text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
import sys
def I():
return sys.stdin.readline().rstrip()
cds = ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" ]
def dist(s, n):
cs = cds[n]
cost = 0
for c, cc in zip(s, cs):
if c != cc:
if c == '1':
return (False, 0)
cost += 1
return (True, cost)
def main():
n, k = map(int, I().split())
dstring = [I() for _ in range(n)]
lens = {n: [0]}
# dp = {(n, 0): ('-', 0)}
dp = [[None] * (k+1) for _ in range(n+1)]
dp[n][0] = ('-', 0)
for i in range(n-1, -1, -1):
idone = set()
lens[i] = set()
for d in range(9, -1, -1):
char = chr(ord('0')+d)
res = dist(dstring[i], d)
if res[0]:
diff = res[1]
if diff in idone:
continue
idone.add(diff)
for ln in lens[i+1]:
ndiff = ln + diff
if ndiff <= k and ndiff not in lens[i]:
lens[i].add(ndiff)
dp[i][ndiff] = (char, diff)
if dp[0][k] != None:
ans = []
for i in range(n):
dpv = dp[i][k]
ans.append(dpv[0])
k -= dpv[1]
print("".join(ans))
else:
print("-1")
main()
```
Yes
| 10,100 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
def a2b(a, b):
if a & (a^b) != 0:
return -1
else:
return sum(d == '1' for d in bin(b & (a^b))[2:])
n, k = map(int, input().split())
arr = []
for _ in range(n):
arr.append(int(input(),2))
num_txt = ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
numbers = list(map(lambda x: int(x, 2), num_txt)) # 0 to 9
cost = [[0] * (10) for _ in range(n)]
for i in range(n):
for d in range(10):
cost[i][d] = a2b(arr[i],numbers[d])
dp = [[0] * (k+1) for _ in range(n+1)]
dp[n][0] = 1
for i in range(n,0,-1):
for j in range(k+1):
if dp[i][j] == 1:
for d in range(10):
if cost[i-1][d] != -1 and cost[i-1][d] <= k-j:
dp[i-1][j+cost[i-1][d]] = 1
if dp[0][k] == 0:
print(-1)
else:
ans = ''
for i in range(n):
for d in range(9,-1,-1):
if 0 <= cost[i][d] <= k and dp[i+1][k-cost[i][d]] == 1:
ans += str(d)
k -= cost[i][d]
break
print(ans)
```
Yes
| 10,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
import sys
readline = sys.stdin.readline
inf = 10**18
digits = [119, 18, 93, 91, 58, 107, 111, 82, 127, 123]
popcount = [
[bin(x - y).count('1') if x & y == x else inf for y in range(2**7)]
for x in range(2**7)
]
N, K = map(int, input().split())
a = [int(readline().strip(), 2) for _ in range(N)]
dp = [[0]*(K+1) for _ in range(N+1)]
dp[0][K] = 1
# rightmost-digit to leftmost-digit
for i, x in zip(range(N), reversed(a)):
for j in range(K, -1, -1):
if not dp[i][j]:
continue
for y in digits:
if popcount[x][y] <= j:
dp[i+1][j - popcount[x][y]] = 1
if not dp[N][0]:
print(-1)
exit()
j = 0
ans = []
# leftmost-digit to rightmost digit
for i, x in zip(range(N, 0, -1), a):
for k, y in zip(range(9, -1, -1), reversed(digits)):
if j + popcount[x][y] <= K and dp[i-1][j + popcount[x][y]]:
ans.append(k)
j += popcount[x][y]
break
print(*ans, sep='')
```
Yes
| 10,102 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
from sys import stderr
m = {
"1110111": 0,
"0010010": 1,
"1011101": 2,
"1011011": 3,
"0111010": 4,
"1101011": 5,
"1101111": 6,
"1010010": 7,
"1111111": 8,
"1111011": 9
}
def diff(s, t):
ans = 0
for i in range(7):
if s[i] < t[i]:
ans += 1
elif s[i] > t[i]:
return - 1
return ans
def debug(*arg):
print(*arg, file=stderr, flush=True)
dp = []
d = []
n, k = 0, 0
def func(i, left):
if i == n:
return 0 if left == 0 else - int(1e9)
if dp[i][left] is not None:
return dp[i][left]
ans = -int(1e9)
for key, value in m.items():
temp = diff(d[i], key)
if temp != -1 and temp <= left:
ans = max(ans, value*(10**(n-i-1))+func(i+1, left-temp))
dp[i][left] = ans
return ans
n, k = list(map(int, input().split()))
dp = n * [None]
for i in range(n):
dp[i] = (k+1)*[None]
d = n * [None]
for i in range(n):
d[i] = input().strip()
temp = func(0, k)
if temp < 0:
print(-1)
else:
print(temp)
```
No
| 10,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
from sys import stdin, stdout
class Param:
inf = int(1e10)
code = ['1110111', '0010010', '1011101', '1011011', '0111010', '1101011', '1101111', '1010010', '1111111', '1111011']
n,k = list(map(int, stdin.readline().split()))
s_list = []
for i in range(n):
s_list.append( str(stdin.readline()) )
pass
D = {}
def move(s, value):
if((s, value) in D):
return D[(s, value)]
pass
x = code[value]
ans = 0
for i in range(7):
if(x[i] == '1' and s[i] == '0'):
ans+=1
elif(x[i] == '1' and s[i] == '1'):
pass
elif(x[i] == '0' and s[i] == '0'):
pass
elif(x[i] == '0' and s[i] == '1'):
ans = Param.inf
break
pass
D[(s, value)] = ans
return ans
#dp = [[False] *(k+1)]*(n+1)
dp = [[False]*(k+1) for i in range(n+1)]
dp[0][0] = True
for pos in range(1, n+1):
for i in range(k+1):
for value in range(10):
need = move(s_list[-pos], value)
if (i-need >= 0 and dp[pos-1][i-need]):
dp[pos][i] = True
print(str(pos) + ' ' + str(i))
break
pass
pass
pass
ans = ''
if(dp[n][k] == False):
ans = '-1'
else:
rest = k
for pos in reversed(range(1, n+1)):
for value in reversed(range(0,10)):
need = move(s_list[-n], value)
if(rest - need >=0 and dp[pos-1][rest - need]):
ans += str(value)
rest = rest - need
break
pass
pass
pass
print(ans)
```
No
| 10,104 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
n,v= map(int,input().split())
w=n
d= ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
l=[]
for _ in range(n):
s= input()
l.append(s)
u=''
for i in l:
m=[]
for j in range(7):
if i[j]=='1':
m.append(j)
p=[]
q=[]
for k in d:
c=0
for o in m:
if k[o]=='1':
c+=1
if c==i.count('1'):
p.append(k)
q.append(d.index(k))
r=[x.count('1')-i.count('1') for x in p]
for t in range(len(p)-1,-1,-1):
if n==1:
if r[t]==v:
v-=r[t]
u+= str(q[t])
n-=1
break
else:
if r[t]<=v:
v-=r[t]
u+= str(q[t])
n-=1
break
if len(u)!=w:
print("-1")
else:
print(u)
```
No
| 10,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 7 segments, which can be turned on or off to display different numbers. The picture shows how all 10 decimal digits are displayed:
<image>
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly k segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly k sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of digits on scoreboard and k (0 ≤ k ≤ 2000) — the number of segments that stopped working.
The next n lines contain one binary string of length 7, the i-th of which encodes the i-th digit of the scoreboard.
Each digit on the scoreboard consists of 7 segments. We number them, as in the picture below, and let the i-th place of the binary string be 0 if the i-th stick is not glowing and 1 if it is glowing. Then a binary string of length 7 will specify which segments are glowing now.
<image>
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 0 to 9 inclusive.
Output
Output a single number consisting of n digits — the maximum number that can be obtained if you turn on exactly k sticks or -1, if it is impossible to turn on exactly k sticks so that a correct number appears on the scoreboard digits.
Examples
Input
1 7
0000000
Output
8
Input
2 5
0010010
0010010
Output
97
Input
3 5
0100001
1001001
1010011
Output
-1
Note
In the first test, we are obliged to include all 7 sticks and get one 8 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 5 of additionally included sticks, you can get the numbers 07, 18, 34, 43, 70, 79, 81 and 97, of which we choose the maximum — 97.
In the third test, it is impossible to turn on exactly 5 sticks so that a sequence of numbers appears on the scoreboard.
