message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3].
Submitted Solution:
```
if __name__=="__main__":
t=int(input())
for _ in range(t):
n=int(input())
ar=list(map(int,input().split()))
# Now create a map
m={}
m[1]=0
m[2]=0
result=0
for i in ar:
if i%3==1:
m[1]+=1
elif i%3==2:
m[2]+=1
elif i%2==0:
result+=1
# Now find the minimum of m[1] and m[2]
minValue=min(m[1],m[2])
result+=minValue
# now see whether the number div by 3 is in m[1]
result+=((m[1]-minValue)//3)
print(result)
``` | instruction | 0 | 60,717 | 22 | 121,434 |
No | output | 1 | 60,717 | 22 | 121,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3].
Submitted Solution:
```
"""
- math -
from math import factorial
from math import gcd
from math import pi
large_p = 10**9 + 7
from itertools import accumulate # 累積和
from operator import mul
- other -
from collections import Counter
from itertools import combinations
from itertools import combinations_with_replacement
from itertools import permutations
from operator import itemgetter
from functools import reduce
- list -
from copy import deepcopy
from collections import deque
from heapq import heapify, heappop, heappush
- output -
p2d = lambda x: print(*x, sep="\n")
p2line = lambda x: print(*x, sep=" ")
print(*list1, sep="\n")
- input -
import sys
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
= map(int, input().split())
= [tuple(map(int, input().split())) for _ in range(n)]
= [tuple(map(int, readline().split())) for _ in range(n)]
m = map(int, read().split()) /ab = zip(m, m) /for a, b in ab:
ps = [line.split() for line in readlines()] # inputの最後まで読んでしまうので注意。
n, *h = map(int, read().split())
a = list(map(int, readlines()))
- run -
sys.setrecursionlimit(10**6)
from time import sleep
"""
# ATC04
import sys
read = sys.stdin.read
readlines = sys.stdin.readlines
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
r = 0
a2 = [i % 3 == 0 for i in a]
r += sum(a2)
a3 = [i for i in a if i < 3]
if a3:
num2 = a3.count(2)
num1 = a3.count(1)
if num2 >= num1:
r += num1
else:
r += num2
num1 -= num2
r += num1 // 3
print(r)
if __name__ == '__main__':
main()
``` | instruction | 0 | 60,719 | 22 | 121,438 |
No | output | 1 | 60,719 | 22 | 121,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ... , a_n.
In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2, 1, 4] you can obtain the following arrays: [3, 4], [1, 6] and [2, 5].
Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
The first line of each query contains one integer n (1 ≤ n ≤ 100).
The second line of each query contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9).
Output
For each query print one integer in a single line — the maximum possible number of elements divisible by 3 that are in the array after performing described operation an arbitrary (possibly, zero) number of times.
Example
Input
2
5
3 1 2 3 1
7
1 1 1 1 1 2 2
Output
3
3
Note
In the first query of the example you can apply the following sequence of operations to obtain 3 elements divisible by 3: [3, 1, 2, 3, 1] → [3, 3, 3, 1].
In the second query you can obtain 3 elements divisible by 3 with the following sequence of operations: [1, 1, 1, 1, 1, 2, 2] → [1, 1, 1, 1, 2, 3] → [1, 1, 1, 3, 3] → [2, 1, 3, 3] → [3, 3, 3].
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
ans = sum(a)
print(ans // 3)
``` | instruction | 0 | 60,720 | 22 | 121,440 |
No | output | 1 | 60,720 | 22 | 121,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In number world, two different numbers are friends if they have a lot in common, but also each one has unique perks.
More precisely, two different numbers a and b are friends if gcd(a,b), (a)/(gcd(a,b)), (b)/(gcd(a,b)) can form sides of a triangle.
Three numbers a, b and c can form sides of a triangle if a + b > c, b + c > a and c + a > b.
In a group of numbers, a number is lonely if it doesn't have any friends in that group.
Given a group of numbers containing all numbers from 1, 2, 3, ..., n, how many numbers in that group are lonely?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) - number of test cases.
On next line there are t numbers, n_i (1 ≤ n_i ≤ 10^6) - meaning that in case i you should solve for numbers 1, 2, 3, ..., n_i.
Output
For each test case, print the answer on separate lines: number of lonely numbers in group 1, 2, 3, ..., n_i.
Example
Input
3
1 5 10
Output
1
3
3
Note
For first test case, 1 is the only number and therefore lonely.
For second test case where n=5, numbers 1, 3 and 5 are lonely.
For third test case where n=10, numbers 1, 5 and 7 are lonely. | instruction | 0 | 60,869 | 22 | 121,738 |
Tags: binary search, math, number theory, two pointers
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
sieve = [0]*(10**6+1)
primes = [0]*(10**6+1)
for i in range(2,10**6+1):
if not sieve[i]:
primes[i] += 1
if i*i <= 10**6:
primes[i*i] -= 1
for x in range(i*i,10**6+1,i):
sieve[x] = 1
primes[1] = 1
for i in range(1,10**6+1):
primes[i] += primes[i-1]
_ = input()
for i in map(int,input().split()):
print(primes[i])
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 60,869 | 22 | 121,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In number world, two different numbers are friends if they have a lot in common, but also each one has unique perks.
More precisely, two different numbers a and b are friends if gcd(a,b), (a)/(gcd(a,b)), (b)/(gcd(a,b)) can form sides of a triangle.
Three numbers a, b and c can form sides of a triangle if a + b > c, b + c > a and c + a > b.
In a group of numbers, a number is lonely if it doesn't have any friends in that group.
Given a group of numbers containing all numbers from 1, 2, 3, ..., n, how many numbers in that group are lonely?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) - number of test cases.
On next line there are t numbers, n_i (1 ≤ n_i ≤ 10^6) - meaning that in case i you should solve for numbers 1, 2, 3, ..., n_i.
Output
For each test case, print the answer on separate lines: number of lonely numbers in group 1, 2, 3, ..., n_i.
Example
Input
3
1 5 10
Output
1
3
3
Note
For first test case, 1 is the only number and therefore lonely.
For second test case where n=5, numbers 1, 3 and 5 are lonely.
For third test case where n=10, numbers 1, 5 and 7 are lonely.
Submitted Solution:
```
from sys import stdout
import math
# gcd = p
# ap,bp where a+b>p
#
# gcd = 7
# 1*7, 7*7
#
# n<p^2 <=> p is lonely
# def gcd(a,b):
# if a == 0:
# return b
# return gcd(b % a, a)
N = 1_000_000
is_prime = [True] * (N+1)
is_prime[0] = False
is_prime[1] = False
for i in range(N+1):
if is_prime[i] and i*i < N+1:
for j in range(i*i, N+1, i):
is_prime[j] = False
primes_until = [0] * (N+1)
for i in range(2,N+1):
if is_prime[i]:
primes_until[i] = primes_until[i-1] + 1
else:
primes_until[i] = primes_until[i-1]
# print(primes_until)
input()
ns = map(int,input().split())
for n in ns:
# lonely_nums = 1
# for i in range(n+1):
# if is_prime[i] and i*i > n:
# lonely_nums += 1
stdout.write(str(primes_until[n]-primes_until[int(math.sqrt(n))]+1) + "\n")
``` | instruction | 0 | 60,875 | 22 | 121,750 |
Yes | output | 1 | 60,875 | 22 | 121,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In number world, two different numbers are friends if they have a lot in common, but also each one has unique perks.
More precisely, two different numbers a and b are friends if gcd(a,b), (a)/(gcd(a,b)), (b)/(gcd(a,b)) can form sides of a triangle.
Three numbers a, b and c can form sides of a triangle if a + b > c, b + c > a and c + a > b.
In a group of numbers, a number is lonely if it doesn't have any friends in that group.
Given a group of numbers containing all numbers from 1, 2, 3, ..., n, how many numbers in that group are lonely?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) - number of test cases.
On next line there are t numbers, n_i (1 ≤ n_i ≤ 10^6) - meaning that in case i you should solve for numbers 1, 2, 3, ..., n_i.
Output
For each test case, print the answer on separate lines: number of lonely numbers in group 1, 2, 3, ..., n_i.
Example
Input
3
1 5 10
Output
1
3
3
Note
For first test case, 1 is the only number and therefore lonely.
For second test case where n=5, numbers 1, 3 and 5 are lonely.
For third test case where n=10, numbers 1, 5 and 7 are lonely.
