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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The Little Elephant loves numbers.
He has a positive integer x. The Little Elephant wants to find the number of positive integers d, such that d is the divisor of x, and x and d have at least one common (the same) digit in their decimal representations.
Help the Little Elephant to find the described number.
Input
A single line contains a single integer x (1 ≤ x ≤ 109).
Output
In a single line print an integer — the answer to the problem.
Examples
Input
1
Output
1
Input
10
Output
2 | instruction | 0 | 92,929 | 22 | 185,858 |
Tags: implementation
Correct Solution:
```
def readln(inp=None): return tuple(map(int, (inp or input()).split()))
x, = readln()
ans = 0
d = 1
dig = set(list(str(x)))
while d * d <= x:
if x % d == 0:
for var in set((d, x // d)):
ans += len(set(list(str(var))).intersection(dig)) > 0
d += 1
print(ans)
``` | output | 1 | 92,929 | 22 | 185,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Little Elephant loves numbers.
He has a positive integer x. The Little Elephant wants to find the number of positive integers d, such that d is the divisor of x, and x and d have at least one common (the same) digit in their decimal representations.
Help the Little Elephant to find the described number.
Input
A single line contains a single integer x (1 ≤ x ≤ 109).
Output
In a single line print an integer — the answer to the problem.
Examples
Input
1
Output
1
Input
10
Output
2 | instruction | 0 | 92,934 | 22 | 185,868 |
Tags: implementation
Correct Solution:
```
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
n = int(input())
list_ = map(str, factors(n))
n = str(n)
cont=0
for el in list_:
flag = False
for i in el:
if i in n:
flag = True
if flag==True:
cont+=1
print(cont)
``` | output | 1 | 92,934 | 22 | 185,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Elephant loves numbers.
He has a positive integer x. The Little Elephant wants to find the number of positive integers d, such that d is the divisor of x, and x and d have at least one common (the same) digit in their decimal representations.
Help the Little Elephant to find the described number.
Input
A single line contains a single integer x (1 ≤ x ≤ 109).
Output
In a single line print an integer — the answer to the problem.
Examples
Input
1
Output
1
Input
10
Output
2
Submitted Solution:
```
import math
import sys
n = int(input())
arr = [0,0,0,0,0,0,0,0,0,0]
number = str(n)
counter = 0
x = math.sqrt(n)+1
divisores = []
for i in range(len(number)):
arr[int(number[i])]=1
for i in range(1,int(x)):
if n%i==0:
d = str(i)
for j in range(len(d)):
if arr[int(d[j])]==1:
divisores.append(d)
break
inv_d = str(n//i)
for j in range(len(inv_d)):
if arr[int(inv_d[j])]==1:
divisores.append(inv_d)
div = list(dict.fromkeys(divisores))
print(len(div))
``` | instruction | 0 | 92,938 | 22 | 185,876 |
Yes | output | 1 | 92,938 | 22 | 185,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Elephant loves numbers.
He has a positive integer x. The Little Elephant wants to find the number of positive integers d, such that d is the divisor of x, and x and d have at least one common (the same) digit in their decimal representations.
Help the Little Elephant to find the described number.
Input
A single line contains a single integer x (1 ≤ x ≤ 109).
Output
In a single line print an integer — the answer to the problem.
Examples
Input
1
Output
1
Input
10
Output
2
Submitted Solution:
```
import math
x = int(input())
divisors = set()
for d in range(1, math.ceil(math.sqrt(x)) + 1):
if x % d == 0:
divisors.add(d)
divisors.add(x / d)
def common_digits(a, b):
a = set(str(a))
b = set(str(b))
return not a.isdisjoint(b)
print(sum(1 for d in divisors if common_digits(d, x)))
``` | instruction | 0 | 92,941 | 22 | 185,882 |
No | output | 1 | 92,941 | 22 | 185,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,975 | 22 | 185,950 |
Tags: brute force
Correct Solution:
```
x, y, xx = map(int, input().split())
cnt = 0
while max(x, y) < xx:
if x < y:
x, y = y, x
l, r, m = 0, 4 * 10 ** 18, 0
while r > l + 1:
m = (l + r) // 2;
if x >= m * x + y:
l = m
else:
r = m
if r == 4e18:
print(-1)
exit(0)
cnt += r
temp = x
x = r * x + y
y = temp
print(cnt)
``` | output | 1 | 92,975 | 22 | 185,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,976 | 22 | 185,952 |
Tags: brute force
Correct Solution:
```
x,y,m = map(int,input().split())
x,y = min(x,y),max(x,y)
if y >= m:
print(0)
elif y <= 0:
print(-1)
else:
cnt=0
if x<0:
cnt += (-x)//y + 1
x = x + cnt*y
while y<m:
x,y = y,x+y
cnt += 1
print(cnt)
``` | output | 1 | 92,976 | 22 | 185,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,977 | 22 | 185,954 |
Tags: brute force
Correct Solution:
```
x, y, m = [int(x) for x in input().split()]
x, y = min(x, y), max(x, y)
if y >= m:
print(0)
exit(0)
if y <= 0:
print(-1)
exit(0)
c = 0
if x < 0 and y > 0:
if m <= 0:
c = (m - x) // y
else:
c = -x // y
x += c * y
while x < m and y < m:
xx = x + y
if xx <= min(x, y):
print(-1)
exit(0)
x, y = xx, max(x, y)
c += 1
print(c)
``` | output | 1 | 92,977 | 22 | 185,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,978 | 22 | 185,956 |
Tags: brute force
Correct Solution:
```
x,y,m = map(int,input().strip().split())
x,y = min(x,y),max(x,y)
if y>= m:
print(0)
elif y<=0:
print(-1)
elif x>=0:
i = 0
while y<m:
x,y = min(x+y,y) , max(x+y,y)
i += 1
print(i)
else:
n = (y-x)//y
if n*y + x < y:
n += 1
if n*y + x >= m:
print(n)
else:
i = n
x,y = y , n*y + x
while y<m:
x,y = min(x+y,y) , max(x+y,y)
i += 1
print(i)
``` | output | 1 | 92,978 | 22 | 185,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,979 | 22 | 185,958 |
Tags: brute force
Correct Solution:
```
x,y,m=map(int,input().split())
x,y=min(x,y),max(x,y)
if y>=m:s=0
elif x+y<=x:s=-1
else:
s=(y-x+y-1)//y
x+=y*s
while max(x,y)<m:
if x<y:x+=y
else:y+=x
s+=1
print(s)
``` | output | 1 | 92,979 | 22 | 185,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,980 | 22 | 185,960 |
Tags: brute force
Correct Solution:
```
import math
a, b, c = map (int, input().split())
if (max (a, b) >= c):
print (0)
raise SystemExit
if (a <= 0 and b <= 0) :
print (-1)
raise SystemExit
tot = 0
if ((a <= 0 and b > 0) or (b <= 0 and a > 0)) :
add = max (a, b)
menor = min (a, b)
adicionar = math.ceil(-menor / add)
tot = adicionar
if (min (a, b) == a) :
a += add * adicionar
else :
b += add * adicionar
times = 500
while (times > 0) :
times -= 1
if (max(a, b) >= c) :
print (tot)
raise SystemExit
tot += 1
add = a + b
if (min (a, b) == a) :
a = add
else :
b = add
print (-1)
``` | output | 1 | 92,980 | 22 | 185,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,981 | 22 | 185,962 |
Tags: brute force
Correct Solution:
```
x, y, m = [int(x) for x in input().split()]
if (x >= m) or (y >= m):
print(0)
exit()
if (x <= 0) and (y <= 0):
print(-1)
exit()
res = 0
if x < 0:
res -= x // y
x %= y
if y < 0:
res -= y // x
y %= x
#print(x, y, res)
while (x < m) and (y < m):
if x < y:
x = x + y
else:
y = x + y
res += 1
print(res)
``` | output | 1 | 92,981 | 22 | 185,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations. | instruction | 0 | 92,982 | 22 | 185,964 |
Tags: brute force
Correct Solution:
```
#_________________ Mukul Mohan Varshney _______________#
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
#define function
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def btod(n):
return int(n,2)
def dtob(n):
return bin(n).replace("0b","")
# Driver Code
def solution():
#for _ in range(Int()):
x,y,m=Mint()
ans=0
if(x>=m or y>=m):
print(0)
elif(x<=0 and y<=0):
print(-1)
else:
if(x>0 and y<0):
ans=(x-y-1)//x
y+=ans*x
elif(y>0 and x<0):
ans=(y-x-1)//y
x+=ans*y
while(x<m and y<m):
t=x+y
if(x<y):
x=t
else:
y=t
ans+=1
print(ans)
#Call the solve function
if __name__ == "__main__":
solution()
``` | output | 1 | 92,982 | 22 | 185,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call a set of positive integers a beautiful if the following condition fulfills: for any prime p, if <image>, then <image>. In other words, if one number from the set is divisible by prime p, then at least half of numbers from the set is divisible by p.
