message stringlengths 2 433k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array [a_1, a_2, ..., a_n] such that 1 ≤ a_i ≤ 10^9. Let S be the sum of all elements of the array a.
Let's call an array b of n integers beautiful if:
* 1 ≤ b_i ≤ 10^9 for each i from 1 to n;
* for every pair of adjacent integers from the array (b_i, b_{i + 1}), either b_i divides b_{i + 1}, or b_{i + 1} divides b_i (or both);
* 2 ∑ _{i = 1}^{n} |a_i - b_i| ≤ S.
Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 ≤ n ≤ 50).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
For each test case, print the beautiful array b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.
Example
Input
4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3
Output
3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3
Submitted Solution:
```
###pyrival template for fast IO
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
t=int(input())
while t>0:
t-=1
n=int(input())
arr=[int(x) for x in input().split()];m=min(arr)
ans=sum(arr)//2
ans=str(ans)
for i in range(n-1):
sys.stdout.write(ans+" ")
sys.stdout.write(ans+"\n")
``` | instruction | 0 | 27,661 | 12 | 55,322 |
No | output | 1 | 27,661 | 12 | 55,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array [a_1, a_2, ..., a_n] such that 1 ≤ a_i ≤ 10^9. Let S be the sum of all elements of the array a.
Let's call an array b of n integers beautiful if:
* 1 ≤ b_i ≤ 10^9 for each i from 1 to n;
* for every pair of adjacent integers from the array (b_i, b_{i + 1}), either b_i divides b_{i + 1}, or b_{i + 1} divides b_i (or both);
* 2 ∑ _{i = 1}^{n} |a_i - b_i| ≤ S.
Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 ≤ n ≤ 50).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
For each test case, print the beautiful array b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.
Example
Input
4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3
Output
3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
lst = list(map(int, input().split()))
even, odd = 0,0
for j in range(n):
if j%2==0:
even += lst[j]
else:
odd += lst[j]
if even>odd:
for j in range(0, n, j+2):
lst[j]=1
else:
for j in range(1, n, j+2):
lst[j]=1
print(*lst)
``` | instruction | 0 | 27,662 | 12 | 55,324 |
No | output | 1 | 27,662 | 12 | 55,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array [a_1, a_2, ..., a_n] such that 1 ≤ a_i ≤ 10^9. Let S be the sum of all elements of the array a.
Let's call an array b of n integers beautiful if:
* 1 ≤ b_i ≤ 10^9 for each i from 1 to n;
* for every pair of adjacent integers from the array (b_i, b_{i + 1}), either b_i divides b_{i + 1}, or b_{i + 1} divides b_i (or both);
* 2 ∑ _{i = 1}^{n} |a_i - b_i| ≤ S.
Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 ≤ n ≤ 50).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
For each test case, print the beautiful array b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.
Example
Input
4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3
Output
3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3
Submitted Solution:
```
def read_int():
return int(input())
def read_ints():
return map(int, input().split(' '))
t = read_int()
for case_num in range(t):
n = read_int()
a = list(read_ints())
a.sort()
median = a[n // 2]
print(' '.join([str(median)] * n))
``` | instruction | 0 | 27,663 | 12 | 55,326 |
No | output | 1 | 27,663 | 12 | 55,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array [a_1, a_2, ..., a_n] such that 1 ≤ a_i ≤ 10^9. Let S be the sum of all elements of the array a.
Let's call an array b of n integers beautiful if:
* 1 ≤ b_i ≤ 10^9 for each i from 1 to n;
* for every pair of adjacent integers from the array (b_i, b_{i + 1}), either b_i divides b_{i + 1}, or b_{i + 1} divides b_i (or both);
* 2 ∑ _{i = 1}^{n} |a_i - b_i| ≤ S.
Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 ≤ n ≤ 50).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
For each test case, print the beautiful array b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.
Example
Input
4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3
Output
3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = [0 for i in range(n)]
mn = min(a)
mx = max(a)
for i in range(n):
if abs(a[i] - mn) < abs(a[i] - mx):
b[i] = mn
else:
b[i] = mx
print(*b)
``` | instruction | 0 | 27,664 | 12 | 55,328 |
No | output | 1 | 27,664 | 12 | 55,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,876 | 12 | 55,752 |
Tags: implementation
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n=Int()
a=array()
steps=0
ans=set()
l=1
r=0
for i in range(1,n-1):
if(a[i-1]==a[i+1] and a[i-1]!=a[i]):
steps+=1
r+=1
else:
if(l<=r): ans.add((l,r))
l=i+1
r=i
steps=0
if(l<=r): ans.add((l,r))
steps=0
b=a
for l,r in ans:
tot=r-l+1
steps=max(tot,steps)
if(tot%2):
b[l:r+1]=[a[l-1]]*(tot)
else:
b[l:l+tot//2]=[a[l-1]]*(tot//2)
b[l+tot//2:r+1]=[a[r+1]]*(tot//2)
print(ceil(steps/2))
print(*b)
``` | output | 1 | 27,876 | 12 | 55,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,877 | 12 | 55,754 |
Tags: implementation
Correct Solution:
```
def main():
n = int(input())
seq = [int(c) for c in input().split()]
alt = [[0, 0]]
prev = seq[0]
for i, e in enumerate(seq):
if e != prev:
alt[-1][1] += 1
else:
alt.append([i, i])
prev = e
alt = [e for e in alt if e[1] - e[0] + 1 > 2]
steps = 0
for a, b in alt:
length = b - a + 1
steps = max(steps, (length + 1) // 2 - 1)
if length % 2 == 1:
for i in range(a, b + 1):
seq[i] = seq[a]
else:
for i in range(a, a + length // 2):
seq[i] = seq[a]
for i in range(a + length // 2, b + 1):
seq[i] = seq[b]
print(steps)
print(*seq)
if __name__ == "__main__":
main()
``` | output | 1 | 27,877 | 12 | 55,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,878 | 12 | 55,756 |
Tags: implementation
Correct Solution:
```
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
#from __future__ import print_function, division #while using python2
def modinv(n,p):
return pow(n,p-2,p)
import math
def main():
# sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
n = int(input())
a = [int(x) for x in input().split()]
# prev = [x for x in a]
# c = 0
# while True:
# b = [-1] * n
# b[0] = prev[0]
# b[n-1] = prev[n-1]
# for i in range(1, n-1):
# b[i] = sorted([prev[i-1], prev[i], prev[i+1]])[1]
# # print(*b)
# if b == prev:
# break
# prev = [x for x in b]
# c += 1
# print(c)
# print(*prev)
# print()
# print(*a)
temp = [1] * n
for i in range(1, n):
if a[i] != a[i-1]:
temp[i] = temp[i-1]+1
