message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer numbers x and y. An array F is called an y-factorization of x iff the following conditions are met:
* There are y elements in F, and all of them are integer numbers;
* <image>.
You have to count the number of pairwise distinct arrays that are y-factorizations of x. Two arrays A and B are considered different iff there exists at least one index i (1 ≤ i ≤ y) such that Ai ≠ Bi. Since the answer can be very large, print it modulo 109 + 7.
Input
The first line contains one integer q (1 ≤ q ≤ 105) — the number of testcases to solve.
Then q lines follow, each containing two integers xi and yi (1 ≤ xi, yi ≤ 106). Each of these lines represents a testcase.
Output
Print q integers. i-th integer has to be equal to the number of yi-factorizations of xi modulo 109 + 7.
Example
Input
2
6 3
4 2
Output
36
6
Note
In the second testcase of the example there are six y-factorizations:
* { - 4, - 1};
* { - 2, - 2};
* { - 1, - 4};
* {1, 4};
* {2, 2};
* {4, 1}.
Submitted Solution:
```
t=int(input())
f=[0]
M=1000000007
for i in range(1,2000001):
f.append(f[i-1]*i%M)
def C(a,b):
return f[a]*pow(f[b],M-2,M)*pow(f[a-b],M-2,M)
for i in range(t):
x,y=map(int,input().split())
y-=1
a=[]
k=2
while k<=x:
c=0
while x%k==0:
c+=1
x//=k
if c:a.append(c)
k+=1
ans=1
for i in a:
ans*=C(i+y,y)
print(ans*pow(2,y,M)%M)
``` | instruction | 0 | 19,074 | 22 | 38,148 |
No | output | 1 | 19,074 | 22 | 38,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
Help Vasya — find the volume disposition with the minimum number of divisors.
Input
The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
Output
Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
Examples
Input
2
Output
2 1
Input
3
Output
1 3 2 | instruction | 0 | 19,805 | 22 | 39,610 |
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
for i in range (2,n+1):
print(i,end=' ')
print(1)
# Made By Mostafa_Khaled
``` | output | 1 | 19,805 | 22 | 39,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
Help Vasya — find the volume disposition with the minimum number of divisors.
Input
The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
Output
Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
Examples
Input
2
Output
2 1
Input
3
Output
1 3 2 | instruction | 0 | 19,806 | 22 | 39,612 |
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
for i in range (2,n+1):
print(i,end=' ')
print(1)
``` | output | 1 | 19,806 | 22 | 39,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
Help Vasya — find the volume disposition with the minimum number of divisors.
Input
The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
Output
Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
Examples
Input
2
Output
2 1
Input
3
Output
1 3 2 | instruction | 0 | 19,807 | 22 | 39,614 |
Tags: constructive algorithms, math
Correct Solution:
```
print(*range(2, int(input())+1), end=' ')
print(1)
``` | output | 1 | 19,807 | 22 | 39,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
Help Vasya — find the volume disposition with the minimum number of divisors.
Input
The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
Output
Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
Examples
Input
2
Output
2 1
Input
3
Output
1 3 2 | instruction | 0 | 19,808 | 22 | 39,616 |
Tags: constructive algorithms, math
Correct Solution:
```
n, ans, be = int(input()), [], 1
if n & 1:
ans.append(1)
be += 1
for i in range(be, n + 1, 2):
ans.extend([i + 1, i])
print(*ans)
``` | output | 1 | 19,808 | 22 | 39,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
Help Vasya — find the volume disposition with the minimum number of divisors.
Input
The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
Output
Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
Examples
Input
2
Output
2 1
Input
3
Output
1 3 2 | instruction | 0 | 19,809 | 22 | 39,618 |
Tags: constructive algorithms, math
Correct Solution:
```
n = int(input())
a = [i+2-i%2*2 for i in range(2*(n//2))]
if n % 2 == 1:
a = [1] + list(map(lambda x: x+1, a))
print(" ".join(map(str, a)))
``` | output | 1 | 19,809 | 22 | 39,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
Help Vasya — find the volume disposition with the minimum number of divisors.
Input
The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
Output
Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
Examples
Input
2
Output
2 1
Input
3
Output
1 3 2 | instruction | 0 | 19,811 | 22 | 39,622 |
Tags: constructive algorithms, math
Correct Solution:
```
#!/usr/bin/env python3
# Consecutive numbers have gcd 1
n = int(input())
answer = [n] + list(range(1,n))
print(" ".join(["%d" % d for d in answer]))
``` | output | 1 | 19,811 | 22 | 39,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
Help Vasya — find the volume disposition with the minimum number of divisors.
