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 |
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Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,725 | 22 | 165,450 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
ans=1
if(n>m):
print(0)
else:
for i in range(n):
for j in range(i+1,n):
ans*=abs(a[i]-a[j])
ans=ans%m
print(ans)
``` | output | 1 | 82,725 | 22 | 165,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,726 | 22 | 165,452 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
import math
n,m=map(int,input().split())
l=list(map(int,input().split()))
if n<=m:
mul=1
for i in range(n-1):
for j in range(i+1,n):
t=(l[i]-l[j])
mul=mul*abs(t)
mul%=m
print(mul%m)
else:
exit(print(0))
``` | output | 1 | 82,726 | 22 | 165,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,727 | 22 | 165,454 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
R=lambda:map(int,input().split())
n, m = R()
a = list(R())
if n > m: print(0)
else:
#a = [x%m for x in a]
ans = 1
for i in range(n):
for j in range(i+1, n):
ans = (ans * abs(a[i] - a[j])) % m
print(ans)
``` | output | 1 | 82,727 | 22 | 165,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,728 | 22 | 165,456 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
n , m = map(int, input().split())
a = list(map(int, input().split()))
ans = 1
if n > m: exit(print(0))
for i in range(n):
for j in range(i+1,n):
ans*=abs(a[i]-a[j])
ans = ans%m
print(ans)
``` | output | 1 | 82,728 | 22 | 165,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,729 | 22 | 165,458 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
from itertools import groupby
n, m = map(int, input().split())
a1 = list(map(int, input().split()))
a = [el for el, _ in groupby(a1)]
if n > m or len(a)!=len(a1): exit(print(0))
ans = 1
for i in range(n):
for j in range(i+1, n):
ans *= abs(a[i] - a[j])
ans %= m
print(ans)
``` | output | 1 | 82,729 | 22 | 165,459 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,730 | 22 | 165,460 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
import sys
input = lambda : sys.stdin.readline()
n,m = map(int,input().split())
a = list(map(int,input().split()))
s = 1
a.sort()
if n>m:
print(0)
exit(0)
for i in range(n):
for j in range(n-1-i):
s = (s*abs(a[i]-a[i+j+1]))%m
if s==0:
print(0)
exit(0)
print(s)
``` | output | 1 | 82,730 | 22 | 165,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,731 | 22 | 165,462 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
# problem 1305 c
N,mod=map(int,input().split())
a=list(map(int,input().split()))
if N>mod:
print(0)
elif N<=mod:
ans=1
for i in range(N):
for j in range(i+1,N):
ans*=abs(a[j]-a[i])
ans%=mod
print(ans)
``` | output | 1 | 82,731 | 22 | 165,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To become the king of Codeforces, Kuroni has to solve the following problem.
He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
Input
The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
Output
Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
1
Note
In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7. | instruction | 0 | 82,732 | 22 | 165,464 |
Tags: brute force, combinatorics, math, number theory
Correct Solution:
```
n, m = map(int, input().split())
li = list(map(int, input().split()))
res = 1
if n > 10000:
print(0)
exit()
for i in range(n):
for j in range(i+1, n):
res = (res * abs(li[i]-li[j])) % m
print(res)
``` | output | 1 | 82,732 | 22 | 165,465 |
Provide a correct Python 3 solution for this coding contest problem.
We ask you to select some number of positive integers, and calculate the sum of them.
It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.
Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.
If the objective is achievable, print `YES`. Otherwise, print `NO`.
Constraints
* 1 ≤ A ≤ 100
* 1 ≤ B ≤ 100
* 0 ≤ C < B
Input
Input is given from Standard Input in the following format:
A B C
Output
Print `YES` or `NO`.
Examples
Input
7 5 1
Output
YES
Input
2 2 1
Output
NO
Input
1 100 97
Output
YES
Input
40 98 58
Output
YES
Input
77 42 36
Output
NO | instruction | 0 | 83,252 | 22 | 166,504 |
"Correct Solution:
```
from fractions import gcd
a,b,c=map(int,input().split())
print("NO"if c%gcd(a,b)!=0else"YES")
``` | output | 1 | 83,252 | 22 | 166,505 |
Provide a correct Python 3 solution for this coding contest problem.
We ask you to select some number of positive integers, and calculate the sum of them.
It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.
Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.
If the objective is achievable, print `YES`. Otherwise, print `NO`.
Constraints
* 1 ≤ A ≤ 100
* 1 ≤ B ≤ 100
* 0 ≤ C < B
Input
Input is given from Standard Input in the following format:
A B C
Output
Print `YES` or `NO`.
Examples
Input
7 5 1
Output
YES
Input
2 2 1
Output
NO
Input
1 100 97
Output
YES
Input
40 98 58
Output
YES
Input
77 42 36
Output
NO | instruction | 0 | 83,253 | 22 | 166,506 |
"Correct Solution:
```
import fractions
a,b,c=map(int,input().split())
if c%fractions.gcd(a,b)==0:
print("YES")
else:
print("NO")
``` | output | 1 | 83,253 | 22 | 166,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We ask you to select some number of positive integers, and calculate the sum of them.
It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.
Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.
If the objective is achievable, print `YES`. Otherwise, print `NO`.
Constraints
* 1 ≤ A ≤ 100
* 1 ≤ B ≤ 100
* 0 ≤ C < B
Input
Input is given from Standard Input in the following format:
A B C
Output
Print `YES` or `NO`.
Examples
Input
7 5 1
Output
YES
Input
2 2 1
Output
NO
Input
1 100 97
Output
YES
Input
40 98 58
Output
YES
Input
77 42 36
Output
NO
Submitted Solution:
```
import fraction
a, b, c = map(int, input().split())
print("YES" if c%(fraction.gcd(a,b)) == 0 else "NO")
``` | instruction | 0 | 83,265 | 22 | 166,530 |
No | output | 1 | 83,265 | 22 | 166,531 |
Provide a correct Python 3 solution for this coding contest problem.
Give you N cards. Only one natural number is written on each card. However, the same number is never written.
From now on, as a question, I will say an appropriate natural number. Please answer the largest remainder you get when you divide the number on the card you have by the number I said.
For example, suppose you have three cards with 9, 3, and 8, respectively. If I say "4", find the remainder of 9 and 3 and 8 divided by 4, respectively. The remainders are 1, 3, and 0, respectively, but the largest remainder is 3, so 3 is the correct answer.
Let's get started. e? Is it hard to have a lot of cards? It can not be helped. Now let's use the computer to find the largest remainder. Create a program that finds the largest of the remainders of the number on the card divided by the number asked. Ask the question many times, not just once, but never ask the same number more than once.
input
The input consists of one dataset. Input data is given in the following format.
N Q
c1 c2 ... cN
q1
q2
::
qQ
The number of cards N (2 ≤ N ≤ 300000) and the number of questions Q (2 ≤ Q ≤ 100000) are given in the first line, separated by one space, and the number ci (1 ≤ 100000) written on the card in the second line. ci ≤ 300000) is given with one space delimiter. The following Q line is given the number qi (1 ≤ qi ≤ 300000) given as a question.
output
Output the maximum remainder on one line for each question.
