message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of all integers from l to r inclusive, l < r, (r - l + 1) ≤ 3 ⋅ 10^5 and (r - l) is always odd.
You want to split these numbers into exactly (r - l + 1)/(2) pairs in such a way that for each pair (i, j) the greatest common divisor of i and j is equal to 1. Each number should appear in exactly one of the pairs.
Print the resulting pairs or output that no solution exists. If there are multiple solutions, print any of them.
Input
The only line contains two integers l and r (1 ≤ l < r ≤ 10^{18}, r - l + 1 ≤ 3 ⋅ 10^5, (r - l) is odd).
Output
If any solution exists, print "YES" in the first line. Each of the next (r - l + 1)/(2) lines should contain some pair of integers. GCD of numbers in each pair should be equal to 1. All (r - l + 1) numbers should be pairwise distinct and should have values from l to r inclusive.
If there are multiple solutions, print any of them.
If there exists no solution, print "NO".
Example
Input
1 8
Output
YES
2 7
4 1
3 8
6 5
Submitted Solution:
```
def main():
l, r = [int(elem) for elem in input().split()]
for i in range(l, r, 2):
print(i, i+1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 52,215 | 22 | 104,430 |
No | output | 1 | 52,215 | 22 | 104,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of all integers from l to r inclusive, l < r, (r - l + 1) ≤ 3 ⋅ 10^5 and (r - l) is always odd.
You want to split these numbers into exactly (r - l + 1)/(2) pairs in such a way that for each pair (i, j) the greatest common divisor of i and j is equal to 1. Each number should appear in exactly one of the pairs.
Print the resulting pairs or output that no solution exists. If there are multiple solutions, print any of them.
Input
The only line contains two integers l and r (1 ≤ l < r ≤ 10^{18}, r - l + 1 ≤ 3 ⋅ 10^5, (r - l) is odd).
Output
If any solution exists, print "YES" in the first line. Each of the next (r - l + 1)/(2) lines should contain some pair of integers. GCD of numbers in each pair should be equal to 1. All (r - l + 1) numbers should be pairwise distinct and should have values from l to r inclusive.
If there are multiple solutions, print any of them.
If there exists no solution, print "NO".
Example
Input
1 8
Output
YES
2 7
4 1
3 8
6 5
Submitted Solution:
```
x=input().split()
(l,r)=(int(x[0]),int(x[1]))
for i in range(l,r+1,2):
print(i,i+1)
``` | instruction | 0 | 52,216 | 22 | 104,432 |
No | output | 1 | 52,216 | 22 | 104,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of all integers from l to r inclusive, l < r, (r - l + 1) ≤ 3 ⋅ 10^5 and (r - l) is always odd.
You want to split these numbers into exactly (r - l + 1)/(2) pairs in such a way that for each pair (i, j) the greatest common divisor of i and j is equal to 1. Each number should appear in exactly one of the pairs.
Print the resulting pairs or output that no solution exists. If there are multiple solutions, print any of them.
Input
The only line contains two integers l and r (1 ≤ l < r ≤ 10^{18}, r - l + 1 ≤ 3 ⋅ 10^5, (r - l) is odd).
Output
If any solution exists, print "YES" in the first line. Each of the next (r - l + 1)/(2) lines should contain some pair of integers. GCD of numbers in each pair should be equal to 1. All (r - l + 1) numbers should be pairwise distinct and should have values from l to r inclusive.
If there are multiple solutions, print any of them.
If there exists no solution, print "NO".
Example
Input
1 8
Output
YES
2 7
4 1
3 8
6 5
Submitted Solution:
```
l,r = map(int,input().split())
print("YES")
for i in range(1,r+1,2):
print(str(i)+' '+str(i+1))
``` | instruction | 0 | 52,217 | 22 | 104,434 |
No | output | 1 | 52,217 | 22 | 104,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,234 | 22 | 104,468 |
Tags: bitmasks, greedy
Correct Solution:
```
n, k = map(int, input().split())
b = bin(n)[2:]
min1 = b.count('1')
if n < k:
print('NO')
elif min1 > k:
print('NO')
else:
print('YES')
i = 0
pows = []
while i < len(b):
idx = len(b) - 1 - i
if b[idx] == '1':
pows.append(2 ** i)
i += 1
while len(pows) != k:
i = len(pows) - 1
while pows[i] == 1 or pows[i] == 0:
i -= 1
#print(pows[i])
while pows[i] != 1 and len(pows) != k:
pows.append(pows[i] // 2)
pows[i] //= 2
#ans = ''
ans = ' '.join(map(str, pows))
'''
for p in pows:
ans += str(p) + ' '
'''
#print(pows)
print(ans)
``` | output | 1 | 52,234 | 22 | 104,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,235 | 22 | 104,470 |
Tags: bitmasks, greedy
Correct Solution:
```
from bisect import bisect_left as bl, bisect_right as br, insort
import sys
import heapq
from math import *
from collections import defaultdict as dd, deque
def data(): return sys.stdin.readline().strip()
def mdata(): return map(int, data().split())
#sys.setrecursionlimit(100000)
n,k=mdata()
d=dd(int)
b=bin(n)[2:]
a=0
for i in range(len(b)):
if b[i]=='1':
a+=1
d[len(b)-1-i]+=1
if k<a or k>n:
print('NO')
else:
while a<k:
for i in d:
if d[i]>0 and i!=0:
d[i]-=1
d[i-1]+=2
a+=1
break
print("YES")
l=[]
