message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,234 | 22 | 178,468 |
"Correct Solution:
```
def gcd(a,b):
if b>a:
a,b = b,a
r = a%b
if r==0:
return b
else:
return gcd(b, a%b)
def solve(a,b):
if a==1:
return 1,0
if b==1:
return 0,1
if a>b:
q = a//b
r = a%b
x, y = solve(r,b)
return x, y-q*x
if a<b:
q = b//a
r = b%a
x, y = solve(a,r)
return x-q*y, y
a,b = map(int, input().split())
g = gcd(a,b)
a//=g
b//=g
x, y = solve(a,b)
def p(n):
return x+b*n
def q(n):
return y-a*n
def f(n):
return abs(p(n))+abs(q(n))
if a>b:
n = y//a
nn = n
m = f(n)
if f(n-1) < m:
nn = n-1
m = f(n-1)
if f(n+1) < m:
nn = n+1
m = f(n+1)
x = p(nn)
y = q(nn)
elif a<b:
n = -x//b
nn = n
m = f(n)
if f(n-1) < m:
nn = n-1
m = f(n-1)
if f(n+1) < m:
nn = n+1
m = f(n+1)
x = p(nn)
y = q(nn)
else:
x = 0
y = 1
print("{} {}".format(x,y))
``` | output | 1 | 89,234 | 22 | 178,469 |
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,235 | 22 | 178,470 |
"Correct Solution:
```
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from fractions import gcd
from bisect import bisect_left
from heapq import heappush, heappop
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
a, b = MAP()
c = gcd(a, b)
def extgcd(a, b, d=0):
g = a
if b == 0:
x, y = 1, 0
else:
x, y, g = extgcd(b, a % b)
x, y = y, x - a // b * y
return x, y, g
print(*extgcd(a, b)[:2])
``` | output | 1 | 89,235 | 22 | 178,471 |
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,236 | 22 | 178,472 |
"Correct Solution:
```
def gcd(a, b):
global queue
r = a % b
if r:
d = a // b
sb = queue.pop()
sa = queue.pop()
queue.append(sb)
queue.append(tuple(map(lambda x, y: x - d * y, sa, sb)))
return gcd(b, r)
else:
return b
a, b = map(int, input().split())
queue = [(1, 0), (0, 1)]
g = gcd(a, b)
print(*queue.pop())
``` | output | 1 | 89,236 | 22 | 178,473 |
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,237 | 22 | 178,474 |
"Correct Solution:
```
#E
a,b = map(int,input().split())
def extgcd(x,y):
c1,c2 = x,y
a1,a2 = 1,0
b1,b2 = 0,1
while c2:
m = c1%c2
q = c1//c2
c1,c2 = c2,m
a1,a2 = a2,(a1-q*a2)
b1,b2 = b2,(b1-q*b2)
return c1,a1,b1
ans = extgcd(a,b)
print(ans[1],ans[2])
``` | output | 1 | 89,237 | 22 | 178,475 |
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,238 | 22 | 178,476 |
"Correct Solution:
```
def extgcd(a, b):
if b == 0:
return a, 1, 0
else:
d, y, x = extgcd(b, a%b)
y -= a // b * x
return d, x, y
a, b = map(int, input().split())
_, x, y = extgcd(a, b)
print(x, y)
``` | output | 1 | 89,238 | 22 | 178,477 |
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,239 | 22 | 178,478 |
"Correct Solution:
```
import math
a, b = map(int, input().split())
c = math.gcd(a, b)
def ex_euclid(x, y):
c0, c1 = x, y
a0, a1 = 1, 0
b0, b1 = 0, 1
while c1 != 0:
m = c0 % c1
q = c0 // c1
c0, c1 = c1, m
a0, a1 = a1, (a0 - q * a1)
b0, b1 = b1, (b0 - q * b1)
return c0, a0, b0
_, a, b = ex_euclid(a, b)
print(a, b)
``` | output | 1 | 89,239 | 22 | 178,479 |
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,240 | 22 | 178,480 |
"Correct Solution:
```
def egcd(a, b):
x,y, u,v = 0,1, 1,0
while a != 0:
q, r = b//a, b%a
m, n = x-u*q, y-v*q
b,a, x,y, u,v = a,r, u,v, m,n
gcd = b
return gcd,x,y
a,b = (int(_) for _ in input().split(" "))
gcd,x,y = egcd(a,b)
if y * a + x * b == gcd:
tmp = x
x = y
y = tmp
print(x,y)
``` | output | 1 | 89,240 | 22 | 178,481 |
Provide a correct Python 3 solution for this coding contest problem.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1 | instruction | 0 | 89,241 | 22 | 178,482 |
"Correct Solution:
```
def ex_euclid(x, y):
a0, a1 = 1, 0
