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.
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least k prime numbers from 2 to n inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
Input
The first line of the input contains two integers n (2 ≤ n ≤ 1000) and k (0 ≤ k ≤ 1000).
Output
Output YES if at least k prime numbers from 2 to n inclusively can be expressed as it was described above. Otherwise output NO.
Examples
Input
27 2
Output
YES
Input
45 7
Output
NO
Note
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
Submitted Solution:
```
import math
def build_sieve(n):
sieve = [True for i in range(n + 1)]
sieve[0] = False
sieve[1] = False
sqrt = math.ceil(math.sqrt(n))
for i in range(2, sqrt):
if sieve[i]:
for j in range(i**2, n, i):
sieve[j] = False
return set(i for i in range(2, n + 1) if sieve[i])
def main():
n, k = [int(x) for x in input().split(" ")]
primes = build_sieve(n)
primes_list = sorted(primes)
for i in range(len(primes_list) - 1):
s = primes_list[i] + primes_list[i + 1] + 1
if s in primes:
k -= 1
if k == 0:
print("YES")
return
print("NO")
main()
``` | instruction | 0 | 52,490 | 22 | 104,980 |
No | output | 1 | 52,490 | 22 | 104,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,560 | 22 | 105,120 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
import math
from collections import defaultdict
MOD = 10**9 + 7
def printDivisors(n) :
i = 1
Factors = []
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
Factors.append(i)
else :
Factors.append(i)
Factors.append(int(n/i))
i = i + 1
Factors.sort(reverse = True)
return Factors
F = defaultdict(list)
n,k = map(int,input().split())
for i in range(1,n+1):
F[i] =printDivisors(i)
dp = [1] * (n+1)
dp[0] = 0
for i in range(1,k):
for j in range(1,n+1):
for f in range(j+j,n+1,j):
dp[j] += dp[f]
dp[j] %= MOD
print(sum(dp)%MOD)
``` | output | 1 | 52,560 | 22 | 105,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
def findDivisors(x):
divisors = []
for i in range(1,((x)**1//2)+2):
if x%i==0:
divisors.append(i)
if x/i!=i:
divisors.append(x//i)
divisors=list(set(divisors))
return divisors
anss = []
n,k = map(int,input().split())
for i in range(1,n+1):
anss.append(findDivisors(i))
dp = [[0 for i in range(n)] for j in range(k)]
for i in range(n):
dp[0][i]=i+1
for j in range(k):
dp[j][0]=1
for i in range(1,k):
for j in range(1,n):
dp[i][j]=1
for x in anss[j]:
dp[i][j]+=dp[i-1][x-1]
print(dp[k-1][n-1])
``` | instruction | 0 | 52,569 | 22 | 105,138 |
No | output | 1 | 52,569 | 22 | 105,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three integers k, pa and pb.
You will construct a sequence with the following algorithm: Initially, start with the empty sequence. Each second, you do the following. With probability pa / (pa + pb), add 'a' to the end of the sequence. Otherwise (with probability pb / (pa + pb)), add 'b' to the end of the sequence.
You stop once there are at least k subsequences that form 'ab'. Determine the expected number of times 'ab' is a subsequence in the resulting sequence. It can be shown that this can be represented by P / Q, where P and Q are coprime integers, and <image>. Print the value of <image>.
Input
The first line will contain three integers integer k, pa, pb (1 ≤ k ≤ 1 000, 1 ≤ pa, pb ≤ 1 000 000).
Output
Print a single integer, the answer to the problem.
Examples
Input
1 1 1
Output
2
Input
3 1 4
Output
370000006
Note
The first sample, we will keep appending to our sequence until we get the subsequence 'ab' at least once. For instance, we get the sequence 'ab' with probability 1/4, 'bbab' with probability 1/16, and 'aab' with probability 1/8. Note, it's impossible for us to end with a sequence like 'aabab', since we would have stopped our algorithm once we had the prefix 'aab'.
The expected amount of times that 'ab' will occur across all valid sequences is 2.
For the second sample, the answer is equal to <image>. | instruction | 0 | 52,790 | 22 | 105,580 |
Tags: dp, math, probabilities
Correct Solution:
```
from collections import defaultdict as di
import sys
sys.setreqursiondepth = 1000000
MOD = int(1e9+7)
def modinvEuler(x,mod):
# if mod is prime
return pow(x, mod-2, mod)
# otherwise exponent should be totient(mod)-1
k,pa,pb = [int(x) for x in input().split()]
Pa = (pa*modinvEuler(pa+pb,MOD))%MOD
Pb = (1-Pa)%MOD
Ea = modinvEuler(Pa,MOD)
Eb = modinvEuler(Pb,MOD)
Pbinv = modinvEuler(Pb,MOD)
mem = di()
def f(na,ns):
#global k,Pa,Pb
if ns>=k:
return ns
if na+ns>=k:
total = ns
total += na
total += Pa*Pbinv
total%= MOD
return total
if (na,ns) not in mem:
mem[(na,ns)] = ( Pa*(f(na+1,ns))+Pb*(f(na,ns+na)) )%MOD
return mem[(na,ns)]
print((f(1,0))%MOD)
``` | output | 1 | 52,790 | 22 | 105,581 |
Provide a correct Python 3 solution for this coding contest problem.