for i in sorted(d.keys()):
l+=[int(pow(2,i))]*d[i]
print(*l)
``` | output | 1 | 52,235 | 22 | 104,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,236 | 22 | 104,472 |
Tags: bitmasks, greedy
Correct Solution:
```
import math
n,k=map(int,input().split())
y=bin(n)
y=y[2:]
mn=0
y=y[::-1]
r=[]
c=0
for i in y:
if int(i)==1:
mn+=1
r+=[2**c]
c+=1
b=0
if mn<=k<=n:
print('YES')
while len(r)!=k:
while r[b]==1:
b+=1
x=r[b]//2
r[b]=x
r.append(x)
print(*r)
else:
print('NO')
``` | output | 1 | 52,236 | 22 | 104,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,237 | 22 | 104,474 |
Tags: bitmasks, greedy
Correct Solution:
```
import heapq
n,k = map(int, input().split())
s = bin(n)[2:]
minval = s.count("1")
maxval = n
if k >= minval and k <= maxval:
print("YES")
L = []
for i,x in enumerate(s[::-1]):
if x == "1":
heapq.heappush(L,-2**i)
while len(L) != k:
current = heapq.heappop(L)
heapq.heappush(L, current // 2)
heapq.heappush(L, current // 2)
print(" ".join([str(-x) for x in L]))
else:
print("NO")
``` | output | 1 | 52,237 | 22 | 104,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,238 | 22 | 104,476 |
Tags: bitmasks, greedy
Correct Solution:
```
n, x = map(int, input().split())
v = n
l = []
i = 0
for k in range(32):
if (n & (1 << k)) >> k:
l.append(pow(2, k))
x -= len(l)
if x < 0:
print("NO")
else:
i = 0
while x > 0:
while i < len(l) and l[i] == 1:
i += 1
if i == len(l):
print("NO")
exit()
l[i] //= 2
l.insert(i, l[i])
x -= 1
l = list(map(str, l))
print("YES")
print(" ".join(l))
``` | output | 1 | 52,238 | 22 | 104,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,239 | 22 | 104,478 |
Tags: bitmasks, greedy
Correct Solution:
```
import heapq
n, k = map(int, input().split())
h = [-x for x in range(30) if n & (1 << x)]
h.reverse()
if len(h) > k or n < k:
print("NO")
quit()
to_do = k - len(h)
while to_do > 0:
x = heapq.heappop(h)
if x == 0:
print("NO")
quit()
for _ in range(2):
heapq.heappush(h, x + 1)
to_do -= 1
print("YES")
ans = [1 << -x for x in h]
for x in reversed(ans):
print(x, end=' ')
``` | output | 1 | 52,239 | 22 | 104,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,240 | 22 | 104,480 |
Tags: bitmasks, greedy
Correct Solution:
```
#/usr/bin/env python3
import sys
a = input().split()
n = int(a[0])
k = int(a[1])
def factor(x):
w = str(bin(x))[2:]
return [len(w) - e[0] - 1 for e in enumerate(w) if int(e[1])]
f = factor(n)
if k < len(f):
print("NO")
sys.exit(0)
w = [2**x for x in f]
if sum(w) < k:
print("NO")
sys.exit(0)
print("YES")
while True:
if len(w) == k:
print(' '.join(map(str, w)))
sys.exit(0)
if w[0] == 2:
w = w[1:] + [1, 1]
elif w[0] < (k - len(w)):
w = w[1:] + [1]*w[0]
else:
w = [w[0]//2, w[0]//2] + w[1:]
``` | output | 1 | 52,240 | 22 | 104,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO | instruction | 0 | 52,241 | 22 | 104,482 |
Tags: bitmasks, greedy
Correct Solution:
```
def main():
n, k = list(map(int, input().split()))
if n < k:
print('NO')
return
if n == k:
print('YES')
print(' '.join(['1', ] * k))
return
powers = []
while n:
x = 1
while x * 2 <= n:
x *= 2
n -= x
powers.append(x)
if len(powers) > k:
print('NO')
return
if len(powers) == k:
print('YES')
print(' '.join(list(map(str, powers))))
return
pd = {}
nums = 0
pm = max(powers)
for e in powers:
pd[e] = pd.get(e, 0) + 1
nums += pd[e]
while nums < k:
x = pm
while x not in pd or pd[x] == 0:
x //= 2
pd[x // 2] = pd.get(x // 2, 0) + 2
pd[x] -= 1
nums += 1
result = []
for k in pd:
result.extend([k, ] * pd[k])
print('YES')
print(' '.join(list(map(str, result))))
return
if __name__ == '__main__':
main()
``` | output | 1 | 52,241 | 22 | 104,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
Submitted Solution:
```
import math
n,k=map(int,input().split())
nn=n
kk=k
g=n-(k-1)
if g<=0:
print("NO")
else:
l=int(math.log(g,2))
c=2**l
li=list()
while n>0 and k>0:
if c+k-1<=n:
li.append(c)
k-=1
n-=c
else:
g=n-(k-1)
if g<=0:
print("NO")
exit(0)
l=int(math.log(g,2))
c=2**l
if sum(li)!=nn:
print("NO")
exit(0)
print("YES")
for i in li:
print(i,end=' ')
``` | instruction | 0 | 52,242 | 22 | 104,484 |
Yes | output | 1 | 52,242 | 22 | 104,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
Submitted Solution:
```
# cook your dish here
# from math import factorial, ceil, pow, sqrt, floor, gcd
from sys import stdin, stdout
# from collections import defaultdict, Counter, deque
# from bisect import bisect_left, bisect_right
# import sympy
# from itertools import permutations
# import numpy as np
# n = int(stdin.readline())
# stdout.write(str())
# s = stdin.readline().strip('\n')
# map(int, stdin.readline().split())
# l = list(map(int, stdin.readline().split()))
a, b = map(int, stdin.readline().split())
shan = [1]*b
a -= b
i = 0
while i < b and a:
while shan[i] <= a:
a -= shan[i]
shan[i] *= 2
i+=1
if a:
print("NO")
else:
print("YES")
print(*shan)
``` | instruction | 0 | 52,243 | 22 | 104,486 |
Yes | output | 1 | 52,243 | 22 | 104,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
Submitted Solution:
```
pt = []
