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.
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor d between two integers a and b that is in a given range from low to high (inclusive), i.e. low ≤ d ≤ high. It is possible that there is no common divisor in the given range.
You will be given the two integers a and b, then n queries. Each query is a range from low to high and you have to answer each query.
Input
The first line contains two integers a and b, the two integers as described above (1 ≤ a, b ≤ 109). The second line contains one integer n, the number of queries (1 ≤ n ≤ 104). Then n lines follow, each line contains one query consisting of two integers, low and high (1 ≤ low ≤ high ≤ 109).
Output
Print n lines. The i-th of them should contain the result of the i-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
Examples
Input
9 27
3
1 5
10 11
9 11
Output
3
-1
9
Submitted Solution:
```
import math,sys
from collections import Counter, defaultdict, deque
from sys import stdin, stdout
input = stdin.readline
li = lambda:list(map(int,input().split()))
def factors(g):
fact=[]
for i in range(1,int(math.sqrt(g))+1):
if(g%i==0):
if(i*i==g):
fact.append(i)
else:
fact.append(i)
fact.append(g//i)
fact.sort(reverse=True)
# print(*fact)
return(fact)
def case():
a,b=li()
g=math.gcd(a,b)
print(g)
fact=factors(g)
q=int(input())
for i in range(q):
l,r=li()
flag=0
for j in range(len(fact)):
if(fact[j]<=r and fact[j]>=l):
print(fact[j])
flag=1
break
if(not flag):
print(-1)
for _ in range(1):
case()
``` | instruction | 0 | 94,911 | 22 | 189,822 |
No | output | 1 | 94,911 | 22 | 189,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor d between two integers a and b that is in a given range from low to high (inclusive), i.e. low ≤ d ≤ high. It is possible that there is no common divisor in the given range.
You will be given the two integers a and b, then n queries. Each query is a range from low to high and you have to answer each query.
Input
The first line contains two integers a and b, the two integers as described above (1 ≤ a, b ≤ 109). The second line contains one integer n, the number of queries (1 ≤ n ≤ 104). Then n lines follow, each line contains one query consisting of two integers, low and high (1 ≤ low ≤ high ≤ 109).
Output
Print n lines. The i-th of them should contain the result of the i-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
Examples
Input
9 27
3
1 5
10 11
9 11
Output
3
-1
9
Submitted Solution:
```
from math import gcd
a,b=map(int,input().split())
c=gcd(a,b)
c=int(c**0.5)
m=[]
for i in range(1,c+1):
if c%i==0:
m.append(i)
m.append(b//i)
m.sort(reverse=True)
n=int(input())
for i in range(n):
c,d=map(int,input().split())
y=0
for i in m:
if i>=c and i<=d:
print(i)
y=1
break
if y==0:
print("-1")
``` | instruction | 0 | 94,912 | 22 | 189,824 |
No | output | 1 | 94,912 | 22 | 189,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor d between two integers a and b that is in a given range from low to high (inclusive), i.e. low ≤ d ≤ high. It is possible that there is no common divisor in the given range.
You will be given the two integers a and b, then n queries. Each query is a range from low to high and you have to answer each query.
Input
The first line contains two integers a and b, the two integers as described above (1 ≤ a, b ≤ 109). The second line contains one integer n, the number of queries (1 ≤ n ≤ 104). Then n lines follow, each line contains one query consisting of two integers, low and high (1 ≤ low ≤ high ≤ 109).