Input
The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
Output
Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
Examples
Input
2
Output
2 1
Input
3
Output
1 3 2 | instruction | 0 | 19,812 | 22 | 39,624 |
Tags: constructive algorithms, math
Correct Solution:
```
print(" ".join([str(x) for x in range(2,int(input())+1)]+["1"]))
``` | output | 1 | 19,812 | 22 | 39,625 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,194 | 22 | 40,388 |
"Correct Solution:
```
while True:
a, b = map(int, input().split())
if a == 0 and b == 0:
break
if a < b:
a, b = b, a
X = a
Y = b
i = 0
while True:
X %= Y
X, Y = Y, X
i += 1
if Y == 0:
break
print(X,i)
``` | output | 1 | 20,194 | 22 | 40,389 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,195 | 22 | 40,390 |
"Correct Solution:
```
while True:
X,Y=map(int,input().split())
if X==0 and Y==0:
break
n=0
while True:
if Y==0:
break
elif X<=Y:
X,Y=Y,X
X%=Y
X,Y=Y,X
n+=1
print(X,n)
``` | output | 1 | 20,195 | 22 | 40,391 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,196 | 22 | 40,392 |
"Correct Solution:
```
while True:
x, y = [int(x) for x in input().split()]
if x == 0 and y == 0:
break
c = 0
if y > x:
x,y = y, x
while True:
c += 1
x = x % y
x,y = y,x
if y == 0:
break
print(x,c)
``` | output | 1 | 20,196 | 22 | 40,393 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,197 | 22 | 40,394 |
"Correct Solution:
```
while True:
n,m = map(int,input().split())
if n==0 and m==0:
break
step = 0
if n >= m:
X = n
Y = m
while True:
X = X % Y
X,Y = Y,X
step +=1
if Y == 0:
ans = X
break
else:
X = m
Y = n
while True:
X = X % Y
X,Y = Y,X
step +=1
if Y == 0 :
ans = X
break
print(ans,step)
``` | output | 1 | 20,197 | 22 | 40,395 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,198 | 22 | 40,396 |
"Correct Solution:
```
while True:
a,b = [int(i) for i in input().split()]
if a == 0 and b == 0:
break
if a < b:
a,b = b,a
cnt = 0
while b > 0:
a, b = b, a % b
cnt += 1
print(a, cnt)
``` | output | 1 | 20,198 | 22 | 40,397 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,199 | 22 | 40,398 |
"Correct Solution:
```
while True:
a,b=map(int,input().split())
if a==0 and b==0:
break
c=0
if a<b:
a,b=b,a
while b!=0:
c+=1
d=a%b
a=b
b=d
print(a,c)
``` | output | 1 | 20,199 | 22 | 40,399 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,200 | 22 | 40,400 |
"Correct Solution:
```
def gozyo(x,y):
w = 0
while True:
if y == 0:
return x,w
x = x % y
z = x
x = y
y = z
w += 1
while True:
a,b = map(int,input().split())
if a == 0 and b == 0:
break
if a < b:
c = a
a = b
b = c
u = gozyo(a,b)
print(u[0],u[1])
``` | output | 1 | 20,200 | 22 | 40,401 |
Provide a correct Python 3 solution for this coding contest problem.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1 | instruction | 0 | 20,201 | 22 | 40,402 |
"Correct Solution:
```
while True:
x,y = map(int, input().split())
if x==0 and y==0:
break
elif x<y:
X=y
Y=x
else:
X=x
Y=y
i = 0
while True:
a = X%Y
X = a
Z = Y
Y = X
X = Z
i += 1
if a==0:
break
print(X, i)
``` | output | 1 | 20,201 | 22 | 40,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1
Submitted Solution:
```
while True:
a, b = map(int,input().split())
if a == 0 and b == 0:
break
if a < b:
a, b = b, a
C = 0
while True:
C = C + 1
a = a % b
if a == 0:
break
a, b = b, a
print(b, C)
``` | instruction | 0 | 20,202 | 22 | 40,404 |
Yes | output | 1 | 20,202 | 22 | 40,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1
Submitted Solution:
```
def euclid_gcd(v1, v2, counter):
if v2 == 0:
return (v1, counter)
else:
return euclid_gcd(v2, v1 % v2, counter + 1)
while True:
a, b = map(int, input().split())
if a == 0:
break
a, b = max(a, b), min(a, b)
print(*euclid_gcd(a, b, 0))
``` | instruction | 0 | 20,203 | 22 | 40,406 |
Yes | output | 1 | 20,203 | 22 | 40,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1
Submitted Solution:
```
# coding: utf-8
# Your code here!
while True:
X,Y=map(int, input().split())
if X==Y==0:
break
i=0
if X<Y:
X,Y=Y,X
while Y!=0:
X=X%Y
X,Y=Y,X
i+=1
print(X, i)
``` | instruction | 0 | 20,204 | 22 | 40,408 |
Yes | output | 1 | 20,204 | 22 | 40,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0197
"""
import sys
from sys import stdin
input = stdin.readline
def solve(a, b):
count = 0
if a > b:
X, Y = a, b
else:
X, Y = b, a
r = 1
while r:
r = X % Y
X = r
X, Y = Y, X
count += 1
return X, count
def main(args):
while True:
a, b = map(int, input().split())
if a == 0 and b == 0:
break
gcd, count = solve(a, b)
print(gcd, count)
if __name__ == '__main__':
main(sys.argv[1:])
``` | instruction | 0 | 20,205 | 22 | 40,410 |
Yes | output | 1 | 20,205 | 22 | 40,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1
Submitted Solution:
```
# Aizu Problem 0197: Greatest Common Divisor: Euclidean Algorithm
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def extended_gcd(a, b):
steps = 0
s, old_s = 0, 1
t, old_t = 1, 0
r, old_r = b, a
while r != 0:
steps += 1
q = old_r // r
old_r, r = r, old_r - q * r
old_s, s = s, old_s - q * s
old_t, t = t, old_t - q * t
return old_r, steps
while True:
a, b = [int(_) for _ in input().split()]
if a == b == 0:
break
g, steps = extended_gcd(a, b)
print(g, steps)
``` | instruction | 0 | 20,206 | 22 | 40,412 |
No | output | 1 | 20,206 | 22 | 40,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The greatest common divisor is an indispensable element in mathematics handled on a computer. Using the greatest common divisor can make a big difference in the efficiency of the calculation. One of the algorithms to find the greatest common divisor is "Euclidean algorithm". The flow of the process is shown below.
<image>
For example, for 1071 and 1029, substitute 1071 for X and 1029 for Y,
The remainder of 1071 ÷ 1029 is 42, and 42 is substituted for X to replace X and Y. (1 step)
The remainder of 1029 ÷ 42 is 21, and 21 is substituted for X to replace X and Y. (2 steps)
The remainder of 42 ÷ 21 is 0, and 0 is substituted for X to replace X and Y. (3 steps)
Since Y has become 0, X at this time is the greatest common divisor. Therefore, the greatest common divisor is 21.
In this way, we were able to find the greatest common divisor of 1071 and 1029 in just three steps. The Euclidean algorithm produces results overwhelmingly faster than the method of comparing divisors.
Create a program that takes two integers as inputs, finds the greatest common divisor using the Euclidean algorithm, and outputs the greatest common divisor and the number of steps required for the calculation.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Two integers a, b (2 ≤ a, b ≤ 231-1) are given on one line for each dataset.
The number of datasets does not exceed 1000.
Output
For each data set, the greatest common divisor of the two input integers and the number of steps of the Euclidean algorithm for calculation are output on one line separated by blanks.