for i in range(31):
pt.append(2**i)
pt = pt[::-1]
n, k = map(int, input().split())
def divide(a, g):
i = 0
while len(a) != g:
if a[i] == 1:
i += 1
else:
e = a.pop(i)
a.append(e//2)
a.append(e//2)
return a
if k > n:
print("NO")
else:
ps = []
for i in pt:
if i <= n:
n -= i
ps.append(i)
if len(ps) > k:
print("NO")
elif len(ps) < k:
print("YES")
print(" ".join(map(str, divide(ps, k))))
else:
print("YES")
print(" ".join(map(str, ps)))
``` | instruction | 0 | 52,245 | 22 | 104,490 |
Yes | output | 1 | 52,245 | 22 | 104,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
Submitted Solution:
```
n, k = map(int, input().split())
if bin(n).count('1') < k:
print ("YES")
n = bin(n)[2:][::-1]
# 13 -> 1101 -> 1011
array = []
for i in range(len(n)):
if n[i] == '1':
array.append(2 ** i)
array = array[::-1]
while len(array) < k:
if array[0] == 1:
array.sort(reverse=True)
array.append(array[0] // 2)
array.append(array[0] // 2)
del array[0]
print (' '.join(map(str, array)))
else:
print ("NO")
``` | instruction | 0 | 52,246 | 22 | 104,492 |
No | output | 1 | 52,246 | 22 | 104,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
Submitted Solution:
```
n=input().split()
k=[]
for i in n:
k.append(int(i))
r=0
if k[0]%2!=0:
k[0]-=1
while k[0]%2==0:
k[0]=k[0]//2
r+=1
if k[0]==1:
if r>=k[1]:
print('YES')
else:
print('NO')
else:
while k[0]%2==0:
k[0]=k[0]//2
r+=1
if k[0]==1:
print('YES')
else:
print('NO')
``` | instruction | 0 | 52,247 | 22 | 104,494 |
No | output | 1 | 52,247 | 22 | 104,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
Submitted Solution:
```
def two_powers_generator(n):
powers = []
i = 0
while(2**i<=n):
powers.append(2**i)
i+=1
return powers
def add_representation(n , k , dic):
if(k == 1 and (n,k) not in dic): return False
if((n,k) in dic) : return True
for i in two_powers_generator(n)[:-1]:
add_representation(n-i,k-1,dic)
if((n-i,k-1) in dic):
a = dic.get((n-i , k-1))
c = a + [i]
dic[(n,k)]= c
return True
return False
def is_representable(n , k , dic):
if((n,k) in dic): return True
for i in two_powers_generator(n):
print(dic)
print(i)
add_representation(n-i , k-1 , dic)
if((n-i,k-1) not in dic) : continue
a = dic.get((n-i , k-1))
c = a+[i]
dic[(n,k)] = c
return True
return False
def C():
inp = input().split()
n , k = int(inp[0]) , int(inp[1])
dic = {}
powers = two_powers_generator(n)
for i in powers:
dic[(i,1)] = [i]
if not is_representable(n,k,dic):
print("NO")
return
print("YES")
a = dic.get((n,k))
a = [str(x) for x in a]
print(" ".join(a))
C()
``` | instruction | 0 | 52,248 | 22 | 104,496 |
No | output | 1 | 52,248 | 22 | 104,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....
You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.
Input
The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).
Output
If it is impossible to represent n as the sum of k powers of two, print NO.
Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.
Examples
Input
9 4
Output
YES
1 2 2 4
Input
8 1
Output
YES
8
Input
5 1
Output
NO
Input
3 7
Output
NO
Submitted Solution:
```
K = []
def isPowerOfTwo(num):
return num & (num - 1) == 0
def divideInParts(_k):
if len(K) == 0:
return False
_n = max(K)
if len(K) != _k:
if _n == 1:
return False
K[K.index(_n)] = int(_n / 2)
K.append(int(_n / 2))
return divideInParts(_k)
return True
# method 1
def finding_k2(_n, _k):
_k = list(bin(_n).replace("0b", ""))
for i in range(len(_k), 0, -1):
if _k[i - 1] == '1':
K.append(pow(2, i - 1))
# method 2
def finding_k(_n, _k):
for i in range(_n, 0, -1):
if isPowerOfTwo(i):
K.append(i)
if _k - 1 != 0:
finding_k(_n - i, _k - 1)
break
if __name__ == '__main__':
n, k = map(int, input().split())
# finding_k(n, k)
finding_k2(n, k)
if divideInParts(k) and sum(K) == n:
print("YES")
print(" ".join(str(i) for i in K))
else:
print("NO")
``` | instruction | 0 | 52,249 | 22 | 104,498 |
No | output | 1 | 52,249 | 22 | 104,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,282 | 22 | 104,564 |
Tags: brute force, math, sortings
Correct Solution:
```
def calc(X, Y):
if len(Y) == 3: return sum(Y)
for x in X:
for y in Y:
if y % x == 0: break
else:
return calc([i for i in X if i != x], sorted(Y+[x])[::-1])
return sum(Y)
for _ in range(int(input())):
N = int(input())
A = sorted(set([int(a) for a in input().split()]))[::-1]
print(max(calc(A, []), calc(A[1:], [])))
``` | output | 1 | 52,282 | 22 | 104,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,283 | 22 | 104,566 |
Tags: brute force, math, sortings
Correct Solution:
```
answer = []
for _ in range(int(input())):
input()
l = sorted(map(int, input().split()))
other = 0
if l[-1] % 30 == 0 and len(set(l) & {l[-1] // 2, l[-1] // 3, l[-1] // 5}) == 3:
other = (31 * l[-1]) // 30
while len(l) >= 2 and l[-1] % l[-2] == 0:
del l[-2]
while len(l) >= 3 and 0 in (l[-1] % l[-3], l[-2] % l[-3]):
del l[-3]
answer.append(str(max(other, sum(l[-3:]))))
print("\n".join(answer))
``` | output | 1 | 52,283 | 22 | 104,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,284 | 22 | 104,568 |
Tags: brute force, math, sortings