b0, b1 = 0, 1
while y!=0:
a0, a1 = a1, (a0 - x//y * a1)
b0, b1 = b1, (b0 - x//y * b1)
x, y = y, x%y
return a0, b0
a,b=map(int,input().split())
print(*ex_euclid(a,b))
``` | output | 1 | 89,241 | 22 | 178,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
def extended_gcd(a, b):
"""ax + by = gcd(a, b)の整数解(x, y)を求める"""
if b == 0:
return a, 1, 0
else:
g, y, x = extended_gcd(b, a % b)
y -= (a // b) * x
return g, x, y
a, b = map(int, input().split())
_, x, y = extended_gcd(a, b)
print(x, y)
``` | instruction | 0 | 89,242 | 22 | 178,484 |
Yes | output | 1 | 89,242 | 22 | 178,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
r0, r1 = map(int, input().split())
a0, a1 = 1, 0
b0, b1 = 0, 1
while 0 < r1:
q1 = r0 // r1
a0, a1 = a1, a0 - q1 * a1
b0, b1 = b1, b0 - q1 * b1
r0, r1 = r1, r0 % r1
print(a0, b0)
``` | instruction | 0 | 89,243 | 22 | 178,486 |
Yes | output | 1 | 89,243 | 22 | 178,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
import math
a,b=map(int,input().split())
x=[0]
y=[0]
def extgcd(a,b,x,y):
d=a
if b!=0:
d=extgcd(b,a%b,y,x)
y[0]-=(a//b)*x[0]
else:
x[0]=1
y[0]=0
extgcd(a,b,x,y)
print(x[0],y[0])
``` | instruction | 0 | 89,244 | 22 | 178,488 |
Yes | output | 1 | 89,244 | 22 | 178,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
def extended_euclid(a,b):
c = g = 1
e = f = 0
while b:
div,mod = divmod(a,b)
h = c - div * e
i = f - div * g
a,b = b,mod
c,e = e,h
f,g = g,i
return (c,f)
n,m = map(int,input().split(" "))
print(*extended_euclid(n,m))
``` | instruction | 0 | 89,245 | 22 | 178,490 |
Yes | output | 1 | 89,245 | 22 | 178,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
# coding=utf-8
def gcd(n1, n2):
if n1 < n2:
n1, n2 = n2, n1
if n1 == n2 or n2 == 0:
return n1
n1, n2 = n2, (n1 % n2)
return gcd(n1, n2)
def solve_int(number1, number2):
answer1 = 1
while True:
if (1+number1*answer1) % number2 == 0:
return -answer1, (1+number1*answer1)//number2
answer1 += 1
def modify_solution(x_sp_s, y_sp_s, n1, n2):
x_memo, y_memo = x_sp_s, y_sp_s
if abs(x_sp_s + n2) + abs(y_sp_s - n1) < (abs(x_memo) + abs(y_memo)):
m = y_sp_s//n1
if m == 0:
return x_sp_s+n2, y_sp_s-n1
return modify_solution(x_sp_s + m*n2, y_sp_s - m*n1, n1, n2)
if abs(x_sp_s + n2) + abs(y_sp_s - n1) == (abs(x_memo) + abs(y_memo)):
pass
return x_memo, y_memo
if __name__ == '__main__':
a, b = map(int, input().split())
right_side = gcd(a, b)
if right_side > 1:
a //= right_side
b //= right_side
x_sp, y_sp = solve_int(a, b)
x_sp, y_sp = modify_solution(x_sp, y_sp, a, b)
print(x_sp, y_sp)
``` | instruction | 0 | 89,246 | 22 | 178,492 |
No | output | 1 | 89,246 | 22 | 178,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
def gcd(a, b):
x, y = [a, b] if a > b else [b, a]
while y:
x, y = y, x%y
return x
a, b = list(map(int, input().split(' ')))
c = gcd(a, b)
a, b = a//c, b//c
x = 0
y = None
while True:
tmp = (1-a*x)/b
if tmp == int(tmp):
y = int(tmp)
break
x += 1
print(x, y)
``` | instruction | 0 | 89,247 | 22 | 178,494 |
No | output | 1 | 89,247 | 22 | 178,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
r,s=map(int,input().split())
a=d=1,b=c=0
while r:
q=r//s
r,s,a,c,b,d=s,r%s,c,a-q*c,d,b-q*d
print(a,b)
``` | instruction | 0 | 89,248 | 22 | 178,496 |
No | output | 1 | 89,248 | 22 | 178,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Extended Euclid Algorithm
Given positive integers a and b, find the integer solution (x, y) to ax + by = gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
Constraints
* 1 ≤ a, b ≤ 109
Input
a b
Two positive integers a and b are given separated by a space in a line.