Output
Print n lines. The i-th of them should contain the result of the i-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
Examples
Input
9 27
3
1 5
10 11
9 11
Output
3
-1
9
Submitted Solution:
```
a, b = map(int, input().split())
import math
g = math.gcd(a, b)
lis = [i for i in range(1, g+1) if g%i == 0]
print(lis)
n = int(input())
for _ in range(n):
a, b = map(int , input().split())
if a > g: print(-1)
else:
bo = 1
for i in range(b, a-1, -1):
if i in lis: print(i); bo = 0
if bo: print(-1)
``` | instruction | 0 | 94,913 | 22 | 189,826 |
No | output | 1 | 94,913 | 22 | 189,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,930 | 22 | 189,860 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
import sys
mod = 10**9 + 7
def solve():
n = int(input())
a = [int(i) for i in input().split()]
cnt = [0]*(10**5 + 1)
pat = [0]*(10**5 + 1)
p2 = [1]*(n + 1)
for i in range(1, n + 1):
p2[i] = (2 * p2[i - 1]) % mod
for ai in a:
cnt[ai] += 1
for i in range(1, 10**5 + 1):
for j in range(2*i, 10**5 + 1, i):
cnt[i] += cnt[j]
pat[i] = p2[cnt[i]] - 1
for i in range(10**5, 0, -1):
for j in range(2*i, 10**5 + 1, i):
pat[i] = (pat[i] - pat[j]) % mod
ans = pat[1] % mod
print(ans)
if __name__ == '__main__':
solve()
``` | output | 1 | 94,930 | 22 | 189,861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,931 | 22 | 189,862 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n = int(input())
#n, k = map(int, input().split())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
a = list(map(int, input().split()))
#b = list(map(int, input().split()))
#s = input()
mod = 10**9 + 7
twopow = [1]*(10**5+69)
for i in range(1, 10**5+69):
twopow[i] = (twopow[i-1] * 2) % mod
count = [0]*100069
for i in a:
count[i] += 1
multiples = [0]*100069
for i in range(1, 10**5+1):
for j in range(i, 10**5+1, i):
multiples[i] += count[j]
gcd_of = [0]*100069
for i in range(10**5, 0, -1):
gcd_of[i] = (twopow[multiples[i]] - 1) % mod
for j in range(2*i, 10**5+1, i):
gcd_of[i] -= gcd_of[j]
print(gcd_of[1] % mod)
``` | output | 1 | 94,931 | 22 | 189,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,932 | 22 | 189,864 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
n=int(input())
r=list(map(int,input().split()))
dp=[0]*(10**5+1)
cnt=[0]*(10**5+1)
tmp=[0]*(10**5+1)
mod=10**9+7
for i in range(n):
cnt[r[i]]+=1
for i in range(1,10**5+1):
for j in range(2*i,10**5+1,i):
cnt[i]+=cnt[j]
tmp[i]=pow(2,cnt[i],mod)-1
for i in range(10**5,0,-1):
for j in range(2*i,10**5+1,i):
tmp[i]=(tmp[i]-tmp[j])%mod
print(tmp[1]%mod)
``` | output | 1 | 94,932 | 22 | 189,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,933 | 22 | 189,866 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
MOD = int( 1e9 ) + 7
N = int( input() )
A = list( map( int, input().split() ) )
pow2 = [ pow( 2, i, MOD ) for i in range( N + 1 ) ]
maxa = max( A )
mcnt = [ 0 for i in range( maxa + 1 ) ]
mans = [ 0 for i in range( maxa + 1 ) ]
for i in range( N ):
mcnt[ A[ i ] ] += 1
for i in range( 1, maxa + 1 ):
for j in range( i + i, maxa + 1, i ):
mcnt[ i ] += mcnt[ j ]
mans[ i ] = pow2[ mcnt[ i ] ] - 1
for i in range( maxa, 0, -1 ):
for j in range( i + i, maxa + 1, i ):
mans[ i ] = ( mans[ i ] - mans[ j ] ) % MOD
print( mans[ 1 ] + ( mans[ 1 ] < 0 ) * MOD )
``` | output | 1 | 94,933 | 22 | 189,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,934 | 22 | 189,868 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
import sys
mod = 10**9 + 7
def solve():
n = int(input())
a = [int(i) for i in input().split()]
cnt = [0]*(10**5 + 1)
pat = [0]*(10**5 + 1)
for ai in a:
cnt[ai] += 1
for i in range(1, 10**5 + 1):
for j in range(2*i, 10**5 + 1, i):
cnt[i] += cnt[j]
pat[i] = (pow(2, cnt[i], mod) - 1)
for i in range(10**5, 0, -1):
for j in range(2*i, 10**5 + 1, i):
pat[i] = (pat[i] - pat[j]) % mod
ans = pat[1] % mod
print(ans)
if __name__ == '__main__':
solve()
``` | output | 1 | 94,934 | 22 | 189,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,935 | 22 | 189,870 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
#Bhargey Mehta (Sophomore)
#DA-IICT, Gandhinagar
import sys, math, queue
#sys.stdin = open("input.txt", "r")
MOD = 10**9+7