else:
temp[i] = 1
print((max(temp) - 1)//2)
# print(*temp)
# print()
# print(*a)
temp2 = [-1] * n
i = 0
while i < n:
# print("i = ", i)
if a[i] == 0:
j = i
while j+2 < n and a[j+1] == 1 and a[j+2] == 0:
j += 2
if j == n-1 or a[j+1] == 0:
while i <= j:
temp2[i] = 0
i += 1
else:
l = j - i + 1
m = i + l//2
while i <= m:
temp2[i] = 0
i += 1
while i <= j:
temp2[i] = 1
i += 1
else:
j = i
while j+2 < n and a[j+1] == 0 and a[j+2] == 1:
j += 2
if j == n-1 or a[j+1] == 1:
while i <= j:
temp2[i] = 1
i += 1
else:
l = j - i + 1
m = i + l//2
while i <= m:
temp2[i] = 1
i += 1
while i <= j:
temp2[i] = 0
i += 1
# i += 1
# print()
print(*temp2)
#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__ == '__main__':
main()
``` | output | 1 | 27,878 | 12 | 55,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,879 | 12 | 55,758 |
Tags: implementation
Correct Solution:
```
import math as mt
import sys,string,bisect
input=sys.stdin.readline
import random
from collections import deque,defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
d=defaultdict(list)
n=I()
a=L()
def solve(l,r):
if(l==r):
return 0
if(a[l]==a[r]):
for i in range(l+1,r):
a[i]=a[l]
return (r-l)//2
else:
m=(l+r+1)//2
for i in range(l+1,m):
a[i]=a[l]
for i in range(m,r):
a[i]=a[r]
return (r-l-1)//2
left=0
ans=0
for i in range(n):
if(i==n-1 or a[i]==a[i+1]):
ans=max(ans,solve(left,i))
left=i+1
print(ans)
print(*a)
``` | output | 1 | 27,879 | 12 | 55,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,880 | 12 | 55,760 |
Tags: implementation
Correct Solution:
```
def main():
input()
seq = [int(c) for c in input().split()]
alt = [[0, 0]]
prev = seq[0]
for i, e in enumerate(seq):
if e != prev:
alt[-1][1] += 1
else:
alt.append([i, i])
prev = e
alt = [e for e in alt if e[1] - e[0] + 1 > 2]
steps = 0
for a, b in alt:
length = b - a + 1
steps = max(steps, (length + 1) // 2 - 1)
for i in range(a, a + length // 2):
seq[i] = seq[a]
for i in range(a + length // 2, b + 1):
seq[i] = seq[b]
print(steps)
print(*seq)
if __name__ == "__main__":
main()
``` | output | 1 | 27,880 | 12 | 55,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,881 | 12 | 55,762 |
Tags: implementation
Correct Solution:
```
import os,io
from sys import stdout
import collections
# import random
import math
# from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import Counter
# from decimal import Decimal
# import heapq
# from functools import lru_cache
# import sys
# sys.setrecursionlimit(10**6)
# from functools import lru_cache
# @lru_cache(maxsize=None)
def primes(n):
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
def binomial_coefficient(n, k):
if 0 <= k <= n:
ntok = 1
ktok = 1
for t in range(1, min(k, n - k) + 1):
ntok *= n
ktok *= t
n -= 1
return ntok // ktok
else:
return 0
def powerOfK(k, max):
if k == 1:
return [1]
if k == -1:
return [-1, 1]
result = []
n = 1
while n <= max:
result.append(n)
n *= k
return result
def prefixSum(arr):
for i in range(1, len(arr)):
arr[i] = arr[i] + arr[i-1]
return arr
def divisors(n):
i = 1
result = []
while i*i <= n:
if n%i == 0:
if n/i == i:
result.append(i)
else:
result.append(i)
result.append(n/i)
i+=1
return result
def kadane(a,size):
max_so_far = 0
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
# @lru_cache(maxsize=None)
def digitsSum(n):
if n == 0:
return 0
r = 0
while n > 0:
r += n % 10
n //= 10
return r
def print_grid(grid):
for line in grid:
print("".join(line))
# INPUTS --------------------------
# s = input().decode('utf-8').strip()
# n = int(input())
# l = list(map(int, input().split()))
# t = int(input())
# for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
i = 1
s = 0
e = None
res = 0
while i <= len(l):
if i == len(l) or l[i] == l[i-1]:
if i-s > 2:
e = i
seq = l[s:e]
if len(seq) % 2 == 0:
res = max(res, (len(seq)//2)-1)
else:
res = max(res, len(seq)//2)
if seq[0] == 1 and seq[-1] == 1:
for k in range(s, e):
l[k] = 1
elif seq[0] == 0 and seq[-1] == 0:
for k in range(s, e):
l[k] = 0
elif seq[0] == 0 and seq[-1] == 1:
for k in range(s, e):
if k < s+(len(seq)//2):
l[k] = 0
else:
l[k] = 1
elif seq[0] == 1 and seq[-1] == 0:
for k in range(s, e):
if k < s+(len(seq)//2):
l[k] = 1
else:
l[k] = 0
s = i
i += 1
print(res)
print(" ".join(map(str, l)))
``` | output | 1 | 27,881 | 12 | 55,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,882 | 12 | 55,764 |
Tags: implementation
Correct Solution:
```
def mp():
return map(int,input().split())
def lt():
return list(map(int,input().split()))
def pt(x):
print(x)
n = int(input())
L = lt()
count = 0
i = 0
c = [None for i in range(n)]
while i < n-1:
j = i
while j < n-1 and L[j] != L[j+1]:
j += 1
span = j-i+1
for k in range(i,i+span//2):
c[k] = L[i]
for k in range(i+span//2,j+1):
c[k] = L[j]
count = max(count,(span-1)//2)
i = j+1
if i == n-1:
c[i] = L[i]
print(count)
print(' '.join([str(r) for r in c]))
# Made By Mostafa_Khaled
``` | output | 1 | 27,882 | 12 | 55,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable. | instruction | 0 | 27,883 | 12 | 55,766 |
Tags: implementation
Correct Solution:
```
input()
seq = list(map(int, input().split()))
def make_ans(seq):
ans = 0
def is_fixed(ind):
#ind != 0 and ind != len(seq) - 1
return seq[ind - 1] == seq[ind] or seq[ind] == seq[ind + 1]
def set_res(left, right):
left_begin = left
if left == right:
return 0
left_res = seq[left - 1]
right_res = seq[right]
right -= 1
while left <= right:
seq[left] = left_res
seq[right] = right_res
left += 1
right -= 1
return left - left_begin
left = 1
right = 1
for i in range(1, len(seq) - 1):
if not is_fixed(i):
right += 1
else:
if left == right:
left += 1
right += 1
else:
ans = max(ans, set_res(left, right))
left = right = i+1
ans = max(ans, set_res(left, right))
return ans, seq
def calc(seq):
#not used
n = 0
def is_fixed(ind):
#ind != 0 and ind != len(seq) - 1
return seq[ind - 1] == seq[ind] or seq[ind] == seq[ind + 1]
def up(seq):
seq_copy = seq[:]
for i in range(1, len(seq) - 1):
if not is_fixed(i):
seq_copy[i] = 1 - seq_copy[i]
return seq_copy
while(up(seq) != seq):
seq = up(seq)
n += 1
return n, seq
ans, seq = make_ans(seq)
print(ans)
print(" ".join(map(str, seq)))
# import random
# for i in range(1000):
# s = [random.randrange(0,2) for x in range(10)]
# x, seq1 = make_ans(s[:])
# y, seq2 = calc(s[:])
# if seq1 != seq2:
# print(s)
# print(seq1)
# print(seq2)
# if x != y:
# print(x,y)
# print(s)
# break
``` | output | 1 | 27,883 | 12 | 55,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=utf-8