We will define Ginkgo numbers and multiplication on Ginkgo numbers.
A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.
The multiplication on Ginkgo numbers is defined by <m, n> * <x, y> = <mx − ny, my + nx>. For example, <1, 1> * <-2, 1> = <-3,-1>.
A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> * <x, y> = <p, q>.
For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.
A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.
The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.
* Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.
* If <m, n> * <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.
Input
The first line of the input contains a single integer, which is the number of datasets.
The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.
Output
For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.
Example
Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Output
C
C
P
C
C
P
P
C | instruction | 0 | 53,009 | 22 | 106,018 |
"Correct Solution:
```
for _ in range(int(input())):
p,q=map(int,input().split())
c=0
for i in range(142):
for j in range(142):
if (i>0 or j>0)and(j*p+i*q)%(j*j+i*i)==0 and (j*q-i*p)%(j*j+i*i)==0:c+=1
print('P' if c<5 else 'C')
``` | output | 1 | 53,009 | 22 | 106,019 |
Provide a correct Python 3 solution for this coding contest problem.
We will define Ginkgo numbers and multiplication on Ginkgo numbers.
A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.
The multiplication on Ginkgo numbers is defined by <m, n> * <x, y> = <mx − ny, my + nx>. For example, <1, 1> * <-2, 1> = <-3,-1>.
A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> * <x, y> = <p, q>.
For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.
A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.
The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.
* Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.
* If <m, n> * <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.
Input
The first line of the input contains a single integer, which is the number of datasets.
The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.
Output
For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.
Example
Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Output
C
C
P
C
C
P
P
C | instruction | 0 | 53,010 | 22 | 106,020 |
"Correct Solution:
```
for i in range(int(input())):
m, n = map(int, input().split())
if m*n < 0:
m, n = -n, m
m = abs(m); n = abs(n)
l = max(abs(m), abs(n))
cnt = 0
for p in range(l+1):
for q in range(l+1):
if p == 0: continue
if (p*m + q*n) % (p**2 + q**2) < 1 and (p*n - q*m) % (p**2 + q**2) < 1:
cnt += 1
print("CP"[cnt == 2])
``` | output | 1 | 53,010 | 22 | 106,021 |
Provide a correct Python 3 solution for this coding contest problem.
We will define Ginkgo numbers and multiplication on Ginkgo numbers.
A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.
The multiplication on Ginkgo numbers is defined by <m, n> * <x, y> = <mx − ny, my + nx>. For example, <1, 1> * <-2, 1> = <-3,-1>.
A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> * <x, y> = <p, q>.
For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.
A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.
The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.
* Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.
* If <m, n> * <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.
Input
The first line of the input contains a single integer, which is the number of datasets.
The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.
Output
For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.
Example
Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Output
C
C
P
C
C
P
P
C | instruction | 0 | 53,011 | 22 | 106,022 |
"Correct Solution:
```
for _ in range(int(input())):
p,q=map(int,input().split())
c=0
for i in range(143):
for j in range(143):
if (i>0 or j>0)and(j*p+i*q)%(j*j+i*i)==0 and (j*q-i*p)%(j*j+i*i)==0:c+=1
print('P' if c<5 else 'C')
``` | output | 1 | 53,011 | 22 | 106,023 |
Provide a correct Python 3 solution for this coding contest problem.
We will define Ginkgo numbers and multiplication on Ginkgo numbers.
A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.
The multiplication on Ginkgo numbers is defined by <m, n> * <x, y> = <mx − ny, my + nx>. For example, <1, 1> * <-2, 1> = <-3,-1>.
A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> * <x, y> = <p, q>.
For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.
A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.
The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.
* Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.
* If <m, n> * <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.
Input
The first line of the input contains a single integer, which is the number of datasets.
The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.
Output
For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.
Example
Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Output
C
C
P
C
C
P
P
C | instruction | 0 | 53,012 | 22 | 106,024 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
M = 20000
s = set()
for i in range(int(M**0.5)+1):
for j in range(int(M**0.5)+1):
t = i**2+j**2
if t > 1:
s.add(t)
n = I()
ni = 0
while ni < n:
ni += 1
a,b = LI()
t = a**2 + b**2
r = 'P'
for c in s:
if t % c == 0 and t // c in s:
r = 'C'
break
rr.append(r)
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 53,012 | 22 | 106,025 |
Provide a correct Python 3 solution for this coding contest problem.
We will define Ginkgo numbers and multiplication on Ginkgo numbers.