Example
Input
1071 1029
5 5
0 0
Output
21 3
5 1
Submitted Solution:
```
def gcd(a,b):
c=0
while b:(a,b),c=(b,a%b),c+1
print(a,c)
while 1:
n,m=map(int,input().split())
if n==0:break
gcd(n,m)
``` | instruction | 0 | 20,207 | 22 | 40,414 |
No | output | 1 | 20,207 | 22 | 40,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,564 | 22 | 41,128 |
Tags: number theory
Correct Solution:
```
n = int(input())
if n == 1:
print(1)
elif n == 2:
print(2)
elif n == 3:
print(6)
elif n%2 == 1:
print(n*(n-1)*(n-2))
else:
if n%3 == 0:
print((n-1)*(n-2)*(n-3))
else:
print(n*(n-1)*(n-3))
``` | output | 1 | 20,564 | 22 | 41,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,565 | 22 | 41,130 |
Tags: number theory
Correct Solution:
```
n=int(input())
if n>2:
if n%2:
print((n)*(n-1)*(n-2))
elif n%3:
print((n)*(n-1)*(n-3))
else:
print((n-1)*(n-2)*(n-3))
else:
print(n)
``` | output | 1 | 20,565 | 22 | 41,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,566 | 22 | 41,132 |
Tags: number theory
Correct Solution:
```
n=int(input())
if(n<3):
print(n)
else:
if(n%2==0):
if(n%3==0):
print((n-1)*(n-3)*(n-2))
else:
print(n*(n-1)*(n-3))
else:
print(n*(n-1)*(n-2))
``` | output | 1 | 20,566 | 22 | 41,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,567 | 22 | 41,134 |
Tags: number theory
Correct Solution:
```
def gcd(x,y):
while y!=0: x,y=y,x%y
return x
def lcm(x,y):
_lcm=(x*y/gcd(x,y))
if int(_lcm)==_lcm: return int(_lcm)
else: return 0
def BiggestLCM(n):
_b=False
k=n-1
LCM=n*k
U=LCM
i=n-2
while gcd(LCM,i)!=1 and i>0:
if lcm(LCM,i)>U: U=lcm(LCM,i)
i-=1
a=1
if i>0:
a=LCM*i
if a>U:return(a)
return(U)
n=int(input())
if n>2 and n%2==0:
print (max([BiggestLCM(n),BiggestLCM(n-1)]))
else: print(BiggestLCM(n))
``` | output | 1 | 20,567 | 22 | 41,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,568 | 22 | 41,136 |
Tags: number theory
Correct Solution:
```
n=int(input())
if n<=2:
print(n)
elif n%6==0:
print((n-1)*(n-2)*(n-3))
elif n%2==0:
print((n)*(n-1)*(n-3))
else:
print((n)*(n-1)*(n-2))
``` | output | 1 | 20,568 | 22 | 41,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,569 | 22 | 41,138 |
Tags: number theory
Correct Solution:
```
n = int(input())
if(n==1):
print(1)
elif(n==2):
print(2)
else:
if(n%2!=0):
print(n*(n-1)*(n-2))
elif(n%6==0):
print((n-1)*(n-2)*(n-3))
else:
print((n)*(n-1)*(n-3))
``` | output | 1 | 20,569 | 22 | 41,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,570 | 22 | 41,140 |
Tags: number theory
Correct Solution:
```
"""
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"""
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def lcm(a, b, c):
l1 = (a * b) // gcd(a, b)
return (l1 * c) // gcd(l1, c)
n = int(data())
if n <= 2:
out(n)
exit()
if n == 3:
out(6)
exit()
answer = 0
for i in range(n, max(1, n-100), -1):
for j in range(i-1, max(1, n-100), -1):
for k in range(j-1, max(1, n-100), -1):
answer = max(answer, lcm(i, j, k))
out(answer)
``` | output | 1 | 20,570 | 22 | 41,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | instruction | 0 | 20,571 | 22 | 41,142 |
Tags: number theory
Correct Solution:
```
from math import *
n, ans = int(input()), 0
if n < 3:
ans = n
elif n % 2:
ans = n * (n - 1) * (n - 2)
else:
a = 0
if gcd(n, n - 3) == 1:
a = n * (n - 1) * (n - 3)
n -= 1
ans = max(n * (n - 1) * (n - 2), a)
print(ans)
``` | output | 1 | 20,571 | 22 | 41,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
#code
import math as m
def lcm(a, b):
return (a*b)//(m.gcd(a, b))
n = int(input())
var = m.gcd(n, n - 1)
if(n == 1):
print(1)
elif(n == 2):
print(2)
elif(n == 3):
print(6)
elif(n == 4):
print(12)
else:
print(max(lcm(n, lcm(n - 1, n - 2)),lcm(n - 3, lcm(n - 1, n - 2)), lcm(n, lcm(n - 1, n - 3))))
``` | instruction | 0 | 20,572 | 22 | 41,144 |
Yes | output | 1 | 20,572 | 22 | 41,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
import math
n = int(input())
ans = n
s = [n, n, n]
for a in range(max(1, n-100), n+1):
for b in range(a, n+1):
m1 = a * b // math.gcd(a, b)
for c in range(b, n+1):
if (m1 * c // math.gcd(c, m1) >= ans):
ans = m1 * c // math.gcd(c, m1)
s = [a, b, c]
print(ans)
# print(s)
``` | instruction | 0 | 20,573 | 22 | 41,146 |
Yes | output | 1 | 20,573 | 22 | 41,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
n = int(input())
if n==1 or n == 2:
ans = n
elif n == 3:
ans = n * 2
elif n % 2 == 0:
if n % 3 == 0:
ans = (n-1) * (n-2) * (n-3)
else:
ans = n * (n-1) * (n-3)
else:
ans = n * (n-1) * (n-2)
print(ans)
``` | instruction | 0 | 20,574 | 22 | 41,148 |
Yes | output | 1 | 20,574 | 22 | 41,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
"""
This template is made by Satwik_Tiwari.
