print(arr2d[f][q])
# To get input of a 2D array given rows and columns
def get_array2d_input(arr_rows, arr_columns):
arr2d = [[0 for f in range(arr_columns)] for q in range(arr_rows)]
for f in range(arr_rows):
line = input()
words = line.split()
for q in range(arr_columns):
arr2d[f][q] = int(words[q])
return arr2d
# *************************************************************************
# FUNCTIONS YOU MAKE GO HERE
def solve():
dummy = 0 # dummy variable
# *************************************************************************
# ******************** MAIN ***********************************
testCases = int(input()) # get total test cases
for test_case in range(1, testCases + 1):
# ********************** SETTING INPUTS ********************************
line1 = input() # First line of a test case
word1 = line1.split() # Splitting line into list of words
N = int(word1[0]) # Getting the value of N
s = input() # Second line of a test case
# ********************** SOLVE HERE ************************************
cost = 0 # To calculate cost
brackets_a = ""
brackets_b = ""
stack_a = []
stack_b = []
possible = True
for i in range(N):
# remaining_str = s[i:]
# remaining_zeros = remaining_str.count('0')
# print("i: " + i)
#
# print("stack_a: " + str(stack_a))
# print("stack_b: " + str(stack_b))
if s[i] == '1':
if len(stack_a) == 0 or len(stack_b) == 0:
brackets_a += '('
brackets_b += '('
stack_a.append('(')
stack_b.append('(')
else:
if i < N - 1:
if len(stack_a) == 1 and len(stack_b) == 1:
if s[i + 1] == '0':
brackets_a += '('
brackets_b += '('
stack_a.append('(')
stack_b.append('(')
else:
if len(stack_a) == 0 or len(stack_b) == 0:
possible = False
break
brackets_a += ')'
brackets_b += ')'
stack_a.pop()
stack_b.pop()
else:
if len(stack_a) == 0 or len(stack_b) == 0:
possible = False
break
brackets_a += ')'
brackets_b += ')'
stack_a.pop()
stack_b.pop()
else:
if len(stack_a) == 0 or len(stack_b) == 0:
possible = False
break
brackets_a += ')'
brackets_b += ')'
stack_a.pop()
stack_b.pop()
else:
if len(stack_a) == 0 and len(stack_b) == 0:
possible = False
break
elif len(stack_a) >= len(stack_b):
if len(stack_a) == 0:
possible = False
break
brackets_a += ')'
brackets_b += '('
stack_a.pop()
stack_b.append('(')
else:
if len(stack_b) == 0:
possible = False
break
brackets_a += '('
brackets_b += ')'
stack_b.pop()
stack_a.append('(')
# print("a: " + brackets_a)
# print("b: " + brackets_b)
# print()
ans = ""
if possible:
if len(stack_a) == 0 and len(stack_b) == 0:
ans = "YES"
else:
ans = "NO"
else:
ans = "NO"
# **********************************************************************
# ************************** OUTPUT ************************************
if ans == "NO":
print("NO")
else:
print("YES")
print(brackets_a)
print(brackets_b)
# *************************************************************************
# ************************** USEFUL CODE SNIPPETS *************************
```
Yes
| 1,092 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
from copy import deepcopy
t = int(input())
for _ in range(t):
n = int(input())
s = input()
count_0 = s.count('0')
if count_0 % 2 != 0:
print("NO")
elif count_0 == 0:
print("YES")
print('(' * (n // 2) + ')' * (n // 2))
print('(' * (n // 2) + ')' * (n // 2))
elif s[0] == '0' or s[-1] == '0':
print("NO")
else:
count_1 = n - count_0
ans = ['0' for _ in range(n)]
counter = 0
for i in range(0, n):
if s[i] == '1':
if counter < count_1 // 2:
ans[i] = '('
else:
ans[i] = ')'
counter += 1
ans_2 = deepcopy(ans)
x,y='(',')'
for i in range(0,n):
if s[i]=='0':
ans[i]=x
ans_2[i]=y
x,y=y,x
print("YES")
print(''.join(ans))
print(''.join(ans_2))
```
Yes
| 1,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
import array
import math
def main() :
tc=1
tc=int(input())
while(tc) :
n=int(input())
s=input()
res=True
cnt=0
for i in range(n) :
if(s[i]=='0') :
if(i==0 or i==n-1) :
res=False
break;
cnt+=1
if(cnt%2==1) :
res=False
if(res) :
c, cnt1, c1, a, b = 0, n-cnt, 0, "", ""
for i in range(n) :
if(s[i]=='1') :
c1+=1
if(c1<=cnt1/2) :
a+="("
b+="("
else :
a+=")"
b+=")"
else :
c+=1
if(c%2) :
a+="("
b+=")"
else :
a+=")"
b+="("
print("YES")
print(a)
print(b)
else :
print("NO")
tc-=1
if __name__=="__main__" :
main()
##### ##### ##### ##### ##### ##### ##### # # # # ##### ######
# # # # # # # # # # # # ## # # # # # #
# # # # # ### ##### ##### ##### # # # ## ##### #
# # # # # # # # # # # # ## # # # # # #
##### ##### ##### ##### # # # # # # # # # # # ####
```
Yes
| 1,094 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
s = input()
s = list(s)
a = ''
b = ''