Your task is to find any beautiful set, where the number of elements is equal to k and each element doesn't exceed 2k2.
Input
The first line contains integer k (10 ≤ k ≤ 5000) that shows how many numbers the required beautiful set should have.
Output
In the first line print k space-separated integers that are a beautiful set. If there are multiple such sets, you are allowed to print any of them.
Examples
Input
10
Output
16 18 24 27 36 48 54 72 108 144
Submitted Solution:
```
n = int(input())
for i in range(n):
print(2, end = " ")
``` | instruction | 0 | 93,007 | 22 | 186,014 |
No | output | 1 | 93,007 | 22 | 186,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call a set of positive integers a beautiful if the following condition fulfills: for any prime p, if <image>, then <image>. In other words, if one number from the set is divisible by prime p, then at least half of numbers from the set is divisible by p.
Your task is to find any beautiful set, where the number of elements is equal to k and each element doesn't exceed 2k2.
Input
The first line contains integer k (10 ≤ k ≤ 5000) that shows how many numbers the required beautiful set should have.
Output
In the first line print k space-separated integers that are a beautiful set. If there are multiple such sets, you are allowed to print any of them.
Examples
Input
10
Output
16 18 24 27 36 48 54 72 108 144
Submitted Solution:
```
primes = [2,3,5,7,11,13,17,19,23,29,31]
def solve(k):
limit = 2 * k * k
cnt = [0 for i in range(len(primes))]
num_by_mask = [[] for i in range(1 << len(primes))]
answer = set()
def good(mask, used):
if len(num_by_mask[mask]) == 0:
return False
for i in range(used + 1):
if mask & (1 << i):
if cnt[i] == 0 or 2 * cnt[i] <= k:
return False
return True
def remove(mask, used):
for i in range(used + 1):
if mask & (1 << i):
cnt[i] -= 1
answer.remove(num_by_mask[mask][0])
num_by_mask[mask].pop(0)
numbers = []
def check_answer():
for x in primes:
cnt = 0
for d in answer:
if d % x == 0:
cnt += 1
if cnt > 0 and cnt * 2 < k:
return False
return True
for used in range(len(primes)):
available_now = []
for i in range(1, 30):
if primes[used] ** i > limit:
break
available_now.append(primes[used] ** i)
for number in numbers:
if number * primes[used] <= limit:
available_now.append(number * primes[used])
for number in available_now:
mask = 0
for i in range(used + 1):
if number % primes[i] == 0:
mask |= 1 << i
cnt[i] += 1
num_by_mask[mask].append(number)
answer.add(number)
numbers.extend(available_now)
masks = [i for i in range(1, 1 << used)]
masks = sorted(masks, key=lambda t: bin(t).count('1'), reverse=True)
change = True
while change and len(answer) > k - (k & 1):
change = False
for mask in masks:
if good(mask, used):
remove(mask, used)
change = True
# print(check_answer(), used, len(answer), answer)
for i in range(used + 1):
if cnt[i] > 0 and cnt[i] * 2 < k:
break
else:
if len(answer) == k:
break
# print(check_answer(), len(answer), k, answer)
assert(check_answer())
return answer, numbers, used
k = int(input())
answer, numbers, used = solve(k - (k & 1))
if k & 1:
cnt = [0 for i in range(used + 1)]
for x in answer:
for i in range(used + 1):
if x % primes[i] == 0:
cnt[i] += 1
need_primes = []
for i in range(used + 1):
if cnt[i] * 2 < k:
need_primes.append(primes[i])
for x in numbers:
if x in answer:
continue
for p in need_primes:
if x % p:
break
else:
answer.add(x)
break
'''
cnt = [0 for i in range(used + 1)]
for x in answer:
for i in range(used + 1):
if x % primes[i] == 0:
cnt[i] += 1
print(cnt)
'''
print(' '.join(map(str, sorted(answer))))
``` | instruction | 0 | 93,008 | 22 | 186,016 |
No | output | 1 | 93,008 | 22 | 186,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call a set of positive integers a beautiful if the following condition fulfills: for any prime p, if <image>, then <image>. In other words, if one number from the set is divisible by prime p, then at least half of numbers from the set is divisible by p.
Your task is to find any beautiful set, where the number of elements is equal to k and each element doesn't exceed 2k2.
Input
The first line contains integer k (10 ≤ k ≤ 5000) that shows how many numbers the required beautiful set should have.
Output
In the first line print k space-separated integers that are a beautiful set. If there are multiple such sets, you are allowed to print any of them.
Examples
Input
10
Output
16 18 24 27 36 48 54 72 108 144
Submitted Solution:
```
k = int(input())
limit = 2 * k * k
primes = [2,3,5,7,11,13,17,19]
cnt = [0 for i in range(len(primes))]
num_by_mask = [[] for i in range(1 << len(primes))]
def good(mask, used):
if len(num_by_mask[mask]) == 0:
return False
for i in range(used + 1):
if mask & (1 << i):
if cnt[i] == 0 or 2 * cnt[i] <= k:
return False
return True
def remove(mask, used):
for i in range(used + 1):
if mask & (1 << i):
cnt[i] -= 1
answer.remove(num_by_mask[mask][0])
num_by_mask[mask].pop(0)
numbers = []
answer = []
for used in range(len(primes)):
available_now = []
for i in range(1, 30):
if primes[used] ** i > limit:
break
available_now.append(primes[used] ** i)
for number in numbers:
if number * primes[used] <= limit:
available_now.append(number * primes[used])
for number in available_now:
mask = 0
for i in range(used + 1):
if number % primes[i] == 0:
mask |= 1 << i
cnt[i] += 1
num_by_mask[mask].append(number)
answer.append(number)
numbers.extend(available_now)
masks = [i for i in range(1, 1 << used)]
masks = sorted(masks, key=lambda t: bin(t).count('1'), reverse=True)
change = True
while change and len(answer) > k:
change = False
for mask in masks:
if good(mask, used):
remove(mask, used)
change = True
# print(used, len(answer), answer)
for i in range(used + 1):
if cnt[i] > 0 and cnt[i] * 2 < k:
break
else:
if len(answer) == k:
break
def check_answer():
for x in primes:
cnt = 0
for d in answer:
if d % x == 0:
cnt += 1
if cnt > 0 and cnt * 2 < k:
return False
return True
#print(check_answer(), len(answer), k, answer)
print(' '.join(map(str, answer)))
``` | instruction | 0 | 93,009 | 22 | 186,018 |
No | output | 1 | 93,009 | 22 | 186,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call a set of positive integers a beautiful if the following condition fulfills: for any prime p, if <image>, then <image>. In other words, if one number from the set is divisible by prime p, then at least half of numbers from the set is divisible by p.
Your task is to find any beautiful set, where the number of elements is equal to k and each element doesn't exceed 2k2.
Input
The first line contains integer k (10 ≤ k ≤ 5000) that shows how many numbers the required beautiful set should have.
Output
In the first line print k space-separated integers that are a beautiful set. If there are multiple such sets, you are allowed to print any of them.