Example
Input
3 3
9 3 8
4
6
5
Output
3
3
4 | instruction | 0 | 83,295 | 22 | 166,590 |
"Correct Solution:
```
# AOJ 0270: Modular Query
# Python3 2018.6.26 bal4u
N, Q = map(int, input().split())
tbl = [0]*300005
nmax, tbl[0] = 0, 1
c = list(map(int, input().split()))
for k in c:
tbl[k] = 1
if k > nmax: nmax = k
tbl[k & 1] = 1
tbl[k & 3] = 1
tbl[k & 7] = 1
for i in range(Q):
q = int(input())
if q > nmax: print(nmax)
else:
f = 0
for k in range(q-1, -1, -1):
for i in range(k, nmax+1, q):
if tbl[i]:
print(k)
f = 1
break
if f: break
``` | output | 1 | 83,295 | 22 | 166,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tuple generator f^{(k)} = (f_1^{(k)}, f_2^{(k)}, ..., f_n^{(k)}), where f_i^{(k)} = (a_i ⋅ f_i^{(k - 1)} + b_i) mod p_i and f^{(0)} = (x_1, x_2, ..., x_n). Here x mod y denotes the remainder of x when divided by y. All p_i are primes.
One can see that with fixed sequences x_i, y_i, a_i the tuples f^{(k)} starting from some index will repeat tuples with smaller indices. Calculate the maximum number of different tuples (from all f^{(k)} for k ≥ 0) that can be produced by this generator, if x_i, a_i, b_i are integers in the range [0, p_i - 1] and can be chosen arbitrary. The answer can be large, so print the remainder it gives when divided by 10^9 + 7
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the tuple.
The second line contains n space separated prime numbers — the modules p_1, p_2, …, p_n (2 ≤ p_i ≤ 2 ⋅ 10^6).
Output
Print one integer — the maximum number of different tuples modulo 10^9 + 7.
Examples
Input
4
2 3 5 7
Output
210
Input
3
5 3 3
Output
30
Note
In the first example we can choose next parameters: a = [1, 1, 1, 1], b = [1, 1, 1, 1], x = [0, 0, 0, 0], then f_i^{(k)} = k mod p_i.
In the second example we can choose next parameters: a = [1, 1, 2], b = [1, 1, 0], x = [0, 0, 1].
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 23 10:17:51 2018
@author: yanni
"""
import math
def factor(N, d=1):
if N == 1:
return 1
for D in range(d+1, int(math.sqrt(N) + 2)):
if (N % D == 0):
count= 0
while (N%D == 0):
N = N//D
count += 1
return (D, count, N)
return (N,1,1)
n = int(input())
p = [int(x) for x in input().split()]
add_one = False
powers = {}
values = {}
for prime in p:
if prime in values:
values[prime] += 1
else:
values[prime] = 1
to_look = sorted(values.keys(), reverse = True)
for pm in values:
if (values[pm]>2):
add_one = True
for pm in values:
unused = values[pm]
if pm not in powers:
powers[pm] = 1
unused -= 1
if (unused > 0):
N = pm-1
d = 1
useful = False
while (N > 1):
d, exp, N = factor(N, d)
if d in powers:
if (exp > powers[d]):
powers[d] = max(exp, powers[d])
useful = True
else:
powers[d] = exp
useful = True
if (useful):
unused -= 1
if (unused > 0):
add_one = True
ans = 1
big_prime = 10**9+7
for pm in powers:
ans *= ((pm**powers[pm]) % big_prime)
ans = ans % big_prime
if (add_one):
ans = (ans+1) % big_prime
print(ans)
``` | instruction | 0 | 83,388 | 22 | 166,776 |
No | output | 1 | 83,388 | 22 | 166,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tuple generator f^{(k)} = (f_1^{(k)}, f_2^{(k)}, ..., f_n^{(k)}), where f_i^{(k)} = (a_i ⋅ f_i^{(k - 1)} + b_i) mod p_i and f^{(0)} = (x_1, x_2, ..., x_n). Here x mod y denotes the remainder of x when divided by y. All p_i are primes.
One can see that with fixed sequences x_i, y_i, a_i the tuples f^{(k)} starting from some index will repeat tuples with smaller indices. Calculate the maximum number of different tuples (from all f^{(k)} for k ≥ 0) that can be produced by this generator, if x_i, a_i, b_i are integers in the range [0, p_i - 1] and can be chosen arbitrary. The answer can be large, so print the remainder it gives when divided by 10^9 + 7
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the tuple.
The second line contains n space separated prime numbers — the modules p_1, p_2, …, p_n (2 ≤ p_i ≤ 2 ⋅ 10^6).
Output
Print one integer — the maximum number of different tuples modulo 10^9 + 7.
Examples
Input
4
2 3 5 7
Output
210
Input
3
5 3 3
Output
30
Note
In the first example we can choose next parameters: a = [1, 1, 1, 1], b = [1, 1, 1, 1], x = [0, 0, 0, 0], then f_i^{(k)} = k mod p_i.
In the second example we can choose next parameters: a = [1, 1, 2], b = [1, 1, 0], x = [0, 0, 1].
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 23 10:17:51 2018
@author: yanni
"""
import math
def factor(N, d=1):
if N == 1:
return 1
for D in range(d+1, int(math.sqrt(N) + 2)):
if (N % D == 0):
count= 0
while (N%D == 0):
N = N//D
count += 1
return (D, count, N)
return (N,1,1)
n = int(input())
p = [int(x) for x in input().split()]
add_one = False
powers = {}
values = {}
for prime in p:
if prime in values:
values[prime] += 1
else:
values[prime] = 1
#print(values)
for pm in values:
if pm not in powers:
powers[pm] = 1
if (values[pm] > 1):
N = pm-1
d = 1
while (N > 1):
#print(factor(N, d))
d, exp, N = factor(N, d)
if d in powers:
#print(d, exp)
powers[d] = max(exp, powers[d])
else:
powers[d] = exp
#print(powers)
#print(powers)
if (values[pm] > 2):
add_one = True
ans = 1
big_prime = 10**9+7
for pm in powers:
ans *= (pm**powers[pm])
ans = ans % big_prime
if (add_one):
ans = (ans+1) % big_prime
print(ans)
``` | instruction | 0 | 83,389 | 22 | 166,778 |
No | output | 1 | 83,389 | 22 | 166,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tuple generator f^{(k)} = (f_1^{(k)}, f_2^{(k)}, ..., f_n^{(k)}), where f_i^{(k)} = (a_i ⋅ f_i^{(k - 1)} + b_i) mod p_i and f^{(0)} = (x_1, x_2, ..., x_n). Here x mod y denotes the remainder of x when divided by y. All p_i are primes.
One can see that with fixed sequences x_i, y_i, a_i the tuples f^{(k)} starting from some index will repeat tuples with smaller indices. Calculate the maximum number of different tuples (from all f^{(k)} for k ≥ 0) that can be produced by this generator, if x_i, a_i, b_i are integers in the range [0, p_i - 1] and can be chosen arbitrary. The answer can be large, so print the remainder it gives when divided by 10^9 + 7
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the tuple.
The second line contains n space separated prime numbers — the modules p_1, p_2, …, p_n (2 ≤ p_i ≤ 2 ⋅ 10^6).