A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.
The multiplication on Ginkgo numbers is defined by <m, n> * <x, y> = <mx − ny, my + nx>. For example, <1, 1> * <-2, 1> = <-3,-1>.
A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> * <x, y> = <p, q>.
For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.
A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.
The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.
* Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.
* If <m, n> * <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.
Input
The first line of the input contains a single integer, which is the number of datasets.
The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.
Output
For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.
Example
Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Output
C
C
P
C
C
P
P
C | instruction | 0 | 53,013 | 22 | 106,026 |
"Correct Solution:
```
def divisor(a):
ret = []
for i in range(1, int(a ** 0.5) + 1):
if not a % i:
ret.append(i)
ret.append(a // i)
return ret
def make_lists():
ret = [[] for i in range(20000)]
for i in range(0, int(20000 ** 0.5) + 1):
for j in range(0, int(20000 ** 0.5) + 1):
if i ** 2 + j ** 2 < 20000:
ret[i ** 2 + j ** 2].append([i, j])
return ret
lists = make_lists()
ans = []
for loop in range(int(input())):
m, n = map(int, input().split())
C = []
for div in divisor(m ** 2 + n ** 2):
for l in lists[div]:
for tmp in [-1, 1]:
for tmp2 in [-1, 1]:
x, y = l[0] * tmp, l[1] * tmp2
if not (m * x + n * y) % (x ** 2 + y ** 2) and not (n * x - m * y) % (x ** 2 + y ** 2):
if [x, y] not in C:
C.append([x, y])
ans.append('P' if len(C) == 8 else 'C')
[print(i) for i in ans]
``` | output | 1 | 53,013 | 22 | 106,027 |
Provide a correct Python 3 solution for this coding contest problem.
We will define Ginkgo numbers and multiplication on Ginkgo numbers.
A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.
The multiplication on Ginkgo numbers is defined by <m, n> * <x, y> = <mx − ny, my + nx>. For example, <1, 1> * <-2, 1> = <-3,-1>.
A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> * <x, y> = <p, q>.
For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.
A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.
The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.
* Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.
* If <m, n> * <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.
Input
The first line of the input contains a single integer, which is the number of datasets.
The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.
Output
For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.
Example
Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Output
C
C
P
C
C
P
P
C | instruction | 0 | 53,014 | 22 | 106,028 |
"Correct Solution:
```
for i in range(int(input())):
m, n = map(int, input().split())
count = 0
for p in range(142):
for q in range(142):
if p == q == 0:
continue
pq = p * p + q * q
if pq > 20000:
break
if (m * p + n * q) % pq == 0 and (n * p - m * q) % pq == 0:
count += 1
if count < 5:
print("P")
else:
print("C")
``` | output | 1 | 53,014 | 22 | 106,029 |
Provide a correct Python 3 solution for this coding contest problem.
We will define Ginkgo numbers and multiplication on Ginkgo numbers.
A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.
The multiplication on Ginkgo numbers is defined by <m, n> * <x, y> = <mx − ny, my + nx>. For example, <1, 1> * <-2, 1> = <-3,-1>.
A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> * <x, y> = <p, q>.
For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.
A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.
The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.
* Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.
* If <m, n> * <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.
Input
The first line of the input contains a single integer, which is the number of datasets.
The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.
Output
For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.
Example
Input
8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1
Output
C
C
P
C
C
P
P
C | instruction | 0 | 53,015 | 22 | 106,030 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
Problems 1325
Problem A: Ginkgo Numbers
"""
for _ in range(int(input())):
m, n = map(int,input().split())
count = 0
for p in range(142):
for q in range(142):
if (p > 0 or q > 0) and ((m*q + n*p) % (p*p + q*q) == 0) and ((n*q - m*p) % (p*p + q*q) == 0):
count += 1
if count < 5:
print('P')
else:
print('C')
``` | output | 1 | 53,015 | 22 | 106,031 |
Provide a correct Python 3 solution for this coding contest problem.
B: Twice as own
problem
You will be given Q queries. Since one positive integer N is given for each query, find the number of positive integers M that satisfy the following two conditions.
* 2 Satisfy \ leq M \ leq N
* Of the divisors of M, excluding M, the total product is more than twice that of M
Input format
The input is given in the following format:
Q
N_1_1
N_2
::
N_Q
* The first line gives the number of queries Q.
* Lines 2 through N + 1 are given one positive integer N given for each query.
Constraint
* 1 \ leq Q \ leq 10 ^ 5
* 2 \ leq N_i \ leq 10 ^ 5 (1 \ leq i \ leq Q)
Output format
Output the number of positive integers that satisfy the above two conditions for each N_i, separated by line breaks.
Input example
3
43
9
twenty four
Output example
11
0
Five
* When N = 24, there are five types of M that can be considered as 12, 16, 18, 20, 24.