sys.setrecursionlimit(1000000)
def getMul(x):
a = 1
for xi in x:
a *= xi
return a
n = int(input())
a = list(map(int, input().split()))
d = {}
for ai in a:
if ai in d: d[ai] += 1
else: d[ai] = 1
f = [[] for i in range(max(a)+10)]
for i in range(1, len(f)):
for j in range(i, len(f), i):
f[j].append(i)
seq = [0 for i in range(max(a)+10)]
for ai in d:
for fi in f[ai]:
seq[fi] += d[ai]
for i in range(len(seq)):
seq[i] = (pow(2, seq[i], MOD) -1 +MOD) % MOD
pf = [[] for i in range(max(a)+10)]
pf[0] = None
pf[1].append(1)
for i in range(2, len(f)):
if len(pf[i]) == 0:
for j in range(i, len(pf), i):
pf[j].append(i)
for i in range(1, len(pf)):
mul = getMul(pf[i])
if mul == i:
if len(pf[i])&1 == 1: pf[i] = -1
else: pf[i] = 1
else:
pf[i] = 0
pf[1] = 1
ans = 0
for i in range(1, len(seq)):
ans += seq[i]*pf[i]
ans = (ans + MOD) % MOD
print(ans)
``` | output | 1 | 94,935 | 22 | 189,871 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,936 | 22 | 189,872 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
# 803F
import math
import collections
def do():
n = int(input())
nums = map(int, input().split(" "))
count = collections.defaultdict(int)
for num in nums:
for i in range(1, int(math.sqrt(num))+1):
cp = num // i
if num % i == 0:
count[i] += 1
if cp != i and num % cp == 0:
count[cp] += 1
maxk = max(count.keys())
freq = {k: (1 << count[k]) - 1 for k in count}
for k in sorted(count.keys(), reverse=True):
for kk in range(k << 1, maxk+1, k):
freq[k] -= freq[kk] if kk in freq else 0
return freq[1] % (10**9 + 7)
print(do())
``` | output | 1 | 94,936 | 22 | 189,873 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime. | instruction | 0 | 94,937 | 22 | 189,874 |
Tags: bitmasks, combinatorics, number theory
Correct Solution:
```
N = 10**5+5
MOD = 10**9+7
freq = [0 for i in range(N)]
# Calculating {power(2,i)%MOD} and storing it at ith pos in p2 arr
p2 = [0 for i in range(N)]
p2[0] = 1
for i in range(1,N):
p2[i] = p2[i-1]*2
p2[i]%=MOD
def Calculate_Mobius(N):
arr = [1 for i in range(N+1)]
prime_count = [0 for i in range(N+1)]
mobius_value = [0 for i in range(N+1)]
for i in range(2,N+1):
if prime_count[i]==0:
for j in range(i,N+1,i):
prime_count[j]+=1
arr[j] = arr[j] * i
for i in range(1, N+1):
if arr[i] == i:
if (prime_count[i] & 1) == 0:
mobius_value[i] = 1
else:
mobius_value[i] = -1
else:
mobius_value[i] = 0
return mobius_value
# Caluclating Mobius values for nos' till 10^5
mobius = Calculate_Mobius(N)
n = int(input())
b = [int(i) for i in input().split()]
# Storing freq of I/p no.s in array
for i in b:
freq[i]+=1
ans = 0
for i in range(1,N):
# Count no of I/p no's which are divisible by i
cnt = 0
for j in range(i,N,i):
cnt += freq[j]
total_subsequences = p2[cnt] - 1
ans = (ans + (mobius[i] * total_subsequences)%MOD)%MOD
# Dealing if ans is -ve due to MOD
ans += MOD
print(ans%MOD)
``` | output | 1 | 94,937 | 22 | 189,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime.
Submitted Solution:
```
N = 10**5+5
MOD = 10**9+7
freq = [0 for i in range(N)]
# Calculating {power(2,i)%MOD} and storing it at ith pos in p2 arr
p2 = [0 for i in range(N)]
p2[0] = 1
for i in range(1,N):
p2[i] = p2[i-1]*2
p2[i]%=MOD
def Calculate_Mobius(N):
arr = [1 for i in range(N+1)]
prime_count = [0 for i in range(N+1)]
mobius_value = [0 for i in range(N+1)]
for i in range(2,N+1):
if prime_count[i]==0:
for j in range(i,N+1,i):
prime_count[j]+=1
arr[j] = arr[j] * i
for i in range(1, N+1):
if arr[i] == i:
if (prime_count[i] & 1) == 0:
mobius_value[i] = 1
else:
mobius_value[i] = -1
else:
mobius_value[i] = 0
return mobius_value
# Caluclating Mobius values for nos' till 10^5
mobius = Calculate_Mobius(N)
b = [int(i) for i in input().split()]
# Storing freq of I/p no.s in array
for i in b:
freq[i]+=1
ans = 0
for i in range(1,N):
# Count no of I/p no's which are divisible by i
cnt = 0
for j in range(i,N,i):
cnt += freq[j]
total_subsequences = p2[cnt] - 1
ans = (ans + (mobius[i] * total_subsequences)%MOD)%MOD
# Dealing if ans is -ve due to MOD
ans += MOD
print(ans%MOD)
``` | instruction | 0 | 94,938 | 22 | 189,876 |
No | output | 1 | 94,938 | 22 | 189,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime.
