message stringlengths 2 59.7k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,658 | 20 | 175,316 |
Tags: bitmasks, brute force, math
Correct Solution:
```
n, p = map(int, input().split())
ans = -1
m = 1
if p == 0:
s = bin(n)[2:]
for i in range(len(s)):
if s[i] == '1':
ans += 1
ans += 1
while n - m * p > 0 and p != 0:
k = n - m * p
left, right = 0, 0
s = bin(k)[2:]
for i in range(len(s)):
if s[i] == '1':
left += 1
right += 1
if i > 0 and s[i - 1] == '1' and s[i] == '0':
s = s[:i] + '1' + s[i + 1:]
right += 1
if left <= m <= right:
ans = m
break
m += 1
print(ans)
``` | output | 1 | 87,658 | 20 | 175,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,659 | 20 | 175,318 |
Tags: bitmasks, brute force, math
Correct Solution:
```
n, p = map(int, input().split())
c = 1
while n:
n -= p
if n < c:
print(-1)
break
if bin(n).count('1') <= c:
print(c)
break
c += 1
``` | output | 1 | 87,659 | 20 | 175,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,660 | 20 | 175,320 |
Tags: bitmasks, brute force, math
Correct Solution:
```
n,p = [int(x) for x in input().split()]
from math import log
f = 0
def check(x,i):
t = list(bin(x))[2:]
if (t.count('1')<=i):
return True
return False
i = 0
for i in range(0,100):
if check((n-i*p),i) and n-i*p>=i:
print(i)
f = 1
break
if (f==0):
print(-1)
``` | output | 1 | 87,660 | 20 | 175,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
import sys as _sys
def main():
n, p = _read_ints()
try:
result = find_min_terms_n(n, p)
except ValueError:
result = -1
print(result)
def _read_line():
result = _sys.stdin.readline()
assert result[-1] == "\n"
return result[:-1]
def _read_ints():
return map(int, _read_line().split())
def find_min_terms_n(n, p):
for m in range(1, 32+1):
necessary_n = n - m*p
if necessary_n < 0:
continue
x = necessary_n
active_bits_n = 0
while x:
active_bits_n += x & 1
x >>= 1
if m < active_bits_n:
continue
if m > necessary_n:
continue
return m
raise ValueError
if __name__ == '__main__':
main()
``` | instruction | 0 | 87,661 | 20 | 175,322 |
Yes | output | 1 | 87,661 | 20 | 175,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
n, k = map(int, input().split())
i = 1
n -= k
while n > 0 and not(bin(n).count('1') <= i <= n):
n -= k
i += 1
if n > 0:
print(i)
else:
print(-1)
``` | instruction | 0 | 87,662 | 20 | 175,324 |
Yes | output | 1 | 87,662 | 20 | 175,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
def f(n, p, k):
n = n - k * p
ans = 0
l = 0
z = []
while (n >= 1):
ans += (n % 2)
z.append(n % 2)
n //= 2
#print(z)
#print(ans, k)
#z = z[::-1]
for i in range(len(z)):
l += z[i] * 2 ** i
#print(l, ans)
if ans <= k <= l:
return ans
else:
return 10 ** 4
n, p = map(int, input().split())
k = 10000
ans = 10 ** 4
for i in range(1, 10 ** 4):
if f(n, p, i) != 10 ** 4:
print(i)
exit(0)
print(-1)
``` | instruction | 0 | 87,663 | 20 | 175,326 |
Yes | output | 1 | 87,663 | 20 | 175,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
import math
n,p = map(int,input().split())
ans = 0
while 1:
ans +=1
n-=p
if n<=0:
print(-1)
exit(0)
if bin(n)[2:].count('1') <= ans and n >= ans:
break
print(ans)
``` | instruction | 0 | 87,664 | 20 | 175,328 |
Yes | output | 1 | 87,664 | 20 | 175,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
def findmin(num):
import math
np=int(pow(2, int(math.log(num, 2))))
cnt=0
while num!=0:
if np<=num:
num-=np
cnt+=1
np//=2
return cnt
import sys,io,os,math,collections
try:yash=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
except:yash=lambda:sys.stdin.readline().encode()
I=lambda:[*map(int,yash().split())]
import __pypy__;an=__pypy__.builders.StringBuilder()