Submitted Solution:
```
import os,io
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = 10**6+1
primes = [1]*n
primes[0],primes[1] = 0,0
v = int(n**0.5)+1
for i in range(2,v):
if primes[i]:
for j in range(i*i,n,i):
primes[j]=0
for i in range(1,n):
primes[i]+=primes[i-1]
cases = int(input())
for t in range(cases):
n1 = list(map(int,input().split()))
for ni in n1:
print(primes[ni]-primes[int(ni**0.5)]+1)
``` | instruction | 0 | 60,876 | 22 | 121,752 |
Yes | output | 1 | 60,876 | 22 | 121,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In number world, two different numbers are friends if they have a lot in common, but also each one has unique perks.
More precisely, two different numbers a and b are friends if gcd(a,b), (a)/(gcd(a,b)), (b)/(gcd(a,b)) can form sides of a triangle.
Three numbers a, b and c can form sides of a triangle if a + b > c, b + c > a and c + a > b.
In a group of numbers, a number is lonely if it doesn't have any friends in that group.
Given a group of numbers containing all numbers from 1, 2, 3, ..., n, how many numbers in that group are lonely?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) - number of test cases.
On next line there are t numbers, n_i (1 ≤ n_i ≤ 10^6) - meaning that in case i you should solve for numbers 1, 2, 3, ..., n_i.
Output
For each test case, print the answer on separate lines: number of lonely numbers in group 1, 2, 3, ..., n_i.
Example
Input
3
1 5 10
Output
1
3
3
Note
For first test case, 1 is the only number and therefore lonely.
For second test case where n=5, numbers 1, 3 and 5 are lonely.
For third test case where n=10, numbers 1, 5 and 7 are lonely.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
sieve = [0]*(10**6+1)
primes = [0]*(10**6+1)
for i in range(2,10**3+1):
if not sieve[i]:
primes[i] += 1
primes[i*i] -= 1
for x in range(i*i,10**6+1,i):
sieve[x] = 1
primes[1] = 1
for i in range(1,10**6+1):
primes[i] += primes[i-1]
_ = input()
for i in map(int,input().split()):
print(primes[i])
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | instruction | 0 | 60,878 | 22 | 121,756 |
No | output | 1 | 60,878 | 22 | 121,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon likes to play with sets, integers and <image>. <image> is defined as the largest positive integer that divides both a and b.
Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, <image>.
Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.
Input
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
Output
On the first line print a single integer — the minimal possible m.
On each of the next n lines print four space separated integers representing the i-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.
Examples
Input
1 1
Output
5
1 2 3 5
Input
2 2
Output
22
2 4 6 22
14 18 10 16
Note
For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <image>. | instruction | 0 | 61,052 | 22 | 122,104 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
'''
Created on Oct 12, 2014
@author: Ismael
'''
#import time
from fractions import gcd
def checkSet(setK,newVal):
for v in setK:
if(gcd(v,newVal) != 1):
return False
return True
def solve(n,k):
j = 1
sets = []
for _ in range(n):
setK = set()
while(len(setK) < 4):
if(checkSet(setK,j) and not(len(setK) == 0 and j%3==0)):
setK.add(j)
j += 1
sets.append(setK)
maxV = 0
for setK in sets:
maxV = max(maxV,max(setK))
print(maxV*k)
for setK in sets:
print(' '.join(map(lambda x:str(x*k),setK)))
#t = time.clock()
n,k = map(int,input().split())
solve(n,k)
#print(time.clock()-t)
``` | output | 1 | 61,052 | 22 | 122,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon likes to play with sets, integers and <image>. <image> is defined as the largest positive integer that divides both a and b.
Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, <image>.
Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.
Input
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
Output
On the first line print a single integer — the minimal possible m.
On each of the next n lines print four space separated integers representing the i-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.
Examples
Input
1 1
Output
5
1 2 3 5
Input
2 2
Output
22
2 4 6 22
14 18 10 16
Note
For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <image>.
Submitted Solution:
```
__author__ = 'neki'
import sys
import math
#sys.stdin = open("sets.in", "r")
global primes, primeDiv
def primality(limit):
global primes, primeDiv
primes = [2, 3]
primeDiv = [[], [], [], []] #according to an abstraction made: 0, 1, 2, 3 have no prime divisors
i = 4
k = 3 #we can use 1, 2 and 3, we need limit prime numbers to be absolutely sure we cannot fail
while k < limit:
primeDiv.append([])
for j in primes:
if 2*j > i:
break
if i % j == 0:
primeDiv[i].append(j)
if not primeDiv[i]:
primes.append(i)
k += 1
i += 1
def gcdPrime(a, b):
if b==0 or a==0:
return 0 #not prime
if b==1 or a==1:
return 1 #prime
if b>a:
return gcdPrime(a, b%a)
return gcdPrime(b, a%b)
def gcdPrimeSet(set, a):
if len(set) >= 4: #full set
return 0
for i in set:
if gcdPrime(i, a) == 0:
return 0
return 1
words = str(input()).split()
n = int(words[0])
k = int(words[1])
#primality(4*n)
#print(primes)
#print(primeDiv)
#print(gcdPrime(8, 12))
sets = []
for i in range(n):
sets.append(set())
m = 0
nEl = 0
iniSet = 0
while nEl < 4*n:
m += 1
#print(m, sets, iniSet)
i = iniSet
while i < iniSet+n:
pointer = i
if i >= n:
pointer = i-n
s = sets[pointer]
if gcdPrimeSet(s, m) == 1:
s.add(m)
nEl += 1
iniSet = 0
break
i += 1
print(k*m)
for s in sets:
for i in range(4):
print(k*s.pop(), end=" ")
print()
``` | instruction | 0 | 61,063 | 22 | 122,126 |
No | output | 1 | 61,063 | 22 | 122,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon likes to play with sets, integers and <image>. <image> is defined as the largest positive integer that divides both a and b.
Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, <image>.
Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.
Input
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
Output
On the first line print a single integer — the minimal possible m.
On each of the next n lines print four space separated integers representing the i-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.
Examples
Input
1 1
Output
5
1 2 3 5
Input
2 2
Output
22
2 4 6 22
14 18 10 16
Note
For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <image>.
Submitted Solution:
```
__author__ = 'neki'
import sys
import math
#sys.stdin = open("sets.in", "r")
global primes, primeDiv
def primality(limit):
global primes, primeDiv
primes = [2, 3]
primeDiv = [[], [], [], []] #according to an abstraction made: 0, 1, 2, 3 have no prime divisors
i = 4
k = 3 #we can use 1, 2 and 3, we need limit prime numbers to be absolutely sure we cannot fail
while k < limit:
primeDiv.append([])
for j in primes:
if 2*j > i:
break
if i % j == 0:
primeDiv[i].append(j)
if not primeDiv[i]:
primes.append(i)
k += 1
i += 1
def gcdPrime(a, b):
if b==0 or a==0:
return 0 #not prime
if b==1 or a==1:
return 1 #prime
if b>a:
return gcdPrime(a, b%a)
return gcdPrime(b, a%b)
def gcdPrimeSet(set, a):
result = []
if len(set) >= 4: #full set
return [x for x in set]
for i in set:
if gcdPrime(i, a) == 0:
result.append(i)
return result
words = str(input()).split()
n = int(words[0])
k = int(words[1])
primality(4*n)
#print(primes)
#print(primeDiv)
#print(gcdPrime(8, 12))
sets = []
for i in range(n):
sets.append(set())
m = 0
nEl = 0
iniSet = 0
while nEl < 4*n:
m += 1
#print(m, sets)
sw = 0
setProblems = [[] for x in sets]
for j in range(iniSet, iniSet + len(sets)):
i = j % len(sets)
setProblems[i] = gcdPrimeSet(sets[i], m)
if len(setProblems[i]) == 0:
sets[i].add(m)
iniSet += 1
nEl += 1
sw = 1
break
if sw == 0:
for i in range(len(setProblems)):
if len(setProblems[i]) == 1 and len(primeDiv[m]) < len(primeDiv[setProblems[i][0]]):
sets[i].remove(setProblems[i][0])
sets[i].add(m)
break
print(k*m)
for s in sets:
for i in range(4):
print(k*s.pop(), end=" ")
print()
``` | instruction | 0 | 61,065 | 22 | 122,130 |
No | output | 1 | 61,065 | 22 | 122,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon likes to play with sets, integers and <image>. <image> is defined as the largest positive integer that divides both a and b.
Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, <image>.
Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.
Input
The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).
Output
On the first line print a single integer — the minimal possible m.
On each of the next n lines print four space separated integers representing the i-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.
Examples
Input
1 1
Output
5
1 2 3 5
Input
2 2
Output
22
2 4 6 22
14 18 10 16
Note
For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <image>.