Submitted Solution:
```
#Bhargey Mehta (Sophomore)
#DA-IICT, Gandhinagar
import sys, math, queue
#sys.stdin = open("input.txt", "r")
MOD = 10**9+7
sys.setrecursionlimit(1000000)
n = int(input())
a = list(map(int, input().split()))
if a[0] == 1:
ans = 1
else:
ans = 0
gs = {a[0]: 1}
for i in range(1, n):
if a[i] == 1:
ans = (ans+1) % MOD
temp = {}
for g in gs:
gx = math.gcd(g, a[i])
if gx == 1:
ans = (ans+gs[g]) % MOD
if gx in gs:
if gx in temp:
temp[gx] = (temp[gx]+gs[g]) % MOD
else:
temp[gx] = (gs[gx]+gs[g]) % MOD
else:
temp[gx] = gs[g]
for k in temp:
gs[k] = temp[k]
if a[i] in gs:
gs[a[i]] += 1
else:
gs[a[i]] = 1
print(ans)
``` | instruction | 0 | 94,939 | 22 | 189,878 |
No | output | 1 | 94,939 | 22 | 189,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime.
Submitted Solution:
```
fact=[1]
temp=1
MOD=10**9+7
for i in range(1,10**5+5):
temp*=i
temp%=MOD
fact+=[temp]
def bino(a,b):
up=fact[a]
down=pow(fact[b]*fact[a-b],MOD-2,MOD)
return (up*down)%MOD
def find(A):
MOD=10**9+7
dp=[0]*(10**5+2)
for x in A:
dp[x]+=1
for i in range(2,len(dp)):
for j in range(2,len(dp)):
if i*j>len(dp)-1:
break
dp[i]+=dp[i*j]
for i in range(2,len(dp)):
dp[i]=(pow(2,dp[i],MOD)-1)%MOD
for i in range(len(dp)-1,1,-1):
for j in range(2,len(dp)):
if i*j>=len(dp):
break
dp[i]-=dp[i*j]
dp[i]%=MOD
ans=0
for i in range(2,len(dp)):
ans+=dp[i]
ans%=MOD
return (pow(2,len(A),MOD)-ans-1)%MOD
print(find(list(map(int,input().strip().split(' ')))))
``` | instruction | 0 | 94,940 | 22 | 189,880 |
No | output | 1 | 94,940 | 22 | 189,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common divisor of all elements of this sequence is equal to 1.
Given an array a consisting of n positive integers, find the number of its coprime subsequences. Since the answer may be very large, print it modulo 109 + 7.
Note that two subsequences are considered different if chosen indices are different. For example, in the array [1, 1] there are 3 different subsequences: [1], [1] and [1, 1].
Input
The first line contains one integer number n (1 ≤ n ≤ 100000).
The second line contains n integer numbers a1, a2... an (1 ≤ ai ≤ 100000).
Output
Print the number of coprime subsequences of a modulo 109 + 7.
Examples
Input
3
1 2 3
Output
5
Input
4
1 1 1 1
Output
15
Input
7
1 3 5 15 3 105 35
Output
100
Note
In the first example coprime subsequences are:
1. 1
2. 1, 2
3. 1, 3
4. 1, 2, 3
5. 2, 3
In the second example all subsequences are coprime.
Submitted Solution:
```
# 803F
import math
import collections
def do():
n = int(input())
nums = map(int, input().split(" "))
count = collections.defaultdict(int)
for num in nums:
for i in range(1, int(math.sqrt(num))+1):
cp = num // i
if num % i == 0:
count[i] += 1
if cp != i and num % cp == 0:
count[cp] += 1
maxk = max(count.keys())
freq = {k: (1 << count[k]) - 1 for k in count}
for k in sorted(count.keys(), reverse=True):
for kk in range(k << 1, maxk+1, k):
freq[k] -= freq[kk] if kk in freq else 0
return freq[1]
print(do())
``` | instruction | 0 | 94,941 | 22 | 189,882 |
No | output | 1 | 94,941 | 22 | 189,883 |
Provide a correct Python 3 solution for this coding contest problem.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3 | instruction | 0 | 95,224 | 22 | 190,448 |
"Correct Solution:
```
from functools import lru_cache
@lru_cache(maxsize=1<<10)
def solve(p,q,a,n):
def _solve(num,dem,d,m,s):
if num==0:
return 1
if d == 0:
return 0
if num * a //m < dem:
return 0
return sum((_solve(num*i-dem, dem*i, d-1, m*i, i) for i in range(s, min(dem*n//num, a//m)+1)),0)
return _solve(p,q,n,1,1)
ans=[]
while True:
p,q,a,n=map(int,input().split())
if p==0:
break
ans.append(solve(p,q,a,n))
print(*ans, sep="\n")
``` | output | 1 | 95,224 | 22 | 190,449 |
Provide a correct Python 3 solution for this coding contest problem.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3 | instruction | 0 | 95,225 | 22 | 190,450 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f(p, q, a, n):
r = 0
fr = fractions.Fraction(p,q)
q = [(fr, 0, 1)]
if fr.numerator == 1 and fr.denominator <= a:
r += 1
n1 = n-1
for i in range(1, 70):
nq = []
ti = fractions.Fraction(1, i)
i2 = (i+1) ** 2
for t, c, m in q:
mik = m
tt = t
for k in range(1, n-c):
mik *= i
tt -= ti
if tt <= 0 or a < mik * i:
break
if tt.numerator == 1 and tt.denominator >= i and mik * tt.denominator <= a:
r += 1
if c+k < n1 and mik * (i ** max(math.ceil(tt / ti), 2)) < a:
nq.append((tt, c+k, mik))
if m * (i ** max(math.ceil(t / ti), 2)) < a:
nq.append((t,c,m))