Example
Input
3
43
9
24
Output
11
0
5 | instruction | 0 | 53,034 | 22 | 106,068 |
"Correct Solution:
```
Q = int(input())
N = list(int(input()) for i in range(Q))
m = [0]*(max(N)+1)
N_max = max(N)
r = 320
for i in range(2,r):
for j in range(2,r):
for k in range(2,100000//(i*j)+1):
if i*j*k < N_max+1 and (i!=j or j!=k or k!=i):
m[i*j*k] = 1
for i in range(1,len(m)):
m[i] = m[i]+m[i-1]
for i in N:
print(m[i])
``` | output | 1 | 53,034 | 22 | 106,069 |
Provide a correct Python 3 solution for this coding contest problem.
B: Twice as own
problem
You will be given Q queries. Since one positive integer N is given for each query, find the number of positive integers M that satisfy the following two conditions.
* 2 Satisfy \ leq M \ leq N
* Of the divisors of M, excluding M, the total product is more than twice that of M
Input format
The input is given in the following format:
Q
N_1_1
N_2
::
N_Q
* The first line gives the number of queries Q.
* Lines 2 through N + 1 are given one positive integer N given for each query.
Constraint
* 1 \ leq Q \ leq 10 ^ 5
* 2 \ leq N_i \ leq 10 ^ 5 (1 \ leq i \ leq Q)
Output format
Output the number of positive integers that satisfy the above two conditions for each N_i, separated by line breaks.
Input example
3
43
9
twenty four
Output example
11
0
Five
* When N = 24, there are five types of M that can be considered as 12, 16, 18, 20, 24.
Example
Input
3
43
9
24
Output
11
0
5 | instruction | 0 | 53,035 | 22 | 106,070 |
"Correct Solution:
```
import math
def get_sieve_of_eratosthenes(n):
if not isinstance(n, int):
raise TypeError('n is int type.')
if n < 2:
raise ValueError('n is more than 2')
prime = []
limit = math.sqrt(n)
data = [i + 1 for i in range(1, n)]
while True:
p = data[0]
if limit <= p:
return prime + data
prime.append(p)
data = [e for e in data if e % p != 0]
n = int(input())
ma = int(math.sqrt(10**5))
lis = get_sieve_of_eratosthenes(ma)
l_b = [False]*(10**5+1)
for item1 in lis:
for item2 in lis:
if item2 < item1:
continue
if item1*item2*3 > 10**5:
break
i = 0
while(item1*item2*(3+i) <= 10**5):
l_b[item1*item2*(3+i)] = True
i += 1
for item in lis:
if item**3 < 10**5:
l_b[item**3] = False
l_count = [0]*(10**5+1)
for i in range(1,10**5+1):
if l_b[i]:
l_count[i] = l_count[i-1] + 1
else:
l_count[i] = l_count[i-1]
for _ in range(n):
num = int(input())
print(l_count[num])
``` | output | 1 | 53,035 | 22 | 106,071 |
Provide a correct Python 3 solution for this coding contest problem.
B: Twice as own
problem
You will be given Q queries. Since one positive integer N is given for each query, find the number of positive integers M that satisfy the following two conditions.
* 2 Satisfy \ leq M \ leq N
* Of the divisors of M, excluding M, the total product is more than twice that of M
Input format
The input is given in the following format:
Q
N_1_1
N_2
::
N_Q
* The first line gives the number of queries Q.
* Lines 2 through N + 1 are given one positive integer N given for each query.
Constraint
* 1 \ leq Q \ leq 10 ^ 5
* 2 \ leq N_i \ leq 10 ^ 5 (1 \ leq i \ leq Q)
Output format
Output the number of positive integers that satisfy the above two conditions for each N_i, separated by line breaks.
Input example
3
43
9
twenty four
Output example
11
0
Five
* When N = 24, there are five types of M that can be considered as 12, 16, 18, 20, 24.
Example
Input
3
43
9
24
Output
11
0
5 | instruction | 0 | 53,036 | 22 | 106,072 |
"Correct Solution:
```
#import sys
#input = sys.stdin.readline
def eratosthenes(N):
from collections import deque
work = [True] * (N+1)
work[0] = False
work[1] = False
ret = []
for i in range(N+1):
if work[i]:
ret.append(i)
for j in range(2* i, N+1, i):
work[j] = False
return ret
from itertools import accumulate
def main():
ANS = [1]*100001
ANS[0] = 0
ANS[1] = 0
Primes = eratosthenes(100000)
for p in Primes:
ANS[p] = 0
if p**2 <= 100000:
ANS[p**2] = 0
if p**3 <= 100000:
ANS[p**3] = 0
for p in Primes:
for q in Primes:
if p*q <= 100000:
ANS[p*q] = 0
else:
break
accANS = list( accumulate(ANS))
Q = int( input())
for _ in range(Q):
print( accANS[ int( input())])
if __name__ == '__main__':
main()
``` | output | 1 | 53,036 | 22 | 106,073 |
Provide a correct Python 3 solution for this coding contest problem.