Submitted Solution:
```
__author__ = 'neki'
import sys
import math
#sys.stdin = open("sets.in", "r")
global primes, primeDiv
def primality(limit):
global primes, primeDiv
primes = [2, 3]
primeDiv = [[], [], [], []] #according to an abstraction made: 0, 1, 2, 3 have no prime divisors
i = 4
k = 3 #we can use 1, 2 and 3, we need limit prime numbers to be absolutely sure we cannot fail
while k < limit:
primeDiv.append([])
for j in primes:
if 2*j > i:
break
if i % j == 0:
primeDiv[i].append(j)
if not primeDiv[i]:
primes.append(i)
k += 1
i += 1
def gcdPrime(a, b):
if b==0 or a==0:
return 0 #not prime
if b==1 or a==1:
return 1 #prime
if b>a:
return gcdPrime(a, b%a)
return gcdPrime(b, a%b)
def gcdPrimeSet(set, a):
result = []
if len(set) >= 4: #full set
return [x for x in set]
for i in set:
if gcdPrime(i, a) == 0:
result.append(i)
return result
words = str(input()).split()
n = int(words[0])
k = int(words[1])
primality(4*n)
#print(primes)
#print(primeDiv)
#print(gcdPrime(8, 12))
sets = []
for i in range(n):
sets.append(set())
m = 0
nEl = 0
iniSet = 0
while nEl < 4*n:
m += 1
#print(m, sets)
sw = 0
setProblems = [[] for x in sets]
for j in range(iniSet, iniSet + len(sets)):
i = j % len(sets)
setProblems[i] = gcdPrimeSet(sets[i], m)
if len(setProblems[i]) == 0:
sets[i].add(m)
iniSet += 1
nEl += 1
sw = 1
break
if sw == 0:
for i in range(len(setProblems)):
if len(setProblems[i]) == 1 and len(primeDiv[m]) < len(primeDiv[setProblems[i][0]]):
sets[i].remove(setProblems[i][0])
sets[i].add(m)
print(k*m)
for s in sets:
for i in range(4):
print(k*s.pop(), end=" ")
print()
``` | instruction | 0 | 61,066 | 22 | 122,132 |
No | output | 1 | 61,066 | 22 | 122,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,147 | 22 | 122,294 |
Tags: math, number theory
Correct Solution:
```
def is_prime(x):
lst = []
nn = x
for i in range(2, round(x**0.5)+1):
while nn % i == 0:
nn //= i
lst.append(i)
if nn > 1:
lst.append(nn)
nn = 1
return lst
n = int(input())
if n == 2:
print(1)
exit(0)
if n == 3:
print(1)
exit(0)
if n == 4:
print(2)
exit(0)
if n == 5:
print(1)
exit(0)
a = is_prime(n)
if len(a) == 1:
print(1)
elif n % 2 == 0:
print(2)
else:
if len(is_prime(n-2)) == 1:
print(2)
else:
print(3)
``` | output | 1 | 61,147 | 22 | 122,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,148 | 22 | 122,296 |
Tags: math, number theory
Correct Solution:
```
def isprime(n):
if(n==1):
return 0
if(n==2 or n==3):
return 1
i=2
while(i*i<=n):
if(n%i==0):
break
i+=1
if(i*i>n):
return 1
else:
return 0
n=int(input())
if(isprime(n)==1):
print(1)
elif(isprime(n-2)==1 or n%2==0):
print(2)
else:
print(3)
``` | output | 1 | 61,148 | 22 | 122,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,149 | 22 | 122,298 |
Tags: math, number theory
Correct Solution:
```
I = lambda: int(input())
IL = lambda: list(map(int, input().split()))
S = [0] * 100001
S[0] = S[1] = 1
for i in range(2, 100001):
if S[i]:
continue
else:
for j in range(2*i, 100001, i):
S[j] = 1
primes = [i for i, s in enumerate(S) if not s]
def is_prime(x):
for p in primes:
if p*p > x:
return True
if x % p == 0:
return False
n = I()
if is_prime(n):
print(1)
elif (n % 2 == 0) or is_prime(n-2):
print(2)
else:
print(3)
``` | output | 1 | 61,149 | 22 | 122,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,150 | 22 | 122,300 |
Tags: math, number theory
Correct Solution:
```
def ok(n):
ok = False
i = int(2)
while i*i <= n:
if n % i == 0:
ok = True
i += 1
return ok == False
n = int(input())
i = int(0)
if(ok(n)):
print(1)
elif(n % 2 == 0):
print(2)
elif(ok(n - 2)):
print(2)
else:
print(3)
``` | output | 1 | 61,150 | 22 | 122,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,151 | 22 | 122,302 |
Tags: math, number theory
Correct Solution:
```
def isPrime(n):
for i in range(2,int(n**0.5+100)):
if n%i==0 and n!=i:
return False
return True
n=int(input())
if isPrime(n):
print(1)
elif n%2==0:
print(2)
elif isPrime(n-2):
print(2)
else:
print(3)
``` | output | 1 | 61,151 | 22 | 122,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,152 | 22 | 122,304 |
Tags: math, number theory
Correct Solution:
```
from itertools import *
from math import *
def isPrime(n):
return n>1and all(n%i for i in islice(count(2),int(sqrt(n)-1)))
b=int(input())
m=isPrime(b)
if m==True:
print(1)
elif b%2==0:
print(2)
elif isPrime(b-2)==True:
print(2)
else:
print(3)
``` | output | 1 | 61,152 | 22 | 122,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,153 | 22 | 122,306 |
Tags: math, number theory
Correct Solution:
```
def snt(n):
for i in range(2, int(n**0.5) + 1):
if n%i == 0: return False
return True
n = int(input())
if n%2 == 0:
if n == 2 : print(1)
else: print(2)
else:
if snt(n):
print('1')
else:
if snt(n-2):
print('2')
else:
print('3')
``` | output | 1 | 61,153 | 22 | 122,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3 | instruction | 0 | 61,154 | 22 | 122,308 |
Tags: math, number theory
Correct Solution:
```
#########################################################################################################\
#########################################################################################################
###################################The_Apurv_Rathore#####################################################
#########################################################################################################
#########################################################################################################
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def power_set(L):
"""
L is a list.
The function returns the power set, but as a list of lists.
"""
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
#the function could stop here closing with
#return powerset
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = 1
# t = int(input())
for _ in range(t):
n = ii()
if (isPrime(n)):
print(1)
continue
if (n%2==0 or (n>2 and isPrime(n-2)==1)):
print(2)
else:
print(3)
``` | output | 1 | 61,154 | 22 | 122,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3
Submitted Solution:
```
from math import sqrt
def is_prime(x):
stop = int(sqrt(x)) + 1
for i in range(2, stop):
if x % i == 0:
return False
return True
n = int(input())
if is_prime(n):
print(1)
else:
if n % 2 == 1:
if is_prime(n - 2):
print(2)
else:
print(3)
else:
print(2)
``` | instruction | 0 | 61,155 | 22 | 122,310 |
Yes | output | 1 | 61,155 | 22 | 122,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3
Submitted Solution:
```
def prime(n):
i=2
while i*i<=n:
if n%i==0:
return False
i+=1
return True
n=int(input())
if prime(n):
print(1)
elif n%2==0 or prime(n-2):
print(2)
else:
print(3)
``` | instruction | 0 | 61,156 | 22 | 122,312 |
Yes | output | 1 | 61,156 | 22 | 122,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3
Submitted Solution:
```
from math import sqrt
def is_prime(n):
for i in range(2, int(sqrt(n))+1):
if n%i == 0:
return False
return True
def tax(n):
if n == 2 or n == 5:
return 1
elif n%2 == 0:
return 2
else:
if is_prime(n):
return 1
elif is_prime(n-2):
return 2
else:
return 3
n = int(input())
print(tax(n))
``` | instruction | 0 | 61,158 | 22 | 122,316 |
Yes | output | 1 | 61,158 | 22 | 122,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (0 ≤ a ≤ 63).
Output
Output a single number.