#
# Copyright © 2016 missingdays <missingdays@missingdays>
#
# Distributed under terms of the MIT license.
"""
"""
n = int(input())
a = [i for i in input().split()]
def change(l, r):
global a
if l == r:
return 0
if a[l] == a[r]:
for i in range(l+1, r):
a[i] = a[l]
return (r-l)//2
else:
m = (l+r+1)//2
for i in range(l+1, m):
a[i] = a[l]
for i in range(m, r):
a[i] = a[r]
return (r-l-1)//2
answ = 0
current = 0
for i in range(n):
if i == n-1 or a[i] == a[i+1]:
answ = max(answ, change(current, i))
current = i+1
print(answ)
print(" ".join(a))
``` | instruction | 0 | 27,884 | 12 | 55,768 |
Yes | output | 1 | 27,884 | 12 | 55,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
def solve():
N = int(input())
A = list(map(int, input().split()))
A.append(A[-1])
ma = 0
s = p = 0
for i in range(1, len(A)):
if A[i] != A[p]:
p = i
continue
else:
e = p
ma = max(ma, (p - s) // 2)
if A[s] == A[e]:
for j in range(s, e):
A[j] = A[s]
else:
l = (p - s + 1) // 2
for j in range(1, l):
A[s + j] = A[s]
A[e - j] = A[e]
s = i
A.pop()
print(ma)
print(' '.join(map(str, A)))
if __name__ == '__main__':
solve()
``` | instruction | 0 | 27,885 | 12 | 55,770 |
Yes | output | 1 | 27,885 | 12 | 55,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
import sys
lines = sys.stdin.readlines()
n = int(lines[0].strip())
arr = list(map(int, lines[1].strip().split(" ")))
arr.append(arr[-1])
res = []
maxChange = 0
left = 0
for i in range(1, n+1):
if arr[i] == arr[i-1]:
L = i - left
res += [arr[left]]*(L//2) + [arr[i-1]]*(L-L//2)
maxChange = max(maxChange, (L-1)//2)
left = i
print(maxChange)
print(" ".join(map(str, res)))
``` | instruction | 0 | 27,886 | 12 | 55,772 |
Yes | output | 1 | 27,886 | 12 | 55,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
r, b, c = 0, [a[0]], 0
for x, y, z in zip(a, a[1:], a[2:]):
if x != y != z:
c += 1
r = max(r, (c+1) // 2)
continue
if c % 2 == 1:
b.extend([y] * (c+1))
else:
b.extend([1-y] * (c//2) + [y] * (c//2+1))
c = 0
y = a[-1]
if c % 2 == 1: b.extend([y] * (c+1))
else: b.extend([1-y] * (c//2) + [y] * (c//2+1))
print(r)
print(*b)
``` | instruction | 0 | 27,887 | 12 | 55,774 |
Yes | output | 1 | 27,887 | 12 | 55,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
input()
seq = list(map(int, input().split()))
def make_ans(seq):
ans = 0
def is_fixed(ind):
#ind != 0 and ind != len(seq) - 1
return seq[ind - 1] == seq[ind] or seq[ind] == seq[ind + 1]
def set_res(left, right):
left_begin = left
if left == right:
return 0
left_res = seq[left - 1]
right_res = seq[right]
while left <= right:
seq[left] = left_res
seq[right] = right_res
left += 1
right -= 1
return left - left_begin
left = 1
right = 1
for i in range(1, len(seq) - 1):
if not is_fixed(i):
right += 1
else:
if left == right:
left += 1
right += 1
else:
ans = max(ans, set_res(left, right))
left = right = i+1
ans = max(ans, set_res(left, right))
return ans, seq
def calc(seq):
#not used
def is_fixed(ind):
#ind != 0 and ind != len(seq) - 1
return seq[ind - 1] == seq[ind] or seq[ind] == seq[ind + 1]
def up(seq):
seq_copy = seq[:]
for i in range(1, len(seq) - 1):
if not is_fixed(i):
seq_copy[i] = 1 - seq_copy[i]
return seq_copy
seq = up(seq)
while(up(seq) != seq):
seq = up(seq)
return seq
ans, seq = make_ans(seq)
print(ans)
print(" ".join(map(str, seq)))
# import random
# for i in range(1000):
# s = [random.randrange(0,2) for x in range(10)]
# x, seq1 = make_ans(s[:])
# seq2 = calc(s[:])
# if seq1 != seq2:
# print(s)
# print(seq1)
# print(seq2)
``` | instruction | 0 | 27,888 | 12 | 55,776 |
No | output | 1 | 27,888 | 12 | 55,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
def solve():
N = int(input())
A = list(map(int, input().split()))
A.append(A[-1])
ma = 0
s = p = 0
for i in range(1, len(A)):
if A[i] != A[p]:
p = i
continue
else:
e = p
if A[s] == A[e]:
ma = max(ma, 1)
for j in range(s, e):
A[j] = A[s]
else:
l = (p - s + 1) // 2
for j in range(1, l):
A[s + j] = A[s]
A[e - j] = A[e]
ma = max(ma, (p - s) // 2)
s = i
A.pop()
print(ma)
print(' '.join(map(str, A)))
if __name__ == '__main__':
solve()
``` | instruction | 0 | 27,889 | 12 | 55,778 |
No | output | 1 | 27,889 | 12 | 55,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
#from __future__ import print_function, division #while using python2
def modinv(n,p):
return pow(n,p-2,p)
import math
def main():
# sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
n = int(input())
a = [int(x) for x in input().split()]
# prev = [x for x in a]
# c = 0
# while True:
# b = [-1] * n
# b[0] = prev[0]
# b[n-1] = prev[n-1]
# for i in range(1, n-1):
# b[i] = sorted([prev[i-1], prev[i], prev[i+1]])[1]
# # print(*b)
# if b == prev:
# break
# prev = [x for x in b]
# c += 1
# print(c)
# print(*prev)
# print()
temp = [1] * n
for i in range(1, n):
if a[i] != a[i-1]:
temp[i] = temp[i-1]+1
else:
temp[i] = 1
print((max(temp) - 1)//2)
# print(*temp)
# print()
# print(*a)
temp2 = [-1] * n
i = 0
while i < n:
# print("i = ", i)
if a[i] == 0:
j = i
while j+2 < n and a[j+1] == 1 and a[j+2] == 0:
j += 2
while i <= j:
temp2[i] = 0
i += 1
else:
j = i
while j+2 < n and a[j+1] == 0 and a[j+2] == 1:
j += 2
while i <= j:
temp2[i] = 1
i += 1
# i += 1
# print()
print(*temp2)
#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__ == '__main__':
main()
``` | instruction | 0 | 27,890 | 12 | 55,780 |
No | output | 1 | 27,890 | 12 | 55,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.
Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, ..., an will result a new sequence b1, b2, ..., bn obtained by the following algorithm:
* b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
* For i = 2, ..., n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.
In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.
Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.
Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.
Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.
The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 1), giving the initial sequence itself.
Output
If the sequence will never become stable, print a single number - 1.
Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.
Examples
Input
4
0 0 1 1
Output
0
0 0 1 1
Input
5
0 1 0 1 0
Output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: <image>, and the sequence 00000 is obviously stable.
Submitted Solution:
```
def solve():
N = int(input())
A = list(map(int, input().split()))
A.append(A[-1])
ma = 0
s = p = 0
for i in range(1, N + 1):
if A[i] != A[p]:
p = i
continue
else:
e = p
l = (p - s + 1) // 2
for j in range(1, l):
A[s + j] = A[s]
A[e - j] = A[e]
ma = max(ma, (p - s) // 2)
s = i
A.pop()
print(ma)
print(' '.join(map(str, A)))
if __name__ == '__main__':
solve()
``` | instruction | 0 | 27,891 | 12 | 55,782 |
No | output | 1 | 27,891 | 12 | 55,783 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,064 | 12 | 56,128 |
"Correct Solution:
```
mod = 10**9 + 7
N, M = map(int, input().split())
S = [None] + list(map(int, input().split()))
T = [None] + list(map(int, input().split()))
dp = [[0 for _ in range(M+1)] for _ in range(N+1)]
for i in range(1, N+1):
for j in range(1, M+1):
if S[i] == T[j]:
dp[i][j] = (dp[i][j-1] + dp[i-1][j] + 1)%mod
else:
dp[i][j] = (dp[i][j-1] + dp[i-1][j] - dp[i-1][j-1])%mod
print((dp[N][M]+1)%mod)
``` | output | 1 | 28,064 | 12 | 56,129 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,065 | 12 | 56,130 |
"Correct Solution:
```
def solve():
N, M = map(int, input().split())
S = list(map(int, input().split()))
T = list(map(int, input().split()))
MOD = 10 ** 9 + 7
dp = [[1] * (M + 1) for _ in range(N + 1)]
for i in range(N):
for j in range(M):
if S[i] == T[j]:
dp[i + 1][j + 1] = (dp[i + 1][j] + dp[i][j + 1]) % MOD
else:
dp[i + 1][j + 1] = (dp[i + 1][j] + dp[i][j + 1] - dp[i][j]) % MOD
print(dp[N][M])
if __name__ == '__main__':
solve()
``` | output | 1 | 28,065 | 12 | 56,131 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,066 | 12 | 56,132 |
"Correct Solution:
```
P = 10**9+7
N, M = map(int, input().split())
S = [int(a) for a in input().split()]
T = [int(a) for a in input().split()]
X = [[1] * (M+1)] + [[1]+[0]*M for _ in range(N)]
for i in range(N):
for j in range(M):
X[i+1][j+1] = (X[i+1][j] + X[i][j+1] - (X[i][j] * (S[i] != T[j])))%P
print(X[N][M])
``` | output | 1 | 28,066 | 12 | 56,133 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,067 | 12 | 56,134 |
"Correct Solution:
```
n,m = map(int,input().split())
s = list(map(int, input().split()))
t = list(map(int, input().split()))
mod = 10**9 + 7
dp = [[0] * (m+1) for _ in range(n+1)]
for i in range(n):
for j in range(m):
if(s[i]==t[j]):
dp[i+1][j+1] = (dp[i][j+1] + dp[i+1][j] + 1) % mod
else:
dp[i+1][j+1] = (dp[i+1][j] + dp[i][j+1] - dp[i][j]) % mod
ans = (dp[-1][-1]+1) % mod
print(ans)
``` | output | 1 | 28,067 | 12 | 56,135 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,068 | 12 | 56,136 |
"Correct Solution:
```
N,M = map(int,input().split())
S = list(map(int,input().split()))
T = list(map(int,input().split()))
dp = [[1]+[0]*(len(T))for i in range(len(S)+1)]
dp[0] = [1]*(len(T)+1)
dp_sum = [[1]+[0]*(len(T))for i in range(len(S)+1)]
dp_sum[0] = [1]*(len(T)+1)
for i in range(len(S)):
for j in range(len(T)):
if S[i] == T[j]:
dp[i+1][j+1] = dp_sum[i][j]
dp_sum[i+1][j+1] = (dp[i+1][j+1]+dp_sum[i+1][j]+\
dp_sum[i][j+1]-dp_sum[i][j])%(10**9+7)
print(dp_sum[-1][-1])
``` | output | 1 | 28,068 | 12 | 56,137 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,069 | 12 | 56,138 |
"Correct Solution:
```
mod=10**9+7
n,m=map(int,input().split())
s=list(map(int,input().split()))
t=list(map(int,input().split()))
dp=[(m+1)*[1]for _ in range(n+1)]
for i in range(n):
for j in range(m):
if s[i]==t[j]:dp[i+1][j+1]=dp[i][j+1]+dp[i+1][j]
else:dp[i+1][j+1]=dp[i][j+1]+dp[i+1][j]-dp[i][j]
dp[i+1][j+1]%=mod
print(dp[-1][-1])
``` | output | 1 | 28,069 | 12 | 56,139 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,070 | 12 | 56,140 |
"Correct Solution:
```
MOD = 10 ** 9 + 7
n, m = map(int, input().split())
s = list(map(int, input().split()))
t = list(map(int, input().split()))
dp = [0] * n
for tj in t:
acc = 0
for i, si in enumerate(s):
acc = (acc + dp[i]) % 1000000007
if si == tj:
dp[i] = (acc + 1) % 1000000007
ans = 1
for x in dp:
ans = (ans + x) % MOD
print(ans)
``` | output | 1 | 28,070 | 12 | 56,141 |
Provide a correct Python 3 solution for this coding contest problem.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861 | instruction | 0 | 28,071 | 12 | 56,142 |
"Correct Solution:
```
N, M = map(int, input().split())
S_array = list(map(int, input().split()))
T_array = list(map(int, input().split()))
mod = 10**9 + 7
add_num_array = [0] * N
ans_array = [0] * (N + 1)
ans_old = [1] * (N + 1)
for t in T_array:
ans_array = [1] * (N+1)
for i, s in enumerate(S_array):
if s == t:
add_num_array[i] = (add_num_array[i] + ans_old[i]) % mod
ans_array[i + 1] = (ans_array[i] + add_num_array[i]) % mod
ans_old = ans_array
print(ans_array[-1])
``` | output | 1 | 28,071 | 12 | 56,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861
Submitted Solution:
```
# ほぼ解説AC
# n,m<2000
# len(s),len(t)<100000
import sys
input = sys.stdin.buffer.readline
n, m = map(int, input().split())
mod = 10 ** 9 + 7
S = list(map(int, input().split()))