B: Twice as own
problem
You will be given Q queries. Since one positive integer N is given for each query, find the number of positive integers M that satisfy the following two conditions.
* 2 Satisfy \ leq M \ leq N
* Of the divisors of M, excluding M, the total product is more than twice that of M
Input format
The input is given in the following format:
Q
N_1_1
N_2
::
N_Q
* The first line gives the number of queries Q.
* Lines 2 through N + 1 are given one positive integer N given for each query.
Constraint
* 1 \ leq Q \ leq 10 ^ 5
* 2 \ leq N_i \ leq 10 ^ 5 (1 \ leq i \ leq Q)
Output format
Output the number of positive integers that satisfy the above two conditions for each N_i, separated by line breaks.
Input example
3
43
9
twenty four
Output example
11
0
Five
* When N = 24, there are five types of M that can be considered as 12, 16, 18, 20, 24.
Example
Input
3
43
9
24
Output
11
0
5 | instruction | 0 | 53,037 | 22 | 106,074 |
"Correct Solution:
```
import bisect
MAX_N=2*10**5
isdmore5=[1 for i in range(MAX_N+1)]
isprime=[1 for i in range(MAX_N+1)]
i=2
isprime[0]=0;isprime[1]=0
isdmore5[0]=0;isdmore5[1]=0
while(i*i<=MAX_N):
if isprime[i]!=0:
for j in range(2*i,MAX_N+1,i):
isprime[j]=0
i+=1
prime=[]
for i in range(MAX_N+1):
if isprime[i]==1:
prime.append(i)
isdmore5[i]=0
#print(prime)
for p in prime:
k=bisect.bisect_left(prime,MAX_N//p)
#print(p,k,prime[k-1]*p,prime[k]*p,prime[k+1]*p)
for i in range(k):
#print(p,prime[i],p*prime[i])
isdmore5[p*prime[i]]=0
i=2
while(i*i*i<=MAX_N):
if isprime[i]==1:
isdmore5[i*i*i]=0
i+=1
ans=[0 for i in range(MAX_N+1)]
for i in range(3,MAX_N+1):
if isdmore5[i]==1:
ans[i]=ans[i-1]+1
else:
ans[i]=ans[i-1]
Q=int(input())
for i in range(Q):
print(ans[int(input())])
``` | output | 1 | 53,037 | 22 | 106,075 |
Provide a correct Python 3 solution for this coding contest problem.
B: Twice as own
problem
You will be given Q queries. Since one positive integer N is given for each query, find the number of positive integers M that satisfy the following two conditions.
* 2 Satisfy \ leq M \ leq N
* Of the divisors of M, excluding M, the total product is more than twice that of M
Input format
The input is given in the following format:
Q
N_1_1
N_2
::
N_Q
* The first line gives the number of queries Q.
* Lines 2 through N + 1 are given one positive integer N given for each query.
Constraint
* 1 \ leq Q \ leq 10 ^ 5
* 2 \ leq N_i \ leq 10 ^ 5 (1 \ leq i \ leq Q)
Output format
Output the number of positive integers that satisfy the above two conditions for each N_i, separated by line breaks.
Input example
3
43
9
twenty four
Output example
11
0
Five
* When N = 24, there are five types of M that can be considered as 12, 16, 18, 20, 24.