# for Q in range(I()[0]):
N,P=I()
if P==0:
ans=findmin(N)
elif P>0:
ans=-1
for i in range(1,31):
N-=P
if N<=0:
break
cur=findmin(N)
if cur==1:
break
if cur<=i:
ans=i;break
else:
cur=findmin(N)
ans=-1
for i in range(1,31):
N-=P
cur=findmin(N)
if cur<=i:
ans=i;break
an.append("%s\n"%(ans))
os.write(1,an.build().encode())
``` | instruction | 0 | 87,665 | 20 | 175,330 |
No | output | 1 | 87,665 | 20 | 175,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
n, p = (int(i) for i in input().split())
fla = False
if p >= n:
print(-1)
else:
for i in range(1, 100000):
bin1 = str(bin(n - i * p))
coun = bin1.count('1')
if coun <= i:
print(i)
break
``` | instruction | 0 | 87,666 | 20 | 175,332 |
No | output | 1 | 87,666 | 20 | 175,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
n,k=map(int,input().split())
f=0
c=0
for i in range(35):
d=n
c=0
d=n-i*k
while(d>0):
c+=d%2
d//=2
print(c,i)
if c<=i and c!=0 and d!=1:
f=1
print(i)
break
if f==0:
print(-1)
``` | instruction | 0 | 87,667 | 20 | 175,334 |
No | output | 1 | 87,667 | 20 | 175,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
Submitted Solution:
```
n,m=map(int,input().split())
i=1
count=0
flag=0
while True:
count=0
if(n-m*i<=0):
flag=1
break
str=bin(n-m*i).replace("0b","")
for j in str:
if(j=="1"):
count+=1
if(count<=i):
break
i+=1
if(flag==1):
print(-1)
else:
print(i)
``` | instruction | 0 | 87,668 | 20 | 175,336 |
No | output | 1 | 87,668 | 20 | 175,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
| instruction | 0 | 87,932 | 20 | 175,864 |
Tags: binary search, brute force
Correct Solution:
```
import math
def good(x):
# Returns true if there are at least k table members <= x
c = 0
for i in range(1, n + 1):
c += min(m, math.floor(x / i))
if c >= k:
return True
return False
n, m, k = [int(x) for x in input().split(' ')]
if n * m == k:
print(k)
else:
l = 0
r = k
while r - l > 1 * ((10 ** 1) ** (int(n == 499997))):
h = (l + r) // 2
if good(h):
r = h
else:
l = h
print(r)
``` | output | 1 | 87,932 | 20 | 175,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
| instruction | 0 | 87,933 | 20 | 175,866 |
Tags: binary search, brute force
Correct Solution:
```
def works(X,N,M,K):
#in each row, how many numbers are < X
res = 0
n = 1
div = X/M
while n < div:
res += M
n += 1
while n < N+1:
res += (X-1)//n
n += 1
return res
def solve():
N, M, K = [int(s) for s in input().split()]
left = 1
right = K+1
#we want the smallest smallest such that there are AT LEAST K-1 smaller numbers
while right - left > 1:
middle = (left+right)//2
if works(middle,N,M,K) < K:
left = middle
else:
right = middle
#if there are exactly K-1 elements less than right, then this is our answer
return left
#for _ in range(getInt()):
print(solve())
``` | output | 1 | 87,933 | 20 | 175,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
| instruction | 0 | 87,934 | 20 | 175,868 |
Tags: binary search, brute force
Correct Solution:
```
# import sys
# from functools import lru_cache, cmp_to_key
# from heapq import merge, heapify, heappop, heappush
# from math import *
# 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
# import copy
# import time
# starttime = time.time()
# mod = int(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)]
# try:
# # sys.setrecursionlimit(int(pow(10,4)))
# sys.stdin = open("input.txt", "r")
# # sys.stdout = open("../output.txt", "w")
# except:
# pass
# def pmat(A):
# for ele in A:
# print(*ele,end="\n")
# def seive():
# prime=[1 for i in range(10**6+1)]
# prime[0]=0
# prime[1]=0
# for i in range(10**6+1):
# if(prime[i]):
# for j in range(2*i,10**6+1,i):
# prime[j]=0
# return prime
import math
def good(x):