Examples
Input
10
Output
16 18 24 27 36 48 54 72 108 144
Submitted Solution:
```
def solve(k):
limit = 2 * k * k
primes = [2,3,5,7,11,13,17,19,23,29,31]
cnt = [0 for i in range(len(primes))]
num_by_mask = [[] for i in range(1 << len(primes))]
answer = set()
def good(mask, used):
if len(num_by_mask[mask]) == 0:
return False
for i in range(used + 1):
if mask & (1 << i):
if cnt[i] == 0 or 2 * cnt[i] <= k:
return False
return True
def remove(mask, used):
for i in range(used + 1):
if mask & (1 << i):
cnt[i] -= 1
answer.remove(num_by_mask[mask][0])
num_by_mask[mask].pop(0)
numbers = []
def check_answer():
for x in primes:
cnt = 0
for d in answer:
if d % x == 0:
cnt += 1
if cnt > 0 and cnt * 2 < k:
return False
return True
for used in range(len(primes)):
available_now = []
for i in range(1, 30):
if primes[used] ** i > limit:
break
available_now.append(primes[used] ** i)
for number in numbers:
if number * primes[used] <= limit:
available_now.append(number * primes[used])
for number in available_now:
mask = 0
for i in range(used + 1):
if number % primes[i] == 0:
mask |= 1 << i
cnt[i] += 1
num_by_mask[mask].append(number)
answer.add(number)
numbers.extend(available_now)
masks = [i for i in range(1, 1 << used)]
masks = sorted(masks, key=lambda t: bin(t).count('1'), reverse=True)
change = True
while change and len(answer) > k - (k & 1):
change = False
for mask in masks:
if good(mask, used):
remove(mask, used)
change = True
# print(check_answer(), used, len(answer), answer)
for i in range(used + 1):
if cnt[i] > 0 and cnt[i] * 2 < k:
break
else:
if len(answer) == k:
break
# print(check_answer(), len(answer), k, answer)
assert(check_answer())
return answer, numbers
k = int(input())
answer, numbers = solve(k - (k & 1))
if k & 1:
for x in numbers:
if x not in answer:
answer.add(x)
break
print(' '.join(map(str, sorted(answer))))
``` | instruction | 0 | 93,010 | 22 | 186,020 |
No | output | 1 | 93,010 | 22 | 186,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4 | instruction | 0 | 93,163 | 22 | 186,326 |
Tags: brute force, implementation, math, number theory
Correct Solution:
```
def solve(n, m, a):
if n == 0:
return 0, 1
if n == 1:
return a[0], 1
d = (a[1]-a[0]) % m
if d < 0: d += m
st = set(a)
cnt = 0
for v in a:
cnt += ((v + d) % m) in st
cnt = n-cnt
d = (d * pow(cnt, m-2, m)) % m
now = a[0]
while (now + m - d) % m in st:
now = (now + m - d) % m
for i in range(n):
if (now + i*d) % m not in st:
return -1, -1
return now, d
m, n = map(int, input().split())
a = list(map(int, input().split()))
if n * 2 > m:
st = set(a)
b = [i for i in range(m) if i not in st]
f, d = solve(len(b), m, b)
f = (f + d * (m-n)) % m
else:
f, d = solve(n, m, a)
if f < 0 or d < 0:
print(-1)
else:
print(f, d)
# Made By Mostafa_Khaled
``` | output | 1 | 93,163 | 22 | 186,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4 | instruction | 0 | 93,164 | 22 | 186,328 |
Tags: brute force, implementation, math, number theory
Correct Solution:
```
from collections import defaultdict
dic = defaultdict(int)
def pow_mod(a,b,c):
n = 1
m = a
v = b
if a == 1:
return 1
while v > 0:
if v%2 == 1:
n *= m%c
n %= c
m *= m
m %= c
v //= 2
return n%c
def pow(a,b):
n = 1
m = a
v = b
if a == 1:
return 1
while v > 0:
if v%2 == 1:
n *= m
m *= m
v //= 2
return n
def tonelli_shanks(a,p):
if a == 0:
return 0
q = p-1
s = 0
while q%2 == 0:
s += 1
q //= 2
z = 2
while pow_mod(z,(p-1)//2,p) != p-1:
z += 1
m = s
c = pow_mod(z,q,p)
t = pow_mod(a,q,p)
r = pow_mod(a,(q+1)//2,p)
while m > 1:
if pow_mod(t,pow(2,m-2),p) == 1:
c = c**2%p
m -= 1
else:
t = (c**2%p)*t%p
r = c*r%p
c = c**2%p
m -= 1
return r%p
m,n = map(int, input().split())
a = list(map(int, input().split()))
for i in a:
dic[i] += 1
if n == 1:
d = 0
x = a[0]
print(x,d)
elif n == m-1:
d = 1
a.sort()
if a[0] != (a[-1]+d)%m:
print(a[0],d)
for i in range(1,n):
if a[i] != (a[i-1]+d)%m:
print(a[i],d)
quit()
elif n == m:
d = 1
x = 0
print(x,d)
elif m == 2:
a.sort()
print(a[0],a[1]-a[0])
elif m == 3:
a.sort()
d = a[1]-a[0]
for i in range(1,n):
if a[i] != a[i-1]+d:
print(-1)
quit()
print(a[0],d)
else:
ave = sum(a)*pow_mod(n,m-2,m)%m
b = [(a[i]-ave)**2%m for i in range(n)]
s = sum(b)*pow_mod(n,m-2,m)%m
k = n**2-1
d2 = s*12*pow_mod(k,m-2,m)%m
d = tonelli_shanks(d2,m)
x = (ave-d*(n-1)*pow_mod(2,m-2,m)%m)%m
if x != int(x):
print(-1)
quit()
x = int(x)
for i in range(n):
if not dic[(x+i*d)%m]:
print(-1)
quit()
print(x,d)
``` | output | 1 | 93,164 | 22 | 186,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4 | instruction | 0 | 93,165 | 22 | 186,330 |
Tags: brute force, implementation, math, number theory
Correct Solution:
```
import sys
def modInverse(x, y, mod):
res = 1
x = x % mod
while(y > 0):
if(y&1):
res = (res * x) % mod
y = y // 2
x = (x * x) % mod
return res
def isEqual(a, b):
for i in range(len(a)):
if (a[i] != b[i]):
return False
return True
m, n = [int(x) for x in input().split()]
arr = [int(x) for x in input().split()]
s = [0, 0]
for x in arr:
s[0] = s[0] + x
s[1] = s[1] + x*x
if (n==1):
print(arr[0], 0)
sys.exit()
if (n==m):
print("0 1")
sys.exit()
arr.sort()
for i in range(1, n):
d = arr[i]-arr[0];
x = (s[0] - (n*(n-1) // 2) * d + m)%m * modInverse(n, m-2, m) % m
Sum = (n*x*x + n*(n-1)*d*x + n*(n-1)*(2*n-1)//6*d*d) % m
if(Sum == (s[1]%m)):
b = [x]
for j in range(1, n):
b.append((b[j-1] + d) % m)
b.sort()
if (isEqual(arr, b)):
print(x, d)
sys.exit()
print("-1")
``` | output | 1 | 93,165 | 22 | 186,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4 | instruction | 0 | 93,166 | 22 | 186,332 |
Tags: brute force, implementation, math, number theory
Correct Solution:
```
def solve(n, m, a):
if n == 0:
return 0, 1
if n == 1:
return a[0], 1
d = (a[1]-a[0]) % m
if d < 0: d += m
st = set(a)
cnt = 0
for v in a:
cnt += ((v + d) % m) in st
cnt = n-cnt
d = (d * pow(cnt, m-2, m)) % m
now = a[0]
while (now + m - d) % m in st:
now = (now + m - d) % m
for i in range(n):
if (now + i*d) % m not in st:
return -1, -1
return now, d
m, n = map(int, input().split())
a = list(map(int, input().split()))
if n * 2 > m:
st = set(a)
b = [i for i in range(m) if i not in st]
f, d = solve(len(b), m, b)
f = (f + d * (m-n)) % m
else:
f, d = solve(n, m, a)
if f < 0 or d < 0:
print(-1)
else:
print(f, d)
``` | output | 1 | 93,166 | 22 | 186,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4 | instruction | 0 | 93,167 | 22 | 186,334 |
Tags: brute force, implementation, math, number theory
Correct Solution:
```
from functools import reduce
def gcd_extended(bigger, less):
if less == 0:
return(bigger, 1, 0)
mod = bigger % less
div = bigger // less
gcd, c_less, c_mod = gcd_extended(less, mod)
#gcd == c_less * less + c_mod * mod
#mod == bigger - div * less
#gcd = (c_less - c_mod * div) * less
# + c_mod * bigger
c_bigger = c_mod
c_less = c_less - c_mod * div
return(gcd, c_bigger, c_less)
def mlp_inverse(x, p):
one, c_p, c_x = gcd_extended(p, x) #one == c_x * x + c_p * p
return (c_x + p) % p
def tests(x, d, p, row):
_row = set(row)
n = len(row)
if d == 0:
return False
for i in range(n):
elem = x + i * d
elem = (elem % p + p) % p
if elem not in _row:
return False
return True
p, n = (int(x) for x in input().split())
row = [int(x) for x in input().split()]
if p == n:
print(1, 1)
exit()
if n == 1:
print(row[0], 0)
exit()
#precounting constants
c1 = reduce(lambda x, y: (x + y) % p, row, 0)
c2 = reduce(lambda x, y: (x + y * y) % p, row, 0)
sum_i = reduce(lambda x, y: (x + y) % p, range(0, n), 0)
sum_i_sq = reduce(lambda x, y: (x + y * y) % p, range(0, n), 0)
inv_sum_i = mlp_inverse(sum_i, p)
inv_sum_i_sq = mlp_inverse(sum_i_sq, p)
#algorythm
for x in row:
#
d = (c1 - n * x) * inv_sum_i
d = (d % p + p) % p
equasion = n * x * x + 2 * d * x * sum_i + d * d * sum_i_sq