Submitted Solution:
```
from sys import stdin, stdout
codes = dict()
def create_codes():
digits = ["1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"]
global codes
for i in range(128):
c = '{:07b}'.format(i)
cnt_dict = dict()
for k,x in enumerate(digits):
cnt = 0
for j in range(7):
if c[j] > x[j]:
j-=1
break
elif c[j] < x[j]:
cnt += 1
if j == 6:
cnt_dict[cnt] = k
codes[c] = cnt_dict
def main():
n,k = list(map(int, stdin.readline().split()))
global codes
create_codes()
mx = [-1] * (k+1)
mx[0] = 0
for _ in range(n):
tmp = [-1] * (k+1)
code = stdin.readline().rstrip()
for i in range(k,-1,-1):
if mx[i] == -1:
continue
for key,val in codes[code].items():
if i + key > k:
continue
if n!= 2000:
tmp[i+key] = max(mx[i] * 10 + val, tmp[i+key])
else:
tmp[i+key] = max(100000000, 911234565)
mx = tmp
if len(str(mx[k])) != n:
stdout.write("-1")
else:
stdout.write(str(mx[k]))
main()
```
No
| 10,106 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
s = set(list(map(int, input().split())))
p=-1
for i in range(1,1025):
for x in s:
if x^i not in s:
break
else:
p=i
if p!=-1:
print(p)
break
else:
print(p)
```
| 10,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
for test in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
b=a.copy()
b.sort()
ans=-1
if b[0]==0:
for i in range(1,n):
temp=[]
for j in range(n):
temp.append(a[j]^b[i])
if set(temp)==set(a):
ans=b[i]
break
print(ans)
else:
m=max(b)
y=0
while 1:
if 2**y>m:
break
y+=1
for i in range(1,2**y+1):
temp=[]
for j in range(n):
temp.append(a[j]^i)
if set(temp)==set(a):
ans=i
break
print(ans)
```
| 10,108 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import ceil, floor, gcd, sqrt, trunc, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
for _ in range(int(data())):
n = int(data())
arr = set(l())
r = False
answer = -1
for i in range(1, 1025):
s = set()
for j in arr:
s.add(j ^ i)
if s == arr:
r = True
answer = i
break
if r:
break
out(answer)
```
| 10,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
import math
t = int(input())
for _ in range(t):
n = int(input())
setList = [int(var) for var in input().split()]
origSet = set(setList)
flag = True
for k in range(1, 1025):
visited = set()
for var in setList:
# print(var, 10^var)
visited.add(k^var)
if len(visited) == len(setList):
if visited==origSet:
print(k)
# print(visited, origSet)
flag = False
break
if flag is True:
print(-1)
```
| 10,110 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
''' ## ## ####### # # ######
## ## ## ## ### ##
## ## ## # # # ##
######### ####### # # ## '''
import sys
import math
# sys.setrecursionlimit(10**6)
def get_ints(): return map(int, input().strip().split())
def get_list(): return list(get_ints())
def printspx(*args): return print(*args, end="")
def printsp(*args): return print(*args, end=" ")
def printchk(*args): return print(*args, end=" \ ")
MODPRIME = int(1e9+7); BABYMODPR = 998244353;
# sys.stdin = open("input.txt","r") # <<< Comment this line >>> #
for _testcases_ in range(int(input())):
n = int(input())
li = get_list()
flag = False
se = set(li)
for i in range(1, 1024):
se2 = set()
for j in li:
se2.add(j ^ i)
if se2 == se:
print(i)
flag = True
break
if not flag:
print(-1)
'''
THE LOGIC AND APPROACH IS BY ME @luctivud ( UDIT GUPTA )
SOME PARTS OF THE CODE HAS BEEN TAKEN FROM WEBSITES LIKE::
(I Own the code if no link is provided here or I may have missed mentioning it)
>>> DO NOT PLAGIARISE.
TESTCASES:
>>> COMMENT THE STDIN !!
'''
```
| 10,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = set(map(int, input().split()))
ans = 0
for i in range(1024):
s = a.copy()
ok = True
while len(s) != 0:
e = s.pop()
if e^i in s:
s.remove(e^i)
else:
ok = False
break
if len(s) == 0 and ok:
ans = i
break
print(ans if ok else -1)
```
| 10,112 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
S = list(map(int, input().split(' ')))
done = False
for k in range(1,1024):
s = set(S)
while len(s) > 0:
x = s.pop()
if (k^x) not in s:
s.add(x)
break
else:
s.remove(k^x)
if len(s) == 0:
print(k)
done = True
break
if not done:
print(-1)
```
| 10,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
t=int(input())
import math
for _ in range(t):
n=int(input())
s=set([int(x) for x in input().split()])
if (n==1):
print(-1)
continue
m=2**(int(math.log2(max(s)))+1)-1
ans = 0
k=1
while (k<=m and ans==0):
s2=set()
for i in s:
s2.add(i^k)
if (s==s2):
ans=k
k+=1
if (ans!=0):
print(ans)
else:
print(-1)
```
| 10,114 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Tags: bitmasks, brute force
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return 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+
# main code
for t in range(ni()):
n=ni()
l=set(li())
f=0
for k in range(1,1025):
s=set()
for i in l:
s.add(i^k)
#print s,l
if s==l:
f=1
break
if f:
pn(k)
else:
pn(-1)
```
| 10,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
t = int(input())
for loop in range(t):
n = int(input())
d = {}
arr = list(map(int,input().split()))
arr.sort()
for i in arr:
d[i] = 1
if(arr[0]==0):
ans =0
for i in range(1,n):
flag = 0
val = arr[i]
for j in arr:
if(val^j not in d):
flag = 1
break
if(flag==0):
print(val)
ans = 1
break
if(ans==0):
print(-1)
else:
ans =0
for i in range(1,1025):
flag = 0
if(i not in d):
for j in arr:
if i^j not in d:
flag = 1
break
else:
continue
if(flag==0):
print(i)
ans = 1
break
if(ans==0):
print(-1)
```
Yes
| 10,116 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
from sys import stdin, stdout
import math,sys
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
import heapq
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(input()))
def In():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def I():return (int(stdin.readline()))
def In():return(map(int,stdin.readline().split()))
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_right(a, x)
if i != len(a):
return i
else:
return -1
def main():
try:
n=I()
l=list(In())
d=dict(l)
ans=-1
for i in range(1,1025):
count=0
for x in l:
temp=x^i
if d.get(temp,-1)==-1:
break
else:
count+=1
if count==n:
ans=i
break
print(ans)
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
for _ in range(I()):main()
#for _ in range(1):main()
```
Yes
| 10,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
t = int(input())
import math
delay = []
for _ in range(t):
n = int(input())
arr = input().split()
arr = [int(a) for a in arr]
res = 1
found = False
while(res <= 1024):
go = True
result = [(res ^ a) for a in arr]
dic = {}
for r in result:
if r in dic:
dic[r] += 1
else:
dic[r] = 1
for a in arr:
if a not in dic:
go = False
break
else:
dic[a] -= 1
if go:
for k in dic:
if dic[k] == 0:
continue
else:
go = False
break
if go:
print(res)
found = True
break
res += 1
if not found:
print(-1)
```
Yes
| 10,118 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
from functools import reduce
for _ in range(int(input())):
n = int(input())
s = [int(i) for i in input().split()]
s.sort()
for j in range(1, 1025):
if j == 1024:
print(-1)
break
t = [j ^ x for x in s]
t.sort()
if all([t[i] == s[i] for i in range(n)]):
print(j)
break
```
Yes
| 10,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
import sys,math
input=sys.stdin.readline
t=int(input())
for r in range(t):
n=int(input())
l=list(map(int,input().split()))
d1 = {}
for i in l:
try:
d1[i]+=1
except:
d1[i]=1
maxi = max(l)
ans = 0
for i in range(1,maxi+1):
temp = []
for j in l:
temp.append(j^i)
d2 = {}
for j in temp:
try:
d2[j]+=1
except:
d2[j]=1
if d1 == d2:
ans = i
break
if ans == 0:
print(-1)
else:
print(ans)
```
No
| 10,120 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
import sys
input=sys.stdin.readline
def N(): return int(input())
def NM():return map(int,input().split())
def L():return list(NM())
def LN(n):return [N() for i in range(n)]
def LL(n):return [L() for i in range(n)]
t=N()
def f():
n=N()
l=set(L())
x=l.pop()
for i in l:
k=x^i
for j in l:
if not (j==i or j^k in l):
break
else:
print(k)
break
else:
print(-1)
for i in range(t):
f()
```
No
| 10,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
def newxtPow(n):
cur = math.log(n,2)
return 2**math.ceil(cur)
import math
for _ in range(int(input())):
n = int(input())
s = set(map(int,input().split()))
l = list(s)
# p = 1/max(l)
if n==1:
print(-1)
continue
ans = -1
if max(l)!=0:
for i in range(1,newxtPow(max(l))+1):
new = set([])
for j in l:
new.add(j^i)
# print(i,new,s)
if new==s:
ans = i
break
print(ans)
```
No
| 10,122 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Among Johnny's numerous hobbies, there are two seemingly harmless ones: applying bitwise operations and sneaking into his dad's office. As it is usually the case with small children, Johnny is unaware that combining these two activities can get him in a lot of trouble.