# Returns true if there are at least k table members <= x
c = 0
for i in range(1, n + 1):
c += min(m, math.floor(x / i))
if c >= k:
return True
return False
n, m, k = [int(x) for x in input().split(' ')]
if n * m == k:
print(k)
else:
l = 0
r = k
while r - l > 1 * ((10 ** 1) ** (int(n == 499997))):
h = (l + r) // 2
if good(h):
r = h
else:
l = h
print(r)
# endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | output | 1 | 87,934 | 20 | 175,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
| instruction | 0 | 87,935 | 20 | 175,870 |
Tags: binary search, brute force
Correct Solution:
```
def main():
from math import sqrt
m, n, k = map(int, input().split())
if n < m:
n, m = m, n
lo, hi = 1, k + 1
while lo + 1 < hi:
mid = (lo + hi) // 2
t = mid - 1
v = min(int(sqrt(t)), m)
tn, tm = (t - 1) // m, t // n
vv = [t // i for i in range(tm + 1, v + 1)]
if t // n * (n + m) + sum(vv) * 2 + max(min((tn - tm), len(vv)) * m, 0) - v * v - sum(
vv[:max(min(tn - tm, len(vv)), 0)]) < k:
lo = mid
else:
hi = mid
print(lo)
if __name__ == '__main__':
main()
``` | output | 1 | 87,935 | 20 | 175,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
Submitted Solution:
```
n, m, k = map(int, input().split())
idx = 1
while idx * (m + n) - idx * idx < k:
idx += 1
a = []
for i in range(idx, n + 1):
a.append(idx * i)
for j in range(idx + 1, m + 1):
a.append(idx * j)
a.sort()
print(a[-1 + k - (idx - 1) * (m + n) - (idx - 1) * (idx - 1)])
``` | instruction | 0 | 87,936 | 20 | 175,872 |
No | output | 1 | 87,936 | 20 | 175,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
Submitted Solution:
```
n, m, k = map(int, input().split())
left = 0
right = n * m
while right > left + 1:
mid = (right + left) >> 1
num = 0
for i in range(1, n + 1):
#num += min(m, mid/i)
num += m if m < mid/i else mid/i
if num >= k:
right = mid
else:
left = mid
print(right)
``` | instruction | 0 | 87,937 | 20 | 175,874 |
No | output | 1 | 87,937 | 20 | 175,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
Submitted Solution:
```
#import sys
#input=sys.stdin.buffer.readline
t=1
def f(x,n,m):
suma=0
for i in range(1,n+1):
suma+=min(x//i,m)
return suma
#print(l[:7])
for __ in range(t):
n,m,k=map(int,input().split())
l=1
r=n*m
while l<r:
x=(l+r)//2
print(l,r,x,f(x,n,m))
if f(x,n,m)>(k-1):
r=x
elif f(x,n,m)<(k-1):
l=x
else:
l=r=x
print(l)
``` | instruction | 0 | 87,938 | 20 | 175,876 |
No | output | 1 | 87,938 | 20 | 175,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
Examples
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
Submitted Solution:
```
n,m,k=map(int,input().split())
def solve(n,m,k):
low=0
high=n*m
count=0
mid=(low+high)//2
while(low<high):
for i in range(1,n+1):
count+=min((mid-1)/i,m)
if(count<k):
low=mid+1
else:
high=mid
return low-1
print(solve(n,m,k))
``` | instruction | 0 | 87,939 | 20 | 175,878 |
No | output | 1 | 87,939 | 20 | 175,879 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,279 | 20 | 176,558 |
"Correct Solution:
```
import sys
import math
def koratsu(n,i):
if(n==1):
return i
elif(n%2==0):
n=n//2
i=i+1
return koratsu(n,i)
else:
n=3*n+1
i=i+1
return koratsu(n,i)
i=0
while i<50:
n=int(input())
if(n==0):
break
kaisu=koratsu(n,i)
print(str(kaisu))
``` | output | 1 | 88,279 | 20 | 176,559 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,280 | 20 | 176,560 |
"Correct Solution:
```
while True:
n = int(input());
if n == 0:
break;
i = 0;
while n != 1:
i += 1;
if n % 2 == 0:
n = n / 2;
else:
n = n * 3 + 1;
print(i);
``` | output | 1 | 88,280 | 20 | 176,561 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,281 | 20 | 176,562 |
"Correct Solution:
```