Examples
Input
2
Output
2
Input
5
Output
24
Input
35
Output
50
Submitted Solution:
```
import sys,os,io
from sys import stdin,stdout
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
# input = stdin.readline
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def power_set(L):
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
def sortByFirstAndSecond(A):
A = sorted(A,key = lambda x:x[0])
A = sorted(A,key = lambda x:x[1])
return list(A)
#__________________________TEMPLATE__________________OVER_______________________________________________________
# if(os.path.exists('input.txt')):
# sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# # else:
# # input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
bi = list(bin(n))[2:]
if(len(bi) < 6):
for i in range(6 - len(bi)):
bi.insert(0,0)
# print(bi)
bi[1],bi[5] = bi[5],bi[1]
bi[2],bi[3] = bi[3],bi[2]
st = ""
for i in bi:
st+= str(i)
print(int(st,2))
``` | instruction | 0 | 61,613 | 22 | 123,226 |
Yes | output | 1 | 61,613 | 22 | 123,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,709 | 22 | 123,418 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
#!/usr/bin/python3
# -*- coding: utf-8 -*-
import sys
import typing
def rl(proc=None) -> typing.Any:
if proc is not None:
return proc(sys.stdin.readline())
else:
return sys.stdin.readline().rstrip()
def srl(proc=None) -> typing.List[typing.Any]:
if proc is not None:
return list(map(proc, rl().split()))
else:
return rl().split()
MOD = 10**9 + 7
def main():
T = rl(int)
for _ in range(1, T+1):
N, M = srl(int)
c0 = 0
for _ in range(N):
c0 += rl().count('0')
if c0:
ret = pow(2, N * M - c0, MOD)
else:
ret = pow(2, N * M, MOD) - 1
if ret < 0:
ret += MOD
print(ret)
if __name__ == '__main__':
main()
``` | output | 1 | 61,709 | 22 | 123,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,710 | 22 | 123,420 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
for i in range(int(input())):
n, m = map(int, input().split())
counta = sum([input().count("#") for i in range(n)])
print(pow(2, counta, int(10**9+7))-(counta == m*n))
``` | output | 1 | 61,710 | 22 | 123,421 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,711 | 22 | 123,422 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
for _ in range(int(input())):
n,m = [int(X) for X in input().split()];aaa = 0
for i in range(n):
x = input();aaa+=x.count('#')
print(pow(2,aaa,int(1e9+7)) - (1 if aaa==n*m else 0))
``` | output | 1 | 61,711 | 22 | 123,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,712 | 22 | 123,424 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
for _ in range(int(input())):
n, m = [int(x) for x in input().split()]
k = 0
for i in range(n):
s = input()
for j in range(m):
if(s[j] != '0'):
k += 1
ans = 2**k
if(n*m == k):
ans -= 1
print(ans % 1000000007)
``` | output | 1 | 61,712 | 22 | 123,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,713 | 22 | 123,426 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
def power(x, y, p):
b = bin(y)[2:]
start = x
answer = 1
for i in range(len(b)-1, -1, -1):
if b[i]=='1':
answer = (answer*start) % p
start = (start*start) % p
return answer
p = 10**9+7
def process(M):
n = len(M)
m = len(M[0])
c = 0
for i in range(n):
for j in range(m):
if M[i][j]=='0':
c+=1
if c==0:
return (power(2, n*m, p)-1) % p
else:
return power(2, n*m-c, p)
t = int(input())
for i in range(t):
n, m = [int(x) for x in input().split()]
M =[]
for j in range(n):
M.append(input())
print(process(M))
``` | output | 1 | 61,713 | 22 | 123,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,714 | 22 | 123,428 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve():
n, m = map(int, input().split())
c = 0
for i in range(n):
c += input().count('#')
MOD = int(1e9+7)
a = pow(2, c, MOD)
if c == n*m:
a = (a - 1) % MOD
print(a)
for i in range(int(input())):
solve()
``` | output | 1 | 61,714 | 22 | 123,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,715 | 22 | 123,430 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
mod = 10**9 + 7
def power(x,e):
res = 1
while e:
if e&1:
res = res * x % mod
x = (x*x) % mod
e >>= 1
return res
for _ in range(int(input())):
n,m = map(int,input().split())
cnt = 0
for _ in range(n):
cnt += input().count('#')
r = 1 if cnt==m*n else 0
res = power(2,cnt) - r % mod
print(res)
``` | output | 1 | 61,715 | 22 | 123,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar's most recent follower, Ajit, has entered the Holy Forest. Ajit realizes that Omkar's forest is an n by m grid (1 ≤ n, m ≤ 2000) of some non-negative integers. Since the forest is blessed by Omkar, it satisfies some special conditions:
1. For any two adjacent (sharing a side) cells, the absolute value of the difference of numbers in them is at most 1.
2. If the number in some cell is strictly larger than 0, it should be strictly greater than the number in at least one of the cells adjacent to it.
Unfortunately, Ajit is not fully worthy of Omkar's powers yet. He sees each cell as a "0" or a "#". If a cell is labeled as "0", then the number in it must equal 0. Otherwise, the number in it can be any nonnegative integer.
Determine how many different assignments of elements exist such that these special conditions are satisfied. Two assignments are considered different if there exists at least one cell such that the numbers written in it in these assignments are different. Since the answer may be enormous, find the answer modulo 10^9+7.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 2000, nm ≥ 2) – the dimensions of the forest.
n lines follow, each consisting of one string of m characters. Each of these characters is either a "0" or a "#".
It is guaranteed that the sum of n over all test cases does not exceed 2000 and the sum of m over all test cases does not exceed 2000.
Output
For each test case, print one integer: the number of valid configurations modulo 10^9+7.
Example
Input
4
3 4
0000
00#0
0000
2 1
#
#
1 2
##
6 29
#############################
#000##0###0##0#0####0####000#
#0#0##00#00##00####0#0###0#0#
#0#0##0#0#0##00###00000##00##
#000##0###0##0#0##0###0##0#0#
#############################
Output
2
3
3
319908071
Note
For the first test case, the two valid assignments are
0000\\\ 0000\\\ 0000
and
0000\\\ 0010\\\ 0000 | instruction | 0 | 61,716 | 22 | 123,432 |
Tags: combinatorics, graphs, math, shortest paths
Correct Solution:
```
"""n = int(input())
for _ in range(n):
m = int(input())
s = input()
if s.count("T") == 2*m/3 and s.count("M") == m/3:
pass
else:
print("NO")
"""
"""n = int(input())
for _ in range(n):
m = int(input())
s = input()
ls = []
for i in range(m):
if i == 0:
ls.append(s[i])
else:
if s[i-1] == s[i]:
pass
else:
ls.append(s[i])
for j in ls:
if ls.count(j) > 1:
print("NO")
break
else:
if j == ls[-1]:
print("YES")
"""
"""
t = int(input())
for i in range(t):
n = int(input())
l = list(map(int, input().split()))
c = 0
l = sorted(l)
q = l.count(2)
for _ in range(q):
l.remove(2)
for _ in range(q):
l.append(2)
for j in l:
if j == 1:
c += 1
if j == 2:
c -= 1
if j == 3:
if c >= 0:
c += 1
if c >= 0:
print(l.count(1)+l.count(3))
else:
print(0)
"""
"""
def sumOfTwoPerfectCubes(N):
# Stores the perfect cubes
# of first N natural numbers
cubes = {}
i = 1
while i * i * i <= N:
cubes[i * i * i] = i
i += 1
# Traverse the map
for itr in cubes:
# Stores first number
firstNumber = itr
# Stores second number
secondNumber = N - itr
# Search the pair for the first
# number to obtain sum N from the Map
if secondNumber in cubes:
print("YES")
return
# If N cannot be represented as
# sum of two positive perfect cubes
print("NO")
n = int(input())
for i in range(n):
sumOfTwoPerfectCubes(int(input()))
"""
"""n = int(input())
for i in range(n):
l = list(map(int, input().split()))
ls = sorted(l)
m = max(l[0], l[1])
n = max(l[2], l[3])
f = [ls[-1], ls[-2]]
if m in f and n in f:
print("YES")
else:
print("NO")
"""
"""from math import gcd
from itertools import permutations
t = int(input())
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
l = sorted(l)
for p in permutations(l):
c = 0
m = 0
for i in range(n-1):
for j in range(i+1, n):
if gcd(p[i], 2*p[j]) > 1:
c += 1
m = max(m, c)
print(m)
"""
"""
from math import gcd
t = int(input())
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
ls = []
for i in l:
if i%2 == 0:
ls.append(i)
for j in l:
if j%2 != 0:
ls.append(j)
m = 0
for i in range(n - 1):
for j in range(i + 1, n):
if gcd(ls[i], 2 * ls[j]) > 1:
m += 1
print(m)
"""
# if 2 come together then unstable
# if can be made then beautiful
# making a substring program
#
# t = int(input())
# for _ in range(t):
# l = input()
# for i in input():
#
"""t = int(input())
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
if len(l) <= 300:
for i in range(n-1):
for j in range(i+1, n):
if abs(l[i]-l[j]) not in l:
l.append(abs(l[i]-l[j]))
print("YES")
else:
print("NO")
"""
"""
def printAllKLength(set, k):
n = len(set)
printAllKLengthRec(set, "", n, k)
def printAllKLengthRec(set, prefix, n, k):
if (k == 0):
return prefix
for i in range(n):
newPrefix = prefix + set[i]
printAllKLengthRec(set, newPrefix, n, k - 1)
set1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
t = int(input())
for _ in range(t):
n = int(input())
s = input()
for k in range(len(s)):
if printAllKLength(set1, k) in s:
"""
"""
from itertools import combinations_with_replacement
set1 = ('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z')
a = list(combinations_with_replacement(set1, 1))
# b = list(combinations(set1, 2))
# c = list(combinations(set1, 3))
#t = int(input())
print(a)
# for _ in range(t):
# n = int(input())
# s = input()
# for i in a:
# if i in s:
# print(i)
# break
"""
"""
for _ in range(int(input())):
n = int(input())
for i in range(2, n):
if n%(2**i-1) == 0:
print(int((n/(2**i-1))))
break
"""
"""
from math import ceil, floor, sqrt
def solution(n):
c = 0
s = N
while s!=0:
if (ceil(sqrt(n)) == floor(sqrt(n))):
N = int(input())
print(solution(N), end = "")
"""
"""
n = int(input())
l = list(map(int, input().split()))
l = l[::-1]
ls = []
for i in l:
if l.count(i)>=1 and i not in ls:
ls.append(i)
print(len(ls))
for j in ls[::-1]:
print(j, end = " ")
"""
"""
n = int(input())
l = input()
while "xxx" in l:
l.replace("xxx", "xx")
print(n-len(l))
"""
MAX = int(1e9+7)
for _ in range(int(input())):
n, m = [int(i) for i in input().split()]
l = []
k = 0
for _ in range(n):
q = input()
l.append(q)
k += q.count("#")
print(pow(2, k, MAX) - (1 if k == n*m else 0))
``` | output | 1 | 61,716 | 22 | 123,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,737 | 22 | 123,474 |
Tags: brute force, dp, number theory
Correct Solution:
```
# # of divisors of p1^e1 p2^e2 ... is (e1 + 1)(e2 + 1) ... (en + 1)
# Try all possible partitions, sort, see which gives the least answer?
# 1: 1
# 2: 2
# 3: 4
# 4: 6, 2^3 = 8, 2 * 3 = 6
# 5: 16
# 6: 12
ans = [0, 1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 14096, 192, 144, 120, 65536, 180, 262144, 240, 576, 3072, 4194304, 360, 1296, 12288, 900, 960, 268435456, 720, 1073741824, 840, 9216, 196608, 5184, 1260, 68719476736, 786432, 36864, 1680, 1099511627776, 2880, 4398046511104, 15360, 3600, 12582912, 70368744177664, 2520, 46656, 6480, 589824, 61440, 4503599627370496, 6300, 82944, 6720, 2359296, 805306368, 288230376151711744, 5040, 1152921504606846976, 3221225472, 14400, 7560, 331776, 46080, 73786976294838206464, 983040, 37748736, 25920, 1180591620717411303424, 10080, 4722366482869645213696, 206158430208, 32400, 3932160, 746496, 184320, 302231454903657293676544, 15120, 44100, 3298534883328, 4835703278458516698824704, 20160, 5308416, 13194139533312, 2415919104, 107520, 309485009821345068724781056, 25200, 2985984, 62914560, 9663676416, 211106232532992, 21233664, 27720, 79228162514264337593543950336, 233280, 230400, 45360, 1267650600228229401496703205376, 2949120, 5070602400912917605986812821504, 430080, 129600, 13510798882111488, 81129638414606681695789005144064, 50400, 324518553658426726783156020576256, 414720, 618475290624, 60480, 5192296858534827628530496329220096, 11796480, 339738624, 4026531840, 921600, 864691128455135232, 47775744, 55440, 60466176, 3458764513820540928, 9895604649984, 16106127360, 810000, 100800, 85070591730234615865843651857942052864, 83160, 39582418599936, 1658880, 1361129467683753853853498429727072845824, 322560, 191102976, 221360928884514619392, 176400, 6881280, 87112285931760246646623899502532662132736, 188743680, 348449143727040986586495598010130648530944, 181440, 633318697598976, 3541774862152233910272, 241864704, 110880, 21743271936, 14167099448608935641088, 1166400, 1030792151040, 356811923176489970264571492362373784095686656, 226800, 1427247692705959881058285969449495136382746624, 27525120, 14745600, 3732480, 86973087744, 1290240, 91343852333181432387730302044767688728495783936, 906694364710971881029632, 40532396646334464, 166320, 3057647616, 352800, 5846006549323611672814739330865132078623730171904, 16492674416640, 2073600, 14507109835375550096474112, 93536104789177786765035829293842113257979682750464, 221760, 2176782336, 26542080, 58982400, 65970697666560, 5986310706507378352962293074805895248510699696029696, 12079595520, 3240000, 967680, 2594073385365405696, 928455029464035206174343168, 383123885216472214589586756787577295904684780545900544, 277200, 1532495540865888858358347027150309183618739122183602176, 14929920, 10376293541461622784, 440401920, 5566277615616, 48318382080, 3869835264, 1055531162664960, 705600, 106168320, 1569275433846670190958947355801916604025588861116008628224, 332640, 6277101735386680763835789423207666416102355444464034512896, 237684487542793012780631851008, 8294400, 1632960, 100433627766186892221372630771322662657637687111424552206336, 1612800, 401734511064747568885490523085290650630550748445698208825344, 498960, 664082786653543858176, 3802951800684688204490109616128, 195689447424, 20643840, 89060441849856, 15211807202738752817960438464512, 943718400, 3870720, 15479341056, 907200, 1645504557321206042154969182557350504982735865633579863348609024, 67553994410557440, 10625324586456701730816, 243388915243820045087367015432192, 356241767399424, 554400, 782757789696, 973555660975280180349468061728768, 42501298345826806923264, 2903040, 34828517376, 3092376453120, 6739986666787659948666753771754907668409286105635143120275902562304, 665280, 1587600, 15576890575604482885591488987660288, 107839786668602559178668060348078522694548577690162289924414440996864, 82575360, 431359146674410236714672241392314090778194310760649159697657763987456, 1698693120, 18662400, 28185722880, 6901746346790563787434755862277025452451108972170386555162524223799296, 6451200, 5699868278390784, 4323455642275676160, 2720083094132915643088896, 238878720, 441711766194596082395824375185729628956870974218904739530401550323154944, 720720, 1766847064778384329583297500742918515827483896875618958121606201292619776, 302330880, 2822400, 17293822569102704640, 29160000, 49478023249920, 139314069504, 112742891520, 43521329506126650289422336, 5670000, 1809251394333065553493296640760748560207343510400633813116524750123642650624, 1108800, 247669456896, 255211775190703847597530955573826158592, 132710400, 1081080, 115792089237316195423570985008687907853269984665640564039457584007913129639936, 197912092999680, 50096498540544, 11612160, 60397977600, 4083388403051261561560495289181218537472, 7410693711188236507108543040556026102609279018600996098525285376506440296955904, 3548160, 364791569817010176, 955514880, 2785365088392105618523029504, 1106804644422573096960, 474284397516047136454946754595585670566993857190463750305618264096412179005177856, 1940400, 1897137590064188545819787018382342682267975428761855001222473056385648716020711424, 61931520, 74649600, 261336857795280739939871698507597986398208, 51840000, 1321205760, 121416805764108066932466369176469931665150427440758720078238275608681517825325531136, 1045347431181122959759486794030391945592832, 241591910400, 1995840, 1942668892225729070919461906823518906642406839052139521251812409738904285205208498176, 3166593487994880, 7770675568902916283677847627294075626569627356208558085007249638955617140820833992704, 17708874310761169551360, 530841600, 1209323520, 801543976648704, 1441440, 2821109907456, 108716359680, 713053462628379038341895553024, 70835497243044678205440, 7957171782556586274486115970349133441607298412757563479047423630290551952200534008528896, 8164800, 23346660468288651264, 7215545057280, 11289600, 1070435769529469910793714477087121352287059968, 2229025112064, 2494800, 3206175906594816, 4281743078117879643174857908348485409148239872, 11408855402054064613470328848384, 247726080, 93386641873154605056, 103219200, 130370302485407109521180524058200202307293977194619920040712988758680403184853549195737432064, 26127360, 45635421608216258453881315393536, 434865438720, 2085924839766513752338888384931203236916703635113918720651407820138886450957656787131798913024, 14192640, 8343699359066055009355553539724812947666814540455674882605631280555545803830627148527195652096, 274031556999544297163190906134303066185487351808, 6350400, 4533471823554859405148160, 133499189745056880149688856635597007162669032647290798121690100488888732861290034376435130433536, 202661983231672320, 15850845241344, 2162160, 730166745731460135262101046296576, 15288238080, 11284439629824, 3880800, 207360000, 17538019647970835018444217992595396235871190515712, 2920666982925840541048404185186304, 115448720916480, 51298814505517056, 14515200, 2187250724783011924372502227117621365353169430893212436425770606409952999199375923223513177023053824, 72535549176877750482370560, 15461882265600, 280608314367533360295107487881526339773939048251392, 5976745079881894723584, 2882880, 139984046386112763159840142535527767382602843577165595931249318810236991948760059086304843329475444736, 10883911680, 46730671726813448656774466962980864, 185794560, 63403380965376, 412876800, 729000000, 461794883665920, 8493465600, 17958932119522135058886879224417685745532099088089088, 143343663499379469475676305956380433799785311823017570233599302461682679755530300504376159569382855409664, 84557168640, 573374653997517877902705223825521735199141247292070280934397209846730719022121202017504638277531421638656, 22680000, 45158400, 10644480, 9173994463960286046443283581208347763186259956673124494950355357547691504353939232280074212440502746218496, 