Correct Solution:
```
import sys
p = int(input())
lines = sys.stdin.readlines()
res = []
def getMax(numset):
n = max(numset)
numset = set([j for j in numset if n % j != 0])
return (n , numset)
for i in range(p):
n = int(lines[2 * i])
m = []
numset = set()
for j in lines[2 * i + 1].split():
numset.add(int(j))
total = max(numset)
n = total
if n // 2 in numset and n // 3 in numset and n // 5 in numset:
total = n // 2 + n // 3 + n // 5
for _ in range(3):
if len(numset) == 0:
break
n, numset = getMax(numset)
m.append(n)
if sum(m) > total:
total = sum(m)
res.append(str(total))
print('\n'.join(res))
``` | output | 1 | 52,284 | 22 | 104,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,285 | 22 | 104,570 |
Tags: brute force, math, sortings
Correct Solution:
```
import sys
q = int(input())
lines = sys.stdin.readlines()
res = []
def getMax(numset):
n = max(numset)
numset = set([j for j in numset if n % j != 0])
return (n , numset)
for i in range(q):
n = int(lines[2 * i])
m = []
numset = set()
for j in lines[2 * i + 1].split():
numset.add(int(j))
total = max(numset)
n = total
if n // 2 in numset and n // 3 in numset and n // 5 in numset:
total = n // 2 + n // 3 + n // 5
for _ in range(3):
if len(numset) == 0:
break
n, numset = getMax(numset)
m.append(n)
if sum(m) > total:
total = sum(m)
res.append(str(total))
print('\n'.join(res))
``` | output | 1 | 52,285 | 22 | 104,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,286 | 22 | 104,572 |
Tags: brute force, math, sortings
Correct Solution:
```
def calcSum(first):
global m
s = m[first]
k = 1
for i in range(first + 1, len(m)):
yes = True
for j in range(first, i):
if m[j] % m[i] == 0:
yes = False
break
if yes:
s += m[i]
k += 1
if k == 3: break
return s
import sys
nnn = int(input())
for _ in range(nnn):
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
if n == 1:
print(a[0])
continue
if n == 2:
if a[0]%a[1] == 0 or a[1]%a[0] == 0:
print(max(a))
else:
print(sum(a))
continue
a.sort(reverse = True)
m = [a[0]]
for i in range(1, len(a)):
if a[i] == a[i-1]: continue
yes = True
for j in range(1, len(m)):
if m[j] % a[i] == 0:
yes = False
break
if yes:
m.append(a[i])
if len(m) >= 10:
break
## print(m)
s1 = calcSum(0)
if len(m) > 1:
s2 = calcSum(1)
else:
s2 = 0
s = max(s1, s2)
print (s)
## if nnn == 16383:
## if _>890:
## print(m, ' - ', a, '-', s)
## else:
## print(s)
``` | output | 1 | 52,286 | 22 | 104,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,287 | 22 | 104,574 |
Tags: brute force, math, sortings
Correct Solution:
```
#def gcd(x,y):
# z=x%y
# while(z!=0):
# x=y;y=z;z=x%y
# return y
T=int(input())
for o in range(T):
n=int(input())
a=list(map(int,input().split()))
a.sort()
b=[]
for i in a:
if len(b)==0 or (len(b)!=0 and b[-1]!=i):b.append(i)
a=b
n=len(b)
if(n==1):print(a[0])
elif(n==2):
if(a[1]%a[0]==0):print(a[1])
else:print(a[0]+a[1])
else:
ans1=a[-1]
for i in reversed(range(0,n-1)):
# print(a)
# print(i)
# print(a[i])
if a[-1]%a[i]!=0:
ans1+=a[i]
for j in reversed(range(0,i)):
if a[-1]%a[j]!=0 and a[i]%a[j]!=0:
ans1+=a[j]
break
break
ans2=a[-2]
for i in reversed(range(0,n-2)):
# print(i)
if a[-2]%a[i]!=0:
ans2+=a[i]
# print(i)
for j in reversed(range(0,i)):
# print(i);print(j);
if a[-2]%a[j]!=0 and a[i]%a[j]!=0:
ans2+=a[j]
break
break
# print(ans2)
print(max(ans1,ans2))
``` | output | 1 | 52,287 | 22 | 104,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,288 | 22 | 104,576 |
Tags: brute force, math, sortings
Correct Solution:
```
from sys import stdin, stdout
q = int(stdin.readline())
for it in range(q):
n = int(stdin.readline())
a = set(map(int, stdin.readline().split()))
M = max(a)
if M % 30 == 0 and M // 2 in a and M // 3 in a and M // 5 in a:
p1 = M * 31 // 30
else:
p1 = 0
ans = [M]
a = sorted(list(a))
idx = len(a)-2
while idx >= 0 and len(ans) < 3:
if any([x % a[idx] == 0 for x in ans]):
idx-=1
continue
ans.append(a[idx])
idx-=1
p2 = sum(ans)
if p1 > p2:
stdout.write(str(p1) + "\n")
else:
stdout.write(str(p2) + "\n")
``` | output | 1 | 52,288 | 22 | 104,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One important contest will take place on the most famous programming platform (Topforces) very soon!
The authors have a pool of n problems and should choose at most three of them into this contest. The prettiness of the i-th problem is a_i. The authors have to compose the most pretty contest (in other words, the cumulative prettinesses of chosen problems should be maximum possible).
But there is one important thing in the contest preparation: because of some superstitions of authors, the prettinesses of problems cannot divide each other. In other words, if the prettinesses of chosen problems are x, y, z, then x should be divisible by neither y, nor z, y should be divisible by neither x, nor z and z should be divisible by neither x, nor y. If the prettinesses of chosen problems are x and y then neither x should be divisible by y nor y should be divisible by x. Any contest composed from one problem is considered good.