python programmers can use this template :)) .
"""
#===============================================================================================
#importing some useful libraries.
import sys
import bisect
import heapq
from math import *
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl #
from bisect import bisect_right as br
from bisect import bisect
#===============================================================================================
#some shortcuts
mod = pow(10, 9) + 7
def inp(): return sys.stdin.readline().strip() #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
#===============================================================================================
# code here ;))
def solve():
n = int(inp())
a = n
b = n-1
ans = a*b
if(n==1):
print(1)
elif(n == 2):
print(2)
elif(n==3):
print(6)
elif(n>20):
ans = 0
for i in range(n-20,n+1):
for j in range(n-20,n+1):
for k in range(n-20,n+1):
ans = max(ans,lcm(lcm(i,j),k))
print(ans)
else:
ans = 0
for i in range(1,n+1):
for j in range(1,n+1):
for k in range(1,n+1):
ans = max(ans,lcm(lcm(i,j),k))
print(ans)
testcase(1)
``` | instruction | 0 | 20,575 | 22 | 41,150 |
Yes | output | 1 | 20,575 | 22 | 41,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
n=int(input())
if n<=2:
print(n)
elif n%2==0:
print((n)*(n-1)*(n-3))
elif n%6==0:
print((n-1)*(n-2)*(n-3))
else:
print((n)*(n-1)*(n-2))
``` | instruction | 0 | 20,576 | 22 | 41,152 |
No | output | 1 | 20,576 | 22 | 41,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
n=int(input())
if n>2:
print (n*(n-1)*(n-2))
elif n==2:
print ('2')
else:
print('1')
``` | instruction | 0 | 20,577 | 22 | 41,154 |
No | output | 1 | 20,577 | 22 | 41,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
n=int(input())
if n>4:
print(n*(n-1)*(n-2))
elif n==4:
print(12)
elif n==3:
print(6)
elif n==2:
print(2)
else:
print(1)
``` | instruction | 0 | 20,578 | 22 | 41,156 |
No | output | 1 | 20,578 | 22 | 41,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?
Input
The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.
Output
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.
Examples
Input
9
Output
504
Input
7
Output
210
Note
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
Submitted Solution:
```
import sys
import math
import itertools
import collections
def getdict(n):
d = {}
if type(n) is list or type(n) is str:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 1
else:
d[t] = 1
return d
def divs(n, start=1):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
def wr(arr): return '\n'.join(map(str, arr))
def revn(n): return int(str(n)[::-1])
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
n = ii()
if n < 3:
print(n)
else:
a, b = n, n - 1
for i in range(n - 2, 0, -1):
if math.gcd(a, i) == 1 and math.gcd(b, i) == 1:
exit(print(a * b * i))
``` | instruction | 0 | 20,579 | 22 | 41,158 |
No | output | 1 | 20,579 | 22 | 41,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,812 | 22 | 41,624 |
Tags: math, number theory
Correct Solution:
```
mod=1e9+7
def primeFactors(n):
factores=[]
if n%2==0:
exp=0
while n%2 ==0:
exp+=1
n//=2
factores.append((2,exp))
i=3
while(i*i<=n):
if(n%i==0):
exp=0
while n%i ==0:
exp+=1
n//=i
factores.append((i,exp))
i+=2
if n > 2:
factores.append((n,1))
return factores
def Phi(n):
phi=1
for fact in primeFactors(n):
phi*=(fact[0]**(fact[1]-1))*(fact[0]-1)
return phi
def f(n):
return Phi(n)
def g(n):
return n
n,k=map(int,input().split())
ans=n
k=(k+1)//2
while k!=0:
ans=Phi(ans)
if ans==1:
break;
k -=1
ans=int(ans%mod)
print(ans)
``` | output | 1 | 20,812 | 22 | 41,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,813 | 22 | 41,626 |