T = list(map(int, input().split()))
dp = [[0] * (m + 2) for _ in range(n + 2)]
sum_dp = [[0] * (m + 2) for _ in range(n + 2)]
# 空集合も要素とみなすのでdp[0][0]は1
dp[0][0] = 0
# i,jを見る時にi,jまでの累積和がほすぃ。
# i,jを見てi+1,j+1を決める。
for i in range(n + 1):
for j in range(m + 1):
# 二次元累積和の計算
sum_dp[i][j] += dp[i][j]
if i - 1 >= 0:
sum_dp[i][j] += sum_dp[i - 1][j]
if j - 1 >= 0:
sum_dp[i][j] += sum_dp[i][j - 1]
if i - 1 >= 0 and j - 1 >= 0:
sum_dp[i][j] -= sum_dp[i - 1][j - 1]
sum_dp[i][j] %= mod
if i < n and j < m:
# 先にsum_dp[i][j]は計算されていないとだめ。↑
# i-1字目、j-1字目まででできるものに、新たに一致したものを加えたもの(sum_dp)と空集合(+1)
if S[i] == T[j]:
dp[i + 1][j + 1] = sum_dp[i][j] + 1
dp[i + 1][j + 1] %= mod
print((sum_dp[n][m] + 1) % mod)
``` | instruction | 0 | 28,077 | 12 | 56,154 |
No | output | 1 | 28,077 | 12 | 56,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integer sequences S and T of length N and M, respectively, both consisting of integers between 1 and 10^5 (inclusive).
In how many pairs of a subsequence of S and a subsequence of T do the two subsequences are the same in content?
Here the subsequence of A is a sequence obtained by removing zero or more elements from A and concatenating the remaining elements without changing the order.
For both S and T, we distinguish two subsequences if the sets of the indices of the removed elements are different, even if the subsequences are the same in content.
Since the answer can be tremendous, print the number modulo 10^9+7.
Constraints
* 1 \leq N, M \leq 2 \times 10^3
* The length of S is N.
* The length of T is M.
* 1 \leq S_i, T_i \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
S_1 S_2 ... S_{N-1} S_{N}
T_1 T_2 ... T_{M-1} T_{M}
Output
Print the number of pairs of a subsequence of S and a subsequence of T such that the subsequences are the same in content, modulo 10^9+7.
Examples
Input
2 2
1 3
3 1
Output
3
Input
2 2
1 1
1 1
Output
6
Input
4 4
3 4 5 6
3 4 5 6
Output
16
Input
10 9
9 6 5 7 5 9 8 5 6 7
8 6 8 5 5 7 9 9 7
Output
191
Input
20 20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Output
846527861
Submitted Solution:
```
# 解説AC
from collections import defaultdict
N,M = map(int, input().split())
S = [int(i) for i in input().split()]
T = [int(i) for i in input().split()]
MOD = 10 ** 9 + 7
# (i, j): S[:i]とT[:j]での共通部分列の個数
dp = defaultdict(int)
# 初期条件
for i in range(N + 1):
dp[(i, 0)] = 1
for j in range(M + 1):
dp[(0, j)] = 1
# 貰うDP
for i, s in enumerate(S):
for j, t in enumerate(T):
dp[(i + 1, j + 1)] = dp[(i, j + 1)] + dp[(i + 1, j)]
if s == t:
dp[(i + 1, j + 1)] += dp[(i, j)]
# ダブルカウントを除く
dp[(i + 1, j + 1)] -= dp[(i, j)]
dp[(i + 1, j + 1)] %= MOD
print(dp[(N, M)])
``` | instruction | 0 | 28,078 | 12 | 56,156 |
No | output | 1 | 28,078 | 12 | 56,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,278 | 12 | 56,556 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
mod = 10**9+7
n = int(input())
a = list(map(int,input().split()))
fact = [[1] for i in range(1000001)]
b = max(a)
for i in range(2,b+1):
for j in range(i,b+1,i):
fact[j].append(i)
dp = [0]*(n+1)
dp[0] = 1
# dp[1] = n
for i in range(n):
for j in range(len(fact[a[i]])-1,-1,-1):
if fact[a[i]][j]<=i+1:
dp[fact[a[i]][j]] = (dp[fact[a[i]][j]]%mod + dp[fact[a[i]][j]-1]%mod)%mod
# print (dp)
# print (count)
# print (dp)
print (sum(dp[1:])%mod)
``` | output | 1 | 28,278 | 12 | 56,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,279 | 12 | 56,558 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
seen=[0]*(n+3)
long=1
for guy in a:
if guy**0.5>long-50:
for i in range(long+1,1,-1):
if guy%i==0:
seen[i]+=seen[i-1]
if i==long+1:
long+=1
else:
facs=[]
bigfacs=[]
for i in range(1,int(guy**0.5)+1):
if guy%i==0:
facs.append(i)
bigfacs.append(guy//i)
if facs[-1]==bigfacs[-1]:
bigfacs.pop()
facs+=bigfacs[::-1]
for guy1 in facs[::-1]:
if guy1<=long+1:
seen[guy1]+=seen[guy1-1]
if guy1==long+1:
long+=1
seen[1]+=1
print(sum(seen)%1000000007)
``` | output | 1 | 28,279 | 12 | 56,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,280 | 12 | 56,560 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
#import numpy as np
import bisect
def main(n,s,mod):
ls = len(s)
L = [[] for i in range(n+1)]
for i,x in enumerate(s):
L[x].append(i)
for i in range(1,ls+1):
for j in range(2*i,n+1,i):
L[i] += L[j]
L[i].sort()
L = L[:ls+1]
counts = [list(range(ls)),[1 for i in range(ls)]]
S = ls
for d in range(2,ls+1):
if not L[d]: break
lcounts = len(counts[0])
sumcounts = [0]
for i in range(1,lcounts+1):
sumcounts.append((sumcounts[i-1] + counts[1][i-1]) % mod)
newcounts = [[],[]]
for x in L[d]:
y = sumcounts[bisect.bisect_left(counts[0],x)]
S = (S + y) % mod
newcounts[0].append(x), newcounts[1].append(y)
counts = newcounts
return S
useless = str(input()).split(' ')
s = [int(i) for i in input().split(' ')]
n = 10**6
mod = 10**9 + 7
#k = 10**5
#s = [0]*(2*k)
#s[0:2*k:2] = list(range(1,k+1))
#s[1:2*k:2] = list(range(1,k+1))
print(main(n,s,mod))
``` | output | 1 | 28,280 | 12 | 56,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,281 | 12 | 56,562 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
import math
def nextDiv(a):
tmp = set()
sq = int(math.sqrt(a))
for i in range(1, sq+1):
if (a%i == 0):
tmp.add(i)
tmp.add(a//i)
return reversed(sorted(tmp))
MOD = int(1e9+7)
def solve(n, lis):
dp = [0] * (max(lis)+1)
dp[0] = 1
for i in lis:
for j in nextDiv(i):
dp[j] += dp[j-1]
dp[j] %= MOD
return (sum(dp)-1) % MOD
###########################
###########################
from sys import stdin
def intRead():
while True:
ln = stdin.readline().strip()
if not ln:
return
for i in map(int, ln.split()):
yield i
if __name__ == "__main__":
ipt = intRead()
n = next(ipt)
lis = [next(ipt) for _ in range(n)]
print(solve(n, lis))
``` | output | 1 | 28,281 | 12 | 56,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,282 | 12 | 56,564 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