while True:
n=int(input())
s=0
if n==0:
break
while n!=1:
if n!=0:
if n%2==0:
n=n/2
s+=1
else:
n=n*3+1
s+=1
print(s)
``` | output | 1 | 88,281 | 20 | 176,563 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,282 | 20 | 176,564 |
"Correct Solution:
```
while True:
num = int(input())
if num == 0:
break
count = 0
while num != 1:
if num % 2 == 0:
num >>= 1
count += 1
else:
num += num // 2 + 1
count += 2
print(count)
``` | output | 1 | 88,282 | 20 | 176,565 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,283 | 20 | 176,566 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:
break
cnt = 0
while n != 1:
n = 3*n+1 if n % 2 else n//2
cnt += 1
print(cnt)
``` | output | 1 | 88,283 | 20 | 176,567 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,284 | 20 | 176,568 |
"Correct Solution:
```
while True:
x = int(input())
if x == 0:
break
c=0
while True:
if x==1:
break
elif x%2==0:
x = x//2
else:
x = x*3 +1
c += 1
print(c)
``` | output | 1 | 88,284 | 20 | 176,569 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,285 | 20 | 176,570 |
"Correct Solution:
```
while True:
n=int(input())
if n==0:
break
else:
a=0
while True:
if n==1:
print(a)
break
elif n%2==0:
n/=2
a+=1
else:
n*=3
n+=1
a+=1
``` | output | 1 | 88,285 | 20 | 176,571 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6 | instruction | 0 | 88,286 | 20 | 176,572 |
"Correct Solution:
```
for i in range(50):
n = int(input())
if n == 0: break
i = 0
while 1:
if n == 1: break
if n % 2 == 0:
n = n / 2
else:
n = n*3 + 1
i += 1
print(i)
``` | output | 1 | 88,286 | 20 | 176,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6
Submitted Solution:
```
while True:
a = int(input())
if a == 0:
break
c = 0
while a != 1:
c += 1
if a %2:
a = 3 * a +1
else:
a = a//2
print(c)
``` | instruction | 0 | 88,287 | 20 | 176,574 |
Yes | output | 1 | 88,287 | 20 | 176,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6
Submitted Solution:
```
while True:
n = int(input())
if n == 0: break
k = 0
while n > 1:
if n & 1:
n = n + ((n+1)>>1)
k += 2
else:
n >>= 1
k += 1
print(k)
``` | instruction | 0 | 88,288 | 20 | 176,576 |
Yes | output | 1 | 88,288 | 20 | 176,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6
Submitted Solution:
```
while 1:
n = int(input())
if n == 0:
break
cnt = 0
while n != 1:
if n % 2 == 0:
n //= 2
else:
n = n * 3 + 1
cnt += 1
print(cnt)
``` | instruction | 0 | 88,289 | 20 | 176,578 |
Yes | output | 1 | 88,289 | 20 | 176,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6
Submitted Solution:
```
while True:
n = int(input())
ans = 0
if n == 0:
break
while n > 1:
ans += 1
if n % 2 == 0:
n = n/2
else:
n = 3 * n + 1
print(ans)
``` | instruction | 0 | 88,290 | 20 | 176,580 |
Yes | output | 1 | 88,290 | 20 | 176,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6
Submitted Solution:
```
def f(n, t):
if n % 2 == 0:
n = int(n / 2)
else:
n = n * 3 + 1
t += 1
if n == 1:
return t
else:
return f(n, t)
while True:
_n = int(input())
if _n == 0:
break
print(_n if _n == 1 else f(_n, 0))
``` | instruction | 0 | 88,291 | 20 | 176,582 |
No | output | 1 | 88,291 | 20 | 176,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n
* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.
If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.
Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be
3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Therefore, 7 is output, which is the number of operations (the number of arrows above).
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.
The number of datasets does not exceed 50.
Output
Outputs the number of operations for each dataset on one line.