if not nq:
break
q = nq
return r
while True:
p, q, a, n = LI()
if p == 0 and q == 0:
break
rr.append(f(p, q, a, n))
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 95,225 | 22 | 190,451 |
Provide a correct Python 3 solution for this coding contest problem.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3 | instruction | 0 | 95,226 | 22 | 190,452 |
"Correct Solution:
```
from fractions import gcd
def solve(p, q, a, n, l=1):
ans = 1 if p==1 and q<=a and q>=l else 0
denom = max(l, q//p)
p_denom = denom*p
while n*q >= p_denom and denom <= a: #n/denom >= p/q:
p_, q_ = p_denom-q, q*denom
if p_ <= 0:
denom += 1
p_denom += p
continue
gcd_ = gcd(p_, q_)
p_ //= gcd_
q_ //= gcd_
if n==2 and p_==1 and q_*denom<=a and q_>=denom:
ans += 1
else:
ans += solve(p_, q_, a//denom, n-1, denom)
denom += 1
p_denom += p
#print(p, q, a, n, l, ans)
return ans
while True:
p, q, a, n = map(int, input().split())
if p==q==a==n==0:
break
gcd_ = gcd(p, q)
p //= gcd_
q //= gcd_
if n==1:
print(1 if p==1 and q<=a else 0)
else:
print(solve(p, q, a, n))
``` | output | 1 | 95,226 | 22 | 190,453 |
Provide a correct Python 3 solution for this coding contest problem.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3 | instruction | 0 | 95,227 | 22 | 190,454 |
"Correct Solution:
```
from functools import lru_cache
@lru_cache(maxsize=1<<10)
def solve(p, q, a, n):
def _solve(num, dem, d, m, s):
if num == 0:
return 1
if d == 0:
return 0
if num * a // m < dem:
return 0
return sum((_solve(num*i-dem, dem*i, d-1, m*i, i)
for i in range(s, min(dem*n//num, a//m)+1)), 0)
return _solve(p, q, n, 1, 1)
if __name__ == "__main__":
ans = []
while True:
p, q, a, n = [int(i) for i in input().split()]
if p == 0:
break
ans.append(solve(p, q, a, n))
print(*ans, sep="\n")
``` | output | 1 | 95,227 | 22 | 190,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3
Submitted Solution:
```
from fractions import Fraction
def frac(n):
return(Fraction(1,n))
def part(s,a,n,m):
if s == 0:
return(1)
k = int(1/s)
if n == 1:
if frac(k) == s and k <= a:
return(1)
else:
return(0)
else:
ans = 0
for i in range(max(k,m),min(a,int(n/s))+1):
t = s-frac(i)
if t == 0:
ans += 1
elif t>0:
ans += part(s-frac(i), a//i, n-1, i)
return(ans)
while(True):
tmp = list(map(int, input().split()))
p,q,a,n = tmp[0],tmp[1],tmp[2],tmp[3]
if p == 0:
quit()
s = Fraction(p,q)
print(part(s,a,n,1))
``` | instruction | 0 | 95,228 | 22 | 190,456 |
No | output | 1 | 95,228 | 22 | 190,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3
Submitted Solution:
```
from fractions import Fraction
def frac(n):
return(Fraction(1,n))
def part(s,a,n,m):
if s == 0:
return(1)
k = int(1/s)
if n == 1:
if frac(k) == s and k <= a:
return(1)
else:
return(0)
else:
ans = 0
for i in range(max(k,m),min(a,int(n/s))+1):
ans += part(s-frac(i), a//i, n-1, i)
return(ans)
while(True):
tmp = list(map(int, input().split()))
p,q,a,n = tmp[0],tmp[1],tmp[2],tmp[3]
if p == 0:
quit()
s = Fraction(p,q)
print(part(s,a,n,1))
``` | instruction | 0 | 95,229 | 22 | 190,458 |
No | output | 1 | 95,229 | 22 | 190,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3
Submitted Solution:
```
from fractions import gcd
def solve(p, q, a, n, l=1):
#if n==1:
# #print(p, q, a, n, l, 1 if p==1 and q<=a and q>=l else 0)
# return 1 if p==1 and q<=a and q>=l else 0
ans = 1 if p==1 and q<=a and q>=l else 0
#denom = l if p==0 else max(l, int(q/p))
for denom in range(max(l, int(q/p)), a+1): #n/denom >= p/q:
if n*q < denom*p: break
p_, q_ = p*denom-q, q*denom
if p_ <= 0:
denom += 1
continue
gcd_ = gcd(p_, q_)
p_ //= gcd_
q_ //= gcd_
if n==2 and p_==1 and q_*denom<=a and q_>=denom:
ans += 1
else:
ans += solve(p_, q_, a//denom, n-1, denom)
denom += 1
#print(p, q, a, n, l, ans)
return ans
while True:
p, q, a, n = map(int, input().split())
if p==q==a==n==0:
break
gcd_ = gcd(p, q)
p //= gcd_
q //= gcd_
if n==1:
print(1 if p==1 and q<=a else 0)
else:
print(solve(p, q, a, n))
``` | instruction | 0 | 95,230 | 22 | 190,460 |
No | output | 1 | 95,230 | 22 | 190,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A fraction whose numerator is 1 and whose denominator is a positive integer is called a unit fraction. A representation of a positive rational number p/q as the sum of finitely many unit fractions is called a partition of p/q into unit fractions. For example, 1/2 + 1/6 is a partition of 2/3 into unit fractions. The difference in the order of addition is disregarded. For example, we do not distinguish 1/6 + 1/2 from 1/2 + 1/6.