import math
n = int(input())
a = list(map(int,input().split()))
dp = [0]*(n+1)
dp[0]=1
mod = int(1e9 +7)
for el in a:
sqrt = int(math.sqrt(el))
for d in range(min(n,sqrt),0,-1):
if el%d==0:
if el/d > sqrt and el/d<=n:
dp[el//d]= (dp[el//d] + dp[el//d-1])%mod
dp[d]= (dp[d]+dp[d-1])%mod
res=0
for i in range(1,n+1):
res = (res+dp[i])%mod
print(res)
``` | output | 1 | 28,282 | 12 | 56,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,283 | 12 | 56,566 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
import math
def factors(n):
ans = list()
sqrt_n = math.floor(math.sqrt(n)) + 1
for i in range(1, sqrt_n):
if n % i == 0:
ans.append(int(i))
ans.append(int(n/i))
return sorted(list(set(ans)), reverse=True)
if __name__ == '__main__':
n = int(input())
ai = [int(e) for e in input().split(' ')]
hash_ = [0] * 1000020
hash_[0] = 1
for a in ai:
f_ai = factors(a)
for f in f_ai:
hash_[f] = (hash_[f] + hash_[f-1]) % 1000000007
ans = sum(hash_[1:n+1]) % 1000000007
print(ans)
``` | output | 1 | 28,283 | 12 | 56,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,284 | 12 | 56,568 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
import math
n=int(input())
arr=list(map(int,input().split()))
# dp=[[0]*(n+1) for _ in range(n+1)]
dp=[0]*(n+1)
dp[1]=1
dp[0]=1
mod=10**9+7
for i in range(2,n+1):
update=[]
for d in range(1,int(math.sqrt(arr[i-1]))+1):
if arr[i-1]%d==0:
update.append(d)
if d*d!=arr[i-1]:
update.append(arr[i-1]//d)
update.sort()
update.reverse()
for j in update:
if j<=n:
dp[j]+=dp[j-1]
dp[j]%=mod
total=-1
for val in dp:
total=(total+val)%mod
print(total)
``` | output | 1 | 28,284 | 12 | 56,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times. | instruction | 0 | 28,285 | 12 | 56,570 |
Tags: data structures, dp, implementation, math, number theory
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
N = 1000005
MOD = 10 ** 9 + 7
dp =[0] * N
dp[0] = 0
for i in range(n):
cur = []
A = int(a[i] ** 0.5) + 1
for j in range(1, A):
if a[i] % j == 0:
cur.append(j)
if a[i] // j != j:
cur.append(a[i] // j)
cur = sorted(cur, reverse=True)
for j in cur:
if j == 1:
dp[j] += 1
else:
dp[j] += dp[j - 1]
ans = 0
for i in dp:
ans += i
ans %= MOD
print(ans)
``` | output | 1 | 28,285 | 12 | 56,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
R = lambda: map(int, input().split())
mod = 10**9 + 7
n = int(input())
arr = list(R())
bs = [list() for i in range(max(arr) + 1)]
for b in range(1, len(bs)):
tmp = b
while tmp < len(bs):
bs[tmp].append(b)
tmp += b
dp = [1] + [0] * n
for x in arr:
for b in bs[x][::-1]:
if b < len(dp):
dp[b] = (dp[b] + dp[b - 1]) % mod
print(sum(dp[1:]) % mod)
``` | instruction | 0 | 28,286 | 12 | 56,572 |
Yes | output | 1 | 28,286 | 12 | 56,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
"""https://codeforces.com/contest/1061/problem/C"""
def divisors(n):
div = []
for i in range(1, int(n**0.5)+1):
if n%i == 0:
if n//i == i:
div.append(i)
else:
div.append(i)
div.append(n//i)
return div
from copy import copy
n = int(input())
arr = list(map(int,input().split()))
mod = int(1e9+7)
# with 1 element
dp = [ 0 for i in range(n+1)]
dp[0] = 1
for i in range(1,n+1):
for j in sorted(divisors(arr[i-1]),reverse=True):
if j < n+1:
dp[j] = (dp[j] + dp[j-1])%mod
print(sum(dp[1:])%mod)
``` | instruction | 0 | 28,287 | 12 | 56,574 |
Yes | output | 1 | 28,287 | 12 | 56,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
import sys
import math
from collections import defaultdict
mod=10**9+7
def get_factors(num,factors):
if factors[num]!=[]:
return
for i in range(1,int(math.sqrt(num)+1)):
if num%i==0:
factors[num].append(i)
if num//i!=i:
factors[num].append(num//i)
n=int(sys.stdin.readline())
arr=list(map(int,sys.stdin.readline().split()))
factors=defaultdict(list)
dp=defaultdict(int)
dp[0]=1
for i in range(n):
get_factors(arr[i],factors)
m=len(factors[arr[i]])
#print(dp,'dp')
for j in range(m-1,-1,-1):
if factors[arr[i]][j]<=i+1:
dp[factors[arr[i]][j]]+=dp[factors[arr[i]][j]-1]
dp[factors[arr[i]][j]]%=mod
#print(dp,'dp')
#print(factors,'factors')
ans=0
for i in dp:
ans+=dp[i]
ans%=mod
ans-=1
ans%=mod
print(ans)
``` | instruction | 0 | 28,288 | 12 | 56,576 |
Yes | output | 1 | 28,288 | 12 | 56,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
import math
def nextDiv(a):
tmp = []
sq = int(math.sqrt(a))
for i in range(1, sq+1):
if (a%i == 0):
j = a//i
yield j
if (i != j):
tmp.append(i)
while tmp:
yield tmp.pop()
MOD = int(1e9+7)
def solve(n, lis):
dp = [0] * (max(lis)+1)
dp[0] = 1
for i in lis:
for j in nextDiv(i):
dp[j] += dp[j-1]
dp[j] %= MOD
return (sum(dp)-1) % MOD
###########################
###########################
from sys import stdin
def intRead():
while True:
ln = stdin.readline().strip()
if not ln:
return
for i in map(int, ln.split()):
yield i
if __name__ == "__main__":
ipt = intRead()
n = next(ipt)
lis = [next(ipt) for _ in range(n)]
print(solve(n, lis))
``` | instruction | 0 | 28,289 | 12 | 56,578 |
Yes | output | 1 | 28,289 | 12 | 56,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
mod = 10**9+7
n = int(input())
a = list(map(int,input().split()))
fact = [[] for i in range(1000001)]
for i in range(2,1000001):
for j in range(i,1000001,i):
fact[j].append(i)
dp = [0]*(n+1)
for i in range(1,n):
for j in range(len(fact[a[i]])-1,-1,-1):
if fact[a[i]][j]<=i+1:
dp[fact[a[i]][j]] = (dp[fact[a[i]][j]]%mod + dp[fact[a[i]][j]-1]%mod)%mod
# print (dp)
# print (count)
print (sum(dp))
``` | instruction | 0 | 28,290 | 12 | 56,580 |
No | output | 1 | 28,290 | 12 | 56,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
def good_sub_num(length, arr):
count = length
sub_arr_prev = list(range(1, length))
sub_arr_curr = [0] * length
k = 2
while True:
curr_count = 0
for j in range(k-1, length):
if arr[j] % k == 0:
sub_arr_curr[j] = sub_arr_prev[j-1]
else:
sub_arr_curr[j] = 0
curr_count += sub_arr_curr[j]