equasion = (equasion % p + p) % p
# print(x, d)
# print(c2, equasion)
if (equasion == c2
and tests(x, d, p, row)
):
print(x, d)
exit()
print(-1)
``` | output | 1 | 93,167 | 22 | 186,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4 | instruction | 0 | 93,168 | 22 | 186,336 |
Tags: brute force, implementation, math, number theory
Correct Solution:
```
import bisect
import copy
import decimal
import fractions
import heapq
import itertools
import math
import random
import sys
from collections import Counter,deque,defaultdict
from functools import lru_cache,reduce
from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
def _heappush_max(heap,item):
heap.append(item)
heapq._siftdown_max(heap, 0, len(heap)-1)
def _heappushpop_max(heap, item):
if heap and item < heap[0]:
item, heap[0] = heap[0], item
heapq._siftup_max(heap, 0)
return item
from math import e, gcd as GCD, modf
read=sys.stdin.read
readline=sys.stdin.readline
readlines=sys.stdin.readlines
def Extended_Euclid(n,m):
stack=[]
while m:
stack.append((n,m))
n,m=m,n%m
if n>=0:
x,y=1,0
else:
x,y=-1,0
for i in range(len(stack)-1,-1,-1):
n,m=stack[i]
x,y=y,x-(n//m)*y
return x,y
class MOD:
def __init__(self,mod):
self.mod=mod
def Pow(self,a,n):
a%=self.mod
if n>=0:
return pow(a,n,self.mod)
else:
assert math.gcd(a,self.mod)==1
x=Extended_Euclid(a,self.mod)[0]
return pow(x,-n,self.mod)
def Build_Fact(self,N):
assert N>=0
self.factorial=[1]
for i in range(1,N+1):
self.factorial.append((self.factorial[-1]*i)%self.mod)
self.factorial_inv=[None]*(N+1)
self.factorial_inv[-1]=self.Pow(self.factorial[-1],-1)
for i in range(N-1,-1,-1):
self.factorial_inv[i]=(self.factorial_inv[i+1]*(i+1))%self.mod
return self.factorial_inv
def Fact(self,N):
return self.factorial[N]
def Fact_Inv(self,N):
return self.factorial_inv[N]
def Comb(self,N,K):
if K<0 or K>N:
return 0
s=self.factorial[N]
s=(s*self.factorial_inv[K])%self.mod
s=(s*self.factorial_inv[N-K])%self.mod
return s
M,N=map(int,readline().split())
A=list(map(int,readline().split()))
set_A=set(A)
MD=MOD(M)
if N==M:
print(0,1)
exit()
if 2*N<M:
if N==1:
print(A[0],0)
exit()
a=(A[0]-A[1])%M
cnt=0
for i in range(N):
if (A[i]+a)%M in set_A:
cnt+=1
d=a*MD.Pow(N-cnt,-1)%M
cnt=0
for i in range(N):
if (A[i]-d)%M in set_A:
cnt+=1
else:
ans=A[i]
if cnt==N-1:
print(ans,d)
exit()
else:
A=[a for a in range(M) if not a in set_A]
N=len(A)
if N==1:
print((A[0]+1)%M,1)
exit()
set_A=set(A)
a=(A[0]-A[1])%M
cnt=0
for i in range(N):
if (A[i]+a)%M in set_A:
cnt+=1
d=a*MD.Pow(N-cnt,-1)%M
cnt=0
for i in range(N):
if (A[i]+d)%M in set_A:
cnt+=1
else:
ans=A[i]
if cnt==N-1:
print((ans+d)%M,d)
exit()
print(-1)
``` | output | 1 | 93,168 | 22 | 186,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4
Submitted Solution:
```
#Soru 1
#n = input()
#lst = list(map(int,n.split()))
#num1 = lst[0]
#num2 = lst[1]
#num3 = lst[2]
#while True:
# if num1 > num2:
# if num1//num2 == num1/num2:
# ebob = num2
# break
# num1 = num1%num2
# else:
# if num2//num1 == num2/num1:
# ebob = num1
# break
# num2 = num2%num1
#
#div = lst[0]*lst[1]/ebob
#res = int(num3/div)
#print(res)
#####################################################
# Soru 2
#def swap(lst,n,i):
# temp = lst[i]
# lst[i] = lst[n-i-1]
# lst[n-i-1] = temp
# return lst
#n = int(input())
#arr = list(map(int,input().split()))
#i = 0
#for i in range(n//2):
# if i%2 == 0:
# arr = swap(arr,n,i)
#print(*arr)
#####################################################
#Soru 3
first_line = list(map(int, input().split()))
prime_m = first_line[0]
seq_n = first_line[1]
seq = list(map(int, input().split()))
if seq_n%2 == 1:
x = seq_n//2+1
for i in seq[seq_n//2+1:]:
seq[x] = i%prime_m - prime_m
x = x + 1
else:
x = seq_n//2
for i in seq[seq_n//2:]:
seq[x] = i%prime_m - prime_m
x = x + 1
seq = sorted(seq)
length = len(seq)
#Tek sayılı sequence ler içindir çift için denenmemiştir.
fark = seq[-1] - seq[0]
progress = int(fark/(len(seq)-1))
if seq[length//2] - seq[length//2 - 1] != progress:
x = seq[length//2]
seq.remove(x)
seq.append(x+prime_m)
fark = seq[-1] - seq[0]
progress = int(fark/(len(seq)-1))
print(seq)
indd = 1
for i in seq[:-1]:
if seq[indd] - i == progress:
indd = indd+1
else:
if indd == len(seq)//2:
if seq[indd]+prime_m - seq[-1] == progress:
indd = indd+1
else:
print(-1)
break
else:
print(-1)
break
if indd == len(seq):
print(seq[0]+prime_m,progress)
``` | instruction | 0 | 93,169 | 22 | 186,338 |
No | output | 1 | 93,169 | 22 | 186,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4
Submitted Solution:
```
import bisect
import copy
import decimal
import fractions
import heapq
import itertools
import math
import random
import sys
from collections import Counter,deque,defaultdict
from functools import lru_cache,reduce
from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
def _heappush_max(heap,item):
heap.append(item)
heapq._siftdown_max(heap, 0, len(heap)-1)
def _heappushpop_max(heap, item):
if heap and item < heap[0]:
item, heap[0] = heap[0], item
heapq._siftup_max(heap, 0)
return item
from math import e, gcd as GCD, modf
read=sys.stdin.read
readline=sys.stdin.readline
readlines=sys.stdin.readlines
def Extended_Euclid(n,m):
stack=[]
while m:
stack.append((n,m))
n,m=m,n%m
if n>=0:
x,y=1,0
else:
x,y=-1,0
for i in range(len(stack)-1,-1,-1):
n,m=stack[i]
x,y=y,x-(n//m)*y
return x,y
class MOD:
def __init__(self,mod):
self.mod=mod
def Pow(self,a,n):
a%=self.mod
if n>=0:
return pow(a,n,self.mod)
else:
assert math.gcd(a,self.mod)==1
x=Extended_Euclid(a,self.mod)[0]
return pow(x,-n,self.mod)
def Build_Fact(self,N):
assert N>=0
self.factorial=[1]
for i in range(1,N+1):
self.factorial.append((self.factorial[-1]*i)%self.mod)
self.factorial_inv=[None]*(N+1)
self.factorial_inv[-1]=self.Pow(self.factorial[-1],-1)
for i in range(N-1,-1,-1):
self.factorial_inv[i]=(self.factorial_inv[i+1]*(i+1))%self.mod
return self.factorial_inv
def Fact(self,N):
return self.factorial[N]
def Fact_Inv(self,N):
return self.factorial_inv[N]
def Comb(self,N,K):
if K<0 or K>N:
return 0
s=self.factorial[N]
s=(s*self.factorial_inv[K])%self.mod
s=(s*self.factorial_inv[N-K])%self.mod
return s
M,N,*A=map(int,read().split())
set_A=set(A)
MD=MOD(M)
if N==1:
print(A[0],0)
exit()
a=(A[0]-A[1])%M
cnt=0
for i in range(N):
if (A[i]+a)%M in set_A:
cnt+=1
d=a*MD.Pow(N-cnt,-1)
cnt=0
for i in range(N):
if (A[i]-d)%M in set_A:
cnt+=1
else:
ans=A[i]
if cnt!=N-1:
print(-1)
else:
print(ans,d)
``` | instruction | 0 | 93,170 | 22 | 186,340 |
No | output | 1 | 93,170 | 22 | 186,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4
Submitted Solution:
```
def solve(n, m, a):
if n == 0:
return 0, 1
if n == 1:
return a[0], 1
d = (m+a[1]-a[0]) % m
st = set(a)
cnt = 0
for v in a:
cnt += ((v + d) % m) in st
cnt = n-cnt
d = (d * pow(cnt, m-2, m)) % m
now = a[0]
while (now + m - d) % m in st:
now = (now + m - d) % m
for i in range(n):
if (now + i*d) % m not in st:
return -1, -1
return now, d
m, n = map(int, input().split())
a = list(map(int, input().split()))
if n * 2 > m:
st = set(a)
b = [i for i in range(m) if i not in st]
f, d = solve(m-n, m, b)
f = (f + d * (m-n)) % m
else:
f, d = solve(n, m, a)
if f < 0:
print(f)
else:
print(f, d)
``` | instruction | 0 | 93,171 | 22 | 186,342 |
No | output | 1 | 93,171 | 22 | 186,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.
One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.
Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.
Input
The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.
The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.
Output
Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.
Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).
If there are multiple answers, print any of them.