Your task is to find out the maximum possible total prettiness of the contest composed of at most three problems from the given pool.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of problems.
The second line of the query contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the prettiness of the i-th problem.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the maximum possible cumulative prettiness of the contest composed of at most three problems from the given pool of problems in the query.
Example
Input
3
4
5 6 15 30
4
10 6 30 15
3
3 4 6
Output
30
31
10 | instruction | 0 | 52,289 | 22 | 104,578 |
Tags: brute force, math, sortings
Correct Solution:
```
q = int(input())
ans = []
for w in range(q):
n = int(input())
a = [int(x) for x in input().split()]
a.sort()
a = a[::-1]
a1 = 0
b1 = 0
c1 = 0
a1 = a[0]
for i in range(n):
if a1 % a[i] != 0:
b1 = a[i]
break
for i in range(n):
if a1 % a[i] != 0 and b1 % a[i] != 0:
c1 = a[i]
break
a2 = 0
b2 = 0
c2 = 0
if a1 % 2 == 0:
for i in range(n):
if a[i] == a1 // 2:
a2 = a[i]
break
if a1 % 3 == 0:
for i in range(n):
if a[i] == a1 // 3:
b2 = a[i]
break
if a1 % 5 == 0:
for i in range(n):
if a[i] == a1 // 5:
c2 = a[i]
break
ans.append(max(a1 + b1 + c1, a2 + b2 + c2))
for i in range(q):
print(ans[i])
``` | output | 1 | 52,289 | 22 | 104,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,423 | 22 | 104,846 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
from collections import defaultdict
def primeFactors(n):
dic = defaultdict(int)
factors = []
while n % 2 == 0:
dic[2] += 1
n = n // 2
i = 3
while i * i <= n:
while n % i == 0:
dic[i] += 1
n = n // i
i += 2
if n > 2:
dic[n] += 1
return dic
def isPrime(n):
if n==1:
return False
i=2
while(i*i<=n):
if n%i==0:
return False
i+=1
return True
def recur(n,prev,prime):
if n <= 1:
return
i = prime
flag = 0
while i * i <= n:
if n % i == 0 and i % prev == 0:
flag = 1
res.append(i)
break
i += 1
if flag:
recur(n // i, i,prime)
else:
if res:
tmp = res.pop()
if n % tmp != 0:
res.append(tmp * n)
else:
res.append(tmp)
res.append(n)
for _ in range(int(input())):
n = int(input())
if isPrime(n):
print(1)
print(n)
continue
factors = primeFactors(n)
ans = -1
fin = []
for prime in factors:
res = []
recur(n, 1, prime)
if len(res) > len(fin):
fin = res[:]
print(len(fin))
print(*fin)
``` | output | 1 | 52,423 | 22 | 104,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,424 | 22 | 104,848 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
def read():
return int(input())
def solve(n):
res = []
for i in range(2, int(n ** 0.5) + 2):
if not (n % i):
cur = [0, i]
while not (n % i):
n //= i
cur[0] += 1
res.append(cur)
if n != 1:
res.append([1, n])
return res
for _ in range(read()):
n = read()
p = solve(n)
p.sort(key=lambda x: x[0], reverse=True)
print(p[0][0])
for i in range(p[0][0] - 1):
print(p[0][1], end=' ')
rem = p[0][1]
for i in range(1, len(p)):
for j in range(p[i][0]):
rem *= p[i][1]
print(rem)
``` | output | 1 | 52,424 | 22 | 104,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,425 | 22 | 104,850 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
from collections import defaultdict
import math
def len(d):
count=0
for jj in d:
count+=1
return count
for _ in range(int(input())):
n=int(input())
no=n
d=defaultdict(int)
i,m,index=2,0,-1
while i * i <= n:
c=0
while n%i == 0:
c+=1
n = n // i
d[i]+=c
if m<d[i]:
index=i
m=d[i]
i = i + 1
if no==2:
print(1)
print(2,end=" ")
elif no==3:
print(1)
print(3,end=" ")
elif no==5:
print(1)
print(5,end=" ")
elif no==4:
print(2)
print(2,2,end=" ")
else:
if d[index]==0:
print(1)
print(no,end=" ")
elif no==index**d[index]:
print(d[index])
for j in range(d[index]):print(int(index),end=" ")
else:
print(d[index])
for j in range(d[index]-1):
print(int(index),end=" ")
print(abs(int(no//(index**(d[index]-1)))),end=" ")
print()
# print((no//(index**(d[index]-1))),len(d))
``` | output | 1 | 52,425 | 22 | 104,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,426 | 22 | 104,852 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import math
def prime_factorize(n):
n2 = n
i = 2
ans = []
while n > 1:
if i > math.sqrt(n2):
if len(ans)==0:
return [(n2, 1)]
else:
ans.append((n,1))
return ans
power = 0
while n % i == 0:
power += 1
n = n//i
if power >= 1:
ans.append((i,power))
i += 1
return ans
def main():
for _ in range(ii()): # codeforces testcases
n=ii()
a = prime_factorize(n)
maxpow = max([j[1] for j in a])
print(maxpow)
for j in range(maxpow):
b=1
for i in range(len(a)):
prime,power = a[i]
if power >= maxpow - j:
b *= prime
print(b)
# shorthands
def mpin(): return map(int, input().split())
def lmpin(): return list(mpin())
def ii(): return int(input())
def gridi(h,w): return [lmpin() for i in range(h)]
def gridstr(h,w): return [list(input()) for i in range(h)]
def zero1(h): return [0 for i in range(h)]
def zero2(h,w): return [[0 for j in range(w)] for i in range(h)]
# 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")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 52,426 | 22 | 104,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,427 | 22 | 104,854 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
import math
global prime
prime = [True for i in range(10**5+1)]
def SieveOfEratosthenes(n):
n2 = int(n)