if curr_count == 0:
return count
else:
count += curr_count
k += 1
sub_arr_prev = sub_arr_curr
n = int(input(""))
second_line = input("")
a = second_line.split()
for i in range(0, n):
a[i] = int(a[i])
print(good_sub_num(n, a))
``` | instruction | 0 | 28,291 | 12 | 56,582 |
No | output | 1 | 28,291 | 12 | 56,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
import math
from collections import defaultdict
n = int(input())
arr = list(map(int,input().strip().split()))
pre = defaultdict(int)
def get_divisors(k, upper):
div = []
for i in range(2, min(upper, int(math.floor(math.sqrt(k)))) + 1):
if k % i == 0:
div.append(i)
if k // i <= upper:
div.append(k // i)
if k <= upper:
div.append(k)
return div
for i in range(n):
div = get_divisors(arr[i], i+1)
for d in div:
if pre[d-1] == 0:
continue
pre[d] += pre[d-1]
pre[1] += 1
print(sum(pre.values()))
``` | instruction | 0 | 28,292 | 12 | 56,584 |
No | output | 1 | 28,292 | 12 | 56,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, …, a_n.
The array b is called to be a subsequence of a if it is possible to remove some elements from a to get b.
Array b_1, b_2, …, b_k is called to be good if it is not empty and for every i (1 ≤ i ≤ k) b_i is divisible by i.
Find the number of good subsequences in a modulo 10^9 + 7.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do not matter in the comparison of subsequences. In particular, the array a has exactly 2^n - 1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of the array a.
The next line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
Output
Print exactly one integer — the number of good subsequences taken modulo 10^9 + 7.
Examples
Input
2
1 2
Output
3
Input
5
2 2 1 22 14
Output
13
Note
In the first example, all three non-empty possible subsequences are good: \{1\}, \{1, 2\}, \{2\}
In the second example, the possible good subsequences are: \{2\}, \{2, 2\}, \{2, 22\}, \{2, 14\}, \{2\}, \{2, 22\}, \{2, 14\}, \{1\}, \{1, 22\}, \{1, 14\}, \{22\}, \{22, 14\}, \{14\}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
dp=[0]*(1000001)
dp[0]=1
for i in range(n):
x=int(a[i]**.5)
for j in range(min(i+1,x),0,-1):
if a[i]%j==0:
dp[j]+=dp[j-1]
if a[i]//j>x and a[i]//j<n:
dp[a[i]//j]+=dp[a[i]//j-1]
dp[1]=i+1
ans=0
for i in range(1,n+1):
ans+=dp[i]
ans%=1000000007
print(ans)
``` | instruction | 0 | 28,293 | 12 | 56,586 |
No | output | 1 | 28,293 | 12 | 56,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.
We define two kinds of functions on sets:
f(S)=min_{u∈ S} u;
g(S)=max_{u∈ S} u.
You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).
Formally, you can perform the following sequence of operations:
* cnt← cnt+1;
* S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).
You are required to obtain some specific sets.
There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.
In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.
Input
The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.
The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.
i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.
Output
It is guaranteed that a solution under given constraints exists.
The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.
cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.
The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.
Please notice the large amount of output.
Examples
Input
3 2
1 3 2
2 3
1 3
Output
6
3 2
1 3
5 2
4 6
Input
2 4
2 1
1 2
1 2
1 2
1 1
Output
5
2 1
2 1
2 1
5 3 3 1
Note
In the first sample:
We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.
In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.
In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.
In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.
For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.
For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6
Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.
Submitted Solution:
```
number_list = [1, 2, 3]
str_list = ['one', 'two', 'three']
# No iterables are passed
result = zip()
# Converting itertor to list
result_list = list(result)
print(result_list)
# Two iterables are passed
result = zip(number_list, str_list)
# Converting itertor to set
result_set = set(result)
print(result_set)
``` | instruction | 0 | 28,450 | 12 | 56,900 |
No | output | 1 | 28,450 | 12 | 56,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.
We define two kinds of functions on sets:
f(S)=min_{u∈ S} u;
g(S)=max_{u∈ S} u.
You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).
Formally, you can perform the following sequence of operations:
* cnt← cnt+1;
* S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).
You are required to obtain some specific sets.
There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.
In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.
Input
The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.
The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.
i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.
Output
It is guaranteed that a solution under given constraints exists.
The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.
cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.
The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.
Please notice the large amount of output.
Examples
Input
3 2
1 3 2
2 3
1 3
Output
6
3 2
1 3
5 2
4 6
Input
2 4
2 1
1 2
1 2
1 2
1 1
Output
5
2 1
2 1
2 1
5 3 3 1
Note
In the first sample:
We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.
In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.
In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.
In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.
For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.
For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6
Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.