There is a set S containing very important numbers on his dad's desk. The minute Johnny heard about it, he decided that it's a good idea to choose a positive integer k and replace each element s of the set S with s ⊕ k (⊕ denotes the [exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Computer_science) operation).
Help him choose such k that Johnny's dad will not see any difference after his son is done playing (i.e. Johnny will get the same set as before playing). It is possible that no such number exists. It is also possible that there are many of them. In such a case, output the smallest one. Note that the order of elements in a set doesn't matter, i.e. set \{1, 2, 3\} equals to set \{2, 1, 3\}.
Formally, find the smallest positive integer k such that \\{s ⊕ k | s ∈ S\} = S or report that there is no such number.
For example, if S = \{1, 3, 4\} and k = 2, new set will be equal to \{3, 1, 6\}. If S = \{0, 1, 2, 3\} and k = 1, after playing set will stay the same.
Input
In the first line of input, there is a single integer t (1 ≤ t ≤ 1024), the number of test cases. In the next lines, t test cases follow. Each of them consists of two lines.
In the first line there is a single integer n (1 ≤ n ≤ 1024) denoting the number of elements in set S. Second line consists of n distinct integers s_i (0 ≤ s_i < 1024), elements of S.
It is guaranteed that the sum of n over all test cases will not exceed 1024.
Output
Print t lines; i-th line should contain the answer to the i-th test case, the minimal positive integer k satisfying the conditions or -1 if no such k exists.
Example
Input
6
4
1 0 2 3
6
10 7 14 8 3 12
2
0 2
3
1 2 3
6
1 4 6 10 11 12
2
0 1023
Output
1
4
2
-1
-1
1023
Note
In the first test case, the answer is 1 because it is a minimum positive integer and it satisfies all the conditions.
Submitted Solution:
```
kl = int(input())
for l in range(kl):
for er in range(1):
pr=1
n = int(input())
a=[int(i) for i in input().split()]
if n%2!=0:
pr=0
break
d, mn1, mn2=0, 1024, 1024
for i in range(n):
d=d^a[i]
if a[i]<mn1:
mn1, mn2 = a[i], mn1
if n%4!=0:
if d==0:
pr=0
break
else:
for i in range(n):
if d^a[i] not in a:
pr=0
break
else:
if d!=0:
pr=0
break
else:
d=mn1^mn2
if pr:
print(d)
else:
print(-1)
```
No
| 10,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
from sys import stdin
input=stdin.readline
def answer():
ans = []
i , j = 0 , n - 1
ind = n - 1
while(i <= j):
if(a[i] == b[ind]):
ans.append(1)
ans.append(ind + 1)
else:
ans.append(ind + 1)
if(i == j):break
ind -= 1
if(a[j] ^ 1 == b[ind]):
ans.append(1)
ans.append(ind + 1)
else:
ans.append(ind + 1)
ind -= 1
i , j = i + 1 , j - 1
print(len(ans) , *ans)
for T in range(int(input())):
n=int(input())
a = [int(i) for i in input().strip()]
b = [int(i) for i in input().strip()]
answer()
```
| 10,124 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys,os,io
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = sys.stdin.readline
for _ in range (int(input())):
n = int(input())
a = input().strip()
b = input().strip()
curr = a[0]
ans = []
for i in range (n):
if a[i]!=curr:
ans.append(i)
curr = a[i]
for i in range (n-1,-1,-1):
if curr != b[i]:
ans.append(i+1)
curr = b[i]
print(len(ans),*ans)
```
| 10,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
str1="First"
str2="Second"
def rev(c,cnt):
if(cnt%2==0):
return c
if(c=='0'):
c='1'
else:
c='0'
return c
def solve():
n=int(input())
a=input()
b=input()
ans=list()
l,r=0,n-1
for i in range(n):
if(i%2==0):
if(rev(a[l],i)==b[n-i-1]):
ans.append(1)
l+=1
else:
if (rev(a[r], i) == b[n - i - 1]):
ans.append(1)
r -= 1
ans.append(n-i)
print(len(ans),sep=' ',end=' ')
for num in ans:
print(num,sep=' ',end=' ')
print()
def main():
t=int(input())
for i in range(t):
solve()
main()
```
| 10,126 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
#------------------------------warmup----------------------------
# *******************************
# * AUTHOR: RAJDEEP GHOSH *
# * NICK : Rajdeep2k *
# * INSTITUTION: IIEST, SHIBPUR *
# *******************************
import os
import sys
from io import BytesIO, IOBase
import math
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now---------------------------------------------------
for _ in range(int(input())):
n = int(input())
s1 = input()
s2 = input()
if(s1 == s2):
print('0')
else:
s1 = list(s1)
s2 = list(s2)
ans=[]
end=n-1
rev=0
fc=s1[0]
while(end>=0):
if rev&1:
fc=chr(1 - (ord(s1[n-(rev//2)-1]) - 48) + 48)
else:
fc=s1[rev//2]
if fc==s2[end]:
ans.append(1)
ans.append(end+1)
rev+=1
end-=1
else:
ans.append(end+1)
rev+=1
end-=1
print(len(ans),end=' ')
print(*ans)
```
| 10,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import math
import sys
#input=sys.stdin.readline
t=int(input())
#t=1
for _ in range(t):
n=int(input())
#n,m=map(int,input().split())
#l1=list(map(int,input().split()))
a=input()
a+='0'
b=input()
b+='0'
ans=[]
ans1=[]
for i in range(n):
if a[i]=='1' and a[i+1]=='0' :
ans.append(i+1)
if a[i]=='0' and a[i+1]=='1':
ans.append(i+1)
if b[i]=='1' and b[i+1]=='0' :
ans1.append(i+1)
if b[i]=='0' and b[i+1]=='1':
ans1.append(i+1)
if len(ans)+len(ans1)==0:
print(0)
else:
print(len(ans)+len(ans1),*ans,*ans1[::-1])
```
| 10,128 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
def f(c):
if c=='1':
return '0'
else:
return '1'
for _ in range(int(input())):
n=int(input())
a=list(input())
b=list(input())
l=[]
i=n-1
j=n-1
am=True
while j>=0:
if am:
if a[i]!=b[j]:
if a[i-j]==b[j]:
l.append(1)
a[i-j]=f(a[i-j])
l.append(j+1)
am=False
i-=j
i+=1
else:
i-=1
else:
if a[i]==b[j]:
if a[i+j]!=b[j]:
l.append(1)
a[i+j]=f(a[i+j])
l.append(j+1)
am=True
i+=j
i-=1
else:
i+=1
j-=1
l1=[len(l)]+l
print(*l1)
```
| 10,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
import sys
t=int(input())
for _ in range(t):
n=int(input())
tempa=input()
tempb=input()
a=[]
for i in tempa:
a.append(int(i))
b=[]
for i in tempb:
b.append(int(i))
count=0
anslist=[]
start=1
rev=0
for i in range(n-1,-1,-1):
if(rev==0):
pos=start+i
else:
pos=start-i
curr=a[pos-1]
if(rev==1):
curr=curr^1
if(curr!=b[i]):
first=a[start-1]
if(rev==1):
first=first^1
if(first!=b[i]):
anslist.append(i+1)
count+=1
else:
anslist.append(1)
anslist.append(i+1)
count+=2
rev=rev^1
start=pos
print(count,end=" ")
for i in anslist:
sys.stdout.write(str(i)+" ")
print()
```
| 10,130 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Tags: constructive algorithms, data structures, implementation, strings, two pointers
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
avl=AvlTree()
#-----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a ,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left)/ 2)
# Check if middle element is
# less than or equal to key
if (arr[mid]<=key):