Output
Print one integer — the maximum number of different tuples modulo 10^9 + 7.
Examples
Input
4
2 3 5 7
Output
210
Input
3
5 3 3
Output
30
Note
In the first example we can choose next parameters: a = [1, 1, 1, 1], b = [1, 1, 1, 1], x = [0, 0, 0, 0], then f_i^{(k)} = k mod p_i.
In the second example we can choose next parameters: a = [1, 1, 2], b = [1, 1, 0], x = [0, 0, 1].
Submitted Solution:
```
mod = 10**9 + 7
_ = input()
l = input().split()
l = list(map(lambda x : int(x), l))
d = {x:x for x in l}
s = set()
for i in l:
if d[i] <= 1: continue
while d[i] in s and d[i] > 1: d[i] -= 1
if d[i] > 1: s.add(d[i])
res = 1
for i in s:
res = (res * i) % mod
print(res)
``` | instruction | 0 | 83,390 | 22 | 166,780 |
No | output | 1 | 83,390 | 22 | 166,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tuple generator f^{(k)} = (f_1^{(k)}, f_2^{(k)}, ..., f_n^{(k)}), where f_i^{(k)} = (a_i ⋅ f_i^{(k - 1)} + b_i) mod p_i and f^{(0)} = (x_1, x_2, ..., x_n). Here x mod y denotes the remainder of x when divided by y. All p_i are primes.
One can see that with fixed sequences x_i, y_i, a_i the tuples f^{(k)} starting from some index will repeat tuples with smaller indices. Calculate the maximum number of different tuples (from all f^{(k)} for k ≥ 0) that can be produced by this generator, if x_i, a_i, b_i are integers in the range [0, p_i - 1] and can be chosen arbitrary. The answer can be large, so print the remainder it gives when divided by 10^9 + 7
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the tuple.
The second line contains n space separated prime numbers — the modules p_1, p_2, …, p_n (2 ≤ p_i ≤ 2 ⋅ 10^6).
Output
Print one integer — the maximum number of different tuples modulo 10^9 + 7.
Examples
Input
4
2 3 5 7
Output
210
Input
3
5 3 3
Output
30
Note
In the first example we can choose next parameters: a = [1, 1, 1, 1], b = [1, 1, 1, 1], x = [0, 0, 0, 0], then f_i^{(k)} = k mod p_i.
In the second example we can choose next parameters: a = [1, 1, 2], b = [1, 1, 0], x = [0, 0, 1].
Submitted Solution:
```
from math import gcd
n = int(input())
primeList = [int(e) for e in input().split(' ')]
uniqueSet = set()
duplicateList = []
pile = 1
for e in primeList:
if e in uniqueSet:
duplicateList.append(e)
else:
uniqueSet.add(e)
pile *= e
fallingEdge = 0
for e in duplicateList:
maxv = 1;
original = e
while e>maxv:
e -= 1;
thegcd = gcd(e,pile);
t = e//thegcd;
if(t>maxv):
maxv = t;
pile *= maxv
if maxv==1 and fallingEdge<original-1:
fallingEdge = original-1
print(pile+fallingEdge)
"""
4
2 3 5 7
-------------------
210
-------------------
3
5 3 3
-------------------
30
"""
``` | instruction | 0 | 83,391 | 22 | 166,782 |
No | output | 1 | 83,391 | 22 | 166,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,541 | 22 | 167,082 |
Tags: data structures, math, number theory
Correct Solution:
```
import math
def pf(n):
l=[]
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return l
n=int(input())
p=[[] for i in range(200005)]
a=list(map(int,input().split()))
ans=1
for i in range(n):
l=pf(a[i])
for j in range(len(l)):
if j==0:
p[l[j]].append(1)
else:
if l[j]==l[j-1]:
p[l[j]][-1]+=1
continue
else:
p[l[j]].append(1)
for i in range(len(p)):
if len(p[i])==(n-1):
for j in range(min(p[i])):
ans*=i
if len(p[i])==n:
p[i].sort()
for j in range(p[i][1]):
ans*=i
print(ans)
``` | output | 1 | 83,541 | 22 | 167,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,542 | 22 | 167,084 |
Tags: data structures, math, number theory
Correct Solution:
```
x=int(input())
s=list(map(int,input().split()))
from math import gcd
def lc(x,y):
return x*y//gcd(x,y)
gc=[0]*x
gc[-1]=s[-1]
for n in range(x-2,-1,-1):
gc[n]=gcd(s[n],gc[n+1])
res=lc(s[0],gc[1])
for n in range(1,x-1):
res=gcd(res,lc(s[n],gc[n+1]))
print(res)
``` | output | 1 | 83,542 | 22 | 167,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,543 | 22 | 167,086 |
Tags: data structures, math, number theory
Correct Solution:
```
import math
n = int(input())
arr = list(map(int, input().split()))
assert len(arr) == n
ans = arr[0]*arr[1]//math.gcd(arr[0],arr[1])
# gcd(t) so far
g = math.gcd(*arr[:2]) # gcd(arr) so far
for i in range(2, n):
g2 = math.gcd(arr[i], g)
ans = math.gcd(ans, arr[i]*g//g2)
g = g2
print(ans)
# import math
# n=int(input())
# a=list(map(int,input().strip().split(' ')))
# ans=(a[0]*a[1])//math.gcd(a[0],a[1])
# g=math.gcd(a[0],a[1])
# for i in a[2:]:
# newg=math.gcd(i,g)
# ans=math.gcd(ans,i*g//newg)
# g=newg
# print(int(ans))
``` | output | 1 | 83,543 | 22 | 167,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,544 | 22 | 167,088 |
Tags: data structures, math, number theory
Correct Solution:
```
def gcd(a,b):
if a==0:
return b
if b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
n=int(input())
li=[int(x) for x in input().split()]
gcd_suffix=[0]*(n)
gcd_suffix[n-1]=li[n-1]
iterator=li[n-1]
for i in range(n-2,-1,-1):
iterator=gcd(iterator,li[i])
gcd_suffix[i]=iterator
for i in range(n-1):
val=int((li[i]*gcd_suffix[i+1])/gcd_suffix[i])
if i==0:
final_gcd=val
else:
final_gcd=gcd(final_gcd,val)
print(final_gcd)
``` | output | 1 | 83,544 | 22 | 167,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,545 | 22 | 167,090 |
Tags: data structures, math, number theory
Correct Solution:
```
from collections import defaultdict
N = 2*(10**5) + 1
def fast_power(a, b):
power = 1
while b:
if b&1: power *= a
a *= a
b >>= 1
return power
def pp():
spf = [i for i in range(N)]
i = 2
while i*i < N:
if spf[i] == i:
j = 2
while i*j < N:
spf[i*j] = min(i, spf[i*j])
j += 1
i += 1
return spf
def f(num, ma, spf):
cnt = {}
while num > 1:
cnt[spf[num]] = cnt.get(spf[num], 0) + 1
num //= spf[num]
for key in cnt: ma[key].append(cnt[key])
def main():
spf = pp()
n = int(input())
arr = list(map(int, input().strip().split()))
if n == 1:
print(arr[0])
return
ma = defaultdict(list)
for num in arr: f(num, ma, spf)
ret = 1
for key in ma:
if len(ma[key]) < (n-1): continue
ma[key].sort()
if len(ma[key]) == n-1: ret *= fast_power(key, ma[key][0])
else: ret *= fast_power(key, ma[key][1])