# Create a boolean array
# "prime[0..n]" and initialize
# all entries it as true.
# A value in prime[i] will
# finally be false if i is
# Not a prime, else true.
p = 2
while (p * p <= n):
# If prime[p] is not
# changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, int(math.sqrt(n))+10, p):
prime[i] = False
p += 1
# Print all prime numbers
def solve(n):
freq_max=0
max_dig = 0
max_prod =0
i=2
n2 = int(n)
# lol = int(math.sqrt(n2))
while (i*i<=n2):
# print(i)
if prime[i]:
n=int(n2)
if n%i==0 :
# print('hi',i)
freq=0
prod = 1
while n%i==0:
# print("hi2")
prod*=i
freq+=1
n/=i
if freq>freq_max :
freq_max=freq
max_dig = i
if n==1: break
i+=1
# print(max_dig,freq_max,n2)
if max_dig==0 :
print(1)
print(n2)
else:
if freq_max==1 :
if (n2//max_dig)%max_dig==0 :
print(2)
print(max_dig, n2//max_dig)
else:
print(1)
print(n2)
else:
z = int(freq_max)
print(z)
while (freq_max!=1):
print(max_dig,end=" ")
freq_max-=1
print(n2//(max_dig**(z-1)))
if __name__=="__main__":
T = int(input())
SieveOfEratosthenes(10**5+1)
for i in range(T):
global n
n = int(input())
solve(n)
``` | output | 1 | 52,427 | 22 | 104,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,428 | 22 | 104,856 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
import sys
LI=lambda:list(map(int, sys.stdin.readline().split()))
MI=lambda:map(int, sys.stdin.readline().split())
SI=lambda:sys.stdin.readline().strip('\n')
II=lambda:int(sys.stdin.readline())
N=1000001
ok=[1]*N
primes=[]
for i in range(2, N):
if ok[i]:
primes.append(i)
for j in range(i+i, N, i):
ok[j]=0
for _ in range(II()):
n=II()
nn=n
num, mx=0, 0
for p in primes:
cnt=0
while n%p==0:
n//=p
cnt+=1
if cnt>mx:
mx=cnt
num=p
if n>1:
if mx<1:
mx=1
num=p
print(mx)
print(*[num]*(mx-1), nn//(num**(mx-1)))
``` | output | 1 | 52,428 | 22 | 104,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,429 | 22 | 104,858 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
# biesect_left is essentially an equivalent of lower_bound function in
# cpp and returns the first index not smaller than x.
from bisect import bisect_left;
from bisect import bisect_right;
from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1;
return x;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k, modulus = 1):
if(k > n - k):
k = n - k;
res = 1;
for i in range(k):
res = res * (n - i);
res = res / (i + 1);
res %= modulus;
return int(res);
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret;
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
for _ in range(int(input())):
n = int(input());
f = primefs(n);
x = 0;
for i in f:
x = max(x,f[i]);
a = None;
b = None;
for i in f:
if f[i] == x and a == None and x > 1:
a = i;
else:
if b == None:
b = (i ** f[i]);
else:
b *= (i ** f[i]);
if not a:
print(1);
print(n);
continue;
ans = [a] * (x - 1);
if b != None:
ans.append(b * a);
else:
ans.append(a);
print(x);
print(*ans);
``` | output | 1 | 52,429 | 22 | 104,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083 | instruction | 0 | 52,430 | 22 | 104,860 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
def sieve_eratosthenes(N):
"Generate list of prime integers from 1 to N (inclusive)"
assert isinstance(N, int), "sieve_eratosthenes(N): Given N non integer " + \
"data types"
if N < 2:
return []
sieve = list(range(3, N+1, 2)) # step=2 bcs only '2' is the prime even number
# rumus indexing: indeks = (angka-3)//2
for i in range(3, N+1, 2):
if (sieve[(i-3)//2] is False): # if sieve[i] is False
continue
# step=2i (j+=2i), bcs j += i will yield an even numbers
for j in range(i*i, N+1, 2*i):
sieve[(j-3)//2] = False
result = [2] # 2 is also a prime
result.extend(filter(None, sieve))
return result
primes = sieve_eratosthenes(10**4)
CONST = 10**5
def decompose(N):
N_before = N
i = 0
_ = primes[i]
__ = len(primes) - 1
factors = []
while _ <= N:
if N % _ == 0:
count = 0
while N%_ == 0:
count += 1
N //= _
factors.append([count, _])
if i < __:
i += 1
_ = primes[i]
else:
break
if N > CONST and False: # asumsi greedy
factors.append([1, N])
elif N > primes[-1] :
i = primes[-1] + 2
while i*i <= N_before:
if N%i == 0:
count = 0
while N%i == 0:
count += 1
N //= i
factors.append([count, i])
i += 2
if N != 1:
factors.append([1, N])
return factors
T = int(input())
for tc in range(T):
N = int(input())
decomposed = decompose(N)
decomposed.sort(reverse=True)
result = []
multiplier = 1
for i in range(len(decomposed)-1):
multiplier *= decomposed[i][1]
while decomposed[i][0] > decomposed[i+1][0]:
decomposed[i][0] -= 1
result.append(multiplier)
multiplier *= decomposed[-1][1]
for i in range(decomposed[-1][0]):
result.append(multiplier)
print(len(result))
print(" ".join(map(str, result)))
``` | output | 1 | 52,430 | 22 | 104,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n > 1).
Your task is to find a sequence of integers a_1, a_2, …, a_k such that:
* each a_i is strictly greater than 1;
* a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n);
* a_{i + 1} is divisible by a_i for each i from 1 to k-1;
* k is the maximum possible (i. e. the length of this sequence is the maximum possible).
If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).
It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).