Submitted Solution:
```
class Node:
l, r = 0, 0
index = -1
lc, rc = None, None
import sys
inpy = [int(x) for x in sys.stdin.read().split()]
n, q = inpy[0], inpy[1]
memo = [0] * (n+1)
for i in range(2, n + 2):
memo[inpy[i]] = i - 1
qq = [(inpy[n + 2 + 2 * i], inpy[n + 3 + i * 2]) for i in range(q)]
cnt = n
res = []
def build(l, r):
global cnt, res
if l == r:
node = Node()
node.l, node.r = l, r
node.index = memo[l]
return node
node = Node()
node.l, node.r = l, r
node.lc = build(l, l + (r - l) // 2)
node.rc = build(l + (r - l) // 2 + 1, r)
res.append((node.lc.index, node.rc.index))
cnt += 1
node.index = cnt
return node
def get(l, r, cur:Node):
global cnt, res
if l == cur.l and r == cur.r:
return cur.index
mid = cur.l + (cur.r - cur.l) // 2
if r <= mid:
return get(l, r, cur.lc)
elif l > mid:
return get(l, r, cur.rc)
a, b = get(l, mid, cur.lc), get(mid + 1, r, cur.rc)
cnt += 1
res.append((a, b))
return cnt
root = build(1, n)
qr = []
for l, r in qq:
x = inpy[1+l: 2+r]
x.sort()
nl, l = x[0], 1
pre = None
# print(x)
for i in range(1, len(x) + 1):
if i < len(x) and x[i] == nl + l:
l += 1
else:
cnt1 = get(nl, nl+l-1, root)
nl, l = i, 1
if pre:
cnt += 1
res.append((pre, cnt1))
cnt1 = cnt
pre = cnt1
else:
pre = cnt1
if i == len(x):
qr.append(cnt1)
print(cnt)
for i, j in res:
print(i, j)
for i in qr:
print(i, end=' ')
``` | instruction | 0 | 28,451 | 12 | 56,902 |
No | output | 1 | 28,451 | 12 | 56,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.
We define two kinds of functions on sets:
f(S)=min_{u∈ S} u;
g(S)=max_{u∈ S} u.
You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).
Formally, you can perform the following sequence of operations:
* cnt← cnt+1;
* S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).
You are required to obtain some specific sets.
There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.
In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.
Input
The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.
The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.
i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.
Output
It is guaranteed that a solution under given constraints exists.
The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.
cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.
The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.
Please notice the large amount of output.
Examples
Input
3 2
1 3 2
2 3
1 3
Output
6
3 2
1 3
5 2
4 6
Input
2 4
2 1
1 2
1 2
1 2
1 1
Output
5
2 1
2 1
2 1
5 3 3 1
Note
In the first sample:
We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.
In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.
In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.
In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.
For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.
For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6
Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.
Submitted Solution:
```
class Node:
l, r = 0, 0
index = -1
lc, rc = None, None
import sys
inpy = [int(x) for x in sys.stdin.read().split()]
n, q = inpy[0], inpy[1]
memo = [0] * (n+1)
for i in range(2, n + 2):
memo[inpy[i]] = i - 1
qq = [(inpy[n + 2 + 2 * i], inpy[n + 3 + i * 2]) for i in range(q)]
cnt = n
res = []
def build(l, r):
global cnt, res
if l == r:
node = Node()
node.l, node.r = l, r
node.index = memo[l]
return node
node = Node()
node.l, node.r = l, r
node.lc = build(l, l + (r - l) // 2)
node.rc = build(l + (r - l) // 2 + 1, r)
res.append((node.lc.index, node.rc.index))
cnt += 1
node.index = cnt
return node
def get(l, r, cur:Node):
global cnt, res
if l == cur.l and r == cur.r:
return cur.index
mid = cur.l + (cur.r - cur.l) // 2
if r <= mid:
return get(l, r, cur.lc)
elif l > mid:
return get(l, r, cur.rc)
a, b = get(l, mid, cur.lc), get(mid + 1, r, cur.rc)
cnt += 1
res.append((a, b))
return cnt
root = build(1, n)
qr = []
m1, m2 = {}, {}
for l, r in qq:
if (l, r) in m1:
qr.append(m1[(l, r)])
continue
x = inpy[1+l: 2+r]
x.sort()
nl, l = x[0], 1
pre = None
# print(x)
for i in range(1, len(x) + 1):
if i < len(x) and x[i] == nl + l:
l += 1
else:
if (nl, nl + l - 1) in m2:
cnt1 = m2[(nl, nl + l - 1)]
else:
cnt1 = get(nl, nl+l-1, root)
m2[(nl, nl + l - 1)] = cnt1
if pre:
cnt += 1
res.append((pre, cnt1))
cnt1 = cnt
pre = cnt1
else:
pre = cnt1
if i == len(x):
qr.append(cnt1)
m1[l, r] = cnt1
else:
nl, l = x[i], 1
print(cnt)
for i, j in res:
print(i, j)
for i in qr:
print(i, end=' ')
``` | instruction | 0 | 28,452 | 12 | 56,904 |
No | output | 1 | 28,452 | 12 | 56,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.
We define two kinds of functions on sets:
f(S)=min_{u∈ S} u;
g(S)=max_{u∈ S} u.
You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).
Formally, you can perform the following sequence of operations:
* cnt← cnt+1;
* S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).
You are required to obtain some specific sets.
There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.
In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.
Input
The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.
The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.
i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.
Output
It is guaranteed that a solution under given constraints exists.
The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.
cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.
The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.
Please notice the large amount of output.
Examples
Input
3 2
1 3 2
2 3
1 3
Output
6
3 2
1 3
5 2
4 6
Input
2 4
2 1
1 2
1 2
1 2
1 1
Output
5
2 1
2 1
2 1
5 3 3 1
Note
In the first sample:
We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.
In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.
In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.
In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.
For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.
For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6
Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.
Submitted Solution:
```
class Node:
l, r = 0, 0
index = -1
lc, rc = None, None
import sys
inpy = [int(x) for x in sys.stdin.read().split()]
n, q = inpy[0], inpy[1]
memo = [0] * (n+1)
for i in range(2, n + 2):
memo[inpy[i]] = i - 1
qq = [(inpy[n + 2 + 2 * i], inpy[n + 3 + i * 2]) for i in range(q)]
cnt = n
res = []
def build(l, r):
global cnt, res
if l == r:
node = Node()
node.l, node.r = l, r
node.index = l
return node
node = Node()
node.l, node.r = l, r
node.lc = build(l, l + (r - l) // 2)
node.rc = build(l + (r - l) // 2 + 1, r)
res.append((node.lc.index, node.rc.index))
cnt += 1
node.index = cnt
return node
def get(l, r, cur:Node):
global cnt, res
if l == cur.l and r == cur.r:
return cur.index
mid = cur.l + (cur.r - cur.l) // 2
if r <= mid:
return get(l, r, cur.lc)
elif l > mid:
return get(l, r, cur.rc)
a, b = get(l, mid, cur.lc), get(mid + 1, r, cur.rc)
cnt += 1
res.append((a, b))
return cnt
root = build(1, n)
qr = []
for l, r in qq:
qr.append(get(l, r, root))
print(cnt)
for i, j in res:
print(i, j)
for i in qr:
print(i, end=' ')
``` | instruction | 0 | 28,453 | 12 | 56,906 |
No | output | 1 | 28,453 | 12 | 56,907 |
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