count = mid+1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def countGreater( arr,n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
for ik in range(int(input())):
n=int(input())
s=list(input())
for i in range(n):
s[i]=int(s[i])
s1=list(input())
for i in range(n):
s1[i]=int(s1[i])
ans=[]
cou=0
lastz=s[0]
su=0
for i in range(n-1,-1,-1):
si=s[su+(-1)**(cou)*i]
if cou%2==1:
si=1-si
if s1[i]!=si:
if lastz==s1[i]:
ans.append(1)
ans.append(i+1)
lastz=s1[i]
if cou % 2 == 1:
su -= i
else:
su += i
cou+=1
#print(s)
print(len(ans),*ans)
#print(*ans)
```
| 10,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
# from collections import Counter
# import math
def tc():
n = int(input())
a = input()
b = input()
segs = []
prev = a[0]
for i, ch in enumerate(a[1:]):
if ch != prev:
segs.append(i + 1)
prev = ch
bsegs = []
bprev = b[0]
for i, ch in enumerate(b[1:]):
if ch != bprev:
bsegs.append(i + 1)
bprev = ch
if a[-1] != b[-1]:
segs.append(n)
segs.extend(bsegs[::-1])
print(len(segs), ' '.join(map(str, segs)))
##################################
T = int(input())
for _ in range(T):
tc()
# tc()
```
Yes
| 10,132 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from sys import stdin
input = lambda: stdin.readline().rstrip("\r\n")
from collections import deque as que, defaultdict as vector
from heapq import*
inin = lambda: int(input())
inar = lambda: list(map(int,input().split()))
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack: return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrappedfunc
Testcase=inin()
for _ in range(Testcase):
n=inin()
a=list(input())
b=list(input())
a=[int(x) for x in a]+[0]
b=[int(x) for x in b]+[0]
aw=[]
bw=[]
for i in range(n):
if a[i]!=a[i+1]:
aw.append(i+1)
if b[i]!=b[i+1]:
bw.append(i+1)
print(len(aw)+len(bw),*(aw),*bw[::-1])
```
Yes
| 10,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def flip(a,flip):
if flip%2==0:
return a
else:
if a==1:
return 0
else:
return 1
for _ in range(int(input())):
n = int(input())
a = list(map(int,list(input())))
b = list(map(int,list(input())))
i = 0
j = n-1
ans = []
flipp = 0
while j>=0:
if flip(a[i],flipp)!=b[j]:
ans.append(j+1)
else:
ans.append(1)
ans.append(j+1)
flipp+=1
if i==0:
a.pop(0)
i = j-1
else:
a.pop()
i = 0
j-=1
ans.insert(0,len(ans))
print(*ans)
```
Yes
| 10,134 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from collections import deque
t = int(input())
for _ in range(t):
n = int(input())
a = deque()
b = deque()
for i in input():
a.append(0 if i=='0' else 1)
for i in input():
b.append(0 if i=='0' else 1)
ans = []
rev = 0
while b:
bb = b.pop()
if rev%2==0:
a0 = a[0]
aa = a[-1]
else:
a0 = 1-a[-1]
aa = 1-a[0]
if aa==bb:
if rev%2==0:
a.pop()
else:
a.popleft()
continue
if a0==bb and b:
ans.append(1)
ans.append(len(b)+1)
rev += 1
if rev%2==0:
a.pop()
else:
a.popleft()
print(len(ans), end=' ')
print(*ans)
```
Yes
| 10,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def helper(s):
ans = []
s = '0' + s
for i in range(1,len(s)):
if s[i]!=s[i-1]:
ans.append(i)
return ans
t = int(input())
for l in range(t):
n = int(input())
a = input()
b = input()
a1 = helper(a)
a2 = helper(b)
a = a1+a2[::-1]
print(len(a),end = " ")
for i in a:
print(i,end=" ")
print()
```
No
| 10,136 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def solve(s):
new = ""
for i in s:
if i=="1":
new += "0"
else:
new += "1"
return new[::-1]
for nt in range(int(input())):
n = int(input())
a = input()
b = input()
ans = []
i = 1
count = 0
while i<n:
if i+1<n:
if a[i]!=b[i] and a[i+1]!=b[i+1]:
if a[i]==a[i+1]:
ans.append(i+2)
ans.append(2)
ans.append(i+2)
else:
ans.append(i+2)
ans.append(1)
ans.append(2)
ans.append(1)
ans.append(i+2)
i += 2
elif a[i]==b[i] and a[i+1]!=b[i+1]:
i += 1
elif a[i]!=b[i] and a[i+1]==b[i+1]:
ans.append(i+1)
ans.append(1)
ans.append(i+1)
i += 2
else:
i += 2
else:
if a[i]!=b[i]:
ans.append(i+1)
ans.append(1)
ans.append(i+1)
break
if a[0]!=b[0]:
ans.append(1)
print (len(ans),end = " ")
print (*ans)
# def solve(s):
# new = ""
# for i in s:
# if i=="1":
# new += "0"
# else:
# new += "1"
# return new[::-1]
# for nt in range(int(input())):
# n = int(input())
# a = input()
# b = input()
# ans = []
# for i in range(n-1,-1,-1):
# print (a,i,a[i])
# if a[i]!=b[i]:
# if a[i]==a[0]:
# ans.append(i+1)
# a = solve(a[0:i+1]) + a[i+1:]
# else:
# ans.append(1)
# ans.append(i+1)
# a = solve(a[i]+a[1:i+1]) + a[i+1:]
# print (len(ans),end = " ")
# print (*ans)
```
No
| 10,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
for _ in range (int(input())):
n = int(input())
a = list(input())
b = list(input())
ans = []
for i in range (n-1,-1,-1):
if b[i]!=a[0]:
ans.append(i+1)
if a[i]=='0':
a[0]='1'
else:
a[0] = '0'
else:
ans.append(1)
if a[0]=='0':
a[0]='1'
else:
a[0] = '0'
ans.append(i+1)
if a[i]=='0':
a[0]='1'
else:
a[0] = '0'
print(len(ans),end=" ")
print(*ans)
```
No
| 10,138 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
"""for _ in range(int(input())):
n,m=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
f=0
for i in range(n):
if l1[i] in l2:
f=1
ans=l1[i]
break
if(f==0):
print("NO")
else:
print("YES")
print(1, ans)"""
"""for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
x=l.count(1)
if(x==0):
print("First")
else:
if(x==n):
if(n%2==0):
print("Second")
else:
print("First")
else:
c=0
for i in range(n):
if(l[i]==1):
c+=1
else:
break
if(c%2==0):
print("First")
else:
print("Second")"""
for _ in range(int(input())):
n=int(input())
la=str(input())
lb=str(input())
ans=n
l=[]
a=list(la)
b=list(lb)
left=0
right=n-1
for i in range(n-1,-1,-1):
temp=n-i+1
if(temp%2==0):
if(a[left]==b[i]):
l.append(1)
left+=1
l.append(i+1)
else:
if(a[right]!=b[i]):
l.append(1)
right-=1
l.append(i+1)
print(len(l),end=" ")
for i in range(len(l)):
print(l[i],end=" ")
print()
```
No
| 10,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Chanek just won the national chess tournament and got a huge chessboard of size N × M. Bored with playing conventional chess, Mr. Chanek now defines a function F(X, Y), which denotes the minimum number of moves to move a knight from square (1, 1) to square (X, Y). It turns out finding F(X, Y) is too simple, so Mr. Chanek defines:
G(X, Y) = ∑_{i=X}^{N} ∑_{j=Y}^{M} F(i, j)
Given X and Y, you are tasked to find G(X, Y).
A knight can move from square (a, b) to square (a', b') if and only if |a - a'| > 0, |b - b'| > 0, and |a - a'| + |b - b'| = 3. Of course, the knight cannot leave the chessboard.
Input
The first line contains an integer T (1 ≤ T ≤ 100), the number of test cases.
Each test case contains a line with four integers X Y N M (3 ≤ X ≤ N ≤ 10^9, 3 ≤ Y ≤ M ≤ 10^9).
Output
For each test case, print a line with the value of G(X, Y) modulo 10^9 + 7.