Output
For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.
If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.
Example
Input
4
2
360
4999999937
4998207083
Output
1
2
3
2 2 90
1
4999999937
1
4998207083
Submitted Solution:
```
import sys
from collections import defaultdict
input = lambda:sys.stdin.readline().strip()
def get_primes(n):
ans = defaultdict(int)
while n%2==0:
ans[2]+=1
n = n//2
for i in range(3,int(n**0.5)+1,2):
while n%i==0:
ans[i]+=1
n = n//i
if n>2:
ans[n]+=1
return ans
t = int(input())
while t:
t-=1
n = int(input())
primes = get_primes(n)
prime,k = max(primes.items(),key=lambda x:x[1])
a = [prime]*k
del primes[prime]
for key in primes:
a[-1]*=key**primes[key]
print(k)
print(*a)
``` | instruction | 0 | 52,433 | 22 | 104,866 |
Yes | output | 1 | 52,433 | 22 | 104,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,475 | 22 | 104,950 |
Tags: brute force, math, number theory
Correct Solution:
```
n,k=map(int,input().split())
if n<13 and k>0:
print("NO")
exit()
if k==0:
print("YES")
exit()
prime=[True for i in range(n+1)]
prime[0]=False
prime[1]=False
p=2
while p*p<=n:
if prime[p]:
for i in range(p*2,n+1,p):
prime[i]=False
p+=1
l=[]
for i in range(n+1):
if prime[i]:
l.append(i)
#print(l)
c=0
for i in range(len(l)-1):
if l[i]+l[i+1]+1 in l:
#print(l[i],l[i+1])
c+=1
if c>=k:
print("YES")
else:
print("NO")
``` | output | 1 | 52,475 | 22 | 104,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,476 | 22 | 104,952 |
Tags: brute force, math, number theory
Correct Solution:
```
n,k = map(int,input().split())
primesList = list()
#function to find prime
for i in range(2,n+1):
isPrime = True
for j in range(2,i):
if i%j==0:
isPrime = False
break
if isPrime:
primesList.append(i)
cnt = 0
for i in range(len(primesList)-1):
if primesList[i]+primesList[i+1]+1 in primesList:
#print(primesList[i]+primesList[i+1]+1)
cnt+=1
if cnt>=k:
print("Yes")
else:
print("No")
``` | output | 1 | 52,476 | 22 | 104,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,477 | 22 | 104,954 |
Tags: brute force, math, number theory
Correct Solution:
```
n,k=map(int,input().split())
prime=[1]*5000
valid=[0]*5000
for i in range(2,32):
if prime[i]==1:
for j in range(2*i,1000,i):
prime[j]=0
lis=list()
for i in range(2,1000):
if prime[i]==1:
lis.append(i)
#print(len(lis))
for i in range(len(lis)-1):
valid[lis[i]+lis[i+1]]=1
c=0
for i in range(2,n+1):
if prime[i] and valid[i-1]:
c+=1
if c>=k:
print("YES")
else :
print("NO")
``` | output | 1 | 52,477 | 22 | 104,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,478 | 22 | 104,956 |
Tags: brute force, math, number theory
Correct Solution:
```
def primenum(k): #functionfor checking if a number is prime
if(k < 2):
return False
i = 2
while(i*i <= k):
if(k%i == 0):
return False
i += 1
return True
n, k = [int(x) for x in input().split()] #input data
#find all prime numbers from 2 to n+1
primes = []
for i in range(1, n+1):
if(primenum(i)):
primes.append(i)
#count how many variants of sums exist
count = 0
for i in range(len(primes)-1):
a = primes[i]
b = primes[i+1]
if(primenum(1+a+b) and 1+a+b<=n):
count += 1
if(count < k): #if number of variants is less than k
print("NO")
else: #if number of variants is greater or equal to k
print("YES")
``` | output | 1 | 52,478 | 22 | 104,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,479 | 22 | 104,958 |
Tags: brute force, math, number theory
Correct Solution:
```
import math as m
def isPrime(n):
if n == 2 or n == 3:
return True
if n == 1 or n%2 == 0 or n % 3 == 0:
return False
i = 1
while (6*i - 1) <= int(m.sqrt(n)):
if n % (6*i + 1) == 0 or n % (6*i - 1) == 0:
return False
i+=1
return True
n, k = list(map(int, input().split()))
sum = 0
result = 'NO'
a = 2
for i in range(3, n+1):
if isPrime(i):
if a+i+1>n:
break
if isPrime(a + i + 1):
sum+=1
a = i
if sum >= k:
result = 'YES'
print(result)
``` | output | 1 | 52,479 | 22 | 104,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,480 | 22 | 104,960 |
Tags: brute force, math, number theory
Correct Solution:
```
from math import sqrt
from math import ceil
def is_prime(number):
if number == 2:
return True
for k in range(2, ceil(sqrt(number)) + 1):
if number % k == 0:
return False
return True
n, k = [int(x) for x in input().split()]
prime_numbers = [x for x in range(2, n +1) if is_prime(x)]
answer = 0
flag = 0
if k == 0:
print('YES')
else:
for i in range(len(prime_numbers) - 2):
if prime_numbers[i] + prime_numbers[i + 1] + 1 in prime_numbers:
answer += 1
if answer == k:
print('YES')
flag = 1
break
if flag == 0:
print('NO')
``` | output | 1 | 52,480 | 22 | 104,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,481 | 22 | 104,962 |
Tags: brute force, math, number theory
Correct Solution:
```
n,k=map(int,input().split())
v=[]
for i in range(2,n+1):
if all(i%j!=0 for j in v):
v.append(i)
c=0
for i in range(len(v)-1):
if v[i]+v[i+1]+1 in v:
c+=1
if c>=k:
print("YES")
else:
print("NO")
``` | output | 1 | 52,481 | 22 | 104,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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. | instruction | 0 | 52,482 | 22 | 104,964 |