print(ret)
if __name__ == "__main__": main()
``` | output | 1 | 83,545 | 22 | 167,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,546 | 22 | 167,092 |
Tags: data structures, math, number theory
Correct Solution:
```
def main():
from array import array
from collections import Counter
from itertools import chain
from math import gcd
from sys import stdin, stdout
n = int(stdin.readline())
if n == 2:
a, b = map(int, stdin.readline().split())
print(a // gcd(a, b) * b)
return
a = array('i', map(int, stdin.readline().split()))
ans = 1
for p in range(2, 444):
pows = [0] * n
for i in range(n):
while not a[i] % p:
a[i] //= p
pows[i] += 1
pows.sort()
ans *= p ** pows[1]
cnt = Counter(a)
nm1 = n - 1
print(ans * next(chain((k for k in cnt.keys() if cnt[k] >= nm1), (1,))))
main()
``` | output | 1 | 83,546 | 22 | 167,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,547 | 22 | 167,094 |
Tags: data structures, math, number theory
Correct Solution:
```
from math import *
a=int(input())
b=list(map(int,input().split()))
l=(b[0]*b[1])//gcd(b[0],b[1])
g=gcd(b[0],b[1])
for i in b[2:]:
l=gcd(l,(i*g)//gcd(i,g))
g=gcd(i,g)
print(l)
``` | output | 1 | 83,547 | 22 | 167,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40. | instruction | 0 | 83,548 | 22 | 167,096 |
Tags: data structures, math, number theory
Correct Solution:
```
from math import gcd
def fun():
n=int(input())
arr=list(map(int,input().split()))
whole=gcd(arr[0],arr[1])
for i in range(2,n):
whole=gcd(whole,arr[i])
for i in range(n):
arr[i]=arr[i]//whole
prefix=[0]*n
suffix=[0]*n
prefix[0]=arr[0]
suffix[-1]=arr[-1]
for i in range(1,n):
prefix[i]=gcd(prefix[i-1],arr[i])
for i in range(n-2,-1,-1):
suffix[i]=gcd(suffix[i+1],arr[i])
ans=whole
for i in range(1,n-1):
ans*=gcd(prefix[i-1],suffix[i+1])
ans*=prefix[n-2]
ans*=suffix[1]
return ans
print(fun())
``` | output | 1 | 83,548 | 22 | 167,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
from math import gcd
def lcm(a,b):
return a*b//gcd(a,b)
input()
a=list(map(int,input().split()))
t=gcd(a[0],a[1])
q=lcm(a[0],a[1])
for i in range(2,len(a)):
q=gcd(q,lcm(a[i],t))
t=gcd(t,a[i])
print(q)
``` | instruction | 0 | 83,549 | 22 | 167,098 |
Yes | output | 1 | 83,549 | 22 | 167,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
import math
#t=int(input())
n=int(input())
suff_gcd=[0]*n
ar=list(map(int,input().split()))
suff_gcd[-1]=ar[-1]
for i in range (n-2,-1,-1):
suff_gcd[i]=math.gcd(suff_gcd[i+1],ar[i])
temp=[]
for i in range (0,n-1):
res=(ar[i]*suff_gcd[i+1])//suff_gcd[i]
temp.append(res)
sol=temp[0]
for i in range (1,len(temp)):
sol=math.gcd(sol,temp[i])
print(sol)
``` | instruction | 0 | 83,550 | 22 | 167,100 |
Yes | output | 1 | 83,550 | 22 | 167,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
d1 = dict()
prost = [2]
for i in range(3, int((10**6+2)**(1/2))+2):
flag = True
for k in prost:
if i % k == 0:
flag = False
break
if flag:
prost.append(i)
for i in a:
d = dict()
x = 0
ind = 0
while prost[ind] <= i**(1/2)+1:
if i % prost[ind] == 0:
if prost[ind] not in d:
d[prost[ind]] = 0
d[prost[ind]] += 1
i //= prost[ind]
else:
ind += 1
if i != 1:
if i in d:
d[i] += 1
else:
d[i] = 1
for c in d:
if c in d1:
d1[c].append(d[c])
else:
d1[c] = []
d1[c].append(d[c])
ans = 1
for i in d1:
if len(d1[i]) <= n-2:
continue
if len(d1[i]) == n-1:
ans *= i**min(d1[i])
continue
if len(d1[i]) == n:
first = 1000000
second = 1000000
for kek in d1[i]:
if kek <= second:
if kek <= first:
second = first
first = kek
else:
second = kek
ans *= i**second
print(ans)
``` | instruction | 0 | 83,551 | 22 | 167,102 |
Yes | output | 1 | 83,551 | 22 | 167,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
MOD = 10 ** 9 + 7
RLIMIT = 1000
DEBUG = 1
def main():
for _ in inputt(1):
n, = inputi()
A = inputl()
if n == 1:
print(A[0])
elif n == 2:
print(A[0] * A[1] // gcd(*A))
else:
D = {d: [-inf, -inf] for d in factordict(A[0]) + factordict(A[1])}
for a in A:
p = factordict(a)
for d in D:
heappushpop(D[d], -p[d])
debug(D)
t = 1
for d, c in D.items():
t *= d ** (-c[0])
print(t)
# region M
# region import
from math import *
from heapq import *
from itertools import *
from functools import reduce, lru_cache, partial
from collections import Counter, defaultdict, deque
import re, copy, operator, cmath
import sys, io, os, builtins
sys.setrecursionlimit(RLIMIT)
# endregion
# region fastio
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
if args:
sys.stdout.write(str(args[0]))
split = kwargs.pop("split", " ")
for arg in args[1:]:
sys.stdout.write(split)
sys.stdout.write(str(arg))
sys.stdout.write(kwargs.pop("end", "\n"))
def debug(*args, **kwargs):
if DEBUG and not __debug__:
print("debug", *args, **kwargs)
sys.stdout.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip()
inputt = lambda t = 0: range(t) if t else range(int(input()))
inputs = lambda: input().split()
inputi = lambda k = int: map(k, inputs())
inputl = lambda t = 0, k = lambda: list(inputi()): [k() for _ in range(t)] if t else list(k())
# endregion
# region bisect
def len(a):
if isinstance(a, range):
return -((a.start - a.stop) // a.step)
return builtins.len(a)
def bisect_left(a, x, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
if key == None: key = do_nothing
while lo < hi:
mid = (lo + hi) // 2
if key(a[mid]) < x: lo = mid + 1
else: hi = mid
return lo
def bisect_right(a, x, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
if key == None: key = do_nothing
while lo < hi:
mid = (lo + hi) // 2
if x < key(a[mid]): hi = mid
else: lo = mid + 1
return lo
def insort_left(a, x, key = None, lo = 0, hi = None):
lo = bisect_left(a, x, key, lo, hi)
a.insert(lo, x)
def insort_right(a, x, key = None, lo = 0, hi = None):
lo = bisect_right(a, x, key, lo, hi)
a.insert(lo, x)
do_nothing = lambda x: x
bisect = bisect_right
insort = insort_right
def index(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
if key == None: key = do_nothing
i = bisect_left(a, x, key, lo, hi)
if lo <= i < hi and key(a[i]) == x: return a[i]
if default != None: return default
raise ValueError