Examples
Input
17 5
0 2 4 13 15
Output
13 2
Input
17 5
0 2 4 13 14
Output
-1
Input
5 3
1 2 3
Output
3 4
Submitted Solution:
```
import bisect
import copy
import decimal
import fractions
import heapq
import itertools
import math
import random
import sys
from collections import Counter,deque,defaultdict
from functools import lru_cache,reduce
from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
def _heappush_max(heap,item):
heap.append(item)
heapq._siftdown_max(heap, 0, len(heap)-1)
def _heappushpop_max(heap, item):
if heap and item < heap[0]:
item, heap[0] = heap[0], item
heapq._siftup_max(heap, 0)
return item
from math import e, gcd as GCD, modf
read=sys.stdin.read
readline=sys.stdin.readline
readlines=sys.stdin.readlines
M,N,*A=map(int,read().split())
set_A=set(A)
if N**2<10**7:
cnt=defaultdict(int)
for i in range(N):
for j in range(N):
if i==j:
continue
cnt[(A[i]-A[j])%M]+=1
else:
cnt=defaultdict(int)
n=int(N**.5)
lst=[[] for i in range(n)]
for i in range(N):
lst[i].append(A[i%n])
for k in range(n):
l=len(lst[n])
for i in range(l):
for j in range(l):
if i==j:
continue
cnt[(A[i]-A[j])%M]+=1
for a,c in cnt.items():
if c==N-1:
for i in range(N):
if not (A[i]-a)%M in set_A:
print(A[i],a)
exit()
print(-1)
``` | instruction | 0 | 93,172 | 22 | 186,344 |
No | output | 1 | 93,172 | 22 | 186,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuro is currently playing an educational game about numbers. The game focuses on the greatest common divisor (GCD), the XOR value, and the sum of two numbers. Kuro loves the game so much that he solves levels by levels day by day.
Sadly, he's going on a vacation for a day, and he isn't able to continue his solving streak on his own. As Katie is a reliable person, Kuro kindly asked her to come to his house on this day to play the game for him.
Initally, there is an empty array a. The game consists of q tasks of two types. The first type asks Katie to add a number u_i to a. The second type asks Katie to find a number v existing in a such that k_i ∣ GCD(x_i, v), x_i + v ≤ s_i, and x_i ⊕ v is maximized, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), GCD(c, d) denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers c and d, and y ∣ x means x is divisible by y, or report -1 if no such numbers are found.
Since you are a programmer, Katie needs you to automatically and accurately perform the tasks in the game to satisfy her dear friend Kuro. Let's help her!
Input
The first line contains one integer q (2 ≤ q ≤ 10^{5}) — the number of tasks the game wants you to perform.
q lines follow, each line begins with an integer t_i — the type of the task:
* If t_i = 1, an integer u_i follow (1 ≤ u_i ≤ 10^{5}) — you have to add u_i to the array a.
* If t_i = 2, three integers x_i, k_i, and s_i follow (1 ≤ x_i, k_i, s_i ≤ 10^{5}) — you must find a number v existing in the array a such that k_i ∣ GCD(x_i, v), x_i + v ≤ s_i, and x_i ⊕ v is maximized, where ⊕ denotes the XOR operation, or report -1 if no such numbers are found.
It is guaranteed that the type of the first task is type 1, and there exists at least one task of type 2.
Output
For each task of type 2, output on one line the desired number v, or -1 if no such numbers are found.
Examples
Input
5
1 1
1 2
2 1 1 3
2 1 1 2
2 1 1 1
Output
2
1
-1
Input
10
1 9
2 9 9 22
2 3 3 18
1 25
2 9 9 20
2 25 25 14
1 20
2 26 26 3
1 14
2 20 20 9
Output
9
9
9
-1
-1
-1
Note
In the first example, there are 5 tasks:
* The first task requires you to add 1 into a. a is now \left\{1\right\}.
* The second task requires you to add 2 into a. a is now \left\{1, 2\right\}.
* The third task asks you a question with x = 1, k = 1 and s = 3. Taking both 1 and 2 as v satisfies 1 ∣ GCD(1, v) and 1 + v ≤ 3. Because 2 ⊕ 1 = 3 > 1 ⊕ 1 = 0, 2 is the answer to this task.
* The fourth task asks you a question with x = 1, k = 1 and s = 2. Only v = 1 satisfies 1 ∣ GCD(1, v) and 1 + v ≤ 2, so 1 is the answer to this task.
* The fifth task asks you a question with x = 1, k = 1 and s = 1. There are no elements in a that satisfy the conditions, so we report -1 as the answer to this task.
Submitted Solution:
```
def Nod(a, b):
while a != 0 and b != 0:
if a > b:
a = a % b
else:
b = b % a
return a + b
q = int(input())
a = set()
for i in range(q):
t = tuple(map(int, input().split()))
if t[0] == 1:
a.add(t[1])
else:
t, x, k, s = t
r = -1
mx = -1
for u in a:
if x + u <= s:
if k % Nod(x, u) == 0:
xu = x ^ u
if xu > mx:
mx = xu
r = u
print(r)
``` | instruction | 0 | 93,232 | 22 | 186,464 |
No | output | 1 | 93,232 | 22 | 186,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuro is currently playing an educational game about numbers. The game focuses on the greatest common divisor (GCD), the XOR value, and the sum of two numbers. Kuro loves the game so much that he solves levels by levels day by day.
Sadly, he's going on a vacation for a day, and he isn't able to continue his solving streak on his own. As Katie is a reliable person, Kuro kindly asked her to come to his house on this day to play the game for him.
Initally, there is an empty array a. The game consists of q tasks of two types. The first type asks Katie to add a number u_i to a. The second type asks Katie to find a number v existing in a such that k_i ∣ GCD(x_i, v), x_i + v ≤ s_i, and x_i ⊕ v is maximized, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), GCD(c, d) denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers c and d, and y ∣ x means x is divisible by y, or report -1 if no such numbers are found.
Since you are a programmer, Katie needs you to automatically and accurately perform the tasks in the game to satisfy her dear friend Kuro. Let's help her!
Input
The first line contains one integer q (2 ≤ q ≤ 10^{5}) — the number of tasks the game wants you to perform.
q lines follow, each line begins with an integer t_i — the type of the task:
* If t_i = 1, an integer u_i follow (1 ≤ u_i ≤ 10^{5}) — you have to add u_i to the array a.
* If t_i = 2, three integers x_i, k_i, and s_i follow (1 ≤ x_i, k_i, s_i ≤ 10^{5}) — you must find a number v existing in the array a such that k_i ∣ GCD(x_i, v), x_i + v ≤ s_i, and x_i ⊕ v is maximized, where ⊕ denotes the XOR operation, or report -1 if no such numbers are found.
It is guaranteed that the type of the first task is type 1, and there exists at least one task of type 2.
Output
For each task of type 2, output on one line the desired number v, or -1 if no such numbers are found.
Examples
Input
5
1 1
1 2
2 1 1 3
2 1 1 2
2 1 1 1
Output
2
1
-1
Input
10
1 9
2 9 9 22
2 3 3 18
1 25
2 9 9 20
2 25 25 14
1 20
2 26 26 3
1 14
2 20 20 9
Output
9
9
9
-1
-1
-1
Note
In the first example, there are 5 tasks:
* The first task requires you to add 1 into a. a is now \left\{1\right\}.
* The second task requires you to add 2 into a. a is now \left\{1, 2\right\}.
* The third task asks you a question with x = 1, k = 1 and s = 3. Taking both 1 and 2 as v satisfies 1 ∣ GCD(1, v) and 1 + v ≤ 3. Because 2 ⊕ 1 = 3 > 1 ⊕ 1 = 0, 2 is the answer to this task.
* The fourth task asks you a question with x = 1, k = 1 and s = 2. Only v = 1 satisfies 1 ∣ GCD(1, v) and 1 + v ≤ 2, so 1 is the answer to this task.
* The fifth task asks you a question with x = 1, k = 1 and s = 1. There are no elements in a that satisfy the conditions, so we report -1 as the answer to this task.
Submitted Solution:
```
def Nod(a, b):
while a != 0 and b != 0:
if a > b:
a = a % b
else:
b = b % a
return a + b
q = int(input())
a = set()
for i in range(q):
t = tuple(map(int, input().split()))
if t[0] == 1:
a.add(t[1])
else:
t, x, k, s = t
r = -1
mx = -1
for u in a:
if x + u <= s:
if k % Nod(x, u) == 0:
xu = x | u
if xu > mx:
mx = xu
r = u
print(r)
``` | instruction | 0 | 93,233 | 22 | 186,466 |
No | output | 1 | 93,233 | 22 | 186,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuro is currently playing an educational game about numbers. The game focuses on the greatest common divisor (GCD), the XOR value, and the sum of two numbers. Kuro loves the game so much that he solves levels by levels day by day.
Sadly, he's going on a vacation for a day, and he isn't able to continue his solving streak on his own. As Katie is a reliable person, Kuro kindly asked her to come to his house on this day to play the game for him.
Initally, there is an empty array a. The game consists of q tasks of two types. The first type asks Katie to add a number u_i to a. The second type asks Katie to find a number v existing in a such that k_i ∣ GCD(x_i, v), x_i + v ≤ s_i, and x_i ⊕ v is maximized, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), GCD(c, d) denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers c and d, and y ∣ x means x is divisible by y, or report -1 if no such numbers are found.