For given four positive integers p, q, a, and n, count the number of partitions of p/q into unit fractions satisfying the following two conditions.
* The partition is the sum of at most n many unit fractions.
* The product of the denominators of the unit fractions in the partition is less than or equal to a.
For example, if (p,q,a,n) = (2,3,120,3), you should report 4 since
2/3 = 1/3 + 1/3 = 1/2 + 1/6 = 1/4 + 1/4 + 1/6 = 1/3 + 1/6 + 1/6
enumerates all of the valid partitions.
Input
The input is a sequence of at most 1000 data sets followed by a terminator.
A data set is a line containing four positive integers p, q, a, and n satisfying p,q <= 800, a <= 12000 and n <= 7. The integers are separated by a space.
The terminator is composed of just one line which contains four zeros separated by a space. It is not a part of the input data but a mark for the end of the input.
Output
The output should be composed of lines each of which contains a single integer. No other characters should appear in the output.
The output integer corresponding to a data set p, q, a, n should be the number of all partitions of p/q into at most n many unit fractions such that the product of the denominators of the unit fractions is less than or equal to a.
Example
Input
2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0
Output
4
7
6
2
42
1
0
9
3
Submitted Solution:
```
from fractions import gcd
def solve(p, q, a, n, l=1):
#if n==1:
# #print(p, q, a, n, l, 1 if p==1 and q<=a and q>=l else 0)
# return 1 if p==1 and q<=a and q>=l else 0
ans = 1 if p==1 and q<=a and q>=l else 0
denom = l if p==0 else max(l, int(q/p))
while n*q >= denom*p and denom <= a: #n/denom >= p/q:
p_, q_ = p*denom-q, q*denom
if p_ <= 0:
denom += 1
continue
gcd_ = gcd(p_, q_)
p_ //= gcd_
q_ //= gcd_
if n==2 and p_==1 and q_*denom<=a and q_>=denom:
ans += 1
else:
ans += solve(p_, q_, a//denom, n-1, denom)
denom += 1
#print(p, q, a, n, l, ans)
return ans
while True:
p, q, a, n = map(int, input().split())
if p==q==a==n==0:
break
gcd_ = gcd(p, q)
p //= gcd_
q //= gcd_
if n==1:
print(1 if p==1 and q<=a else 0)
else:
print(solve(p, q, a, n))
``` | instruction | 0 | 95,231 | 22 | 190,462 |
No | output | 1 | 95,231 | 22 | 190,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,109 | 22 | 192,218 |
Tags: brute force, constructive algorithms
Correct Solution:
```
x = int(input())
if x == 1: print('-1')
elif x == 2 or x == 3: print('2 2')
elif x == 4: print('4 2')
elif x%2 == 0:
print(str(x-2)+' 2')
else:
print(str(x-1)+' 2')
``` | output | 1 | 96,109 | 22 | 192,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,110 | 22 | 192,220 |
Tags: brute force, constructive algorithms
Correct Solution:
```
x = int(input())
a, b = None, None
for i in range(1, x + 1):
for j in range(1, x + 1):
if(i % j == 0 and i*j > x and i//j < x):
a = i
b = j
break
if(a != None or b != None):
break
if(a == None or b == None):
print(-1)
else:
print(a, b)
``` | output | 1 | 96,110 | 22 | 192,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,111 | 22 | 192,222 |
Tags: brute force, constructive algorithms
Correct Solution:
```
x = int(input())
if x == 1:print(-1)
elif x%2 == 0:
a,b = x,2
print(a,b)
else:
a,b = x-1,2
print(a,b)
``` | output | 1 | 96,111 | 22 | 192,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,112 | 22 | 192,224 |
Tags: brute force, constructive algorithms
Correct Solution:
```
x = int(input())
print(-1) if x == 1 else print(x - x % 2, 2)
``` | output | 1 | 96,112 | 22 | 192,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,113 | 22 | 192,226 |
Tags: brute force, constructive algorithms