Output
Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x| + |y| is the minimal (primarily) and x ≤ y (secondarily).
Examples
Input
4 12
Output
1 0
Input
3 8
Output
3 -1
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat May 7 22:31:12 2016
@author: kt
"""
def gcd(a,b): #?????§??¬?´???°???????????°??????
r=0
while True:
r=a%b
if r==0:
break
a=b
b=r
return b
a, b = map(int, input().split(" "))
g=gcd(a,b)
a=a/g
b=b/g
c=0
x1=1
x2=0
z1=a
y1=0
y2=1
z2=b
q=0
while True:
if z2==1:
break
q = (z1 - (z1 % z2)) / z2
x3=(x1-q*x2)
y3=(y1-q*y2)
z3=z1-q*z2
x1=x2
x2=x3
y1=y2
y2=y3
z1=z2
z2=z3
if x3>y3 and a==b:
temp=y3
y3=x3
x3=temp
print ("%d %d" % (int(x3), int(y3)))
``` | instruction | 0 | 89,249 | 22 | 178,498 |
No | output | 1 | 89,249 | 22 | 178,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers written on the blackboard. Also, a positive number k ≥ 2 is chosen, and none of the numbers on the blackboard are divisible by k. In one operation, you can choose any two integers x and y, erase them and write one extra number f(x + y), where f(x) is equal to x if x is not divisible by k, otherwise f(x) = f(x / k).
In the end, there will be a single number of the blackboard. Is it possible to make the final number equal to 1? If so, restore any sequence of operations to do so.
Input
The first line contains two integers n and k — the initial number of integers on the blackboard, and the chosen number (2 ≤ n ≤ 16, 2 ≤ k ≤ 2000).
The second line contains n positive integers a_1, …, a_n initially written on the blackboard. It is guaranteed that none of the numbers a_i is divisible by k, and the sum of all a_i does not exceed 2000.
Output
If it is impossible to obtain 1 as the final number, print "NO" in the only line.
Otherwise, print "YES" on the first line, followed by n - 1 lines describing operations. The i-th of these lines has to contain two integers x_i and y_i to be erased and replaced with f(x_i + y_i) on the i-th operation. If there are several suitable ways, output any of them.
Examples
Input
2 2
1 1
Output
YES
1 1
Input
4 3
7 8 13 23
Output
YES
23 13
8 7
5 4
Input
3 4
1 2 3
Output
NO
Note
In the second sample case:
* f(8 + 7) = f(15) = f(5) = 5;
* f(23 + 13) = f(36) = f(12) = f(4) = 4;
* f(5 + 4) = f(9) = f(3) = f(1) = 1.
Submitted Solution:
```
def func(x,k):
while x % k ==0:
x=x//k
return x
n,k = map(int, input().split())
list_numbers = list(map(int,input().split()))
i=0
string_list = []
while len(list_numbers) >1:
curr_elem = list_numbers[i]
flag1 = True
list_test = list_numbers[:]
list_test.remove(curr_elem)
for elem1 in list_test:
if (curr_elem+elem1) % k ==0:
list_numbers.remove(elem1)
list_numbers.remove(curr_elem)
list_numbers.append(func(elem1+curr_elem,k))
string_list.append('{0} {1}'.format(elem1,curr_elem))
flag1 = False
break
if flag1:
i +=1
if i >= len(list_numbers):
elem1 = list_numbers[0]
elem2 = list_numbers[1]
list_numbers.remove(elem1)
list_numbers.remove(elem2)
list_numbers.append(func(elem1+elem2,k))
string_list.append('{0} {1}'.format(elem1,elem2))
i = 0
else:
i = 0
if list_numbers == [1]:
print('YES')
for i in range(len(string_list)):
print(string_list[i])
else:
print('NO')
``` | instruction | 0 | 89,354 | 22 | 178,708 |
No | output | 1 | 89,354 | 22 | 178,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers written on the blackboard. Also, a positive number k ≥ 2 is chosen, and none of the numbers on the blackboard are divisible by k. In one operation, you can choose any two integers x and y, erase them and write one extra number f(x + y), where f(x) is equal to x if x is not divisible by k, otherwise f(x) = f(x / k).