Tags: math, number theory
Correct Solution:
```
phi = [0 for i in range(100100)]
phi[1] = 1
for i in range(2, 100050):
if (phi[i] == 0):
for j in range(i, 100051, i):
if (phi[j] == 0):
phi[j]=j
phi[j] = phi[j]//i*(i-1)
def Fenjie(n):
k = {}
if (n == 1):
return {}
a = 2
while (n >= 2):
b = n%a
if (a*a > n):
if (n in k):
k[n] += 1
else:
k[n] = 1
break
if (b == 0):
if (a in k):
k[a] += 1
else:
k[a] = 1
n = n//a
else:
a += 1
return k
def Euler(k):
if (len(k) == 0):
return 1
ans = 1
for i in k:
s = k[i]
ans *= pow(i, s)
ans = ans//i*(i-1)
return ans
n, k = map(int, input().split())
k = (k+1)//2
while (k):
k -= 1
n = Euler(Fenjie(n))
if (n==1):
break
print(n%(10**9+7))
``` | output | 1 | 20,813 | 22 | 41,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,814 | 22 | 41,628 |
Tags: math, number theory
Correct Solution:
```
import math
def phi(n):
res = n
for i in range(2, int(math.sqrt(n)) + 1):
if(n % i == 0):
while(n % i == 0):
n /= i
res -= res/i
if(n > 1):
res -= int(res / n)
return res
n, k = map(int, input().split())
res = n
k = int((k + 1) / 2)
while(res != 1 and k != 0):
res = phi(res)
k -= 1
print("{}".format(int(res % (1e9 + 7))))
``` | output | 1 | 20,814 | 22 | 41,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,815 | 22 | 41,630 |
Tags: math, number theory
Correct Solution:
```
import math
def phi(n):
res = n
for i in range(2, int(math.sqrt(n)) + 1):
if(n % i == 0):
while n % i == 0:
n /= i
res -= res/i
if(n>1):
res -= int(res/n)
return int(res)
def F(n, k):
k = int((k + 1)/2)
res = n
while k != 0:
res = phi(res)
if res == 1:
break
k -=1
return int(res % (10 ** 9 + 7))
n, k = [int(i) for i in input().split(' ')]
print(F(n, k))
``` | output | 1 | 20,815 | 22 | 41,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,816 | 22 | 41,632 |
Tags: math, number theory
Correct Solution:
```
MOD = 1000000007
def phi(n):
res = n
for i in range(2,int(n**(0.5)+1)):
if n % i == 0:
while n % i == 0:
n = n//i
res -= res//i
if n > 1:
res -= res//n
return res
n,k = map(int,input().split())
k = (k+1)//2
ans = n
for _ in range(k):
if ans > 1:
ans = phi(ans)
else:
break
print(ans % MOD)
``` | output | 1 | 20,816 | 22 | 41,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,817 | 22 | 41,634 |
Tags: math, number theory
Correct Solution:
```
import math
def f_k_n(n,k):
k = int((k+1)/2)
while n>1 and k:
n = phi_improved(n)
k -= 1
return n
def phi(n):
p = 1
for i in range(2,n):
if gcd(i,n) == 1:
p += 1
return p
def phi_improved(n):
prime_fact = find_prime_factors(n)
result = n
for p in prime_fact:
result *= 1 - (1/p)
return int(result)
def find_prime_factors(n):
l = []
temp_n = n
if temp_n % 2 == 0:
l.append(2)
#mientras sea divisible por 2 lo divido hasta que sea impar
while temp_n % 2 == 0:
temp_n = temp_n / 2
i = 3
last_i = 0
#busco hasta la raiz los factores de la descomposicion en primos
while i <= math.sqrt(n):
if temp_n % i == 0:
if last_i != i:
l.append(i)
last_i = i
temp_n = temp_n / i
else:
i += 2
# ya no tiene mas divisores entonces es primo y no se ha agregado a la lista
if temp_n > 2:
l.append(int(temp_n))
return l
def gcd(x,y):
r = x % y
while r != 0:
x = y
y = r
r = x % y
return y
def find_dividers(n):
d = 2
dividers = [1]
while d < math.sqrt(n):
if n % d == 0:
dividers.append(d)
d += 1
return dividers
def main():
n,k = [int(i) for i in input().split()]
print(str(f_k_n(n,k) % 1000000007))
main()
``` | output | 1 | 20,817 | 22 | 41,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,818 | 22 | 41,636 |
Tags: math, number theory
Correct Solution:
```
mod=1e9+7 # la respuesta te piden que se de modulo este numero
def primeFactors(n): #Algoritmo para la descomposicion de n en factores primos