def find_lt(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_left(a, x, key, lo, hi)
if lo < i <= hi: return a[i - 1]
if default != None: return default
raise ValueError
def find_le(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_right(a, x, key, lo, hi)
if lo < i <= hi: return a[i - 1]
if default != None: return default
raise ValueError
def find_gt(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_right(a, x, key, lo, hi)
if lo <= i < hi: return a[i]
if default != None: return default
raise ValueError
def find_ge(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_left(a, x, key, lo, hi)
if lo <= i < hi: return a[i]
if default != None: return default
raise ValueError
# endregion
# region csgraph
# TODO
class Tree:
def __init__(n):
self._n = n
self._conn = [[] for _ in range(n)]
self._list = [0] * n
def connect(a, b):
pass
# endregion
# region ntheory
class Sieve:
def __init__(self):
self._n = 6
self._list = [2, 3, 5, 7, 11, 13]
def extend(self, n):
if n <= self._list[-1]: return
maxbase = int(n ** 0.5) + 1
self.extend(maxbase)
begin = self._list[-1] + 1
newsieve = [i for i in range(begin, n + 1)]
for p in self.primerange(2, maxbase):
for i in range(-begin % p, len(newsieve), p):
newsieve[i] = 0
self._list.extend([x for x in newsieve if x])
def extend_to_no(self, i):
while len(self._list) < i:
self.extend(int(self._list[-1] * 1.5))
def primerange(self, a, b):
a = max(2, a)
if a >= b: return
self.extend(b)
i = self.search(a)[1]
maxi = len(self._list) + 1
while i < maxi:
p = self._list[i - 1]
if p < b:
yield p
i += 1
else: return
def search(self, n):
if n < 2: raise ValueError
if n > self._list[-1]: self.extend(n)
b = bisect(self._list, n)
if self._list[b - 1] == n: return b, b
else: return b, b + 1
def __contains__(self, n):
if n < 2: raise ValueError
if not n % 2: return n == 2
a, b = self.search(n)
return a == b
def __getitem__(self, n):
if isinstance(n, slice):
self.extend_to_no(n.stop + 1)
return self._list[n.start: n.stop: n.step]
else:
self.extend_to_no(n + 1)
return self._list[n]
sieve = Sieve()
def isprime(n):
if n <= sieve._list[-1]:
return n in sieve
for i in sieve:
if not n % i: return False
if n < i * i: return True
prime = sieve.__getitem__
primerange = lambda a, b = 0: sieve.primerange(a, b) if b else sieve.primerange(2, a)
def factorint(n):
factors = []
for i in sieve:
if n < i * i: break
while not n % i:
factors.append(i)
n //= i
if n != 1: factors.append(n)
return factors
factordict = lambda n: Counter(factorint(n))
# endregion
# region main
if __name__ == "__main__":
main()
# endregion
# endregion
``` | instruction | 0 | 83,552 | 22 | 167,104 |
Yes | output | 1 | 83,552 | 22 | 167,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
# Contest No.: 641
# Problem No.: C
# Solver: JEMINI
# Date: 20200512
##### FAST INPUT #####
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##### END OF FAST INPUT #####
#import sys
def main():
n = int(input())
nums = list(map(int, input().split()))
primeList = [True] * (10 ** 5 + 1)
primeList[0] = False
primeList[1] = False
prime = []
for i in range(2, 10 ** 5 + 1):
if primeList[i]:
cnt = 2 * i
prime.append(i)
while cnt <= 10 ** 5:
primeList[cnt] = False
cnt += i
ans = 1
for i in prime:
noCnt = 0
bigCnt = 0
tempList = []
for j in nums:
if j < i:
bigCnt += 1
if j % i or j < i:
noCnt += 1
else:
tempj = j
cnt = 0
while not tempj % i:
temp = 1
while not tempj % (i ** (temp * 2)):
temp <<= 1
cnt += temp
tempj = tempj // (i ** temp)
if len(tempList) < 2:
tempList.append(cnt)
tempList.sort()
else:
if tempList[0] <= cnt <= tempList[1]:
tempList[1] = cnt
elif cnt < tempList[0]:
tempList[1] = tempList[0]
tempList[0] = cnt
if noCnt >= 2:
break
if bigCnt == n:
break
if noCnt == 0:
ans *= i ** tempList[1]
elif noCnt == 1:
ans *= i ** tempList[0]
print(ans)
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 83,553 | 22 | 167,106 |
No | output | 1 | 83,553 | 22 | 167,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
from itertools import combinations
def gcd(a,b):
if b == 0:
return a
return gcd(b,a%b)
n = int(input())
lis = list(map(int, input().split()))
temp = lis[0]
for x in range(1,len(lis)):
if temp != lis[x]:
arr = list(set(lis))
z = list(combinations(arr,2))
lcm = []
count = 0
gcdes = 0
if len(z) >= 2:
for x in z:
a = x[0]
b = x[1]
lcm.append(a*b//(gcd(a,b)))
if count == 1:
gcdes = gcd(lcm[count-1],lcm[count])
#print(g )
elif count > 1:
gcdes = gcd(gcdes,lcm[count])
count += 1
print(gcdes)
else:
print(gcd(z[0][0],z[0][1]))
break
elif x == len(lis)-1:
print(temp)
``` | instruction | 0 | 83,554 | 22 | 167,108 |
No | output | 1 | 83,554 | 22 | 167,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
from collections import defaultdict, Counter
n = int(input())
a = [int(x) for x in input().split()]
def fact(n):
gaps = [1,2,2,4,2,4,2,4,6,2,6]
length, cycle = 11, 3
f, fs, nxt = 2, [], 0
while f * f <= n:
while n % f == 0:
fs.append(f)
n //= f
f += gaps[nxt]
nxt += 1
if nxt == length:
nxt = cycle
if n > 1: fs.append(n)
return Counter(fs)
# ~ print(fact(24))
min_p = defaultdict(list)
cnt = defaultdict(int)
for i in range(n):
fa = fact(a[i])
for p in fa:
cnt[p]+=1
act = fa[p]
l = min_p[p]
l.append(act)
l.sort()
min_p[p] = l[:2]
ans = 1
# ~ print(min_p)
for p in min_p:
l = min_p[p]
l.sort()
if cnt[p] <= n-2:
exp = 0
if cnt[p] == n-1:
exp = l[0]
else:
exp = l[-1]
ans *= p**exp
print(ans)
``` | instruction | 0 | 83,555 | 22 | 167,110 |
No | output | 1 | 83,555 | 22 | 167,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the multiset of positive integers s=\\{s_1,s_2,...,s_k\}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:
* \gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.
* lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, \gcd(\{8,12\})=4,\gcd(\{12,18,6\})=6 and lcm(\{4,6\})=12. Note that for any positive integer x, \gcd(\\{x\})=lcm(\\{x\})=x.
Orac has a sequence a with length n. He come up with the multiset t=\{lcm(\\{a_i,a_j\})\ |\ i<j\}, and asked you to find the value of \gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.