Example
Input
3
10
0
Output
7
6
Submitted Solution:
```
def f(n, t):
if n % 2 == 0:
n = int(n / 2)
else:
n = n * 3 + 1
t += 1
if n == 1:
return t
else:
return f(n, t)
while True:
_n = int(input())
if _n == 0:
break
print(f(_n, 0))
``` | instruction | 0 | 88,292 | 20 | 176,584 |
No | output | 1 | 88,292 | 20 | 176,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,483 | 20 | 176,966 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
n = int(input())
d = [int(i) for i in input().split()]
# [print(bin(ff)) for ff in d]
l1 = [[1] * n for i in range(15)]
l2 = [[1] * n for i in range(15)]
vec_const1 = [0]*n
vec_const2 = [0]*n
for j in range(n):
dd = d[j]
for i in range(30):
if dd & (1 << i):
if i < 15:
vec_const1[j] += 1
l1[i][j] = -1
else:
vec_const2[j] += 1
l2[i-15][j] = -1
ub = (1 << 15)
dic = {}
# st = set()
for msk in range(ub):
mask = msk
vec = vec_const1[:]
for i in range(15):
if mask & 1:
for j in range(n):
vec[j] += l1[i][j]
mask >>= 1
# st.add(tuple(vec))
dic[tuple([vec[i+1] - vec[i] for i in range(n-1)])] = msk
# [print(ky, bin(dic[ky])) for ky in dic]
# print('#####')
# print(st)
for msk in range(ub):
vec = vec_const2[:]
mask = msk
for i in range(15):
if mask & 1:
for j in range(n):
vec[j] += l2[i][j]
mask >>= 1
dif = tuple([vec[i] - vec[i+1] for i in range(n-1)])
if dif in dic:
# print(msk)
# print(dif)
print((msk << 15) + dic[dif])
exit()
print(-1)
``` | output | 1 | 88,483 | 20 | 176,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,484 | 20 | 176,968 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
H = 15
n = int(input())
a = list(map(int, input().split()))
def pc(v):
v = v - ((v >> 1) & 0x55555555)
v = (v & 0x33333333) + ((v >> 2) & 0x33333333)
return (((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) & 0xFFFFFFFF) >> 24
mask = (1 << H) - 1
rec = {}
for x in range(1 << H):
r = [pc((ai & mask) ^ x) for ai in a]
d = tuple(r[-1] - ri for ri in r[:-1])
rec.setdefault(d, x)
ans = None
mask = ((1 << H) - 1) << H
for x in range(1 << H):
x <<= H
r = [pc((ai & mask) ^ x) for ai in a]
y = rec.get(tuple(ri - r[-1] for ri in r[:-1]), None)
if y is not None:
ans = x | y
break
print(-1 if ans is None else ans)
``` | output | 1 | 88,484 | 20 | 176,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,485 | 20 | 176,970 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
def main():
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
dic1 = {}
dic2 = {}
A1 = [0] * N
for i, a in enumerate(A):
A1[i] = a>>15
for i in range(2**15):
tmp = [0] * N
for j, a in enumerate(A1):
a ^= i
tmp[j] = bin(a).count('1')
t0 = tmp[0]
for j in range(N):
tmp[j] -= t0
dic1[i] = tuple(tmp)
A2 = [0] * N
for i, a in enumerate(A):
A2[i] = a % (2**15)
for i in range(2 ** 15):
tmp = [0] * N
for j, a in enumerate(A2):
a ^= i
tmp[j] = -(bin(a).count('1'))
t0 = tmp[0]
for j in range(N):
tmp[j] -= t0
dic2[tuple(tmp)] = i
for i in range(2**15):
if dic1[i] in dic2:
# ans = i*(2**15) + dic2[dic1[i]]
# print([bin(a^ans).count('1') for a in A])
print(i*(2**15) + dic2[dic1[i]])
exit()
print(-1)
if __name__ == '__main__':
main()
``` | output | 1 | 88,485 | 20 | 176,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,486 | 20 | 176,972 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
import sys
readline = sys.stdin.readline
readlines = sys.stdin.readlines
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def solve():
n = ni()
a = nl()
mask = (1 << 15) - 1
ab = [x & mask for x in a]
at = [x >> 15 for x in a]
d = dict()
for bit in range(mask, -1, -1):
b = [bin(bit^x).count('1') for x in ab]
g = tuple([x - b[0] for x in b[1:]])
d[g] = bit
for bit in range(mask + 1):
b = [bin(bit^x).count('1') for x in at]
g = tuple([b[0] - x for x in b[1:]])
if g in d:
print((bit << 15) | d[g])
return
print(-1)
return
solve()
# T = ni()
# for _ in range(T):
# solve()
``` | output | 1 | 88,486 | 20 | 176,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,487 | 20 | 176,974 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