Example
Input
2
3 4 5 6
5 5 8 8
Output
27
70
Submitted Solution:
```
import itertools
def f(x):
scores = [0, 0]
for i in itertools.cycle([0, 1]):
if x & 1:
scores[i] += 1
x -= 1
elif x == 0:
return scores[0]
elif x & 0b10:
x >>= 1
scores[i] += x
else:
x -= 1
scores[i] += 1
N = int(input())
results = []
import sys
for n in map(f, map(int, sys.stdin.read().split())):
results.append(n)
print('\n'.join(map(str, results)))
```
No
| 10,140 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Chanek just won the national chess tournament and got a huge chessboard of size N × M. Bored with playing conventional chess, Mr. Chanek now defines a function F(X, Y), which denotes the minimum number of moves to move a knight from square (1, 1) to square (X, Y). It turns out finding F(X, Y) is too simple, so Mr. Chanek defines:
G(X, Y) = ∑_{i=X}^{N} ∑_{j=Y}^{M} F(i, j)
Given X and Y, you are tasked to find G(X, Y).
A knight can move from square (a, b) to square (a', b') if and only if |a - a'| > 0, |b - b'| > 0, and |a - a'| + |b - b'| = 3. Of course, the knight cannot leave the chessboard.
Input
The first line contains an integer T (1 ≤ T ≤ 100), the number of test cases.
Each test case contains a line with four integers X Y N M (3 ≤ X ≤ N ≤ 10^9, 3 ≤ Y ≤ M ≤ 10^9).
Output
For each test case, print a line with the value of G(X, Y) modulo 10^9 + 7.
Example
Input
2
3 4 5 6
5 5 8 8
Output
27
70
Submitted Solution:
```
def sum_triangle(x,y,len):
if(len==0):
return 0
ans=0
if((x+y)%3==0):
a=len//3
ans+=(a*(a+1)*(2*a+1)*9)/6-(3*a*(a+1))/2
ans+=(3*a*(a+1))/2-a
if(len%3==1):
ans+=(len)*(a+1)
if(len%3==2):
ans+=(len)*(a+1)+(len-1)*(a+1)
ans+=len
ans+=((len*(len+1))/2)*((x+y)/3 -1)
return ans
if((x+y)%3==2):
ans+=0
a=(len-1)//3
ans+=(a*(a+1)*(2*a+1)*9)//6
ans+=(3*a*(a+1))//2
if((len-1)%3==1):
ans+=(len)*(a+1)
if((len-1)%3==2):
ans+=(len)*(a+1)+(len-1)*(a+1)
ans+=len
ans+=((len*(len+1))//2)*((x+y)//3)
return ans
if((x+y)%3==1):
ans+=0;
a=(len-2)//3
ans+=(a*(a+1)*(2*a+1)*9)//6+(3*a*(a+1))//2
ans+=1
ans+=(3*a*(a+1))//2+a
if((len-2)%3==1):
ans+=(len)*(a+1)
if((len-2)%3==2):
ans+=(len)*(a+1)+(len-1)*(a+1)
ans+=len;
ans+=((len*(len+1))//2)*((x+y)//3)
return ans
t=int(input())
for you in range(t):
l=input().split()
x=int(l[0])
y=int(l[1])
n=int(l[2])
m=int(l[3])
a=sum_triangle(x,y,(m-y+1)+(n-x+1))
b=sum_triangle(x,m+1,(n-x+1))
c=sum_triangle(n+1,y,(m-y+1))
if(x==3 and y==3):
a+=2
print(a-b-c)
```
No
| 10,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Chanek just won the national chess tournament and got a huge chessboard of size N × M. Bored with playing conventional chess, Mr. Chanek now defines a function F(X, Y), which denotes the minimum number of moves to move a knight from square (1, 1) to square (X, Y). It turns out finding F(X, Y) is too simple, so Mr. Chanek defines:
G(X, Y) = ∑_{i=X}^{N} ∑_{j=Y}^{M} F(i, j)
Given X and Y, you are tasked to find G(X, Y).
A knight can move from square (a, b) to square (a', b') if and only if |a - a'| > 0, |b - b'| > 0, and |a - a'| + |b - b'| = 3. Of course, the knight cannot leave the chessboard.
Input
The first line contains an integer T (1 ≤ T ≤ 100), the number of test cases.
Each test case contains a line with four integers X Y N M (3 ≤ X ≤ N ≤ 10^9, 3 ≤ Y ≤ M ≤ 10^9).
Output
For each test case, print a line with the value of G(X, Y) modulo 10^9 + 7.
Example
Input
2
3 4 5 6
5 5 8 8
Output
27
70
Submitted Solution:
```
def sum_triangle(x,y,len):
if(len==0):
return 0
ans=0
if((x+y)%3==0):
a=len//3
ans+=(a*(a+1)*(2*a+1)*9)/6-(3*a*(a+1))/2
ans+=(3*a*(a+1))/2-a
if(len%3==1):
ans+=(len)*(a+1)
if(len%3==2):
ans+=(len)*(a+1)+(len-1)*(a+1)
ans+=len
ans+=((len*(len+1))/2)*((x+y)/3 -1)
return ans
if((x+y)%3==2):
ans+=0
a=(len-1)//3
ans+=(a*(a+1)*(2*a+1)*9)//6
ans+=(3*a*(a+1))//2
if((len-1)%3==1):
ans+=(len)*(a+1)
if((len-1)%3==2):
ans+=(len)*(a+1)+(len-1)*(a+1)
ans+=len
ans+=((len*(len+1))//2)*((x+y)//3)
return ans
if((x+y)%3==1):
ans+=0;
a=(len-2)//3
ans+=(a*(a+1)*(2*a+1)*9)//6+(3*a*(a+1))//2
ans+=1
ans+=(3*a*(a+1))//2+a
if((len-2)%3==1):
ans+=(len)*(a+1)
if((len-2)%3==2):
ans+=(len)*(a+1)+(len-1)*(a+1)
ans+=len;
ans+=((len*(len+1))//2)*((x+y)//3)
return ans
t=int(input())
for you in range(t):
l=input().split()
x=int(l[0])
y=int(l[1])
n=int(l[2])
m=int(l[3])
a=sum_triangle(x,y,(m-y+1)+(n-x+1))
b=sum_triangle(x,m+1,(n-x+1))
c=sum_triangle(n+1,y,(m-y+1))
if(x==3 and y==3):
a+=2
print((a-b-c)%1000000007)
```
No
| 10,142 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
for _ in range(int(input())):
n = int(input())
print(n)
print(*range(1, n + 1))
```
| 10,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=[]
for i in range(1,n+1):
a.append(i)
print(n)
print(*a)
```
| 10,144 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
t = int(input().strip())
for _ in range(t):
n = int(input().strip())
ans = [i for i in range(2,n + 1)]
print(len(ans))
print(' '.join(str(i) for i in ans))
```
| 10,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
print(n-1)
j=n
for i in range(2,n+1):
print(i,end=" ")
```
| 10,146 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=[x+1 for x in range(n)]
print(n)
print(*l)
```
| 10,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
#bsdwalon aasan question diya karo
for _ in range(int(input())):
n = int(input())
print(n-1)
temp = []
for i in range(2, n+1):
temp.append(i)
print(*temp)
```
| 10,148 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
a=int(input())
for i in range(a):
b=int(input())
arr=[str(i) for i in range(b,1,-1)]
arr=arr[::-1]
print(len(arr))
print(' '.join(arr))
```
| 10,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Tags: constructive algorithms, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
print(n)
a = [i+1 for i in range(n)]
print(*a)
```
| 10,150 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
#########
def solve():
a = int(input())
print(a)
lst = []
for i in range(1,a+1):
lst.append(i)
print(*lst)
#########
def main():
testcases = 1
testcases = int(input())
for _ in range(testcases):
solve()
# 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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
Yes
| 10,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
T = int(input())
for _ in range(T):
N = int(input())
print(N)
print(*range(1, N + 1))
```
Yes
| 10,152 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
print(n)
for i in range(1, n + 1):
print(i, end=' ')
```
Yes
| 10,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
print(n)
for i in range(n):
print(i+1,end=" ")
print()
```
Yes
| 10,154 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
import math
def lcm(a, b):
m = a * b
while a != 0 and b != 0:
if a > b:
a %= b
else:
b %= a
return m // (a + b)
def dtod(a):
s=''
while a!=0:
s+=str(a%2)
a=int(a/2)
return s[::-1]
n = int(input())
for i in range(n):
a = int(input())
print(a - 1)
b = list(range(a, 1, -1))
for item in b:
print((str(item) + ' ') * (item - 2) + str(item), end = ' ')
```
No
| 10,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
print(n)
for i in range(n, 0, -1):
print(i, end = " ")
print()
```
No
| 10,156 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
import sys
import math
from collections import Counter
from collections import OrderedDict
from collections import defaultdict
from functools import reduce
#from itertools import groupby
sys.setrecursionlimit(10**6)
def inputt():
return sys.stdin.readline().strip()
def printt(n):
sys.stdout.write(str(n)+'\n')
def listt():
return [int(i) for i in inputt().split()]
def gcd(a,b):
return math.gcd(a,b)
def lcm(a,b):
return (a*b) // gcd(a,b)
def factors(n):
step = 2 if n%2 else 1
return set(reduce(list.__add__,([i, n//i] for i in range(1, int(math.sqrt(n))+1, step) if n % i == 0)))
def comb(n,k):
factn=math.factorial(n)
factk=math.factorial(k)
fact=math.factorial(n-k)
ans=factn//(factk*fact)
return ans
def is_prime(n):
if n <= 1:
return False
if n == 2:
return True
if n > 2 and n % 2 == 0:
return False
max_div = math.floor(math.sqrt(n))
for i in range(3, 1 + max_div, 2):
if n % i == 0:
return False
return True
def maxpower(n,x):
B_max = int(math.log(n, x)) + 1 #tells upto what power of x n is less than it like 1024->5^4
return B_max
t=int(input())
for _ in range(t):
n=int(inputt())
print(n)
for i in range(n,0,-1):
print(i,end=" ")
print()
```
No
| 10,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 ≤ m ≤ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 ≤ n≤ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].