Since you are a programmer, Katie needs you to automatically and accurately perform the tasks in the game to satisfy her dear friend Kuro. Let's help her!
Input
The first line contains one integer q (2 ≤ q ≤ 10^{5}) — the number of tasks the game wants you to perform.
q lines follow, each line begins with an integer t_i — the type of the task:
* If t_i = 1, an integer u_i follow (1 ≤ u_i ≤ 10^{5}) — you have to add u_i to the array a.
* If t_i = 2, three integers x_i, k_i, and s_i follow (1 ≤ x_i, k_i, s_i ≤ 10^{5}) — you must find a number v existing in the array a such that k_i ∣ GCD(x_i, v), x_i + v ≤ s_i, and x_i ⊕ v is maximized, where ⊕ denotes the XOR operation, or report -1 if no such numbers are found.
It is guaranteed that the type of the first task is type 1, and there exists at least one task of type 2.
Output
For each task of type 2, output on one line the desired number v, or -1 if no such numbers are found.
Examples
Input
5
1 1
1 2
2 1 1 3
2 1 1 2
2 1 1 1
Output
2
1
-1
Input
10
1 9
2 9 9 22
2 3 3 18
1 25
2 9 9 20
2 25 25 14
1 20
2 26 26 3
1 14
2 20 20 9
Output
9
9
9
-1
-1
-1
Note
In the first example, there are 5 tasks:
* The first task requires you to add 1 into a. a is now \left\{1\right\}.
* The second task requires you to add 2 into a. a is now \left\{1, 2\right\}.
* The third task asks you a question with x = 1, k = 1 and s = 3. Taking both 1 and 2 as v satisfies 1 ∣ GCD(1, v) and 1 + v ≤ 3. Because 2 ⊕ 1 = 3 > 1 ⊕ 1 = 0, 2 is the answer to this task.
* The fourth task asks you a question with x = 1, k = 1 and s = 2. Only v = 1 satisfies 1 ∣ GCD(1, v) and 1 + v ≤ 2, so 1 is the answer to this task.
* The fifth task asks you a question with x = 1, k = 1 and s = 1. There are no elements in a that satisfy the conditions, so we report -1 as the answer to this task.
Submitted Solution:
```
from heapq import heappush as push, heappop as pop, heapify
def gcd(a, b):
a, b = max(a, b), min(a, b)
while a % b != 0:
a, b = a % b, a
return b
def main():
q = int(input())
a = []
for i in range(q):
task = [int(_) for _ in input().split()]
# print(task)
if task[0] == 1:
a.append(task[1])
else:
x, k, s = task[1:]
max_xor = -1
# print(a)
for v in a:
# print(k, gcd(x, v), s, x + v, x ^ v, max_xor)
if k % gcd(x, v) == 0 and x + v <= s:
xor = x ^ v
if xor > max_xor:
max_xor = xor
v_ = v
if max_xor == -1:
print(-1)
else:
print(v_)
if __name__ == '__main__':
main()
``` | instruction | 0 | 93,234 | 22 | 186,468 |
No | output | 1 | 93,234 | 22 | 186,469 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,439 | 22 | 186,878 |
"Correct Solution:
```
def ms_number(n):
count = 1
msnumber_list = [6]
while msnumber_list[-1] < n:
msnumber_list.append(7 * count + 1)
msnumber_list.append(7 * count + 6)
count += 1
return msnumber_list[:-1]
def ms_prime(n, msnumber_list):
msprime_list = []
msprime_flag = [0 for i in range(n + 1)]
for i in msnumber_list:
if msprime_flag[i] == 0:
msprime_list.append(i)
for j in msnumber_list:
if i * j > n:
break
msprime_flag[i * j] = 1
return msprime_list
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5) + 1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n // i)
divisors.sort()
return divisors
msnumber_list = ms_number(300000)
msprime_list = ms_prime(300000, msnumber_list)
answer = []
input_list = []
while True:
n = int(input())
if n == 1:
break
input_list.append(n)
divisors = make_divisors(n)
answer.append([d for d in divisors if d in msprime_list])
for i in range(len(input_list)):
print(str(input_list[i]) + ": ", end="")
print(*answer[i])
``` | output | 1 | 93,439 | 22 | 186,879 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,440 | 22 | 186,880 |
"Correct Solution:
```
def modifiedSieve():
L = 300001
msPrimes = []
isMSPrime = [ True for _ in range(L) ]
for i in range(6):
isMSPrime[i] = False
for i in range(6, L):
if isMSPrime[i]:
if i % 7 == 1 or i % 7 == 6:
msPrimes.append(i)
for j in range(i * i, L, i):
isMSPrime[j] = False
else:
isMSPrime[i] = False
return msPrimes
def getMSFactors(num, msPrimes):
factors = set()
for msp in msPrimes:
if msp > num:
break
if num % msp == 0:
factors.add(msp)
if num in msPrimes:
factors.add(num)
return sorted(list(factors))
if __name__ == '__main__':
msPrimes = modifiedSieve()
while True:
num = int(input())
if num == 1:
break
factors = getMSFactors(num, msPrimes)
print("{}: {}".format(num, " ".join(list(map(str, factors)))))
``` | output | 1 | 93,440 | 22 | 186,881 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,441 | 22 | 186,882 |
"Correct Solution:
```
a=[0]*6+[1,0,1,0,0,0,0]*42857
p=[]
for i in range(6,300000,7):
if a[i]:
p+=[i]
for j in range(i**2,300000,i):
a[j]=0
if a[i+2]:
p+=[i+2]
for j in range((i+2)**2,300000,i+2):
a[j]=0
while 1:
n=int(input())
if n==1:break
b=[x for x in p if n%x==0]
print('{}: {}'.format(n, ' '.join(map(str, b))))
``` | output | 1 | 93,441 | 22 | 186,883 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,442 | 22 | 186,884 |
"Correct Solution:
```
def main():
while True:
data = int(input())
if data == 1: break
i = 1
factlist = []
while True:
item = 7 * i - 1
if item > int(data / 6) + 1: break
if data % item == 0:
quo = int(data / item)
if isms(quo) and ismsp(quo):
factlist.append(quo)
item = 7 * i + 1
if item > int(data / 6) + 1:break
if data % item == 0:
quo = int(data / item)
if isms(quo) and ismsp(quo):
factlist.append(quo)
i += 1
factlist.sort()
if factlist == []: factlist.append(data)
print(str(data) + ":", end = "")
for item in factlist:
print(" " + str(item), end = "")
print("")
def isms(num):
if num % 7 == 1 or num % 7 == 6: return True
else: return False
def ismsp(num):
i = 1
while True:
item = 7 * i - 1
if item > int(num / 6) + 1: break
if num % item == 0: return False
item = 7 * i + 1
if item > int(num / 6) + 1: break
if num % item == 0: return False
i += 1
return True
main()
``` | output | 1 | 93,442 | 22 | 186,885 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,443 | 22 | 186,886 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
class Prime():
def __init__(self, n):
self.M = m = int(math.sqrt(n)) + 10
self.A = a = [True] * m
a[0] = a[1] = False
self.T = t = []
for i in range(2, m):
if not a[i]:
continue
if not (i % 7 == 1 or i % 7 == 6):
continue
t.append(i)
for j in range(i*i,m,i):
a[j] = False
def is_prime(self, n):
return self.A[n]
def division(self, n):
d = []
for c in self.T:
if n < c:
break
if n % c == 0:
d.append(c)
return d
def main():
rr = []
pr = Prime(300000**2)
while True:
n = I()
if n == 1:
break
d = pr.division(n)
rr.append('{}: {}'.format(n,' '.join(map(str,d))))
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 93,443 | 22 | 186,887 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,444 | 22 | 186,888 |
"Correct Solution:
```
N = 300000
a = [((i % 7 ==1)or(i % 7 ==6)) for i in range(N)]
#print(a[11])
a[1]=0
p =[]
for i in range(6,N):
if a[i]:
p+=[i]
for j in range(i * i ,N,i):
a[j] = 0
while True:
n = int(input())
if n == 1:
break
print(n,end = ":")
for x in p:
if n % x ==0:
print(" "+str(x),end = "")
print()
``` | output | 1 | 93,444 | 22 | 186,889 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,445 | 22 | 186,890 |
"Correct Solution:
```
from math import floor
# 約数全列挙(ソート済み)
# n = 24 -> [1,2,3,4,6,8,12,24]
def divisor(n):
left = [1]
right = [n]
sup = floor(pow(n,1/2))
for p in range(2,sup+1):
if n % p == 0:
if p == n//p:
left.append(p)
continue
left.append(p)
right.append(n//p)
res = left + right[::-1]
return res
ans_list = []
def main():
while True:
ans = solve()
if ans == "end":
break
ans_list.append(ans)
for ans in ans_list:
print(ans)
def solve():
n = int(input())
if n == 1:
return "end"
cand = [c for c in divisor(n) if c % 7 in (1,6)]
ans = "{}:".format(n)
for c in cand:
flag = True
for p in [p for p in divisor(c) if p % 7 in (1,6)]:
if p not in (1,c):
flag = False
break
if c > 1 and flag is True:
ans += " {}".format(c)
return ans
main()
``` | output | 1 | 93,445 | 22 | 186,891 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998 | instruction | 0 | 93,446 | 22 | 186,892 |
"Correct Solution:
```
p = [1]*300010
p[0]=p[1]=0
for i in range(2, 300010):
if not(i%7==1 or i%7==6):
p[i] = 0
for i in range(2, 300010):
if p[i]:
for j in range(i*2, 300010, i):
p[j] = 0
doyososu = []
for i in range(2, 300010):
if p[i]:
doyososu.append(i)
while True:
n = int(input())
if n==1:
break
ans = []
for e in doyososu:
if n%e==0:
ans.append(e)
print(n, ":", sep="", end=" ")
print(*ans)
``` | output | 1 | 93,446 | 22 | 186,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
def monsat(num):
L = [False for i in range(num+1)]
for i in range(0,num,7):
if i == 1 or i == 0:
continue
for d in [-1,1]:
if L[i+d] == False:
L[i+d] = True
for j in range((i+d)*2,num,i+d):
if L[j] is False:
L[j] = [i+d]
else:
L[j].append(i+d)
return L
L = monsat(3*10**5+1)
while True:
n = int(input())
if n == 1:
break
print(str(n) + ':',n if L[n] is True else ' '.join(map(str,L[n])))
``` | instruction | 0 | 93,447 | 22 | 186,894 |
Yes | output | 1 | 93,447 | 22 | 186,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
sieve = [0] * 300001
for i in range(6, 300000, 7):
sieve[i] = sieve[i+2] = 1
MSprimes = []
for i in range(6, 300000, 7):
if sieve[i] == 1:
MSprimes.append(i)
for j in range(2*i, 300000, i):
sieve[j] = 0
if sieve[i+2] == 1:
MSprimes.append(i+2)
for j in range(2*(i+2), 300000, i+2):
sieve[j] = 0
while True:
N = int(input())
if N == 1: break
ansp = [x for x in MSprimes if N % x == 0]
print('{}: {}'.format(N, ' '.join(map(str, ansp))))
``` | instruction | 0 | 93,448 | 22 | 186,896 |
Yes | output | 1 | 93,448 | 22 | 186,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
# coding: utf-8
isgdprime=[1 for i in range(300001)]
gdprimes=[]
for i in range(6):
isgdprime[i]=0
for i in range(300001):
if isgdprime[i]==1 and (i%7==1 or i%7==6):
gdprimes.append(i)
j=i*2
while j<300001:
isgdprime[j]=0
j+=i
else:
isgdprime[i]=0
while 1:
n=int(input())
if n==1:
break
print(n,end=':')
for i in range(len(gdprimes)):
if n%gdprimes[i]==0:
print('',gdprimes[i],end='')
print()
``` | instruction | 0 | 93,449 | 22 | 186,898 |
Yes | output | 1 | 93,449 | 22 | 186,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
from array import array
size = 300000
s = array('i', range(size + 1))
for i in range(size+1):
if i % 7 != 1 and i % 7 != 6:
s[i] = 0
s[1] = 0
sqrtn = int(round(size ** 0.5))
for i in range(6, sqrtn + 1):
if s[i]:
s[i * i: size + 1: i] = array('i', [0] * len(range(i*i, size+1, i)))
primes = list(filter(None, s))
while True:
n = int(input())
if n == 1: break
i = 0
ls = []
for p in primes:
if p > n: break
if n % p == 0:
ls.append(p)
print(n, end="")
print(':', end=" ")
print(*ls)
``` | instruction | 0 | 93,450 | 22 | 186,900 |
Yes | output | 1 | 93,450 | 22 | 186,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
this_prime_num = [0 for i in range(300001)]
for i in range(300001):
if i % 7 == 1 or i % 7 == 6:
this_prime_num[i] = 1
for i in range(2, 300001):
j = i * 2
while this_prime_num[i] == 1 and j < len(this_prime_num):
#print(j)
this_prime_num[j] = 0
j += i
this_primenum_list = []
for i in range(2, len(this_prime_num)):
if this_prime_num[i]:
this_primenum_list.append(i)
print(this_primenum_list[:30])
ans_list = []
while True:
n = int(input())
if n == 1:
break
i = 0
ans = [n]
while this_primenum_list[i] <= n:
#print(i)
if n % this_primenum_list[i] == 0:
ans.append(this_primenum_list[i])
i += 1
if i >= len(this_primenum_list):
break
ans_list.append(ans)
for i in ans_list:
print(str(i[0]) + ":", end = "")
for j in i[1:]:
print("", end = " ")
print(j, end = "")
print()
``` | instruction | 0 | 93,451 | 22 | 186,902 |
No | output | 1 | 93,451 | 22 | 186,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
def factors(n):
sieve = [1]*(n+10)
m = 1
while m*7-1 <= n:
for p in [m*7-1,m*7+1]:
if sieve[p] == 0: continue
for pp in range(2*p,n+1,p):
sieve[pp] = 0
m += 1
ret = []
m = 1
while m*7-1 <= n:
for p in [m*7-1,m*7+1]:
if sieve[p] and n % p == 0:
ret.append(p)
m += 1
return(ret)
while True:
N = int(input())
if N == 1: break
ans = ' '.join(list(map(str,factors(N))))
print('{0}: {1}'.format(N,ans))
``` | instruction | 0 | 93,452 | 22 | 186,904 |
No | output | 1 | 93,452 | 22 | 186,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
def factors(n):
sieve = [1]*(n+10)
ret = []
m = 1
while m*7-1 <= n:
for p in [m*7-1,m*7+1]:
if sieve[p] == 0: continue
if n%p == 0:
ret.append(p)
for pp in range(2*p,n+1,p):
sieve[pp] = 0
m += 1
return(ret)
while True:
N = int(input())
if N == 1: break
ans = ' '.join(list(map(str,factors(N))))
print('{0}: {1}'.format(N,ans))
``` | instruction | 0 | 93,453 | 22 | 186,906 |
No | output | 1 | 93,453 | 22 | 186,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
S = [0 for i in range(300000)]
A = [-1, 1]
for i in range(1, int(300000 / 7) + 1):
d = 7 * i
for a in A:
da = d + a
if da < 300000 and S[da] == 0:
S[da] = 1
for j in range(2, int(300000 / 2)):
if da * j >= 300000:break
S[da * j] = -1
while 1:
n = int(input())
if n == 1:break
res = [str(n) + ':']
for i in range(2, n+1):
if S[i] == 1 and n % i == 0:
res.append(str(i))
print(' '.join(res))
``` | instruction | 0 | 93,454 | 22 | 186,908 |
No | output | 1 | 93,454 | 22 | 186,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties:
* for all x <image> the following inequality holds l ≤ x ≤ r;
* 1 ≤ |S| ≤ k;
* lets denote the i-th element of the set S as si; value <image> must be as small as possible.
Help Victor find the described set.
Input
The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)).
Output
Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order.
If there are multiple optimal sets, you can print any of them.
Examples
Input
8 15 3
Output
1
2
10 11
Input
8 30 7
Output
0
5
14 9 28 11 16
Note
Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. | instruction | 0 | 93,885 | 22 | 187,770 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
def solve(L,R,k):
#print(*[(bin(x)[2:]).zfill(max([len(bin(y)[2:]) for y in [L,R]])) for x in [L,R]],sep='\n')
if k < 5:
if k==1:
return [L]
elif k==2:
if R-L+1==2:
if L < (L^R):
return [L]
else:
return [L,R]
else:
if L&1:
return [L+1,L+2]
else:
return [L,L+1]
else:
b = 60
while (L&(1<<b))==0 and (R&(1<<b))==0:
b -= 1
if L==1:
return [1,2,3]
elif k==3:
if (L&(1<<b))==(R&(1<<b)):
return solve(L,R,2)
else:
r = b
while (L&(1<<r))==0:
r -= 1
if (1<<b)+(1<<r) <= R:
return [L,(1<<b)+(1<<r),(1<<b)+((1<<r)^L)]
else:
return solve(L,R,2)
else: #k==4
if (L^1)&1:
return [L,L+1,L+2,L+3]
elif L+4 <= R:
return [L+1,L+2,L+3,L+4]
else:
return solve(L,R,3)
else:
if L&1:
return [L+1,L+2,L+3,L+4]
else:
return [L,L+1,L+2,L+3]
ans = solve(*map(int,input().split()))
xor = 0
for x in ans:
xor ^= x
print(xor)
print(len(ans))
print(*ans)
``` | output | 1 | 93,885 | 22 | 187,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties:
* for all x <image> the following inequality holds l ≤ x ≤ r;
* 1 ≤ |S| ≤ k;
* lets denote the i-th element of the set S as si; value <image> must be as small as possible.