12970366926827028480, 95627921278110315577344, 4642275147320176030871715840, 1194393600, 1149371655649416643768760270362731887714054341637701632, 587135645693458306972370149197334256843920637227079967676822742883052256278652110865924749596192175757983744, 3603600, 101559956668416, 4597486622597666575075041081450927550856217366550806528, 1511654400, 104509440, 382511685112441262309376, 51881467707308113920, 150306725297525326584926758194517569752043683130132471725266622178061377607334940381676735896625196994043838464, 3963617280, 247390116249600, 27831388078080, 3283124128353091584, 338228674560, 9619630419041620901435312524449124464130795720328478190417063819395928166869436184427311097384012607618805661696, 19349176320, 39690000, 7388718138654720, 142657607172096, 7761600, 615656346818663737691860001564743965704370926101022604186692084441339402679643915803347910232576806887603562348544, 743178240, 765635325572111542792592866721478475776, 4707826301540010572876842067405749812076766583348025884672, 9850501549098619803069760025035903451269934817616361666987073351061430442874302652853566563721228910201656997576704, 4324320, 466560000, 18831305206160042291507368269622999248307066333392103538688, 989560464998400, 1188422437713965063903159255040, 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638668899051110339730039268171788341141584403339346413280815117579907805398068501051186548957643424348589582990141598478071180156285949971436534975451146054502401815344347416485612070862196155898449074158254055677865859345638678524444433685903374130402278690200241933325554718473710083176923136, 263070294719562525276663269888930943538067857735680, 18274780344940604963129596575744, 2245320000, 73016674573146013526210104629657600, 9590011546570728134804266689924606891259693494266791375439330179969141981654319985009580348611047335402988332404961145534816181286773612460923420672, 40874809539271061742722513162994453833061401813718170449972167525114099545476384067275939133289179158309733311369062302596555530002300798171938238428873347488153716182038234655079172535180553977500740746128259563383414998120875425564443755897815944345745836172815483732835501982317445323323080704, 3809807790243840, 8135123849061138269931183092477135675268652656025388728713216, 57776330904698880, 1282470362637926400, 4875992432640, 138015359728109104398336, 1117670400, 10463951242053391806136963369726580181263718864311851635192874886429209483641954321222640418122029864527291727710479949464718215680589004332016189037791576956967351342601788071700268169006221818240189631008834448226154239518944108944497601509840881752510934060240763835605888507473266002770708660224, 11161546260480, 19685256523047107319352520044058592288178524878038911927831935457689576992794383309011618593207484416, 4303256457514964200980480, 32540495396244553079724732369908542701074610624101554914852864, 1523246532714432760129781760, 669692879491417075592765655662501131600878007315958504652343992731469406953085076558248986759809911329746670573470716765741965803557696277249036098418660925245910485926514436588817162816398196367372136384565404686473871329212422972447846496629816432160699779855408885478776864478289024177325354254336, 2455042955922106402509892272620699364162481534532298592112468526072100347303505916162452569244428117863165013095670053256912942409414044789996395692032, 757632231014400, 810810000]
print(ans[int(input())])
``` | output | 1 | 61,737 | 22 | 123,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,738 | 22 | 123,476 |
Tags: brute force, dp, number theory
Correct Solution:
```
P = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
def power(a, b):
res = 1
while (b):
if (b & 1): res *= a
a *= a
b >>= 1
return res
ans = []
ans.append(1e30)
def solve(pos, n, res):
if (n == 1):
ans[0] = min(ans[0], res)
for i in range(2, 62):
if (n % i == 0):
solve(pos + 1, n / i, res * power(P[pos], i - 1))
n = int(input())
solve(0, n, 1)
print(ans[0])
``` | output | 1 | 61,738 | 22 | 123,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,739 | 22 | 123,478 |
Tags: brute force, dp, number theory
Correct Solution:
```
import math
N = 1003
n = int(input())
dp = [[1000000000000000000 for i in range(N)] for j in range(62)]
primes = []
for i in range(2, 1000):
if(len(primes) > 62):
break
flag = True
for j in range(2, i):
if(i % j == 0):
flag = False
break
if(flag):
primes.append(i)
for i in range(1, n + 1):
dp[0][i] = min(dp[0][i], 2**(i - 1))
primepowers = [[1 for i in range(n)] for j in range(62)]
for i in range(1, 62):
for j in range(1, n):
primepowers[i][j] = primes[i]*primepowers[i][j - 1]
for i in range(1, 62):
for j in range(1, n + 1):
dp[i][j] = dp[i - 1][j]
for k in range(1, int(math.sqrt(j)) + 1):
if(j % k != 0):
continue
dp[i][j] = min(dp[i][j], dp[i - 1][j//k]*primepowers[i][k - 1])
dp[i][j] = min(dp[i][j], dp[i - 1][k]*primepowers[i][j//k - 1])
# for k in range(0, j + 1):
# if(j % (k + 1) != 0):
# continue
# dp[i][j] = min(dp[i][j], dp[i - 1][j//(k + 1)]*primes[i]**k)
print(dp[61][n])
``` | output | 1 | 61,739 | 22 | 123,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,740 | 22 | 123,480 |
Tags: brute force, dp, number theory
Correct Solution:
```
def fact(n):
p = {}
d = 2
while d * d <= n:
while n % d == 0:
p[d-1] = p.get(d-1,0)+1
n //= d
d += 1
if n > 1:
p[n-1] = p.get(n-1,0) + 1
return p
def isPrime(n):
if n % 2 == 0:
return n == 2
d = 3
while d * d <= n and n % d != 0:
d += 2
return d * d > n
m = int(input())
g = fact(m)
#print(g)
z = []
for i,j in g.items():
z.extend([i]*(j))
z.sort(reverse=True)
#print(z)
res = 1
pros = 2
for i in z :
res *= pros**i
pros += 1
while not isPrime(pros) :
pros += 1
x = list(g)
if m > 1 :
y = g[x[0]] - 3
#print(x,y)
if len(g) == 1 and x[0] == 1 and m > 4 :
res = 24
kkk = 5
while y > 0 :
res *= kkk
y -= 1
kkk += 2
print(res)
``` | output | 1 | 61,740 | 22 | 123,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,741 | 22 | 123,482 |
Tags: brute force, dp, number theory
Correct Solution:
```
def main():
def f(x, i, aa):
if (x, i) not in cache:
bb = [q for q in aa if not x % q]
p = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43)[i]
cache[x, i] = min(p ** (d - 1) * f(x // d, i + 1, bb) for d in bb)
return cache[x, i]
n, cache = int(input()), dict.fromkeys(((1, i) for i in range(14)), 1)
print(f(n, 0, [d for d in range(2, n + 1) if not n % d]))
if __name__ == '__main__':
main()
``` | output | 1 | 61,741 | 22 | 123,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,742 | 22 | 123,484 |
Tags: brute force, dp, number theory
Correct Solution:
```
import math
p = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43]
def solve(n, pos):
ans = [1, 10 ** 18][n > 1] # si n>1 ans = 10**18 sino ans =1 dado que solo se necesita 1 divisor y solo 1 cumple tener un unico divisor
for exp in range(2, n + 1):
if n % exp == 0: # analizamos los posibles exponentes como se calculo matematicamente no tiene sentido pregunta por un exponente que no cumpla que (exp+1)|n
ans = min(ans, p[pos] ** (exp - 1) * solve(n // exp, pos + 1)) # ans queda siendo la menor de las respuestas hasta ahora o la nueva respuesta
# calculada mediante el primo que se esta analizando elevado al exponente -1
# * solve en los divisores restantes n/exp y cambiando la pos para analiar el primo siguiente
return ans
inp = int(input())
print(solve(inp, 0))
``` | output | 1 | 61,742 | 22 | 123,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,743 | 22 | 123,486 |
Tags: brute force, dp, number theory
Correct Solution:
```
def delit(a) :
res = []
i = 1
while i * i < a + 1 :
if a % i == 0 :
temp = [i, a // i]
res.append(temp)
i += 1
return res
n = int(input())