Submitted Solution:
```
"""
Code of Ayush Tiwari
Codeforces: servermonk
Codechef: ayush572000
"""
import sys
input = sys.stdin.buffer.readline
def solution():
# This is the main code
n=int(input())
print(n)
l=[]
for i in range(n,0,-1):
l.append(i)
print(*l)
t=int(input())
for _ in range(t):
solution()
```
No
| 10,158 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
def main():
t = int(input())
for _ in range(t):
n, x = map(int, input().split())
a = list(map(int, input().split()))
l = []; mx = 0
for i in range(n):
e = a[i]
p = 0
while e % x == 0:
p += 1
e //= x
l.append(p + 1)
if l[i] < l[mx]:
mx = i
#print(l, mx)
suma = sum(a)
total = suma * l[mx]
for j in range(mx):
total += a[j]
print(total)
main()
```
| 10,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
def strangeList(a,x):
i=0
n=len(a)
count=0
flag=0
while True:
j=0
while j<n:
if a[j]%(x**i)==0:
count+=a[j]
else:
return count
j+=1
i+=1
return count
t=int(input())
for i in range(t):
n,x=map(int,input().split())
a=list(map(int,input().split()))
print(strangeList(a,x))
```
| 10,160 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n,x=map(int,input().split())
arr=list(map(int,input().split()))
res=0
tmp=arr.copy()
new_arr=[0]*n
for i in range(n):
cnt=0
while arr[i]%x==0:
cnt+=1
arr[i]=arr[i]//x
new_arr[i]=cnt
idx=new_arr.index(min(new_arr))
min_value=min(new_arr)
res=(min_value+1)*sum(tmp)+sum(tmp[:idx])
print(res)
```
| 10,161 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
# Code Submission
#
# Author : GuptaSir
# Date : 05:01:2021
# Time : 20:14:45
#
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
for _ in range(int(input())):
n, x = map(int, input().split())
a = [*map(int, input().split())]
Ma = max(a)
px = [None for i in range(35)]
px[0] = 1
for i in range(1, 35):
px[i] = px[i - 1] * x
if px[i] > 10 ** 19:
break
px = [i for i in px if i != None]
lpx = len(px)
ac = [None for i in range(n)]
for i in range(n):
j = 1
while True:
if a[i] % px[j] == 0:
j += 1
else:
break
ac[i] = j - 1
ans = 0
mac = min(ac)
mac_reached = False
for i in range(n):
if ac[i] == mac:
mac_reached = True
if mac_reached == False:
ans += (mac + 2) * a[i]
else:
ans += (mac + 1) * a[i]
print(ans)
```
| 10,162 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
for i in range(int(input())):
n, x = map(int, input().split())
a = list(map(int, input().split()))
ans, mi, midx = 0, 10**10, n - 1
for i in range(n):
y = a[i]
cnt = 0
while(True):
if(y % x):
break
else:
cnt += 1;
y /= x
if(cnt < mi):
mi = cnt
midx = i
for i in a:
ans += (mi + 1) * i
for i in range(midx):
ans += a[i]
print(ans)
```
| 10,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
###pyrival template for fast IO
import os
import sys
from io import BytesIO, IOBase
# 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")
t=int(input())
while t>0:
t-=1
n,x=[int(x) for x in input().split()]
arr=[int(x) for x in input().split()]
count=[0 for x in range(n)]
for i in range(n):
val=arr[i]
c=1
while True:
if val%x==0:
val=val//x
c+=1
else:
count[i]=c
break
s=sum(arr)
index=0;val=count[0]
for i in range(1,n):
if count[i]<val:
index=i
val=count[i]
total=s*val
for i in range(index):
total+=arr[i]
print(total)
```
| 10,164 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
t=int(input())
for _ in range(t):
l=list(map(int,input().split()))
n,x=l[0],l[1]
l=list(map(int,input().split()))
arr=[]
for i in range(n):
k=0
p=l[i]
while(p%x==0):
k+=1
p=p//x
arr.append(k)
b=min(arr)
f=arr.index(min(arr))
s=0
for i in range(n):
if(i<=f):
d=l[i]
su=l[i]
k=0
while(d%x==0 and k<b+1):
d=d//x
su+=l[i]
k+=1
else:
d=l[i]
su=l[i]
k=0
while(d%x==0 and k<b):
d=d//x
su+=l[i]
k+=1
s+=su
print(s)
```
| 10,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Tags: brute force, greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n, x = [int(i) for i in input().split(' ')]
a = [int(i) for i in input().split(' ')]
m = 0
summ = 0
stop = 0
while True:
for i in a:
if i % x**m == 0:
summ += i
else:
stop = 1
break
m += 1
if stop == 1:
break
print(summ)
```
| 10,166 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n,k=[int(x) for x in input().split()]
arr=[int(x) for x in input().split()]
ans=0
ind=-1
res=1000000000000
for i in range(n):
now=1
curr=arr[i]
while curr%k==0:
now+=1
curr=curr//k
if now<res:
res=now
ind=i
for i in range(n):
if i<ind:
ans+=(res+1)*arr[i]
else:
ans+=res*arr[i]
print(ans)
```
Yes
| 10,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
import sys
import math
def ans(A, x):
pivot = None
numSplits = math.inf
i = len(A) - 1
while i >= 0:
splits = 0
curr = A[i]
while curr%x==0:
splits += 1
curr //= x
if splits > numSplits:
break
if splits <= numSplits:
pivot = i
numSplits = splits
i -= 1
ans = 0
for i in range(0, pivot):
ans += A[i]*(numSplits+2)
for i in range(pivot, len(A)):
ans += A[i]*(numSplits+1)
return ans
def main():
l1 = []
l2 = []
for i, line in enumerate(sys.stdin):
if i == 0:
continue
idx = i - 1
if idx%2 == 0:
l1 = [int(j) for j in line.split(' ')]
else:
l2 = [int(j) for j in line.split(' ')]
b = ans(l2, l1[1])
print(b)
if __name__ == '__main__':
main()
```
Yes
| 10,168 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, x = map(int, input().split())
A = list(map(int, input().split()))
B = [0]*n
flag = 1
for i in range(n):
temp = A[i]
while(temp%x==0):
temp = temp//x
B[i]+=1
cut = min(B)
#print(B, cut)
ans = sum(A)
for i in range(n):
if flag == 1:
ans+=A[i]*min(cut+1, B[i])
if cut==B[i]:
flag =0
else:
ans+=A[i]*(min(B[i], cut))
print(ans)
```
Yes
| 10,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
tc=int(input())
for k in range(tc):
n,k=list(map(int,input().split()))
l=list(map(int,input().split()))
div=1
ans=0
while(1):
flag=1
for i in l:
if(i%div==0):
ans+=i
else:
flag=0
break
if(flag==0):
break
div=div*k
print(ans)
```
Yes
| 10,170 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
t=int(input())
while(t!=0):
t-=1
n,x=map(int,input().split())
fg=list(map(int,input().split()))
for i in range(0,len(fg)):
if(fg[i]%2==0):
k=0
while(k!=x):
fg.append(fg[i]//2)
k+=1
else:
break
f=sum(fg)
print(f)
```
No
| 10,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
for t in range(int(input())):
n,x=map(int,input().split())
l=list(map(int,input().split()))
l1=[[]for i in range(n)]
a=x
z=0
ans=0
for j in range(n):
if l[j]%x:
a=0
ans+=l[j]
break
ans+=l[j]
c=0
b=l[j]
while b%x==0:
b//=x
l1[j].append(b)
c+=1
l1[j].append(b)
if a>c:
a=c
z=j
for k in range(a):
for j in range(n):
ans+=(x**(k+1))*l1[j][k]
for j in range(z):
ans+=(x**(a+1))*l1[j][a]
print(ans)
```
No
| 10,172 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
def findpow(n, x):
t = 0
while n%x==0:
n //= x
t += 1
return t
def solve(arr, x):
pows = [findpow(i, x) for i in arr]
return arr[0]*(pows[0]+1) + sum(arr[i]*pows[i] for i in range(1, len(arr)))
t = int(input())
ans = []
for i in range(t):
n, x = map(int, input().split())
arr = [int(i) for i in input().split()]
ans.append(solve(arr, x))
for i in ans:
print(i)
```
No
| 10,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.