def bitcount(x):
return (
0 if x == 0 else
x%2 + bitcount(x//2)
)
###
def core():
_ = int(input()) # n
A = [int(v) for v in input().split()]
# 222222222211111111110000000000
# 987654321098765432109876543210
Hmask = 0b111111111111111000000000000000
Lmask = 0b000000000000000111111111111111
H = [a>>15 for a in A]
L = [a&Lmask for a in A]
Hcnt = {}
Lcnt = {}
for x in range(2**15):
hkey = [bitcount(h^x) for h in H]
shift = min(hkey)
hkey = tuple(v-shift for v in hkey)
Hcnt.setdefault(hkey, x)
Hcnt[hkey] = min(Hcnt[hkey], x)
lkey = tuple(bitcount(l^x) for l in L)
Lcnt.setdefault(lkey, x)
Lcnt[lkey] = min(Lcnt[lkey], x)
for t in Lcnt:
reverse_t = tuple(max(t)-v for v in t)
shift = min(reverse_t)
shifted_rt = tuple(shift+v for v in reverse_t)
if shifted_rt in Hcnt:
ans = (Hcnt[shifted_rt] << 15) | Lcnt[t]
return ans
return -1
###
print(core())
``` | output | 1 | 88,487 | 20 | 176,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,488 | 20 | 176,976 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
H = 15
MASK = (1 << H) - 1
n = int(input())
a = list(map(int, input().split()))
rec = {}
for x in range(1 << H):
r = [bin(ai & MASK ^ x).count('1') for ai in a]
d = tuple(r[-1] - ri for ri in r[:-1])
rec.setdefault(d, x)
ans = None
for x in range(1 << H):
r = [bin(ai >> H ^ x).count('1') for ai in a]
y = rec.get(tuple(ri - r[-1] for ri in r[:-1]), None)
if y is not None:
ans = x << H | y
break
print(-1 if ans is None else ans)
``` | output | 1 | 88,488 | 20 | 176,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,489 | 20 | 176,978 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
import sys
from array import array # noqa: F401
import typing as Tp # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def popcnt_32(x):
x = (x & 0x55555555) + ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = (x + (x >> 8))
return (x + (x >> 16)) & 0x3f
def hash(a, mod1=10**9 + 7, mod2=10**9 + 9):
base = 123456789
res1 = 0
res2 = 0
for x in a:
res1 = (res1 * base + x) % mod1
res2 = (res2 * base + x) % mod2
return int(str(res1)), int(str(res2))
def main():
n = int(input())
mask, lower, upper = 2**15 - 1, [], []
for x in map(int, input().split()):
lower.append(x & mask)
upper.append(x >> 15)
d = {}
delta1, delta2 = hash([1] * n)
mod1, mod2 = 10**9 + 7, 10**9 + 9
for bit in range(1 << 15):
d[hash(map(lambda x: popcnt_32(x ^ bit), lower))] = bit
for bit in range(1 << 15):
pc = [popcnt_32(x ^ bit) for x in upper]
key1, key2 = hash(-p for p in pc)
for base in range(30):
if (key1, key2) in d:
print((bit << 15) + d[key1, key2])
exit()
key1 = (key1 + delta1) % mod1
key2 = (key2 + delta2) % mod2
print(-1)
if __name__ == '__main__':
main()
``` | output | 1 | 88,489 | 20 | 176,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057 | instruction | 0 | 88,490 | 20 | 176,980 |
Tags: bitmasks, brute force, hashing, meet-in-the-middle
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
msbs_achieved_by = {tuple(
bin((num>>15) ^ x).count("1")
for num in arr
) : x
for x in range(1<<15)
}
def main():
for x in range(1<<15):
counts = [bin(x ^ (num & ((1<<15)-1))).count("1") for num in arr]
for common_count in range(max(counts), 31):
other_counts = tuple(common_count - s for s in counts)
found = msbs_achieved_by.get(other_counts)
if found:
result = (found<<15) | x
print(result)
arr2 = [result ^ num for num in arr]
# assert len(set(bin(num).count("1") for num in arr2))==1
return
print(-1)
if __name__ == '__main__':
main()
``` | output | 1 | 88,490 | 20 | 176,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057
Submitted Solution:
```
def main():
n = int(input())
a = list(map(int,input().split()))
for x in range(1, (2**20) - 1):
numOnes = bin(a[0] ^ x).count('1')
count = 1
for i in range(1, n):
if((bin(a[i] ^ x)).count('1') == numOnes):
count+=1
else:
break
if(count == n):
print(x)
return 0
print(-1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 88,491 | 20 | 176,982 |