In the end, there will be a single number of the blackboard. Is it possible to make the final number equal to 1? If so, restore any sequence of operations to do so.
Input
The first line contains two integers n and k — the initial number of integers on the blackboard, and the chosen number (2 ≤ n ≤ 16, 2 ≤ k ≤ 2000).
The second line contains n positive integers a_1, …, a_n initially written on the blackboard. It is guaranteed that none of the numbers a_i is divisible by k, and the sum of all a_i does not exceed 2000.
Output
If it is impossible to obtain 1 as the final number, print "NO" in the only line.
Otherwise, print "YES" on the first line, followed by n - 1 lines describing operations. The i-th of these lines has to contain two integers x_i and y_i to be erased and replaced with f(x_i + y_i) on the i-th operation. If there are several suitable ways, output any of them.
Examples
Input
2 2
1 1
Output
YES
1 1
Input
4 3
7 8 13 23
Output
YES
23 13
8 7
5 4
Input
3 4
1 2 3
Output
NO
Note
In the second sample case:
* f(8 + 7) = f(15) = f(5) = 5;
* f(23 + 13) = f(36) = f(12) = f(4) = 4;
* f(5 + 4) = f(9) = f(3) = f(1) = 1.
Submitted Solution:
```
def func(x,k):
while x % k ==0:
x=x//k
return x
n,k = map(int, input().split())
list_numbers = list(map(int,input().split()))
i=0
string_list = []
while len(list_numbers) >1:
curr_elem = list_numbers[i]
flag1 = True
list_test = list_numbers[:]
list_test.remove(curr_elem)
for elem1 in list_test:
if (curr_elem+elem1) % k ==0:
list_numbers.remove(elem1)
list_numbers.remove(curr_elem)
list_numbers.append(func(elem1+curr_elem,k))
string_list.append('{0} {1}'.format(elem1,curr_elem))
flag1 = False
break
if flag1:
i +=1
if i >= len(list_numbers):
break
else:
i = 0
if list_numbers == [1]:
print('YES')
for i in range(len(string_list)):
print(string_list[i])
else:
print('NO')
``` | instruction | 0 | 89,355 | 22 | 178,710 |
No | output | 1 | 89,355 | 22 | 178,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers written on the blackboard. Also, a positive number k ≥ 2 is chosen, and none of the numbers on the blackboard are divisible by k. In one operation, you can choose any two integers x and y, erase them and write one extra number f(x + y), where f(x) is equal to x if x is not divisible by k, otherwise f(x) = f(x / k).
In the end, there will be a single number of the blackboard. Is it possible to make the final number equal to 1? If so, restore any sequence of operations to do so.
Input
The first line contains two integers n and k — the initial number of integers on the blackboard, and the chosen number (2 ≤ n ≤ 16, 2 ≤ k ≤ 2000).
The second line contains n positive integers a_1, …, a_n initially written on the blackboard. It is guaranteed that none of the numbers a_i is divisible by k, and the sum of all a_i does not exceed 2000.
Output
If it is impossible to obtain 1 as the final number, print "NO" in the only line.
Otherwise, print "YES" on the first line, followed by n - 1 lines describing operations. The i-th of these lines has to contain two integers x_i and y_i to be erased and replaced with f(x_i + y_i) on the i-th operation. If there are several suitable ways, output any of them.
Examples
Input
2 2
1 1
Output
YES
1 1
Input
4 3
7 8 13 23
Output
YES
23 13
8 7
5 4
Input
3 4
1 2 3
Output
NO
Note
In the second sample case:
* f(8 + 7) = f(15) = f(5) = 5;
* f(23 + 13) = f(36) = f(12) = f(4) = 4;
* f(5 + 4) = f(9) = f(3) = f(1) = 1.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(lambda x: (int(x), int(x) % k), input().split()))
c = 0
for i in a:
c += i[1]
if c % k != 0:
print('NO')
else:
print('YES')
while len(a) > 1:
k1 = a.pop(0)[0]
k2 = a.pop(0)[0]
print(k1, k2)
s = k1 + k2
while s % k == 0:
s //= k
a.append((s, s % k))
``` | instruction | 0 | 89,356 | 22 | 178,712 |
No | output | 1 | 89,356 | 22 | 178,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers written on the blackboard. Also, a positive number k ≥ 2 is chosen, and none of the numbers on the blackboard are divisible by k. In one operation, you can choose any two integers x and y, erase them and write one extra number f(x + y), where f(x) is equal to x if x is not divisible by k, otherwise f(x) = f(x / k).