Please determine the sum of all values of the array at the end of the process.
Input
The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.
The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.
It is guaranteed that the sum of values n over all test cases does not exceed 10^5.
Output
For each test case output one integer — the sum of all elements at the end of the process.
Example
Input
2
1 2
12
4 2
4 6 8 2
Output
36
44
Note
In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.
In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].
Submitted Solution:
```
"""
n=int(z())
for _ in range(int(z())):
x=int(z())
l=list(map(int,z().split()))
n=int(z())
l=sorted(list(map(int,z().split())))[::-1]
a,b=map(int,z().split())
l=set(map(int,z().split()))
led=(6,2,5,5,4,5,6,3,7,6)
vowel={'a':0,'e':0,'i':0,'o':0,'u':0}
color4=["G", "GB", "YGB", "YGBI", "OYGBI" ,"OYGBIV",'ROYGBIV' ]
"""
#!/usr/bin/env python
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def powofx(n, x, m, flag):
c = 0
t = n
if t % x == 0 and flag:
while t % x == 0:
c += 1
t //= x
if c == m and m != 0:
break
return n*(c+1), c
else:
return n*(c+1), False
def main():
try:
from math import ceil
for _ in range(inp()):
n, x = invr()
a = inlt()
s = 0
m = 10**5
f = True
for i in range(n):
temp = a[i]
c = 0
if temp % x != 0:
f = False
if f:
while temp % x == 0:
c += 1
temp //= x
if c == m:
break
m = min(c, m)
s += (c+1)*a[i]
else:
s += a[i]
print(s)
except Exception as e:
print(e)
# 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()
```
No
| 10,174 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
import sys,math
input=sys.stdin.readline
from collections import defaultdict
for _ in range(int(input())):
n,w=map(int,input().split())
ans=0
v=list(map(int,input().split()))
d=defaultdict(int)
vis=[]
for i in v:
d[i]+=1
if d[i]==1:
vis.append(i)
vis.sort(reverse=True)
cnt=0
while cnt<n:
cur=0
i=0
while i<len(vis):
if d[vis[i]]>0:
if cur+vis[i]<=w:
cnt+=1
cur+=vis[i]
d[vis[i]]-=1
i-=1
i+=1
ans+=1
print(ans)
```
| 10,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
from heapq import heappop, heappush
T = int(input())
for _ in range(T):
N, W = map(int, input().split())
A = list(map(int, input().split()))
A.sort(reverse=True)
lst = [0]
idx = 0
for a in A:
flag = False
v = heappop(lst)
if v + a <= W:
heappush(lst, v + a)
else:
heappush(lst, v)
heappush(lst, a)
# for idx in range(len(lst)):
# if flag:
# break
# if lst[idx] + a <= W:
# lst[idx] += a
# flag = True
# if not flag:
# lst.append(a)
# print(lst)
print(len(lst))
```
| 10,176 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
def main():
from sys import stdin, stdout
from collections import Counter
rl = stdin.readline
wl = stdout.write
for _ in range(int(rl())):
n, w = map(int, rl().split())
a = Counter(int(x) for x in rl().split())
a = dict(a.most_common())
keys = sorted(a.keys(), reverse=True)
h = -1
while True:
cur = 0
h += 1
for key in keys:
while a[key] != 0 and cur + key <= w:
cur += key
a[key] -= 1
if cur == 0:
break
wl(str(h) + '\n')
main()
```
| 10,177 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
import sys
from os import path
if(path.exists("inp.txt")):
sys.stdin = open("inp.txt",'r')
sys.stdout = open("out.txt",'w')
from collections import Counter
t=int(input())
while(t):
t-=1
n,w=map(int,input().split())
li=list(map(int,input().split()))
li.sort()
omp=Counter(li)
r=0
while(omp!={}):
r+=1
h=w
for j in sorted(omp.keys(),reverse=1):
while(omp[j]>0 and h>=j):
h-=j
omp[j]-=1
if omp[j]==0:
del(omp[j])
break
print(r)
```
| 10,178 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
import sys
import heapq
input = sys.stdin.readline
for _ in range(int(input())):
n, w = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
heap = []
ans = 1
heapq.heappush(heap, -(w-a[n-1]))
for i in range(n-2, -1, -1):
tmp = -heapq.heappop(heap)
if tmp>=a[i]:
heapq.heappush(heap, -(tmp-a[i]))
else:
ans+=1
heapq.heappush(heap, -tmp)
heapq.heappush(heap, -(w-a[i]))
print(ans)
```
| 10,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
t=int(input())
for _ in range(t):
n,y=map(int,input().split())
l=list(map(int,input().split()))
d={}
ans=0
for i in l:
d[i]=d.get(i,0)+1
while d:
w=y
for j in reversed(sorted(d.keys())):
#print(x)
#print(d.keys())
x=min(w//j,d[j])
d[j]-=x
w-=x*j
if d[j]==0:del d[j]
if w==0:break
ans+=1
print(ans)
```
| 10,180 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
from collections import *
for _ in range(int(input())):
n, w = map(int, input().split())
l = list(map(int, input().split()))
C = Counter(l)
height = 0
while len(C) > 0:
height += 1
rest = w
for i in range(30, -1, -1):
while (1 << i) in C and rest >= (1 << i):
rest -= (1 << i)
C[(1 << i)] -= 1
if C[1 << i] == 0:
del C[1 << i]
print(height)
```
| 10,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Tags: binary search, bitmasks, data structures, greedy
Correct Solution:
```
# cp template by varun
from math import ceil
from math import floor
from random import random
from math import gcd
def ar ():
return [int(x) for x in input().split()]
# -----
T = 1
T = int(input())
size = 2**20 + 7
dp = [0]*size
for t in range (T):
N,W = ar()
A = ar()
for i in A:
dp[i]+=1
ans = 0
while N > 0:
curr = W
i = 2**20
while i>0:
while dp[i]>0 and curr>=i:
dp[i]-=1
curr-=i
N-=1
i//=2
ans+=1
print (ans)
```
| 10,182 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
from math import log2
t = int(input())
while t > 0:
t -= 1
def solve():
n, w = list(map(int, input().split()))
widths = list(map(int, input().split()))
counts = [0 for i in range(20)]
for width in widths:
counts[int(log2(width))] += 1
space = w
height = 1
for i in range(n):
largest = -1
for size, count_width in list(enumerate(counts))[::-1]:
if counts[size] > 0 and (2 ** size) <= space:
largest = size
break
if largest == -1:
space = w
height += 1
for size, count_width in list(enumerate(counts))[::-1]:
if counts[size] > 0 and (2 ** size) <= space:
largest = size
break
largest != -1
counts[largest] -= 1
space -= 2 ** largest
print(height)
solve()
```
Yes
| 10,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
import math
T = int(input())
def process(tab, nb_piece, width):
tab = sorted(tab)
data = [0] * (int(math.log(tab[-1], 2)) +1)
power = 1
index = 0
for elem in tab:
while (elem > power):
power *= 2
index += 1
if elem == power:
data[index] += 1
#print(data)
#print(tab, nb_piece, width)
stage = 1
trous = width
#cumulate_trous = width
while nb_piece > 0:
#print("process", stage, data, trous, nb_piece)
for index in range(len(data)-1, -1, -1):
elem = 2 ** index
while (data[index] > 0 and elem <= trous):
data[index] -= 1
trous -= elem
nb_piece -= 1
#print(elem, trous, data[index], index)
stage += 1
trous = width
if trous == width:
stage -= 1
return stage
#return sum(tab) // width
for _ in range(T):
data = input().split()
n, width = int(data[0]), int(data[1])
data = input().split()
tab = [int(elem) for elem in data]
res = process(tab, n, width)
print(res)
```
Yes
| 10,184 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
from collections import Counter
for _ in range(int(input())):
n,w=map(int,input().split())
w1=list(map(int,input().split()))
w1.sort(reverse=True)
d=dict(Counter(w1))
h=0
while n>0:
k=w
for i in d:
while i<=k and d[i]>0:
n-=1
k-=i
d[i]-=1
h+=1
print(h)
```
Yes
| 10,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
t = int(input())
pow2 = {i:2**i for i in range(31)}
log2 = {v:k for k, v in pow2.items()}
for _ in range(t):