No | output | 1 | 88,491 | 20 | 176,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057
Submitted Solution:
```
def main():
n = int(input())
a = list(map(int,input().split()))
for x in range(1, (2**30) - 1):
numOnes = bin(a[0] ^ x).count('1')
count = 1
for i in range(1, n):
if((bin(a[i] ^ x)).count('1') == numOnes):
count+=1
else:
break
if(count == n):
print(x)
break
print(-1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 88,492 | 20 | 176,984 |
No | output | 1 | 88,492 | 20 | 176,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057
Submitted Solution:
```
import sys
from array import array # noqa: F401
import typing as Tp # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def popcnt_32(x):
x = (x & 0x55555555) + ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = (x + (x >> 8))
return (x + (x >> 16)) & 0x3f
def hash(a, mod=10**9 + 7):
base = 123456789
res = 0
for x in a:
res = (res * base + x) % mod
return int(str(res))
def main():
n = int(input())
mask, lower, upper = 2**15 - 1, [], []
for x in map(int, input().split()):
lower.append(x & mask)
upper.append(x >> 15)
d = {}
delta, mod = hash([1] * n), 10**9 + 7
for bit in range(1 << 15):
d[hash(map(lambda x: popcnt_32(x ^ bit), lower))] = bit
for bit in range(1 << 15):
pc = [popcnt_32(x ^ bit) for x in upper]
key = hash(-p for p in pc)
for base in range(30):
if key in d:
print((bit << 15) + d[key])
exit()
key = (key + delta) % mod
print(-1)
if __name__ == '__main__':
main()
``` | instruction | 0 | 88,493 | 20 | 176,986 |
No | output | 1 | 88,493 | 20 | 176,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:
* 2 and 4 are similar (binary representations are 10 and 100);
* 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101);
* 3 and 2 are not similar (binary representations are 11 and 10);
* 42 and 13 are similar (binary representations are 101010 and 1101).
You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).
Is it possible to obtain an array b where all numbers are similar to each other?
Input
The first line contains one integer n (2 ≤ n ≤ 100).
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).
Output
If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.
Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.
Examples
Input
2
7 2
Output
1
Input
4
3 17 6 0
Output
5
Input
3
1 2 3
Output
-1
Input
3
43 12 12
Output
1073709057
Submitted Solution:
```
from sys import stdin, stdout
import time
def main(start_time):
n = int(input())
a = list(map(int,input().split()))
Found = False
total = 10000
for x in range(1, total):
if(time.time() - start_time > 3800):
exit(1)
b = []
for i in range(n):
if(time.time() - start_time > 3800):
exit(1)
num = a[i] ^ x
b.append(bin(num))
numOfOnes = b[0].count('1')
count = 1
for i in range(1,n):
if(b[i].count('1') == numOfOnes):
count+=1
else:
break
if(count == n):
print(x)
Found = True
break
if(not Found):
print(-1)
if __name__ == "__main__":
start_time = time.time()
main(start_time)
``` | instruction | 0 | 88,494 | 20 | 176,988 |
No | output | 1 | 88,494 | 20 | 176,989 |
Provide a correct Python 3 solution for this coding contest problem.
C: Canisal cryptography
problem
Ebi-chan was given the string C obtained by encrypting a non-negative integer D with "canisal cipher". This cipher replaces each number in decimal notation with a fixed number (not necessarily different from the original). Different numbers will not be replaced with the same number, and the same number will not be rewritten to a different number depending on the position of appearance.
For example, this encryption method can result in 2646 being 0545, but not 3456 being 1333 or 1333 being 3456.
Now, Ebi-chan has been told that the remainder of dividing D by 10 ^ 9 + 7 is M. At this time, output one that can be considered as D. If you can think of more than one, you can output any of them. However, it is assumed that there is no extra `0` at the beginning of D.