c1 = c2 = 0
c = 0
d = 0
for i in range(n):
if s[i]=='1':
c+=1
for i in range(n):
if s[0]=='0' or s[-1]=='0' or c%2==1 :
print('NO')
d = 1
break
if s[i]=='1':
c1+=1
if c1<n//2:
a=a+'('
b+='('
else:
a+=')'
b+=')'
else:
if c2==0:
a+='('
b+=')'
if c2==1:
a+=')'
b+='('
if c2==0:
c2=1
elif c2==1:
c2=0
if d==0:
print('YES')
print(a)
print(b)
```
No
| 1,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
st=input()
if(st[0]=="0" or st[-1]=="0"):
print("NO")
else:
s1=[]
s2=[]
o1,o2=0,0
f=1
for i in range(len(st)):
if(st[i]=="1"):
if(o1>0 and o2>0):
o1-=1
o2-=1
s1.append(")")
s2.append(")")
elif(o1==0 or o2==0):
o1+=1
o2+=1
s1.append("(")
s2.append("(")
elif(o1<0 or o2<0):
print("NO")
f=0
break
else:
if(o1>0 and o2>0):
if(o1>=o2):
o1-=1
o2+=1
s1.append(")")
s2.append("(")
else:
o2-=1
o1+=1
s1.append(")")
s2.append("(")
elif(o1>0 and o2==0):
o1-=1
o2+=1
s1.append(")")
s2.append("(")
elif(o2>0 and o1==0):
o1+=1
o2-=1
s1.append("(")
s2.append(")")
elif((o2< 0 or o1<0) or (o1==0 and o2==0)):
print("NO")
f=0
break
if(f==1 and o1==0 and o2==0):
print("YES")
print("".join(s1))
print("".join(s2))
```
No
| 1,096 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
t = int(input())
while(t>0):
n = int(input())
s = input()
a = ""
b = ""
one = 0
zero = 0
for i in s:
if i is "1":
one+=1
elif i is "0":
zero+=1
if(zero%2==1 or one%2==1 or s[0]=="0" or s[-1] =="0"):
print("NO")
else:
for i in s:
if(i=="1"):
if(one):
a+="("
b+="("
one-=2
else:
a+=")"
b+=")"
else:
if(zero%2==0):
a+=")"
b+="("
zero-=1
else:
a+="("
b+=")"
print("YES")
print(a)
print(b)
t-=1
```
No
| 1,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
def TestLine(sLine, nLen):
x = 0
for i in range(nLen):
if sLine[i] == "(":
x = x + 1
else:
x = x - 1
if x < 0:
return False
if x != 0:
return False
return True
nTests = int(input())
for i in range(nTests):
nLen = int(input())
if nLen//2*2 != nLen:
print("NO")
break
sMask = input()
a = ""
for j in range(nLen//2):
a = a + "("
for j in range(nLen//2):
a = a + ")"
b = ""
for j in range(nLen):
if sMask[j] == "1":
b = b + a[j]
elif a[j] == "(":
b = b + ")"
else:
b = b + "("
if TestLine(b, nLen):
print("YES")
print(a)
print(b)
else:
print("NO")
# print(a)
# print(b)
```
No
| 1,098 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time, Oolimry saw a suffix array. He wondered how many strings can produce this suffix array.
More formally, given a suffix array of length n and having an alphabet size k, count the number of strings that produce such a suffix array.
Let s be a string of length n. Then the i-th suffix of s is the substring s[i … n-1]. A suffix array is the array of integers that represent the starting indexes of all the suffixes of a given string, after the suffixes are sorted in the lexicographic order. For example, the suffix array of oolimry is [3,2,4,1,0,5,6] as the array of sorted suffixes is [imry,limry,mry,olimry,oolimry,ry,y].
A string x is lexicographically smaller than string y, if either x is a prefix of y (and x≠ y), or there exists such i that x_i < y_i, and for any 1≤ j < i , x_j = y_j.
Input
The first line contain 2 integers n and k (1 ≤ n ≤ 200000,1 ≤ k ≤ 200000) — the length of the suffix array and the alphabet size respectively.
The second line contains n integers s_0, s_1, s_2, …, s_{n-1} (0 ≤ s_i ≤ n-1) where s_i is the i-th element of the suffix array i.e. the starting position of the i-th lexicographically smallest suffix. It is guaranteed that for all 0 ≤ i< j ≤ n-1, s_i ≠ s_j.
Output
Print how many strings produce such a suffix array. Since the number can be very large, print the answer modulo 998244353.
Examples
Input
3 2
0 2 1
Output
1
Input
5 1
0 1 2 3 4
Output
0
Input
6 200000
0 1 2 3 4 5
Output
822243495
Input
7 6
3 2 4 1 0 5 6
Output
36
Note
In the first test case, "abb" is the only possible solution.
In the second test case, it can be easily shown no possible strings exist as all the letters have to be equal.
In the fourth test case, one possible string is "ddbacef".
Please remember to print your answers modulo 998244353.
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
import sys
input = sys.stdin.readline
mod = 998244353
n, k = map(int, input().split())
A = list(map(int, input().split()))
rank = [0] * (n + 1)
for i, a in enumerate(A):
rank[a] = i
rank[n] = -1
cnt = 0
for i in range(n - 1):
if rank[A[i] + 1] > rank[A[i + 1] + 1]: cnt += 1
x, y = 1, 1
for i in range(n):
x = x * (n - 1 + k - cnt - i) % mod
y = y * (i + 1) % mod
print(x * pow(y, mod - 2, mod) % mod)
```
| 1,099 |
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