Example
Input
3
43
9
24
Output
11
0
5 | instruction | 0 | 53,038 | 22 | 106,076 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
from itertools import product, accumulate
from math import gcd
from bisect import bisect
import sys
from sys import setrecursionlimit
setrecursionlimit(10**9)
input = sys.stdin.readline
def inpl(): return list(map(int, input().split()))
def primes(N):
sieve = [1]*(N+1)
sieve[:2] = [0, 0]
P = []
for i in range(2, N+1):
if sieve[i]:
P.append(i)
for j in range(2*i, N+1, i):
sieve[j] = 0
return P
C = [1]*(10**5+1)
C[0] = 0
C[1] = 0
P = primes(10**5)
for p in P:
C[p] = 0
for p in P:
if p**3 > 100000:
break
C[p**3] = 0
for i in range(len(P)):
for j in range(i, bisect(P, 10**5//P[i])):
C[P[i] * P[j]] = 0
S = list(accumulate(C))
for _ in range(int(input())):
print(S[int(input())])
``` | output | 1 | 53,038 | 22 | 106,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,539 | 22 | 107,078 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
n = int(input())
if n&1:
print((n-3)//2+1)
n-=3
for i in range(n//2):
print(2,end=' ')
print(3)
else:
print(n//2)
for i in range(n//2):
print(2,end=' ')
``` | output | 1 | 53,539 | 22 | 107,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,540 | 22 | 107,080 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
n = int(input())
if n % 2 == 0:
print(n // 2)
s = ''
for i in range(n // 2 - 1):
s += '2 '
s += '2'
print(s)
else:
print(n // 2)
s = ''
for i in range(n // 2 - 1):
s += '2 '
s += '3'
print(s)
``` | output | 1 | 53,540 | 22 | 107,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,541 | 22 | 107,082 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
n = int(input())
print(n//2)
if n%2:
print(3,end= ' ')
n -= 3
while n:
print(2,end=' ')
n -= 2
``` | output | 1 | 53,541 | 22 | 107,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,542 | 22 | 107,084 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
n=int(input())
if n%2==0:
print(int(n/2))
for i in range(int(n/2)):
print(2)
else:
print(int((n-3)/2+1))
for i in range(int((n-3)/2)):
print(2)
print(3)
``` | output | 1 | 53,542 | 22 | 107,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,543 | 22 | 107,086 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
if n%2 == 0:
ans1 = n//2
ans2 = [2]*ans1
else:
ans1=n//2
ans2 = [2]*(ans1-1)
ans2.append(3)
print(ans1)
print(*ans2)
``` | output | 1 | 53,543 | 22 | 107,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,544 | 22 | 107,088 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
x = int(input())
print(int(x/2))
for i in range(int(x/2)-1):
print(2, end=" ")
print([2,3][x%2])
``` | output | 1 | 53,544 | 22 | 107,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,545 | 22 | 107,090 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
n = int(input())
if n % 2 == 0:
print(n // 2)
print('2 ' * (n // 2))
else:
print((n - 3) // 2 + 1)
print('3 ' + (n - 3) // 2 * '2 ')
``` | output | 1 | 53,545 | 22 | 107,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2 | instruction | 0 | 53,546 | 22 | 107,092 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
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")
#-------------------game starts now-----------------------------------------------------
n=int(input())
r=0
a=[]
if n%2==1:
n-=3
r+=1+n//2
a=[3]+[2]*(n//2)
else:
r+=n//2
a=[2]*r
print(r)
print(*a)
``` | output | 1 | 53,546 | 22 | 107,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
n = int(input())
print(n // 2)
if n % 2 == 1:
print("3", end = " ")
n -= 3
while n > 0:
print("2", end = " ")
n -= 2
``` | instruction | 0 | 53,547 | 22 | 107,094 |
Yes | output | 1 | 53,547 | 22 | 107,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
n=int(input())
two=n//2
if(n%2!=0):
two-=1
three=(n-two*2)//3
while(two*2+three*3!=n):
three+=1
two-=1
print(two+three)
print('2 '*two+'3 '*three)
``` | instruction | 0 | 53,548 | 22 | 107,096 |
Yes | output | 1 | 53,548 | 22 | 107,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
n = int(input())
k = n//2
l=[]
if not n%2:
for i in range(k):
l.append(n//k)
else:
for i in range(k-1):
l.append(n//k)
l.append(3)
print(k)
print(" ".join(map(str, l)))
``` | instruction | 0 | 53,549 | 22 | 107,098 |
Yes | output | 1 | 53,549 | 22 | 107,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
n = int(input())
if ( n % 2 == 1):
print(1+((n-3)//2))
if (n % 2 == 0):
print((n//2) )
if ( n % 2 == 1):
print("3 ",end="")
n -= 3
if (n % 2 == 0):
print("2 "*(n//2) )
``` | instruction | 0 | 53,550 | 22 | 107,100 |
Yes | output | 1 | 53,550 | 22 | 107,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