factores=[]
if n%2==0: # todo numero de la forma 2k tiene a 2 en su descomposicion en primos por lo que al ser bastante probable que aparezca lo analisamos como un caso aparte
exp=0
while n%2 ==0: # mientras todavia tenga a 2 en su descomposicion 2
exp+=1
n//=2
factores.append((2,exp)) #al terminar me quedo con un n resultado de quitar tanta veces como aparezca 2 en su descomposicion
i=3 #Al no tener que considerar los numeros de la forma 2k pues como son multiplos de 2 no pueden ser primos empiezo en 3 y avanzo de 2 en 2
while(i*i<=n): #solamente se analiza hasta sqrt(n) pues el unico primo mayor que la raiz que puede estar en la descomposicion es el propio n en caso que n sea primo
if(n%i==0): # en caso que i divida a n es porque i es un factor primo de n pues de lo contrario el se dividiria por otro primo hallado previamente y eso no puede ser pues esos primos fueron quitados de n
exp=0
while n%i ==0: # tantas veces como aparezca ese primo en la descomposicion lo voy quitando de n
exp+=1
n//=i
factores.append((i,exp))
i+=2
if n > 2: # en caso que el n que me quede sea primo entonces el tambien forma parte de la descomposicion
factores.append((n,1))
return factores # me quedo con la descomposicion en primos de n en una lista que tiene para la posicion i una tupla <primo i,exponente i >
def Phi(n): # Calcula Phi de n
phi=1
for fact in primeFactors(n): #Recorro cada primo que esta en la descomposicion de n
phi*=(fact[0]**(fact[1]-1))*(fact[0]-1) # voy actualizando phi usando la propiedad que phi(n)=p1^{e1-1}*(p1-1)+p2^{e2-1}*(p2-1)+...+pk^{ek-1}*(pk-1)
return phi #devuevlo el valor de Phi(n)
n,k=map(int,input().split()) #recibo los enteros n y k de entrada al problema
ans=n # variable que va a ir guardando la salida de cada computo de phi(n) mientras queden iteraciones por hacerse
k=(k+1)//2 # numero de iteraciones que voy a ejecutar Phi
while k!=0:
ans=Phi(ans) # en cada paso calculo Phi del resultado previo de haber computado Phi o n la primera vez
if ans==1: # Cuando ans=1 Phi(1) es 1 por lo que no seria eficiente seguir iterando por el while pues nos daria el mismo resultado y nos pudieramos quitar esas iteraciones
break;
k -=1 # voy disminuyendo el contador en 1 hasta hacer las |_k+1/2_| iteraciones necesarias para el computo de F(k,n)
ans=int(ans%mod) # Devuelvo la respuesta modulo 10^9 +7 pues asi es como nos los piden en el problema ya que nos pueden entrar valores de n y k bastantes grandes
print(ans) # imprimo la respuesta al problema F(n,k)
``` | output | 1 | 20,818 | 22 | 41,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6. | instruction | 0 | 20,819 | 22 | 41,638 |
Tags: math, number theory
Correct Solution:
```
import math
def f_k_n(n,k):
k = int((k+1)/2)
while n>1 and k:
n = phi_improved(n)
k -= 1
return n
def phi_improved(n):
prime_fact = find_prime_factors(n)
result = n
for p in prime_fact:
result *= 1 - (1/p)
return int(result)
def find_prime_factors(n):
l = []
temp_n = n
if temp_n % 2 == 0:
l.append(2)
#mientras sea divisible por 2 lo divido hasta que sea impar
while temp_n % 2 == 0:
temp_n = temp_n / 2
i = 3
last_i = 0
#busco hasta la raiz los factores de la descomposicion en primos
while i <= math.sqrt(n):
if temp_n % i == 0:
if last_i != i:
l.append(i)
last_i = i
temp_n = temp_n / i
else:
i += 2
# ya no tiene mas divisores entonces es primo y no se ha agregado a la lista
if temp_n > 2:
l.append(int(temp_n))
return l
def main():
n,k = [int(i) for i in input().split()]
print(str(f_k_n(n,k) % 1000000007))
main()
``` | output | 1 | 20,819 | 22 | 41,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6.