npr = [2,3,5,7,11,13,17,19,23,29]
fact = delit(n)
flen = len(fact)
while True :
for j in range(flen) :
zero = 1
if not (1 in fact[j]) :
zero = 0
n1 = list(fact[j])
factj = delit(n1.pop())
for k in range(len(factj)) :
factj[k].extend(n1)
factj[k].sort()
fact.append(factj[k])
flen = len(fact)
if zero == 1:
break
for ik in fact :
ik.sort(reverse=True)
if ik[-1] == 1 and len(ik) > 2:
ik.pop()
j =[0]
resj = []
for i in fact :
if i != j :
resj.append(i)
j=i
minc = int(1e18)
for i in resj :
j = 0
ss = 1
for ii in i:
ss *= npr[j] ** (ii-1)
if ss > int(1e18) :
ss = int(1e18) + 1
break
j += 1
#print(ss)
minc = min(minc,ss)
#print(*resj)
print(minc)
``` | output | 1 | 61,743 | 22 | 123,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12 | instruction | 0 | 61,744 | 22 | 123,488 |
Tags: brute force, dp, number theory
Correct Solution:
```
def isPrime(i) -> bool:
t = 2
while t * t <= i:
if i % t == 0:
return False
t += 1
return i > 1
MAX_ANS = 10 ** 18
d = int(input())
bound = 25
p = [-1]
for i in range(100000):
if(isPrime(i)):
p.append(i)
if len(p) > bound:
break
dp = [[MAX_ANS + 1]*(1 + bound) for _ in range(1 + d)]
for i in range(1 + bound):
dp[1][i] = 1
for i in range(2, 1 + d):
for k in range(1, 1 + bound):
dp[i][k] = dp[1][k - 1] * p[k]**(i - 1)
t = 2
while t * t <= i:
if i % t == 0:
for k in range(1, 1 + bound):
dp[i][k] = min(dp[i][k], dp[i//t][k - 1] * p[k]**(t - 1), dp[t][k - 1] * p[k]**(i//t - 1))
t += 1
print(min(dp[d]))
``` | output | 1 | 61,744 | 22 | 123,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
a = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
def rec(n, pos):
if n == 1:
return 1
cur = 10**20
for i in range(2, n + 1):
if n % i == 0:
cur = min(cur, a[pos] ** (i - 1) * rec(n // i, pos + 1))
return cur
n = int(input())
print(rec(n, 0))
``` | instruction | 0 | 61,745 | 22 | 123,490 |
Yes | output | 1 | 61,745 | 22 | 123,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
def delit(a) :
res = []
i = 1
while i * i < a + 1 :
if a % i == 0 :
temp = [i, a // i]
res.append(temp)
i += 1
return res
n = int(input())
npr = [2,3,5,7,11,13,17,19,23,29]
# первичное разложение на парные делители
fact = delit(n)
flen = len(fact)
while True :
for j in range(flen) :
zero = 1
# если в очередном списке делителей
#есть "1", пропустить
if 1 not in fact[j] :
zero = 0
n1 = list(fact[j])
# "вытаскиваем" последний делитель из списка
# и ищем все попарные разложения на множители
factj = delit(n1.pop())
# каждую полученную пару
#объединяем со списком
for k in range(len(factj)) :
factj[k].extend(n1)
factj[k].sort()
fact.append(factj[k])
flen = len(fact)
# если zero не изменилось во всех списках,
# все варианты найдены
if zero == 1:
break
# удаляем 1 из всех вариантов
for ik in fact :
ik.sort(reverse=True)
if ik[-1] == 1 :
ik.pop()
fact.sort()
# удаляем одинаковые списки
j =[0]
resj = []
for i in fact :
if i != j :
resj.append(i)
j=i
#print(*resj)
######################################
minc = int(1e18)
for i in resj :
j = 0
ss = 1
for ii in i:
ss *= npr[j] ** (ii-1)
if ss > int(1e18) :
ss = int(1e18) + 1
break
j += 1
minc = min(minc,ss)
print(minc)
``` | instruction | 0 | 61,746 | 22 | 123,492 |
Yes | output | 1 | 61,746 | 22 | 123,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
def is_prime( n ):
i = 2
while i * i <= n:
if n % i == 0:
return ( False )
i += 1
return ( True )
m = {}
p = []
i = 2
while len(p) < 12:
if is_prime(i):
p.append(i)
i += 1
INF = 10**18
m = {}
ans = INF
def f( a, b, div ):
global ans
if b >= len(p):
return
if div == n:
ans = min( ans, a )
return
for j in range( 1, 60 ):
if a <= ans//p[b]:
a *= p[b]
f( a, b+1, div*(j+1) )
else:
break
n = int( input() )
#for i in range( len(p) ):
# f( p[i], i )
f( 1, 0, 1 )
print( ans )
``` | instruction | 0 | 61,747 | 22 | 123,494 |
Yes | output | 1 | 61,747 | 22 | 123,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
y=[2,3,5,7,11,13,17,23,29,31]
def dfs(n,pos):
v=[1,10**18][n>1]
for i in range(2,n+1):
if n%i==0:
v=min(v,y[pos]**(i-1)*dfs(n//i,pos+1))
return v
print(dfs(n,0))
``` | instruction | 0 | 61,748 | 22 | 123,496 |
Yes | output | 1 | 61,748 | 22 | 123,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
def get_factors(n):
# Find all prime factors of n, including multiples
result = []
factor = 2
while (n != 1):
if ((n % factor) == 0):
n = n // factor
result.append(factor)
else:
factor += 1
return result
factors = get_factors(n)
primes = [2,3,5,7,11,13,17,19,23,29]
answer = 1
for (factor, prime) in zip(reversed(factors), primes):
answer *= prime**(factor-1)
print(answer)
``` | instruction | 0 | 61,749 | 22 | 123,498 |
No | output | 1 | 61,749 | 22 | 123,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
p = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43]
it = 0
def solve(n, pos):
global it
it+=1
# si n>1 ans = 10**18 sino ans =1 dado que solo se necesita 1 divisor y
# solo 1 cumple tener un unico divisor
ans = 10 ** 18 if n > 1 else 1
for exp in range(2, n + 1):
# analizamos los posibles exponentes como se calculo matematicamente no
# tiene sentido pregunta por un exponente que no cumpla que (exp+1)|n
if n % exp == 0:
# ans queda siendo la menor de las respuestas hasta ahora o la
# nueva respuesta se calcula mediante el primo que se esta
# analizando elevado al exponente -1 * solve en los divisores
# restantes n/exp cambiando la pos para analiar el primo siguiente
ans = min(ans, p[pos] ** (exp - 1) * solve(n // exp, pos + 1))
return ans
inp = int(input())
print(solve(inp, 0))
print(it)
``` | instruction | 0 | 61,750 | 22 | 123,500 |
No | output | 1 | 61,750 | 22 | 123,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
from functools import reduce
from operator import mul
primes = [2,3,5,7,11,13,17,19,23,29,31,37]
s, c = int(input()), []
for p in range(2, s):
while s % p == 0:
c.append(p)
s //= p
if p*p > s:
if s > 1: c.append(s)
break
print(sorted(c))
print(reduce(lambda p, e: p*e[0]**(e[1]-1), zip(primes, sorted(c, reverse=True)), 1))
``` | instruction | 0 | 61,751 | 22 | 123,502 |
No | output | 1 | 61,751 | 22 | 123,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000).
Output
Output the smallest positive integer with exactly n divisors.
Examples
Input
4
Output
6
Input
6
Output
12
Submitted Solution:
```
from itertools import combinations
#import numpy as np
def prime_divisors(n):
a = []
i = 2
while i*i <= n:
if n % i == 0:
a.append(i)
n //= i
else:
i += 1
a.append(n)
a.sort(reverse=True)
return a
n = int(input())
npr = [2,3,5,7,11,13,17,19,23,29]
# поиск всех списков-вариантов множителей "n"
res =[]
# разложение на простые делители "n"
delit = prime_divisors(n)
for i in range(2,len(delit)) :
# все комбинации перебора делителей,
# начиная с двух делителей до всех делителей
data = list(combinations(delit, i))
for j in data:
jj = list(j)
# здесь пытался применить prod от numpy,
# но ниже при нахождении "ss" шло
# переполнение (????)
temp = 1
for k in jj:
temp *= k
temp1 = n//temp
# добавление к набору делителей
# частного от "n" к произведению делителей
jj.append(temp1)
jj.sort(reverse=True)
# добавление набора из произведения
# делителей и частного от "n" к произведению
# и удаление одинаковых списков-наборов
if jj not in res:
res.append(jj)
ttemp = sorted([temp,temp1],reverse=True)
if ttemp not in res:
res.append(ttemp)
tt =[n,1]
res.append(tt)
######################################
minc = int(1e18) + 1
for i in res :
j = 0
ss = 1
for ii in i:
ss *= npr[j] ** (ii-1)
if ss > int(1e18) + 1 :
ss = int(1e18) + 1
break
j += 1
minc = min(minc,ss)
print(minc)
``` | instruction | 0 | 61,752 | 22 | 123,504 |
No | output | 1 | 61,752 | 22 | 123,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m, and a list of n distinct integers between 0 and m - 1.