n, w = map(int, input().split())
arr = [0 for _ in range(31)]
nums = list(map(int, input().split()))
for num in nums:
arr[log2[num]] += 1
pw = f'{bin(w)[2:]:0>31}'[::-1]
ans=0
while True:
changed = False
curr2 = -1
currleft = 0
for i in range(30, -1, -1):
curr2 = i
currleft += (pw[i] == '1')
if not currleft:
continue
if arr[i]:
if arr[i] >= currleft:
arr[i] -= currleft
currleft = 0
changed = True
else:
currleft -= arr[i]
arr[i] = 0
changed = True
currleft *= 2
if not changed:
break
else:
ans += 1
print(ans)
```
Yes
| 10,186 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
def bs(e,target,top):
low = 0
mid = (top - 0) // 2
while low <= top:
if e[mid] <= target:
return mid,e[mid]
else:
return bs(e,target,mid-1)
return -1,-1
t = int(input())
for i in range(t):
n,w = map(int,input().split())
e = list(map(int,input().split()))
e.sort()
counter = 0
target = w - e[0]
del e[0]
index = 0
while len(e) > 0:
target = w - e[0]
del e[0]
#print(e)
while target != 0:
length = len(e)
index,num = bs(e,target,length-1)
if index == -1:
counter += 1
break
else:
target -= num
del e[index]
else:
counter += 1
print(counter)
```
No
| 10,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
for _ in range(int(input())):
n, w = map(int, input().split())
l = sum(list(map(int, input().split())))
if l % w == 0:
print(l // w)
else:
print((l // w) + 1)
```
No
| 10,188 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
for i in range(int(input())):
n,w=list(map(int,input().split()))
lst=list(map(int,input().split()))
lst.sort(reverse=True)
dict1={}.fromkeys(lst,0)
for key in lst:
dict1[key]+=1
#print(dict1)
for key in dict1:
print(dict1[key])
break
```
No
| 10,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n rectangles, each of height 1. Each rectangle's width is a power of 2 (i. e. it can be represented as 2^x for some non-negative integer x).
You are also given a two-dimensional box of width W. Note that W may or may not be a power of 2. Moreover, W is at least as large as the width of the largest rectangle.
You have to find the smallest height of this box, such that it is able to fit all the given rectangles. It is allowed to have some empty space left in this box after fitting all the rectangles.
You cannot rotate the given rectangles to make them fit into the box. Moreover, any two distinct rectangles must not overlap, i. e., any two distinct rectangles must have zero intersection area.
See notes for visual explanation of sample input.
Input
The first line of input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^3) — the number of test cases. Each test case consists of two lines.
For each test case:
* the first line contains two integers n (1 ≤ n ≤ 10^5) and W (1 ≤ W ≤ 10^9);
* the second line contains n integers w_1, w_2, ..., w_n (1 ≤ w_i ≤ 10^6), where w_i is the width of the i-th rectangle. Each w_i is a power of 2;
* additionally, max_{i=1}^{n} w_i ≤ W.
The sum of n over all test cases does not exceed 10^5.
Output
Output t integers. The i-th integer should be equal to the answer to the i-th test case — the smallest height of the box.
Example
Input
2
5 16
1 2 8 4 8
6 10
2 8 8 2 2 8
Output
2
3
Note
For the first test case in the sample input, the following figure shows one way to fit the given five rectangles into the 2D box with minimum height:
<image>
In the figure above, the number inside each rectangle is its width. The width of the 2D box is 16 (indicated with arrow below). The minimum height required for the 2D box in this case is 2 (indicated on the left).
In the second test case, you can have a minimum height of three by keeping two blocks (one each of widths eight and two) on each of the three levels.
Submitted Solution:
```
import math
for _ in range(int(input())):
n,k=map(int,input().split())
a=list(map(int,input().split()))
if sum(a)<k:
print(1)
continue
d={}
for i in a:
if i not in d:
d[i]=1
else:
d[i]+=1
m=-1
for i in d:
if d[i]>m:
m=d[i]
if m==1:
print(1)
else:
print(m)
```
No
| 10,190 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
e = int(input())
task = []
for i in range(e):
day = int(input())
string = input()
task.append([day,string])
for x in task:
temp = []
flag = True
# temp = list(set(x[1]))
# ctr = x[1][0]
temp.append(x[1][0])
for i in x[1]:
if (i != temp[-1]):
temp.append(i)
if (len(temp) == len(list(set(temp)))):
print("yes")
else:
print("no")
```
| 10,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
t = int(input())
while t:
n = int(input())
st1 = input()
st2 = {}
st2 = set(st2)
f = 0
st2.add(st1[0])
for i in range(1, n):
if st1[i] != st1[i-1]:
if st1[i] not in st2:
st2.add(st1[i])
else:
f = 1
break
if f == 1:
print("NO")
else:
print("YES")
t -= 1
```
| 10,192 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
from collections import Counter
for _ in range(int(input())):
n=int(input())
s=input()
x=Counter(s)
string=''
for m in x:
string+=m*(x[m])
if string==s:
print("YES")
else:
print("NO")
```
| 10,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
s = input()
letters = set()
letters.add(s[0])
for i in range(1, n):
if s[i] != s[i-1]:
if s[i] in letters:
print("NO")
break
else:
letters.add(s[i])
else:
print("YES")
```
| 10,194 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
n=int(input())
for i in range(n):
l=int(input())
str=input()
li=[]
c=0
for j in str:
if j not in li:
li.append(j)
else:
if li[len(li)-1]==j:
continue
else:
c=1
print("NO")
break
if c!=1:
print("YES")
```
| 10,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
# cook your dish here
try:
for t in range(int(input())):
n = int(input())
#l = list(map(int,input().split()))
s = input()
l = []
l.append(s[0])
c = 'YES'
for j in range(1,n):
if(s[j]==s[j-1]):
pass
elif(s[j] in l):
c = 'NO'
else:
l.append(s[j])
print(c)
except EOFError as e:
pass
```
| 10,196 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
for x in range(int(input())):
n=int(input())
s=input()
d={}
for i in range(n):
if s[i] not in d:
d[s[i]]=[i]
else:
d[s[i]].append(i)
flag=0
for i in d:
for j in range(len(d[i])-1):
if(d[i][j+1]-d[i][j]==1):
pass
else:
flag=1
break
if(flag==1):
print("No")
else:
print("Yes")
```
| 10,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Tags: brute force, implementation
Correct Solution:
```
def solve(n,char_str):
checked = []
checking = char_str[0]
for i in char_str:
if i != checking:
if i in checked:
return 'NO'
checked.append(checking)
checking = i
return 'YES'
t = int(input())
for i in range(t):
n = int(input())
char_str = input()
print(solve(n,char_str))
```
| 10,198 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.
The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.
Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.
For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".
If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".
Help Polycarp find out if his teacher might be suspicious.
Input
The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.
The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.
Output
For each test case output:
* "YES", if the teacher cannot be suspicious;
* "NO", otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB
Output
NO
NO
YES
YES
YES
Submitted Solution:
```
# A: Do Not be Distracted!
#
t = int(input())
for elee in range(t):
n = int(input())
st = input()
lst = list(st)
for ind, ite in enumerate(st):
if st[ind-1]!=ite and ite in lst[:ind]:
print("NO")
break
else:
print("YES")
```
Yes
| 10,199 |
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