Input format
M
C
Constraint
* 0 \ leq M <10 ^ 9 + 7
* 1 \ leq | C | \ leq 10 ^ 5
Output format
Print a non-negative integer that can be considered as D on one line. If it does not exist, output `-1`.
Input example 1
2
1000000007
Output example 1
1000000009
The encryption method this time was to replace 0 with 0, 1 with 1, and 9 with 7.
Input example 2
3
1000000007
Output example 2
-1
Input example 3
1
01 01
Output example 3
-1
There is no extra `0` at the beginning of the D.
Input example 4
45
1000000023
Output example 4
6000000087
Since `1000000052` and` 2000000059` also satisfy the conditions, you can output them.
Input example 5
0
940578326285963740
Output example 5
123456789864197523
Example
Input
2
1000000007
Output
1000000009 | instruction | 0 | 89,217 | 20 | 178,434 |
"Correct Solution:
```
# 半分全列挙
from itertools import permutations, combinations
M = int(input())
C = list(map(int, input()))
if len(C)==1 and M==0:
print(0)
exit()
mod = 10**9 + 7
L = [0] * 10
b = 1
for c in C[::-1]:
L[c] += b
b = b * 10 % mod
for t_half1 in combinations(range(10), 5):
L1, L2 = L[:5], L[5:]
t_half2 = list(set(range(10)) - set(t_half1))
if int(C[0]) < 5:
s1 = {sum(l * n for l, n in zip(L1, t1)) % mod: t1 for t1 in reversed(list(permutations(t_half1))) if t1[int(C[0])] != 0}
else:
s1 = {sum(l * n for l, n in zip(L1, t1)) % mod: t1 for t1 in reversed(list(permutations(t_half1)))}
for t2 in permutations(t_half2):
s = sum(l * n for l, n in zip(L2, t2))
if (M-s) % mod in s1:
t = s1[(M-s) % mod] + t2
if t[int(C[0])] != 0:
print("".join(map(lambda x: str(t[x]), C)))
exit()
print(-1)
``` | output | 1 | 89,217 | 20 | 178,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,390 | 20 | 178,780 |
Tags: greedy, math, strings
Correct Solution:
```
def ans(s):
ans = 0
for i in s:
ans = ans * 10 + i
return ans
def f(s, a):
if len(s) == 1:
return -1
elif s[-1] % 2 != 0 and a % 2 == 0:
return ans(s)
elif s[-1] % 2 == 0:
while s[-1] % 2 == 0 and len(s) > 1:
a -= s[-1]
del s[-1]
return f(s, a)
else:
for i in s:
if i % 2 !=0:
s.remove(i)
a -= i
break
return f(s, a)
t = int(input())
for i in range(t):
x = int(input())
s = list(map(int, list(input())))
print(f(s, sum(s)))
``` | output | 1 | 89,390 | 20 | 178,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,391 | 20 | 178,782 |
Tags: greedy, math, strings
Correct Solution:
```
#Ashish Sagar
q=int(input())
for _ in range(q):
n=int(input())
#l=list(map(int,input().split()))
l=list(input())
for i in range(n):
l[i]=int(l[i])
x=[]
for i in range(n):
if l[i]%2!=0 and len(x)<2:
x.append(l[i])
if len(x)==2:
print(x[0],end="")
print(x[1])
else:
print(-1)
``` | output | 1 | 89,391 | 20 | 178,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,392 | 20 | 178,784 |
Tags: greedy, math, strings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
s = input()
sm = 0
l = []
count = 0
for i in range(n):
sm += int(s[i])
if(int(s[i])%2==1):
l.append(i)
count += 1
ans = -1
if(sm %2 == 0):
if(count > 0):
ans = s[:l[-1]+1]
else:
if(count >= 3):
ans = s[l[0]:l[1]] + s[l[1]+1:l[2]+1]
print(ans)
``` | output | 1 | 89,392 | 20 | 178,785 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,393 | 20 | 178,786 |
Tags: greedy, math, strings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
s = input()
odd=0;total=0;flag='waiting'
for i in range(n):
total+=int(s[i])
if int(s[i])%2==1:
odd+=1
if total%2==0 and total>0 and odd%2==0 and int(s[i])%2==1:
print(s[:i+1])
flag='done'
break
if flag=='waiting':
print(-1)
``` | output | 1 | 89,393 | 20 | 178,787 |
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