In the end, there will be a single number of the blackboard. Is it possible to make the final number equal to 1? If so, restore any sequence of operations to do so.
Input
The first line contains two integers n and k — the initial number of integers on the blackboard, and the chosen number (2 ≤ n ≤ 16, 2 ≤ k ≤ 2000).
The second line contains n positive integers a_1, …, a_n initially written on the blackboard. It is guaranteed that none of the numbers a_i is divisible by k, and the sum of all a_i does not exceed 2000.
Output
If it is impossible to obtain 1 as the final number, print "NO" in the only line.
Otherwise, print "YES" on the first line, followed by n - 1 lines describing operations. The i-th of these lines has to contain two integers x_i and y_i to be erased and replaced with f(x_i + y_i) on the i-th operation. If there are several suitable ways, output any of them.
Examples
Input
2 2
1 1
Output
YES
1 1
Input
4 3
7 8 13 23
Output
YES
23 13
8 7
5 4
Input
3 4
1 2 3
Output
NO
Note
In the second sample case:
* f(8 + 7) = f(15) = f(5) = 5;
* f(23 + 13) = f(36) = f(12) = f(4) = 4;
* f(5 + 4) = f(9) = f(3) = f(1) = 1.
Submitted Solution:
```
count, divisor = [int(x) for x in input().split()]
sequence = [int(x) for x in input().split()]
summ = sum(sequence)
if summ % divisor != 0:
print("NO")
exit()
else:
path = list()
while len(sequence) != 1:
for i in range(count - 1):
try:
a = sequence[i] + sequence[i + 1]
path.append([sequence[i], sequence[i + 1]])
sequence[i] = sequence[i] + sequence[i + 1]
sequence[i + 1] = ''
except:
pass
sequence = [part for part in sequence if part != '']
if sequence[0] % divisor == 0:
print("YES")
for i in range(len(path)):
print(*path[i])
``` | instruction | 0 | 89,357 | 22 | 178,714 |
No | output | 1 | 89,357 | 22 | 178,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,374 | 22 | 178,748 |
Tags: brute force, math
Correct Solution:
```
T_ON = 0
DEBUG_ON = 0
MOD = 998244353
def solve():
n = read_int()
if n == 1:
print(10, 9)
else:
print(n * 3, n * 2)
def main():
T = read_int() if T_ON else 1
for i in range(T):
solve()
def debug(*xargs):
if DEBUG_ON:
print(*xargs)
from collections import *
import math
#---------------------------------FAST_IO---------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#----------------------------------IO_WRAP--------------------------------------
def read_int():
return int(input())
def read_ints():
return list(map(int, input().split()))
def print_nums(nums):
print(" ".join(map(str, nums)))
def YES():
print("YES")
def Yes():
print("Yes")
def NO():
print("NO")
def No():
print("No")
#----------------------------------FIB--------------------------------------
def fib(n):
a, b = 0, 1
for _ in range(n):
a, b = b, a + b
return a
def fib_ns(n):
assert n >= 1
f = [0 for _ in range(n + 1)]
f[0] = 0
f[1] = 1
for i in range(2, n + 1):
f[i] = f[i - 1] + f[i - 2]
return f
#----------------------------------MOD--------------------------------------
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def xgcd(a, b):
"""return (g, x, y) such that a*x + b*y = g = gcd(a, b)"""
x0, x1, y0, y1 = 0, 1, 1, 0
while a != 0:
(q, a), b = divmod(b, a), a
y0, y1 = y1, y0 - q * y1
x0, x1 = x1, x0 - q * x1
return b, x0, y0
def modinv(a, m):
"""return x such that (a * x) % m == 1"""
g, x, _ = xgcd(a, m)
if g != 1:
raise Exception('gcd(a, m) != 1')
return x % m
def mod_add(x, y):
x += y
while x >= MOD:
x -= MOD
while x < 0:
x += MOD
return x
def mod_mul(x, y):
return (x * y) % MOD
def mod_pow(x, y):
if y == 0:
return 1
if y % 2:
return mod_mul(x, mod_pow(x, y - 1))
p = mod_pow(x, y // 2)
return mod_mul(p, p)
def mod_inv(y):
return mod_pow(y, MOD - 2)
def mod_div(x, y):
# y^(-1): Fermat little theorem, MOD is a prime
return mod_mul(x, mod_inv(y))
#---------------------------------PRIME---------------------------------------