n = int(input())
print(n // 2)
if n % 2 == 0:
for i in range(n // 2):
print(2, end=' ')
else:
for i in range(n // 2):
print(2, end=' ')
print(3)
``` | instruction | 0 | 53,551 | 22 | 107,102 |
No | output | 1 | 53,551 | 22 | 107,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
n = int(input())
if n%2 == 0:
print(n//2)
print((str(2) + " ")*(n//2))
else:
print(n//2 + 1)
print((str(2) + " ")*(n//2), end = "")
print(3)
``` | instruction | 0 | 53,552 | 22 | 107,104 |
No | output | 1 | 53,552 | 22 | 107,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
n = int(input())
if(n%2)==0:
print(int(n/2))
print(("2 ")*(int(n/2)))
else:
c = int(n//2)
print(("2 ")*(c-1),3)
``` | instruction | 0 | 53,553 | 22 | 107,106 |
No | output | 1 | 53,553 | 22 | 107,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
Output
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
Examples
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
Submitted Solution:
```
def isPrime(n):
if n > 1:
for e in range(2, n//2 + 1):
if not n % e:
return False
return True
else:
return False
n = int(input())
primes = []
for x in range(n+1):
if isPrime(x):
primes.append(x)
y = []
for prime in primes:
while True:
if n - prime >= 0 and not n % prime:
n -= prime
y.append(str(prime))
else:
break
if n == 0:
break
print(len(y))
print(' '.join(y))
``` | instruction | 0 | 53,554 | 22 | 107,108 |
No | output | 1 | 53,554 | 22 | 107,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,621 | 22 | 107,242 |
Tags: math
Correct Solution:
```
p, k = map(int, input().split())
if p < k:
print(1)
print(p)
else:
ans = [p]
cur = p
while cur < 0 or cur >= k:
cur = - (cur // k)
ans.append(cur)
arr = []
for i in range(1, len(ans)):
arr.append(str(k*ans[i]+ans[i-1]))
arr.append(str(ans[-1]))
print(len(arr))
print(' '.join(arr))
``` | output | 1 | 53,621 | 22 | 107,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,622 | 22 | 107,244 |
Tags: math
Correct Solution:
```
p,k = map(int,input().split())
a = []
while p != 0:
t = p % k
a.append(t)
p = -((p-t)//k)
print(len(a))
a = list(map(str,a))
print(' '.join(a))
``` | output | 1 | 53,622 | 22 | 107,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,623 | 22 | 107,246 |
Tags: math
Correct Solution:
```
def solve(n, k):
if n == 0:
return []
x = n%k
return [x] + solve(-(n-x)//k, k)
n, k = map(int, input().split())
a = solve(n, k)
print(len(a))
print(*a)
``` | output | 1 | 53,623 | 22 | 107,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,624 | 22 | 107,248 |
Tags: math
Correct Solution:
```
s = input()
# s = '46 2'
p, k = list(map(int, s.split()))
res = []
f = True
while p != 0:
if f:
n = p % k
m = (p - n) // k
f = False
else:
n = (-p) % k
m = (p + n) // k
f = True
res.append(n)
p = m
print(len(res))
print(' '.join(list(map(str, res))))
``` | output | 1 | 53,624 | 22 | 107,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,625 | 22 | 107,250 |
Tags: math
Correct Solution:
```
l = []
def f(p, k):
if p == 0:
return
r = p % k
q = (p-r)//(-k)
l.append(r)
f(q, k)
p, k = map(int, input().split())
f(p, k)
print(len(l))
print(*l)
``` | output | 1 | 53,625 | 22 | 107,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,626 | 22 | 107,252 |
Tags: math
Correct Solution:
```
p, k = map(int, input().split())
a = ''
cnt = 0
while p != 0:
cnt += 1
a += str(p % k) + ' '
p -= p % k
p //= -k
print(cnt)
print(a)
``` | output | 1 | 53,626 | 22 | 107,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,627 | 22 | 107,254 |
Tags: math
Correct Solution:
```
p, k = map(int, input().split())
r = []
while p:
r.append(p % k)
p = -(p // k)
print(len(r))
print(*r, sep=' ')
# Made By Mostafa_Khaled
``` | output | 1 | 53,627 | 22 | 107,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018. | instruction | 0 | 53,628 | 22 | 107,256 |
Tags: math
Correct Solution:
```
p, k = map(int, input().split())
t = []
while p:
t.append(p % k)
p = -(p // k)
print(len(t), *t)
``` | output | 1 | 53,628 | 22 | 107,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
p,k = map(int,input().split())
num = [0] * 64
sum = 0
c = 0
power = k-1
while(sum < p):
sum += power
power *= k*k
c += 2
#print(power / k / k << endl << sum << endl;
print(c-1)
i = 0
p = sum - p
while(p > 0):
num[i] = p % k
p //= k
i += 1
#cout << sum << endl;
for i in range(c-1):
#cout << num[i] << " ";
if(i % 2 == 0) :print(k - 1 - num[i],end = ' ')
else :print(num[i],end = ' ')
``` | instruction | 0 | 53,629 | 22 | 107,258 |
Yes | output | 1 | 53,629 | 22 | 107,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