Input
The first line contains one integer n\ (2≤ n≤ 100 000).
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 200 000).
Output
Print one integer: \gcd(\{lcm(\\{a_i,a_j\})\ |\ i<j\}).
Examples
Input
2
1 1
Output
1
Input
4
10 24 40 80
Output
40
Input
10
540 648 810 648 720 540 594 864 972 648
Output
54
Note
For the first example, t=\{lcm(\{1,1\})\}=\{1\}, so \gcd(t)=1.
For the second example, t=\{120,40,80,120,240,80\}, and it's not hard to see that \gcd(t)=40.
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
from math import gcd
import heapq
def primeFactors(n):
# Returns a counter of prime factors of n
# e.g., 12 returns {2: 2, 3: 1}
assert n >= 1
factors = Counter()
while n % 2 == 0:
factors[2] += 1
n //= 2
x = 3
while x * x <= n:
while n % x == 0:
factors[x] += 1
n //= x
x += 2
if n != 1:
factors[n] += 1
return factors
def lcm(a, b):
return (a * b) // gcd(a, b)
def solve(N, A):
A = list(set(A))
N = len(A)
if len(A) == 1:
return A[0]
if len(A) <= 2:
l = []
for i in range(N):
for j in range(i + 1, N):
l.append(lcm(A[i], A[j]))
g = 0
for x in l:
g = gcd(g, x)
return g
pToCount = defaultdict(list)
for x in A:
for p, count in primeFactors(x).items():
pToCount[p].append(count)
g = 1
for p, counts in pToCount.items():
if N - len(counts) >= 2:
continue
g *= p ** max(heapq.nsmallest(2, counts))
return g
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
(N,) = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
ans = solve(N, A)
print(ans)
``` | instruction | 0 | 83,556 | 22 | 167,112 |
No | output | 1 | 83,556 | 22 | 167,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One quite ordinary day Valera went to school (there's nowhere else he should go on a week day). In a maths lesson his favorite teacher Ms. Evans told students about divisors. Despite the fact that Valera loved math, he didn't find this particular topic interesting. Even more, it seemed so boring that he fell asleep in the middle of a lesson. And only a loud ringing of a school bell could interrupt his sweet dream.
Of course, the valuable material and the teacher's explanations were lost. However, Valera will one way or another have to do the homework. As he does not know the new material absolutely, he cannot do the job himself. That's why he asked you to help. You're his best friend after all, you just cannot refuse to help.
Valera's home task has only one problem, which, though formulated in a very simple way, has not a trivial solution. Its statement looks as follows: if we consider all positive integers in the interval [a;b] then it is required to count the amount of such numbers in this interval that their smallest divisor will be a certain integer k (you do not have to consider divisor equal to one). In other words, you should count the amount of such numbers from the interval [a;b], that are not divisible by any number between 2 and k - 1 and yet are divisible by k.
Input
The first and only line contains three positive integers a, b, k (1 ≤ a ≤ b ≤ 2·109, 2 ≤ k ≤ 2·109).
Output
Print on a single line the answer to the given problem.
Examples
Input
1 10 2
Output
5
Input
12 23 3
Output
2
Input
6 19 5
Output
0
Note
Comments to the samples from the statement:
In the first sample the answer is numbers 2, 4, 6, 8, 10.
In the second one — 15, 21
In the third one there are no such numbers. | instruction | 0 | 83,902 | 22 | 167,804 |
Tags: dp, math, number theory
Correct Solution:
```
import math
def prime(x):
for i in range(2, int(math.sqrt(x))+1):
if x % i == 0:
return 0
return 1
def cal(n, k):
if not prime(k):
return 0
elif k >= n:
return k == n
res = n // k
for i in range(2, min(k, n // k +1)):
res -= cal(n // k, i)
return res
a, b, k = input().split()
a = int(a)
b = int(b)
k = int(k)
print(cal(b, k) - cal(a - 1, k))
``` | output | 1 | 83,902 | 22 | 167,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One quite ordinary day Valera went to school (there's nowhere else he should go on a week day). In a maths lesson his favorite teacher Ms. Evans told students about divisors. Despite the fact that Valera loved math, he didn't find this particular topic interesting. Even more, it seemed so boring that he fell asleep in the middle of a lesson. And only a loud ringing of a school bell could interrupt his sweet dream.
Of course, the valuable material and the teacher's explanations were lost. However, Valera will one way or another have to do the homework. As he does not know the new material absolutely, he cannot do the job himself. That's why he asked you to help. You're his best friend after all, you just cannot refuse to help.
Valera's home task has only one problem, which, though formulated in a very simple way, has not a trivial solution. Its statement looks as follows: if we consider all positive integers in the interval [a;b] then it is required to count the amount of such numbers in this interval that their smallest divisor will be a certain integer k (you do not have to consider divisor equal to one). In other words, you should count the amount of such numbers from the interval [a;b], that are not divisible by any number between 2 and k - 1 and yet are divisible by k.
Input
The first and only line contains three positive integers a, b, k (1 ≤ a ≤ b ≤ 2·109, 2 ≤ k ≤ 2·109).
Output
Print on a single line the answer to the given problem.
Examples
Input
1 10 2
Output
5
Input
12 23 3
Output
2
Input
6 19 5
Output
0
Note
Comments to the samples from the statement:
In the first sample the answer is numbers 2, 4, 6, 8, 10.
In the second one — 15, 21
In the third one there are no such numbers. | instruction | 0 | 83,903 | 22 | 167,806 |
Tags: dp, math, number theory
Correct Solution:
```
def pr(x):
d = 2
while d * d <= x:
if x % d == 0:
return 0
d += 1
return 1
def cnt(n, k):
if not pr(k) or n < k: return 0
n1 = n // k
return n1 - sum(cnt(n1, i) for i in range(2, min(k, n1 + 1)))
a, b, k = map(int, input().split())
ans = cnt(b, k) - cnt(a - 1, k)
print(ans)
``` | output | 1 | 83,903 | 22 | 167,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One quite ordinary day Valera went to school (there's nowhere else he should go on a week day). In a maths lesson his favorite teacher Ms. Evans told students about divisors. Despite the fact that Valera loved math, he didn't find this particular topic interesting. Even more, it seemed so boring that he fell asleep in the middle of a lesson. And only a loud ringing of a school bell could interrupt his sweet dream.
Of course, the valuable material and the teacher's explanations were lost. However, Valera will one way or another have to do the homework. As he does not know the new material absolutely, he cannot do the job himself. That's why he asked you to help. You're his best friend after all, you just cannot refuse to help.
Valera's home task has only one problem, which, though formulated in a very simple way, has not a trivial solution. Its statement looks as follows: if we consider all positive integers in the interval [a;b] then it is required to count the amount of such numbers in this interval that their smallest divisor will be a certain integer k (you do not have to consider divisor equal to one). In other words, you should count the amount of such numbers from the interval [a;b], that are not divisible by any number between 2 and k - 1 and yet are divisible by k.
Input
The first and only line contains three positive integers a, b, k (1 ≤ a ≤ b ≤ 2·109, 2 ≤ k ≤ 2·109).
Output
Print on a single line the answer to the given problem.