Correct Solution:
```
t=1
while t>0:
t-=1
n=int(input())
if n==1:
print(-1)
else:
print(n,n)
``` | output | 1 | 96,113 | 22 | 192,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,114 | 22 | 192,228 |
Tags: brute force, constructive algorithms
Correct Solution:
```
x = int(input())
if x > 1:
num = str(x) +" "+ str(x)
print (num)
else:
print (-1)
``` | output | 1 | 96,114 | 22 | 192,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,115 | 22 | 192,230 |
Tags: brute force, constructive algorithms
Correct Solution:
```
x=int(input())
if(x!=1):
print(x," ",x)
else:
print(-1)
``` | output | 1 | 96,115 | 22 | 192,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1 | instruction | 0 | 96,116 | 22 | 192,232 |
Tags: brute force, constructive algorithms
Correct Solution:
```
x = int(input())
if x == 1:
print(-1)
else:
for b in range(1, x+1):
for a in range(b, x+1):
if a % b == 0 and a * b > x and a / b < x:
print(a, b)
break
else:
continue
break
``` | output | 1 | 96,116 | 22 | 192,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
x = int(input())
if x == 1 :
print(-1)
elif x == 2 or x == 3:
print(2,2)
else:
print(2*(x//2),x//2)
``` | instruction | 0 | 96,117 | 22 | 192,234 |
Yes | output | 1 | 96,117 | 22 | 192,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
x=int(input())
flag=False
h=0
g=0
for k in range(1,x+1):
for j in range(1,x+1):
if (k%j)==0 and (k//j)<x and (j*k)>x:
h=j
g=k
flag=True
break
break
if flag==True:
print('{} {}'.format(g,h))
else:
print("-1")
``` | instruction | 0 | 96,118 | 22 | 192,236 |
Yes | output | 1 | 96,118 | 22 | 192,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
a=int(input())
if a==1:
print(-1)
else:
if a%2==0:
print(a,2)
else:
print(a-1,2)
``` | instruction | 0 | 96,119 | 22 | 192,238 |
Yes | output | 1 | 96,119 | 22 | 192,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
x=int(input())
a,b=0,0
for i in range(x+1):
for j in range(x+1):
if(i*j>x and i%j==0 and i//j<x):
a=i
b=j
break
if(a!=0 and b!=0):
break
if(a!=0 and b!=0):
print(a,b)
else:
print("-1")
``` | instruction | 0 | 96,120 | 22 | 192,240 |
Yes | output | 1 | 96,120 | 22 | 192,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
x=int(input())
flag=False
h=0
g=0
for k in range(1,x+1):
for j in range(1,x+1):
if j//k<x and j*k>x:
h=j
g=k
flag=True
break
break
if flag==True:
print('{} {}'.format(h,g))
else:
print("-1")
``` | instruction | 0 | 96,121 | 22 | 192,242 |
No | output | 1 | 96,121 | 22 | 192,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
import math
def search(x):
if(x == 1):
print(-1)
else:
if (math.floor(x/2)+1 % 2== 0):
print(math.floor(x/2)+1, math.floor(x/2)+1)
else:
print(math.floor(x/2)+2, math.floor(x/2)+2)
search(int(input()))
``` | instruction | 0 | 96,122 | 22 | 192,244 |
No | output | 1 | 96,122 | 22 | 192,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
import sys
import os.path
from collections import *
import math
import bisect
if (os.path.exists('input.txt')):
sys.stdin = open("input.txt", "r")
sys.stdout = open("output.txt", "w")
else:
input = sys.stdin.readline
############## Code starts here ##########################
n = int(input())
if n == 1 or n == 2 or n == 3:
print(-1)
else:
x = (n // 2) * 2
y = (n // 2)
print(x,y)
############## Code ends here ############################
``` | instruction | 0 | 96,123 | 22 | 192,246 |
No | output | 1 | 96,123 | 22 | 192,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer x, find 2 integers a and b such that:
* 1 ≤ a,b ≤ x
* b divides a (a is divisible by b).
* a ⋅ b>x.
* a/b<x.
Input
The only line contains the integer x (1 ≤ x ≤ 100).