def solve(n,p,k):
# print(n,p,k)
P=p
cf=1
a=[0]*n
for i in range(n):
if i&1:
p+=cf*(k-1)
a[i]-=k-1
cf*=k
# print(p)
for i in range(n):
a[i]+=p%k
p//=k
# print(n,a)
if p:
return
for i in range(n):
if i&1:
a[i]*=-1
cf=1
p=P
for i in range(n):
if a[i]<0 or a[i]>=k:
return
if i&1:
p+=a[i]*cf
else:
p-=a[i]*cf
cf*=k
if p:
return
print(len(a))
print(*a)
exit(0)
p,k=map(int,input().split())
for i in range(100):
if k**i>1<<100:
break
solve(i,p,k)
print(-1)
``` | instruction | 0 | 53,630 | 22 | 107,260 |
Yes | output | 1 | 53,630 | 22 | 107,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
def fu(a,b):
r=a%b
if r<0:
r=a%b-b
return [(a-r)//b,r]
p,k=list(map(int,input().split()))
s=[]
while 1>0:
a=fu(p,-k)
if a[0]==0:
if a[1]!=0:
s.append(a[1])
break
else:
s.append(a[1])
p=a[0]
print(len(s))
print(*s)
``` | instruction | 0 | 53,631 | 22 | 107,262 |
Yes | output | 1 | 53,631 | 22 | 107,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
p, k = map(int, input().split())
ans = []
while p != 0:
ans.append(p % k)
p //= k
p *= -1
print(len(ans))
print(*ans)
``` | instruction | 0 | 53,632 | 22 | 107,264 |
Yes | output | 1 | 53,632 | 22 | 107,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
p, k = map(int, input().split())
print('3\n1', k - p, p - p * k)
``` | instruction | 0 | 53,633 | 22 | 107,266 |
No | output | 1 | 53,633 | 22 | 107,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
p,k=map(int,input().split())
f=1
t=0
ms=0
a=[]
while p+ms*k>=k or p+ms*k<0:
if t%2==0:
while p+ms*k>=k:
ms-=1
else:
while p+ms*k<0:
ms+=1
a.append(p+ms*k)
t+=1
p=ms
ms=0
print(len(a))
for i in a:
print(i,end=' ')
print(1)
``` | instruction | 0 | 53,634 | 22 | 107,268 |
No | output | 1 | 53,634 | 22 | 107,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
p, k = map(int, input().split())
k = -k
a = []
while p != 0:
r = p % k
p //= k
if r < 0:
r += -k
p += 1
a.append(r)
a.reverse()
print(len(a))
print(" ".join(map(str, a)))
``` | instruction | 0 | 53,635 | 22 | 107,270 |
No | output | 1 | 53,635 | 22 | 107,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
Input
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
Output
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial <image> fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
Examples
Input
46 2
Output
7
0 1 0 0 1 1 1
Input
2018 214
Output
3
92 205 1
Note
In the first example, f(x) = x6 + x5 + x4 + x = (x5 - x4 + 3x3 - 6x2 + 12x - 23)·(x + 2) + 46.
In the second example, f(x) = x2 + 205x + 92 = (x - 9)·(x + 214) + 2018.
Submitted Solution:
```
p, k = map(int, input().split())
if p < k:
print(1)
print(p)
else:
ans = [p]
cur = p
while cur < 0 or cur >= k:
cur = - (cur // k)
ans.append(cur)
arr = []
for i in range(1, len(ans)):
arr.append(str(k*ans[i]+ans[i-1]))
arr.append(str(ans[-1]))
print(' '.join(arr))
``` | instruction | 0 | 53,636 | 22 | 107,272 |
No | output | 1 | 53,636 | 22 | 107,273 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.
Input
Input consists of several datasets. Each dataset has an integer n (1 ≤ n ≤ 999,999) in a line.
The number of datasets is less than or equal to 30.
Output
For each dataset, prints the number of prime numbers.
Example
Input
10
3
11
Output
4
2
5 | instruction | 0 | 53,805 | 22 | 107,610 |
"Correct Solution:
```
import sys
import math
M = 1000000
B = int(math.sqrt(M)+1)
N = M//6
N1 = N+1 if M % 6 == 5 else N
N2 = N+1 if M % 6 >= 1 else N
l = [1] * N1
m = [1] * N2
for i in range(5,B,6):
p = (i-5)//6
ini = (i*5-1)//6
if l[p] == 1:
l[p+i::i] = [0] * len(l[p+i::i])
m[ini-1::i] = [0] * len(m[ini-1::i])
if m[p] == 1:
m[p+i+2::i+2] = [0] * len(m[p+i+2::i+2])
l[ini+1::i+2] = [0] * len(l[ini+1::i+2])
for i in sys.stdin:
n = int(i)
r = n-1 if n < 3 else sum(l[0:(n+1)//6])+sum(m[0:(n-1)//6])+2
print(r)
``` | output | 1 | 53,805 | 22 | 107,611 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.
Input
Input consists of several datasets. Each dataset has an integer n (1 ≤ n ≤ 999,999) in a line.
The number of datasets is less than or equal to 30.
Output
For each dataset, prints the number of prime numbers.
Example
Input
10
3
11
Output
4
2
5 | instruction | 0 | 53,806 | 22 | 107,612 |
"Correct Solution:
```
import sys
def primes(n):
is_prime = [True] * (n + 1)
is_prime[0] = False
is_prime[1] = False
for i in range(2, int(n**0.5) + 1):
if not is_prime[i]:
continue
for j in range(i * 2, n + 1, i):
is_prime[j] = False
return [i for i in range(n + 1) if is_prime[i]]
for e in sys.stdin:
n = int(e)
print(len(primes(n)))
``` | output | 1 | 53,806 | 22 | 107,613 |
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