Examples
Input
1 10 2
Output
5
Input
12 23 3
Output
2
Input
6 19 5
Output
0
Note
Comments to the samples from the statement:
In the first sample the answer is numbers 2, 4, 6, 8, 10.
In the second one — 15, 21
In the third one there are no such numbers.
Submitted Solution:
```
import math
def P(n):
if n <= 1:
return 0
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return 0
return 1
def q(x,y,w):
return x//w-y//w+1
x=[int(i) for i in input().split()]
k=x[2]
c=int((x[0]-0.5)//k+1)
d=x[1]//k
if P(k)==0:
print(0)
raise SystemExit
if k==2:
print(d-c+1)
raise SystemExit
if k==3:
print(d-c+1-q(c,d,2))
raise SystemExit
if k==5:
print(d-c+1-q(c,d,2)-q(c,d,3)+q(c,d,6))
raise SystemExit
kp=[]
j=3
while j<k:
if P(j)==1:
kp.append(j)
j=j+2
s=0
i=c+int(c%2==0)
while i <=d:
t=1
for j in kp:
if i%j==0:
t=0
break
s=s+t
i=i+2
print(s)
``` | instruction | 0 | 83,904 | 22 | 167,808 |
No | output | 1 | 83,904 | 22 | 167,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One quite ordinary day Valera went to school (there's nowhere else he should go on a week day). In a maths lesson his favorite teacher Ms. Evans told students about divisors. Despite the fact that Valera loved math, he didn't find this particular topic interesting. Even more, it seemed so boring that he fell asleep in the middle of a lesson. And only a loud ringing of a school bell could interrupt his sweet dream.
Of course, the valuable material and the teacher's explanations were lost. However, Valera will one way or another have to do the homework. As he does not know the new material absolutely, he cannot do the job himself. That's why he asked you to help. You're his best friend after all, you just cannot refuse to help.
Valera's home task has only one problem, which, though formulated in a very simple way, has not a trivial solution. Its statement looks as follows: if we consider all positive integers in the interval [a;b] then it is required to count the amount of such numbers in this interval that their smallest divisor will be a certain integer k (you do not have to consider divisor equal to one). In other words, you should count the amount of such numbers from the interval [a;b], that are not divisible by any number between 2 and k - 1 and yet are divisible by k.
Input
The first and only line contains three positive integers a, b, k (1 ≤ a ≤ b ≤ 2·109, 2 ≤ k ≤ 2·109).
Output
Print on a single line the answer to the given problem.
Examples
Input
1 10 2
Output
5
Input
12 23 3
Output
2
Input
6 19 5
Output
0
Note
Comments to the samples from the statement:
In the first sample the answer is numbers 2, 4, 6, 8, 10.
In the second one — 15, 21
In the third one there are no such numbers.
Submitted Solution:
```
import math
def P(n):
if n <= 1:
return 0
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return 0
return 1
def q(x,y,w):
return x//w-y//w+1
x=[int(i) for i in input().split()]
k=x[2]
c=int((x[0]-0.5)//k+1)
d=x[1]//k
if P(k)==0:
print(0)
raise SystemExit
if k==2:
print(d-c+1)
raise SystemExit
if k==3:
print(q(c,d,3)-q(c,d,2))
raise SystemExit
if k==5:
print(q(c,d,5)-q(c,d,2)-q(c,d,3)+q(c,d,6))
raise SystemExit
kp=[]
j=3
while j<k:
if P(j)==1:
kp.append(j)
j=j+2
s=0
i=c+int(c%2==0)
while i <=d:
t=1
for j in kp:
if i%j==0:
t=0
break
s=s+t
i=i+2
print(s)
``` | instruction | 0 | 83,905 | 22 | 167,810 |
No | output | 1 | 83,905 | 22 | 167,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One quite ordinary day Valera went to school (there's nowhere else he should go on a week day). In a maths lesson his favorite teacher Ms. Evans told students about divisors. Despite the fact that Valera loved math, he didn't find this particular topic interesting. Even more, it seemed so boring that he fell asleep in the middle of a lesson. And only a loud ringing of a school bell could interrupt his sweet dream.
Of course, the valuable material and the teacher's explanations were lost. However, Valera will one way or another have to do the homework. As he does not know the new material absolutely, he cannot do the job himself. That's why he asked you to help. You're his best friend after all, you just cannot refuse to help.
Valera's home task has only one problem, which, though formulated in a very simple way, has not a trivial solution. Its statement looks as follows: if we consider all positive integers in the interval [a;b] then it is required to count the amount of such numbers in this interval that their smallest divisor will be a certain integer k (you do not have to consider divisor equal to one). In other words, you should count the amount of such numbers from the interval [a;b], that are not divisible by any number between 2 and k - 1 and yet are divisible by k.
Input
The first and only line contains three positive integers a, b, k (1 ≤ a ≤ b ≤ 2·109, 2 ≤ k ≤ 2·109).
Output
Print on a single line the answer to the given problem.
Examples
Input
1 10 2
Output
5
Input
12 23 3
Output
2
Input
6 19 5
Output
0
Note
Comments to the samples from the statement:
In the first sample the answer is numbers 2, 4, 6, 8, 10.
In the second one — 15, 21
In the third one there are no such numbers.
Submitted Solution:
```
import math
def P(n):
if n <= 1:
return 0
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return 0
return 1
def q(x,y,w):
return y//w-x//w+1
x=[int(i) for i in input().split()]
k=x[2]
c=int((x[0]-0.5)//k+1)
d=x[1]//k
if P(k)==0:
print(0)
raise SystemExit
if k==2:
print(d-c+1)
raise SystemExit
if k==3:
print(d-c+1-q(c,d,2))
raise SystemExit
if k==5:
print(d-c+1-q(c,d,2)-q(c,d,3)+q(c,d,6))
raise SystemExit
kp=[]
j=3
while j<k:
if P(j)==1:
kp.append(j)
j=j+2
s=0
i=c+int(c%2==0)
while i <=d:
t=1
for j in kp:
if i%j==0:
t=0
break
s=s+t
i=i+2
print(s)
``` | instruction | 0 | 83,906 | 22 | 167,812 |
No | output | 1 | 83,906 | 22 | 167,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One quite ordinary day Valera went to school (there's nowhere else he should go on a week day). In a maths lesson his favorite teacher Ms. Evans told students about divisors. Despite the fact that Valera loved math, he didn't find this particular topic interesting. Even more, it seemed so boring that he fell asleep in the middle of a lesson. And only a loud ringing of a school bell could interrupt his sweet dream.
Of course, the valuable material and the teacher's explanations were lost. However, Valera will one way or another have to do the homework. As he does not know the new material absolutely, he cannot do the job himself. That's why he asked you to help. You're his best friend after all, you just cannot refuse to help.
Valera's home task has only one problem, which, though formulated in a very simple way, has not a trivial solution. Its statement looks as follows: if we consider all positive integers in the interval [a;b] then it is required to count the amount of such numbers in this interval that their smallest divisor will be a certain integer k (you do not have to consider divisor equal to one). In other words, you should count the amount of such numbers from the interval [a;b], that are not divisible by any number between 2 and k - 1 and yet are divisible by k.