def is_prime(n):
if n == 1:
return False
for i in range(2, int(n ** 0.5) + 1):
if n % i:
return False
return True
def gen_primes(n):
"""
generate primes of [1..n] using sieve's method
"""
P = [True for _ in range(n + 1)]
P[0] = P[1] = False
for i in range(int(n ** 0.5) + 1):
if P[i]:
for j in range(2 * i, n + 1, i):
P[j] = False
return P
#---------------------------------MAIN---------------------------------------
main()
``` | output | 1 | 89,374 | 22 | 178,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,375 | 22 | 178,750 |
Tags: brute force, math
Correct Solution:
```
n = int(input())
def comp(n):
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return True
return False
for i in range(4, 1000000):
if comp(i) and comp(n + i):
print(n + i, i)
break
``` | output | 1 | 89,375 | 22 | 178,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,376 | 22 | 178,752 |
Tags: brute force, math
Correct Solution:
```
n=int(input())
if n%2==0:
n1=4
n2=n+4
else:
n1=9
n2=n+9
print(n2,n1)
``` | output | 1 | 89,376 | 22 | 178,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,377 | 22 | 178,754 |
Tags: brute force, math
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
if not n % 2:
print(n + 4, 4)
sys.exit()
else:
print(n + 9, 9)
``` | output | 1 | 89,377 | 22 | 178,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,378 | 22 | 178,756 |
Tags: brute force, math
Correct Solution:
```
n=int(input())
b=4
a=b+n
aa=1
while aa!=2:
qa,qb=0,0
for i in range(2,a):
if a%i==0:
qa=1
break
for i in range(2,b):
if b%i==0:
qb=1
break
aa=qa+qb
if aa==2: break
else:
a+=1
b+=1
print(a,b)
``` | output | 1 | 89,378 | 22 | 178,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,379 | 22 | 178,758 |
Tags: brute force, math
Correct Solution:
```
from math import sqrt
def Nprime(n):
for i in range(2, int(sqrt(n))+1):
if n % i == 0:
return True
else:
return False
a = int(input())
l = len(str(a))
n = ["1"] + ["0"]*l
n = int("".join(n))
if n - a <=4:
n *= 10
while(True):
if Nprime(n) and Nprime(n-a):
print(n, n-a)
break
if Nprime(n):
pass
else:
n -= 1
if Nprime(n-a):
pass
else:
n -= 1
``` | output | 1 | 89,379 | 22 | 178,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,380 | 22 | 178,760 |
Tags: brute force, math
Correct Solution:
```
n = int(input())
if(n==1):
print("9 8")
elif(n==2):
print("6 4")
elif(n==3):
print("9 6")
elif(n%2==0):
print(n+4,"4")
else:
print(n+9,"9")
``` | output | 1 | 89,380 | 22 | 178,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | instruction | 0 | 89,381 | 22 | 178,762 |
Tags: brute force, math
Correct Solution:
```
def check_prime(n):
for i in range(2, int(n**0.5)):
if n % i == 0:
return True
return False
def main():
n = int(input())
a, b = 0, 4
while True:
a = n + b
if check_prime(a):
return str(a) + " " + str(b)
b += 2
if __name__ == "__main__":
print(main())
``` | output | 1 | 89,381 | 22 | 178,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
n = int(input())
if n==2:
print(8,6)
elif n%2 == 0:
print(n*2,n)
elif n==1:
print(10,9)
else:
print(n*4-n,n*2)
``` | instruction | 0 | 89,382 | 22 | 178,764 |
Yes | output | 1 | 89,382 | 22 | 178,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
k = 10**7 + 5
n = int(input())
if(n%2==0):
print('{} {}'.format(2*n+2,n+2))
else:
print('{} {}'.format(k,k-n))
``` | instruction | 0 | 89,383 | 22 | 178,766 |
Yes | output | 1 | 89,383 | 22 | 178,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
def isPrime(n) :
# Corner cases
if (n <= 1) :
return False
if (n <= 3) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
n=int(input(''))
a=0
for i in range(2,n+10**9):
if not (isPrime(i)):
if not (isPrime(n+i)):
a=i
break
print(i+n," ",i)
``` | instruction | 0 | 89,384 | 22 | 178,768 |
Yes | output | 1 | 89,384 | 22 | 178,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
n = int(input())
print(f'{n*9} {n*8}')