Help Victor find the described set.
Input
The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)).
Output
Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order.
If there are multiple optimal sets, you can print any of them.
Examples
Input
8 15 3
Output
1
2
10 11
Input
8 30 7
Output
0
5
14 9 28 11 16
Note
Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. | instruction | 0 | 93,886 | 22 | 187,772 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
import random
l, r, k = map(int, input().split(' '))
if k == 1:
print(l)
print(1)
print(l)
quit()
if k == 2:
if r == l+1:
a = l
b = l^r
if a <= b:
print(a)
print(1)
print(l)
quit()
else:
print(b)
print(2)
print(l, l+1)
quit()
for i in range(l, r+1):
if i%2==0:
print(1)
print(2)
print(i, i+1)
quit()
if k == 3:
if abs(l-r) <= 10000:
for i in range(l, r+1):
for j in range(i+1, r+1):
if l <= i^j <= r:
print(0)
print(3)
print(i, j, i^j)
quit()
print(1)
print(2)
if (l%2==0):
print(l, l+1)
quit()
else:
print(l+1, l+2)
quit()
if abs(l-r) <= 1000000:
for x in range(l, r+1):
if l <= x^(x+1) <= r:
print(0)
print(3)
print(x, x+1, x^(x+1))
quit()
for i in range(5):
if i == 1:
x = l
y = l+1
elif i == 2:
x = r
y = r-1
else:
x = random.randint(l, r)
y = random.randint(l, r)
if x == y:
continue
if l <= x^y<= r:
print(0)
print(3)
print(x, y, x^y)
quit()
print(1)
print(2)
if (l%2==0):
print(l, l+1)
quit()
else:
print(l+1, l+2)
quit()
if k == 4:
if r == l+3:
if l%2 == 0:
print(0)
print(4)
print(l, l+1, l+2, l+3)
quit()
a = l; b = l+1; c = l+2; d = l+3;
if a^b^c == 0:
print(0)
print(3)
print(a, b, c)
quit()
if a^b^d == 0:
print(0)
print(3)
print(a, b, d)
quit()
if a^c^d == 0:
print(0)
print(3)
print(a, c, d)
quit()
if b^c^d == 0:
print(0)
print(3)
print(b, c, d)
quit()
if a^b == 1:
print(1)
print(2)
print(a, b)
quit()
print(1)
print(2)
print(b, c)
quit()
for i in range(l, r+1):
if i%2 == 0:
print(0)
print(4)
print(i, i+1, i+2, i+3)
quit()
if k >= 5:
for i in range(l, r+1):
if i%2 == 0:
print(0)
print(4)
print(i, i+1, i+2, i+3)
quit()
``` | output | 1 | 93,886 | 22 | 187,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given Q queries of the form (L, R).
For each query you have to find the number of such x that L ≤ x ≤ R and there exist integer numbers a > 0, p > 1 such that x = ap.
Input
The first line contains the number of queries Q (1 ≤ Q ≤ 105).
The next Q lines contains two integers L, R each (1 ≤ L ≤ R ≤ 1018).
Output
Output Q lines — the answers to the queries.
Example
Input
6
1 4
9 9
5 7
12 29
137 591
1 1000000
Output
2
1
0
3
17
1111
Note
In query one the suitable numbers are 1 and 4. | instruction | 0 | 94,120 | 22 | 188,240 |
Tags: binary search, math, number theory
Correct Solution:
```
import math
import bisect
import sys
def flrt(exp,x):
l=max(0,math.floor(x**(1/exp))-3)
r= math.floor(x**(1/exp))+3
while l<r:
mid=(l+r)//2
if mid**exp<=x:
l=mid+1
else:
r=mid
return l-1
def c1(r):
ans=set()
pr = [5,7,11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61]
for i in pr:
x=2
while x**i <= r:
val=x**i
b2=flrt(2,val)
b3=flrt(3,val)
if b2**2 != val and b3**3 != val:
ans.add(val)
x+=1
return ans
def solve(r, pc):
if r==0:
return 0
a=[2,3]
ans=0
for i in range(1,2**len(a)):
tot=0
mult=1
for j,x in enumerate(a):
if i&(1<<j):
mult*=x
tot+=1
d= flrt(mult,r)
ans+= d if tot%2 else -d
return ans + bisect.bisect_right(pc,r)
lp=0
rp=len(pc)
while lp<rp:
mid = (lp+rp)//2
if pc[mid] <= r:
lp = mid+1
else:
rp = mid
return ans + lp
q = int(input())
pc= c1(1e18+1)
pca=list(pc)
pca.sort()
for i in range(q):
l,r = [int(x) for x in sys.stdin.readline().split()]
ans=solve(r,pca)-solve(l-1,pca)
print(ans)
``` | output | 1 | 94,120 | 22 | 188,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given Q queries of the form (L, R).
For each query you have to find the number of such x that L ≤ x ≤ R and there exist integer numbers a > 0, p > 1 such that x = ap.
Input
The first line contains the number of queries Q (1 ≤ Q ≤ 105).
The next Q lines contains two integers L, R each (1 ≤ L ≤ R ≤ 1018).
Output
Output Q lines — the answers to the queries.
Example
Input
6
1 4
9 9
5 7
12 29
137 591
1 1000000
Output
2
1
0
3
17
1111
Note
In query one the suitable numbers are 1 and 4. | instruction | 0 | 94,121 | 22 | 188,242 |
Tags: binary search, math, number theory
Correct Solution:
```
import sys
readline = sys.stdin.buffer.readline
J = set()
for i in range(2000):
J.add(i**2)
J.add(i**3)
Ri = set()
for p in range(5, 61, 2):
if p%3 == 0:
continue
for base in range(2, 10**9):
if base in J:
continue
bp = pow(base, p)
if bp > 10**18:
break
Ri.add(bp)
Ri = list(Ri)
Ri.sort()
LL = len(Ri)
def calc(x):
res = 0
ng = 10**9+1
ok = 0
while abs(ok-ng) > 1:
med = (ok+ng)//2
if med*med <= x:
ok = med
else:
ng = med
res += ok
ng = 10**6+1
ok = 0
while abs(ok-ng) > 1:
med = (ok+ng)//2
if med*med*med <= x:
ok = med
else:
ng = med
res += ok
ng = 10**3+1
ok = 0
while abs(ok-ng) > 1:
med = (ok+ng)//2
if med*med*med*med*med*med <= x:
ok = med
else:
ng = med
res -= ok
ng = LL
ok = -1
while abs(ok-ng) > 1:
med = (ok+ng)//2
if Ri[med] <= x:
ok = med
else:
ng = med
res += ok+1
return res
Q = int(readline())
Ans = [None]*Q
for qu in range(Q):
L, R = map(int, readline().split())
Ans[qu] = calc(R) - calc(L-1)
print('\n'.join(map(str, Ans)))
``` | output | 1 | 94,121 | 22 | 188,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given Q queries of the form (L, R).
For each query you have to find the number of such x that L ≤ x ≤ R and there exist integer numbers a > 0, p > 1 such that x = ap.
Input
The first line contains the number of queries Q (1 ≤ Q ≤ 105).
The next Q lines contains two integers L, R each (1 ≤ L ≤ R ≤ 1018).
Output
Output Q lines — the answers to the queries.
Example
Input
6
1 4
9 9
5 7
12 29
137 591
1 1000000
Output
2
1
0
3
17
1111
Note
In query one the suitable numbers are 1 and 4. | instruction | 0 | 94,122 | 22 | 188,244 |
Tags: binary search, math, number theory
Correct Solution:
```
import math
import bisect
import sys
def flrt(exp,x):
l=max(0,math.floor(x**(1/exp))-3)
r= math.floor(x**(1/exp))+3
while l<r:
mid=(l+r)//2
if mid**exp<=x:
l=mid+1
else:
r=mid
return l-1
def c1(r):
ans=set()
pr = [5,7,11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61]
for i in pr:
x=2
while x**i <= r:
val=x**i
b2=flrt(2,val)
b3=flrt(3,val)
if b2**2 != val and b3**3 != val:
ans.add(val)
x+=1
return ans
def solve(r, pc):
if r==0:
return 0
a=[2,3]
ans=0
for i in range(1,2**len(a)):
tot=0
mult=1
for j,x in enumerate(a):
if i&(1<<j):
mult*=x
tot+=1
d= flrt(mult,r)
ans+= d if tot%2 else -d
return ans + bisect.bisect_right(pc,r)
lp=0
rp=len(pc)
while lp<rp:
mid = (lp+rp)//2
if pc[mid] <= r:
lp = mid+1
else:
rp = mid
return ans + lp
q = int(input())
pc= c1(1e18+1)
pca=list(pc)
pca.sort()
for i in range(q):
l,r = [int(x) for x in sys.stdin.readline().split()]
ans=solve(r,pca)-solve(l-1,pca)
sys.stdout.write(str(ans)+'\n')
``` | output | 1 | 94,122 | 22 | 188,245 |
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