Input
The first and only line contains three positive integers a, b, k (1 ≤ a ≤ b ≤ 2·109, 2 ≤ k ≤ 2·109).
Output
Print on a single line the answer to the given problem.
Examples
Input
1 10 2
Output
5
Input
12 23 3
Output
2
Input
6 19 5
Output
0
Note
Comments to the samples from the statement:
In the first sample the answer is numbers 2, 4, 6, 8, 10.
In the second one — 15, 21
In the third one there are no such numbers.
Submitted Solution:
```
def pr(x):
d = 2
while d * d <= x:
if x % d == 0:
return 0
d += 1
return 1
def cnt(n, k):
if not pr(n) or n < k: return 0
n1 = n // k
return n1 - sum(cnt(n1, i) for i in range(2, min(k, n1 + 1)))
a, b, k = map(int, input().split())
ans = cnt(b, k) - cnt(a - 1, k)
print(ans)
``` | instruction | 0 | 83,907 | 22 | 167,814 |
No | output | 1 | 83,907 | 22 | 167,815 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4 | instruction | 0 | 84,158 | 22 | 168,316 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n = int(input())
i = 1
num = [0] * 13
while True:
cnt = 0
for j in range(1, i+1):
if i % j == 0:
cnt += 1
if cnt > 12:
i += 1
continue
if num[cnt] == 0:
num[cnt] = i
if num[n] > 0:
ans = num[n]
break
i += 1
print(ans)
``` | output | 1 | 84,158 | 22 | 168,317 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4 | instruction | 0 | 84,159 | 22 | 168,318 |
"Correct Solution:
```
print([0,1,2,4,6,16,12,64,24,36,48,1024,60][int(input())])
``` | output | 1 | 84,159 | 22 | 168,319 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4 | instruction | 0 | 84,160 | 22 | 168,320 |
"Correct Solution:
```
# AOJ 1562: Divisor
# Python3 2018.7.13 bal4u
ans = [0, 1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60]
print(ans[int(input())])
``` | output | 1 | 84,160 | 22 | 168,321 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4 | instruction | 0 | 84,161 | 22 | 168,322 |
"Correct Solution:
```
def solve(n):
for i in range(1,10000):
cnt=0
for j in range(1,i+1):
if i%j==0:
cnt+=1
if cnt==n:
return(i)
while True:
try:
n=int(input())
print(solve(n))
except EOFError:
break
``` | output | 1 | 84,161 | 22 | 168,323 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4 | instruction | 0 | 84,162 | 22 | 168,324 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def f(n):
ret = 0
for a in range(1, n+1):
if n % a == 0:
ret += 1
return ret
N = int(input())
n = 1
while True:
if f(n) == N:
print(n)
break
n += 1
``` | output | 1 | 84,162 | 22 | 168,325 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4 | instruction | 0 | 84,163 | 22 | 168,326 |
"Correct Solution:
```
print([1,2,4,6,16,12,64,24,36,48,1024,60][int(input())-1])
``` | output | 1 | 84,163 | 22 | 168,327 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4 | instruction | 0 | 84,164 | 22 | 168,328 |
"Correct Solution:
```
N = int(input())
target = 1
while True:
count = 0
for i in range(1, target+1):
if target % i == 0:
count += 1
if count == N:
print(target)
break
target += 1
``` | output | 1 | 84,164 | 22 | 168,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4
Submitted Solution:
```
def solve(n):
for i in range(1,100):
cnt=0
for j in range(1,i+1):
if i%j==0:
cnt+=1
if cnt==n:
return(i)
n=int(input())
print(solve(n))
``` | instruction | 0 | 84,165 | 22 | 168,330 |
No | output | 1 | 84,165 | 22 | 168,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n = int(input())
i = 1
num = [0] * 13 #num[i] -> ?´???°???i???????????°??§????°??????°
while True:
cnt = 0
for j in range(1, i+1):
if i % j == 0:
cnt += 1
if num[cnt] == 0:
num[cnt] = i
if num[n] > 0:
ans = num[n]
break
i += 1
print(ans)
``` | instruction | 0 | 84,166 | 22 | 168,332 |
No | output | 1 | 84,166 | 22 | 168,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4
Submitted Solution:
```
def solve(n):
for i in range(1,100):
cnt=0
for j in range(1,i+1):
if i%j==0:
cnt+=1
if cnt==n:
return(i)
while True:
try:
n=int(input())
print(solve(n))
except EOFError:
break
``` | instruction | 0 | 84,167 | 22 | 168,334 |
No | output | 1 | 84,167 | 22 | 168,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Given a natural number N less than or equal to 12, find the smallest natural number such that the number of divisors is exactly N.
Constraints
* 1 ≤ N ≤ 12
Input
One natural number N is given in one line.
Output
Output the smallest natural number on a line so that the number of divisors is exactly N.
Examples
Input
1
Output
1
Input
2
Output
2
Input
3
Output
4
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n = int(input())
i = 1
num = [0] * 13
while True:
cnt = 0
for j in range(1, i+1):
if i % j == 0:
cnt += 1
if cnt >= 12:
i += 1
continue
if num[cnt] == 0:
num[cnt] = i
if num[n] > 0:
ans = num[n]
break
i += 1
print(ans)
``` | instruction | 0 | 84,168 | 22 | 168,336 |
No | output | 1 | 84,168 | 22 | 168,337 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given a positive integer n.
Find a sequence of fractions (a_i)/(b_i), i = 1 … k (where a_i and b_i are positive integers) for some k such that:
$$$ \begin{cases} $b_i$ divides $n$, $1 < b_i < n$ for $i = 1 … k$ \\\ $1 ≤ a_i < b_i$ for $i = 1 … k$ \\\ \text{$∑_{i=1}^k (a_i)/(b_i) = 1 - 1/n$} \end{cases} $$$
Input
The input consists of a single integer n (2 ≤ n ≤ 10^9).
Output
In the first line print "YES" if there exists such a sequence of fractions or "NO" otherwise.
If there exists such a sequence, next lines should contain a description of the sequence in the following format.
The second line should contain integer k (1 ≤ k ≤ 100 000) — the number of elements in the sequence. It is guaranteed that if such a sequence exists, then there exists a sequence of length at most 100 000.
Next k lines should contain fractions of the sequence with two integers a_i and b_i on each line.
Examples
Input
2
Output
NO
Input
6
Output
YES
2
1 2
1 3
Note
In the second example there is a sequence 1/2, 1/3 such that 1/2 + 1/3 = 1 - 1/6. | instruction | 0 | 84,259 | 22 | 168,518 |
Tags: math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
from fractions import Fraction
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
n=input()
arr=[]
i=2
N=n
while i*i<=n:
if n%i==0:
#print i,n
n/=i
arr.append(i)
while n%i==0:
n/=i
arr[-1]*=i
i+=1
if n>1:
arr.append(n)
if len(arr)>1:
for i in range(1,arr[0]):
f=Fraction(N-1,N)-Fraction(i,arr[0])
if f.denominator<N:
pr('YES\n')
pr_num(2)
pr_arr([i,arr[0]])
pr_arr([f.numerator,f.denominator])
exit()
pr('NO')
``` | output | 1 | 84,259 | 22 | 168,519 |
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