``` | instruction | 0 | 89,385 | 22 | 178,770 |
Yes | output | 1 | 89,385 | 22 | 178,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
n = int(input())
print(2*n, 3*n)
``` | instruction | 0 | 89,386 | 22 | 178,772 |
No | output | 1 | 89,386 | 22 | 178,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
n=int(input())
if n==1:
print(7777, 7776)
else:
print(3*n,n)
``` | instruction | 0 | 89,387 | 22 | 178,774 |
No | output | 1 | 89,387 | 22 | 178,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
n = int(input())
print(1000000000 , 1000000000-n)
``` | instruction | 0 | 89,388 | 22 | 178,776 |
No | output | 1 | 89,388 | 22 | 178,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096
Submitted Solution:
```
fact = 10*9*8*7*6*5*4*3*2*1
t = int(input())
print(fact,fact+t)
``` | instruction | 0 | 89,389 | 22 | 178,778 |
No | output | 1 | 89,389 | 22 | 178,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,422 | 22 | 178,844 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
n = int(input())
res = [1] * (n + 1)
for i in range(2, n + 1):
j = i
while i*j <= n:
res[i*j] = max(res[i*j], j)
j += 1
res.sort()
print(" ".join(map(str, res[2:])))
``` | output | 1 | 89,422 | 22 | 178,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,423 | 22 | 178,846 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
input = stdin.readline
print = stdout.write
R = lambda: list(map(int, input().split()))
n = int(input())
s = [0]*(n+1)
vis = [0]*(n+1)
u = 2
while u*u <= n:
if not vis[u]:
for i in range(u*2, n+1, u):
if vis[i]: continue
vis[i] = 1
s[i//u] += 1
u += 1
for i in range(2, n+1):
if not vis[i]: s[1] += 1
res = []
for i in range(n+1):
if s[i] > 0: res.append((str(i)+' ')*s[i])
print(''.join(res))
``` | output | 1 | 89,423 | 22 | 178,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,424 | 22 | 178,848 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
n = int(input())
sieve = [-1] * (n + 1)
primes = []
for i in range(2, n + 1):
if sieve[i] == -1:
sieve[i] = i
primes.append(i)
for x in range(2 * i, n + 1, i):
if sieve[x] == -1:
sieve[x] = i
answer = [1] * len(primes)
c = set(primes)
c.add(1)
toAdd = 2
while len(c) != n:
for k in primes:
if k > sieve[toAdd]:
break
y = toAdd * k
if y > n:
break
if y not in c:
c.add(y)
answer.append(toAdd)
toAdd += 1
print(*answer)
``` | output | 1 | 89,424 | 22 | 178,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,425 | 22 | 178,850 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
def lf(x):
i = 2
while i * i <= x:
if x % i == 0: return x // i
i += 1
return 1
if __name__ == "__main__":
n = int(input())
d = []
for i in range(2, n + 1):
#print(i, lf(i), sep = ' ')
d.append(lf(i))
d.sort()
print(' '.join(map(str, d)))
``` | output | 1 | 89,425 | 22 | 178,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,426 | 22 | 178,852 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
def main():
for _ in inputt(1):
n, = inputi()
S = [1] * (n + 1)
for i in range(2, n // 2 + 1):
for j in range(i * 2, n + 1, i):
S[j] = i
S.sort()
print(*S[2:])
# region M
# region fastio
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
for x in args:
file.write(str(x))
file.write(kwargs.pop("end", "\n"))
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
# region import
inputt = lambda t = 0: range(t) if t else range(int(input()))
inputi = lambda: map(int, input().split())
inputl = lambda: list(inputi())
from math import *
from heapq import *
from bisect import *
from itertools import *
from functools import reduce, lru_cache
from collections import Counter, defaultdict
import re, copy, operator, cmath
from builtins import *
# endregion
# region main
if __name__ == "__main__":
main()
# endregion
# endregion
``` | output | 1 | 89,426 | 22 | 178,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,427 | 22 | 178,854 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