Tags: brute force, math, number theory
Correct Solution:
```
N, K = map(int, input().split())
primes = [2, 3]
for n in range(5, N + 1, 2):
if all((n % x != 0 for x in primes)):
primes.append(n)
count = 0
for index in range(len(primes) - 1):
if (primes[index] + primes[index + 1] + 1) in primes:
count += 1
if count >= K:
print('YES')
else:
print('NO')
``` | output | 1 | 52,482 | 22 | 104,965 |
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:
```
def prime_number(n):
l=[]
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 1
prime[0]= False
prime[1]= False
for p in range(n + 1):
if prime[p]:
l.append(p)
return l
n,k=map(int,input().split())
l=prime_number(n)
cnt=0
for i in range((len(l)-1)):
if l[i]+l[i+1]+1 in l:
cnt+=1
if cnt>=k:
print("Yes")
else:
print("NO")
``` | instruction | 0 | 52,483 | 22 | 104,966 |
Yes | output | 1 | 52,483 | 22 | 104,967 |
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 sys
import math
def primes(number):
size = number + 1
lst = [True] * (number + 1)
primeList = []
for i in range(2, number):
if lst[i] == False : continue
for j in range(i*2, size, i):
if j < size and lst[j] == True and j % i == 0:
lst[j] = False
for i in range(size-1, 0, -1):
if lst[i] == True :
primeList.append(i)
primeList.sort()
return primeList[1:]
def solve(firstLine):
n ,k = firstLine[0], firstLine[1]
primeList = primes(n * 2 + 1)
sol = 0
for i, prime in enumerate(primeList):
if primeList[i + 1] > n:
break
num = primeList[i + 1] + primeList[i] + 1
if num <= n and num in primeList:
#print(primeList[i + 1] ,primeList[i], num)
sol += 1
# for prime in primeList:
# lst[prime] += 1
# for i in range(prime * 2, n + 1 , prime):
# lst[i] = max([lst[i], lst[i-prime]]) + 1
#print("aa", sol)
if sol >= k:
print("YES")
else :
print("NO")
return
def main():
firstLine = sys.stdin.readline().split()
firstLine = list(map(int, firstLine))
solve(firstLine)
if __name__ == "__main__":
main()
``` | instruction | 0 | 52,484 | 22 | 104,968 |
Yes | output | 1 | 52,484 | 22 | 104,969 |
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:
```
from decimal import Decimal
import math
def primo(n):
for i in range(2, int(math.sqrt(n)+1)):
if n % i == 0:
return False;
return n>1;
l1 = input().split()
#l2 = input().split()
#l3 = input().split()
l1 = [int(i) for i in l1]
#l2 = [int(i) for i in l2]
l = l1[0]
v = l1[1]
primos = [x for x in range(l+1) if primo(x)]
total = 0
for k in range(2,len(primos)):
for m in range (k-1):
if(primos[m] + primos[m + 1] + 1 == primos[k]):
total+=1
if(total >= v):
print("YES")
else:
print("NO")
``` | instruction | 0 | 52,485 | 22 | 104,970 |
Yes | output | 1 | 52,485 | 22 | 104,971 |
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 isprime(n):
if n % 2 == 0 and n > 2:
return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
n, k = map(int, input().split())
primes, somas = [], []
aux = 0
for i in range(2, n + 1):
if isprime(i):
primes.append(i)
for i in range(1, len(primes)):
somas.append(primes[i] + primes[i - 1])
for i in primes:
index = primes.index(i)
for j in somas:
if somas.index(j) < index and (j + 1 == i):
k -= 1
if k <= 0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 52,486 | 22 | 104,972 |
Yes | output | 1 | 52,486 | 22 | 104,973 |
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:
```
#!/usr/bin/env python3
def is_prime(p):
for i in range(3, p):
if p % i == 0:
return False
return True
def next_prime(k):
result = k+1
while not is_prime(result):
result += 1
return result
def check_noldbach(n, k):
count = 0
p1 = 3
p2 = 5
while (p1 + p2 + 1 <= n):
print('Checking pair: ' + str(p1) + ' ' + str(p2), end = ' ')
if is_prime(p1 + p2 + 1):
print('Good pair')
count += 1
else:
print('')
p1 = p2
p2 = next_prime(p2)
if count >= k:
return True
else:
return False
def main():
n, k = [int(j) for j in input().split(' ')]
if check_noldbach(n, k):
print('YES')
else:
print('NO')
if __name__ == '__main__':
main()
``` | instruction | 0 | 52,487 | 22 | 104,974 |
No | output | 1 | 52,487 | 22 | 104,975 |
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:
```
def isPrime(x):
i=int(3)
while i*i<=x:
if x%i==0:
return False
i+=1
return True
l=[]
for i in range(3,1009,2):
if isPrime(i):
l.append(i)
# print(i)
n,k=map(int,input().split())
for i in range(len(l)-1):
if l[i]+l[i+1]+1 in l:
k-=1
if l[i]+l[i+1]+1>n or k==0:
break
print("N0" if k else "YES")
``` | instruction | 0 | 52,488 | 22 | 104,976 |
No | output | 1 | 52,488 | 22 | 104,977 |
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:
```
s=[int(n) for n in input().split()]
x=sum(s)
fl='YES'
pr=[2,3,5]
if x==2 or x==3 or x==5:
fl='YES'
elif x%2==0 or x%3==0 or x%5==0 or x==1:
fl='NO'
elif (x**.5)==int(x**.5) or (x**(1/3))==int(x**(1/3)) :
fl='NO'
else:
for n in range(2,int(x**.5)):
if n not in pr:
if x%n==0:
fl='NO'
else:
fl='YES'
print(fl)
``` | instruction | 0 | 52,489 | 22 | 104,978 |
No | output | 1 | 52,489 | 22 | 104,979 |
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