Submitted Solution:
```
mod=1e9+7
def Phi(n):
phi=n
i=2
while i*i<= n:
if n%i==0:
while n%i==0:
n//=i
phi-=phi//i
i+=1
if n > 1:
phi-=phi//n
return phi
def f(n):
return Phi(n)
def g(n):
return n
n,k=map(int,input().split())
ans=n
k=(k+1)//2
while k!=0:
ans=Phi(ans)
if ans==1:
break;
k -=1
ans=int(ans%mod)
print(ans)
``` | instruction | 0 | 20,820 | 22 | 41,640 |
Yes | output | 1 | 20,820 | 22 | 41,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6.
Submitted Solution:
```
import math
def f_k_n(n,k):
k = int((k+1)/2)
while n>1 and k:
n = phi_improved(n)
k -= 1
return n
def f(n):
return phi_improved(n)
def phi_improved(n):
prime_fact = find_prime_factors(n)
result = n
for p in prime_fact:
result *= 1 - (1/p)
return int(result)
def find_prime_factors(n):
l = []
temp_n = n
if temp_n % 2 == 0:
l.append(2)
while temp_n % 2 == 0:#mientras sea divisible por 2 lo divido hasta que sea impar
temp_n = temp_n / 2
i = 3
last_i = 0
while i <= math.sqrt(n):#busco hasta la raiz los factores de la descomposicion en primos
if temp_n % i == 0:
if last_i != i:
l.append(i)
last_i = i
temp_n = temp_n / i
else:
i += 2
if temp_n > 2:# ya no tiene mas divisores entonces es primo y no se ha agregado a la lista
l.append(int(temp_n))
#print(l)
return l
def main():
n,k = [int(i) for i in input().split()]
print(str(f_k_n(n,k) % 1000000007))
main()
``` | instruction | 0 | 20,821 | 22 | 41,642 |
Yes | output | 1 | 20,821 | 22 | 41,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6.
Submitted Solution:
```
import math
def f_k_n(n,k):
k = int((k+1)/2)
while n>1 and k:
n = phi_improved(n)
k =k-1
return n
def f(n):
return phi_improved(n)
def phi_improved(n):
prime_fact = find_prime_factors(n)
result = n
for p in prime_fact:
result *= 1 - (1/p)
return int(result)
def find_prime_factors(n):
l = []
temp_n = n
if temp_n % 2 == 0:
l.append(2)
while temp_n % 2 == 0:#mientras sea divisible por 2 lo divido hasta que sea impar
temp_n = temp_n / 2
i = 3
last_i = 0
while i <= math.sqrt(n):#busco hasta la raiz los factores de la descomposicion en primos
if temp_n % i == 0:
if last_i != i:
l.append(i)
last_i = i
temp_n = temp_n / i
else:
i += 2
if temp_n > 2:# ya no tiene mas divisores entonces es primo y no se ha agregado a la lista
l.append(int(temp_n))
return l
def main():
n,k = [int(i) for i in input().split()]
print(str(f_k_n(n,k) % 1000000007))
main()
``` | instruction | 0 | 20,822 | 22 | 41,644 |
Yes | output | 1 | 20,822 | 22 | 41,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Holmes children are fighting over who amongst them is the cleverest.
Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.
Sherlock said that solving this was child's play and asked Mycroft to instead get the value of <image>. Summation is done over all positive integers d that divide n.
Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.
She defined a k-composite function Fk(n) recursively as follows:
<image>
She wants them to tell the value of Fk(n) modulo 1000000007.
Input
A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.
Output
Output a single integer — the value of Fk(n) modulo 1000000007.
Examples
Input
7 1
Output
6
Input
10 2
Output
4
Note
In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6.
Submitted Solution:
```
import fractions
def phi(n):
amount = 0
for k in range(1, n + 1):
if fractions.gcd(n, k) == 1:
amount += 1
return amount
n,k=map(int,input().split())
if n==557056 and k==12:
print(8192)
elif int((k-1)/2)>n:
u1=1
else:
u1=phi(n)%1000000007
for j in range(1,int((k-1)/2)+1):
if u1==1:
break
u1=phi(u1)%1000000007
print(u1)
``` | instruction | 0 | 20,823 | 22 | 41,646 |
No | output | 1 | 20,823 | 22 | 41,647 |
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