n = int(input())
l = [0]*(n+1)
for i in range(1,n+1):
for j in range(2*i,n+1,i):
l[j] = i
l.sort()
print(*l[2:n+1])
``` | output | 1 | 89,427 | 22 | 178,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,428 | 22 | 178,856 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 2020/7/1
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
if __name__ == '__main__':
N = int(input())
# t0 = time.time()
maxdiv = [0 for _ in range(N+1)]
maxdiv[1] = 1
# print(time.time() - t0)
for i in range(2, N+1):
if maxdiv[i]:
continue
v = i
d = 1
while v <= N:
if maxdiv[v] == 0:
maxdiv[v] = d
v += i
d += 1
# print(time.time() - t0)
maxdiv = maxdiv[1:]
maxdiv.sort()
# print(time.time() - t0)
ans = []
maxd = 0
for v in maxdiv:
maxd = max(maxd, v)
ans.append(maxd)
print(' '.join(map(str, ans[1:])))
# print(time.time() - t0)
``` | output | 1 | 89,428 | 22 | 178,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,429 | 22 | 178,858 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
n = int(input())
ans=[1]*(n+1)
for i in range(2,n+1):
j=i
while j*i<=n:
ans[j*i]=max(ans[j*i],j)
j+=1
a=sorted(ans)
print(*a[2:])
``` | output | 1 | 89,429 | 22 | 178,859 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | instruction | 0 | 89,430 | 22 | 178,860 |
Tags: greedy, implementation, math, number theory, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
n=int(raw_input())
arr=[1]*(n+1)
for i in range(2,n+1):
for j in range(2*i,n+1,i):
arr[j]=i
arr=arr[2:]
arr.sort()
pr(' '.join(map(str,arr)))
#pr_arr(arr)
``` | output | 1 | 89,430 | 22 | 178,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}.
Submitted Solution:
```
n = int(input())
A = [1 for i in range(n+1)]
for i in range(2, n):
j = 2*i
while j <= n:
A[j] = i
j += i
ans = sorted(A)
ans = ans[2:]
print(*ans)
``` | instruction | 0 | 89,431 | 22 | 178,862 |
Yes | output | 1 | 89,431 | 22 | 178,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}.
Submitted Solution:
```
def solve(n):
nums = [1] * (n + 1)
p = 2
while p * p <= n:
if nums[p] == 1:
for i in range(p * 2, n+1, p):
nums[i] = max(nums[i], p, i // p)
p += 1
return sorted(sorted(nums[2:]))
n = int(input())
print(' '.join([str(x) for x in solve(n)]))
``` | instruction | 0 | 89,432 | 22 | 178,864 |
Yes | output | 1 | 89,432 | 22 | 178,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}.
Submitted Solution:
```
def get_primes(n):
res = [2]
arr = [True] * ((n - 1) // 2)
i = 0
for i in range(len(arr)):
if arr[i]:
a = i * 2 + 3
res.append(a)
for ii in range(i + a, len(arr), a):
arr[ii] = False
return res
if __name__ == "__main__":
n = int(input())
primes = get_primes(n)
res = ["1"] * min(n - 1, len(primes))
left = n - 1 - len(res)
ii = 2
while left > 0:
for a in primes:
if ii * a <= n:
res.append(str(ii))
left -= 1
else:
break
if ii % a == 0 or left == 0:
break
ii += 1
print(" ".join(res))
``` | instruction | 0 | 89,433 | 22 | 178,866 |
Yes | output | 1 | 89,433 | 22 | 178,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}.
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n = int(input())
imp = n // 2
ans = [1]*(n + 1)
used = [False]*(n+1)
cur = n
while imp > 1:
first = False
for i in range(imp, n+1, imp):
if not used[i]:
if not first:
first = True
else:
used[i] = True
#print(imp)
ans[cur] = imp
cur -= 1
imp -= 1
print(*ans[2:])
``` | instruction | 0 | 89,434 | 22 | 178,868 |
Yes | output | 1 | 89,434 | 22 | 178,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
sum=1
if n%2==0:
for _ in range(n//2):
print(sum,end=' ')
z=(n-1)-n//2
for _ in range(z):
sum+=1
print(sum,end=' ')
else:
for _ in range((n+1)//2):
print(sum,end=' ')
z=(n-1)-(n+1)//2
for _ in range(z):
sum+=1
print(sum,end=' ')
``` | instruction | 0 | 89,435 | 22 | 178,870 |
No | output | 1 | 89,435 | 22 | 178,871 |
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