You would like to construct a sequence satisfying the properties:
* Each element is an integer between 0 and m - 1, inclusive.
* All prefix products of the sequence modulo m are distinct.
* No prefix product modulo m appears as an element of the input list.
* The length of the sequence is maximized.
Construct any sequence satisfying the properties above.
Input
The first line of input contains two integers n and m (0 ≤ n < m ≤ 200 000) — the number of forbidden prefix products and the modulus.
If n is non-zero, the next line of input contains n distinct integers between 0 and m - 1, the forbidden prefix products. If n is zero, this line doesn't exist.
Output
On the first line, print the number k, denoting the length of your sequence.
On the second line, print k space separated integers, denoting your sequence.
Examples
Input
0 5
Output
5
1 2 4 3 0
Input
3 10
2 9 1
Output
6
3 9 2 9 8 0
Note
For the first case, the prefix products of this sequence modulo m are [1, 2, 3, 4, 0].
For the second case, the prefix products of this sequence modulo m are [3, 7, 4, 6, 8, 0]. | instruction | 0 | 61,948 | 22 | 123,896 |
Tags: constructive algorithms, dp, graphs, math, number theory
Correct Solution:
```
import math
def gcdExtended(a, b):
# Base Case
if a == 0 :
return b, 0, 1
gcd, x1, y1 = gcdExtended(b%a, a)
# Update x and y using results of recursive
# call
x = y1 - (b//a) * x1
y = x1
return gcd, x, y
def rev_elem(x, m):
return (gcdExtended(x, m)[1] % m + m) % m
n, m = map(int, input().split())
a = []
if n > 0:
a = [int(i) for i in input().split()]
banned = [False] * (m + 5)
for i in a:
banned[i] = True
cycle = [[] for i in range(m + 5)]
d, dp, p = [], [], []
for i in range(m):
cycle[math.gcd(m, i)].append(i)
cycle = [[i for i in j if not banned[i]] for j in cycle]
d = [i for i in range(1, m + 1) if m % i == 0]
dp = [len(cycle[i]) for i in d]
p = [-1 for i in d]
ans, lst = -1, -1
for i in range(len(d)):
if dp[i] > ans:
ans, lst = dp[i], i
for j in range(i + 1, len(d)):
if d[j] % d[i] != 0 or dp[j] > dp[i] + len(cycle[d[j]]):
continue
dp[j] = dp[i] + len(cycle[d[j]])
p[j] = i
print(ans)
pos, dpos, pref = [], [], []
cur = lst
while cur != -1:
dpos.append(d[cur])
cur = p[cur]
dpos.reverse()
for i in dpos:
pref += cycle[i]
cur = 1
for i in pref:
ad = 1
if math.gcd(i, m) != math.gcd(cur, m):
ad = ((cur * math.gcd(i, m) // math.gcd(cur, math.gcd(i, m))) // cur) % m
ncur = (cur * ad) % m
ad *= i // math.gcd(ncur, m) * (rev_elem(ncur // math.gcd(ncur, m), m // math.gcd(ncur, m)))
ad %= m
cur = (cur * ad) % m
pos.append(ad)
print(*pos)
``` | output | 1 | 61,948 | 22 | 123,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m, and a list of n distinct integers between 0 and m - 1.
You would like to construct a sequence satisfying the properties:
* Each element is an integer between 0 and m - 1, inclusive.
* All prefix products of the sequence modulo m are distinct.
* No prefix product modulo m appears as an element of the input list.
* The length of the sequence is maximized.
Construct any sequence satisfying the properties above.
Input
The first line of input contains two integers n and m (0 ≤ n < m ≤ 200 000) — the number of forbidden prefix products and the modulus.
If n is non-zero, the next line of input contains n distinct integers between 0 and m - 1, the forbidden prefix products. If n is zero, this line doesn't exist.
Output
On the first line, print the number k, denoting the length of your sequence.
On the second line, print k space separated integers, denoting your sequence.
Examples
Input
0 5
Output
5
1 2 4 3 0
Input
3 10
2 9 1
Output
6
3 9 2 9 8 0
Note
For the first case, the prefix products of this sequence modulo m are [1, 2, 3, 4, 0].
For the second case, the prefix products of this sequence modulo m are [3, 7, 4, 6, 8, 0]. | instruction | 0 | 61,949 | 22 | 123,898 |
Tags: constructive algorithms, dp, graphs, math, number theory
Correct Solution:
```
maxn = 200500
g = [[] for i in range(maxn)] # Array de array g[i][j] en el cual la posicion i representa gcd(a,m)
# y en el array [j] se guardan todos los ¨a¨ con gcd comunes
dp = [0 for i in range(maxn)] # guarda la profundidad mayor para cada nodo
ban = [0 for i in range(maxn)] # Array donde se guardan los productos prefijos mod m prohibidos
nxt = [0 for i in range(maxn)] # Array que contiene la proxima posicion ¨i¨ a leer en el array g
m = 0 #
def gcd(a, b): #maximo comun divisor de a y b
if (b == 0):
return a
return gcd(b, a % b)
def exgcd(a, b): # Algoritmo de Euclides extendido (recursivo),
if b == 0: # el cual devuelve los menores x, y
x = 1 # tales que se cumple que
y = 0 # a * x + b * y = gcd(a, b)
return a,x,y #
ret, x, y = exgcd(b, a % b) #
tmp = x #
x = y #
y = tmp -int(a / b) * y #
return ret, x, y #
def f(a, b): # resuelve ecuacion de congruencia lineal
r1, x, y = exgcd(a, m) # haciendo uso del algoritmo de Euclides extendido
if b % r1 != 0: # si !=0
return -1 # la ecuacion no tiene solucion
x = ((x + m) % m * int(b / r1)) % m # solucion de la ecuacion
return x
def dfs(p): # Algoritmo DFS (Depth First Search) utilizado para encontrar
if dp[p]!=0: # la secuencia de mayor longitud
return #
res = 0 #
for i in range(p*2,m,p): # recorrido por los nodos
if (len(g[i])>0): #
dfs(i) #
if (dp[i] > res): #
res = dp[i] #
nxt[p] = i # guardar indice i para despues recorrer desde la raiz (1) todos los nodos que
if len(g[p])>0: # se encuentran en el camino mas largo e ir accediendo a los elementos asociados a i
res+= len(g[p]) #
dp[p] = res #
#print("teclee n y m, dar enter e introducir 2da linea listado separados por coma ")
a, b = input().split(' ')
n = int(a)
m = int(b)
if n>0:
lista = input().split(' ')
for i in range(1,n+1):
x = int(lista[i - 1])
ban[x] = 1 #marcar elem x como prohibido
for i in range(1,m):
if ban[i] == 0:
d = gcd(i, m)
g[d].append(i) #guardando los numeros con gcd =d si no es un elemento prohibido
dfs(1)
ban0 = 0
if ban[0] == ban0: #verificar si 0 no es elemento prohibido, en ese caso se añade a la secuencia
ban0=1
print(dp[1] + ban0) #en dp[1] se encuentra la longitud mayor, [1] (origen del recorrido)
u = 1
x = 1
sec = ""
while (u != 0): # recorrer nxt para obtener los elementos que estan en la misma clase modulo m
for i in range(0,len(g[u])): #
v = g[u][i]
print(str(f(x, v)) + " ", end=" ") #solucion de la ecuacion
x = v
u = nxt[u] # obtener siguiente indice u para recorrer todos los elementos i asociados a u (gcd(i,m)=u)
if ban0==1 :
print("0")
``` | output | 1 | 61,949 | 22 | 123,899 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N.
We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.
* Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.
In how many choices of K will N become 1 in the end?
Constraints
* 2 \leq N \leq 10^{12}
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of choices of K in which N becomes 1 in the end.
Examples
Input
6
Output
3
Input
3141
Output
13
Input
314159265358
Output
9 | instruction | 0 | 62,045 | 22 | 124,090 |
"Correct Solution:
```
def divisor(n):
res = []
i = 1
while i*i <= n:
if not n % i:
res.append(i)
if (i*i != n): res.append(n//i)
i += 1
return res
N = int(input())
ans = 0
for d in divisor(N):
if d == 1: continue
n = N
while not n % d:
n //= d
if n%d == 1: ans += 1
print(ans + len(divisor(N-1))-1)
``` | output | 1 | 62,045 | 22 | 124,091 |
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