Output
You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3
Input
1
Output
-1
Submitted Solution:
```
x=int(input())
n=x//2
if n%2==0:
a=(n+2)
b=a//2
if a*b>x and (a%b==0) and a//b<x:
print(a,b)
else:
a=(n+1)
b=a//2
print(a,b)
``` | instruction | 0 | 96,124 | 22 | 192,248 |
No | output | 1 | 96,124 | 22 | 192,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,237 | 22 | 192,474 |
Tags: greedy, number theory
Correct Solution:
```
number = int(input())
for i in range(0,number):
n = int(input())
arr = [int(x) for x in input().split()]
uni = {}
cnt = []
arr.sort()
j = 0
while(j<n):
if(arr[j]%2!=0):
arr.pop(j)
n= n-1
elif(j!=0 and arr[j]==arr[j-1]):
arr.pop(j)
n= n-1
else: j+=1
n = len(arr)
for j in range(0,n):
k = 0
while(arr[j]%2==0):
arr[j]=arr[j]/2
k+=1
if(k>0):
uni[arr[j]] = max(k, uni.get(arr[j], 0))
#print(uni)
#print(cnt)
summ = 0
for j in uni.values():
summ+=j
print (summ)
``` | output | 1 | 96,237 | 22 | 192,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,238 | 22 | 192,476 |
Tags: greedy, number theory
Correct Solution:
```
# lET's tRy ThIS...
import math
import os
import sys
#-------------------BOLT------------------#
#-------Genius----Billionare----Playboy----Philanthropist----NOT ME:D----#
input = lambda: sys.stdin.readline().strip("\r\n")
def cin(): return sys.stdin.readline().strip("\r\n")
def fora(): return list(map(int, sys.stdin.readline().strip().split()))
def string(): return sys.stdin.readline().strip()
def cout(ans): sys.stdout.write(str(ans))
def endl(): sys.stdout.write(str("\n"))
def ende(): sys.stdout.write(str(" "))
#---------ND-I-AM-IRON-MAN------------------#
def main():
for _ in range(int(input())):
#LET's sPill the BEANS
n=int(cin())
a=fora()
b=[]
cnt=0
p=0
for i in a:
j=0
while((i%2)==0):
i/=2
j+=1
b.append(j)
a[p]=math.trunc(i)
p+=1
dict={}
dict=dict.fromkeys(a,0)
for i in range(n):
dict[a[i]]=max(dict[a[i]],b[i])
for i in dict.values():
cnt+=i
# cout(dict)
# endl()
cout(cnt)
endl()
if __name__ == "__main__":
main()
``` | output | 1 | 96,238 | 22 | 192,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,239 | 22 | 192,478 |
Tags: greedy, number theory
Correct Solution:
```
def odd(n,m):
if len(m)!=n:
return "incorrect input"
t=set()
f={}
for i in range(len(m)):
if int(m[i])%2==0:
t.add(int(m[i]))
# Предлагается поддерживать t -- множество всевозможных посещаемых чётных чисел.
for i in t:
while i%2!=1:
if i in f:
i = i // 2
else:
f[i]=i//2
i=i//2
return len(f)
"""
#2 attempt
for i in range(len(m)):
if int(m[i])%2==0:
if int(m[i]) in t:
t[int(m[i])]+=1
else:
t[int(m[i])] = 1
count=0
while len(t)!=0:
print(t)
f=max(t)
count+=1
if (f//2)%2==0:
if f//2 in t:
t[f//2]+=t[f]
else:
t[f//2]=t[f]
del t[f]
#1 attempt
while t:
#g=max(m,key=lambda f:[f%2==0,f])
#m.sort(key=lambda f: [f % 2 == 1, -f])
print(g)
f=m[0]
for i in range(len(m)):
if m[i]==f:
m[i]//=2
else:
break
#t = find(m)
count+=1
"""
for i in range(int(input().strip())):
n=int(input().strip())
m=input().strip().split()
print(odd(n,m))
``` | output | 1 | 96,239 | 22 | 192,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,240 | 22 | 192,480 |
Tags: greedy, number theory
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
arr=list(map(int,input().split()))
arr1=[]
for a in arr:
if a%2==0:
arr1.append(a)
arr1=list(set(arr1))
arr1.sort()
i=0
ca={}
while i<len(arr1):
b=arr1[i]
count=0
while b%2==0:
count+=1
b=b//2
ca[b]=count
i+=1
print(sum(ca.values()))
``` | output | 1 | 96,240 | 22 | 192,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,241 | 22 | 192,482 |
Tags: greedy, number theory
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
s=set()
ans=0
for x in a:
while(x%2==0):
s.add(x)
x/=2
print(len(s))
``` | output | 1 | 96,241 | 22 | 192,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,242 | 22 | 192,484 |
Tags: greedy, number theory
Correct Solution:
```
def fil(i):
if int(i)%2==0:
return True
else:
return False
def mapf(i):
s=bin(int(i))[2:]
x=s.rindex("1")
y=len(s)-1-x
return (y,int(s[0:x+1],2))
def sortf(i):
return i[1]
def func(l,n):
prev=l[0]
sum=0
for i in range(n):
if l[i][1]!=prev[1]:
sum=sum+prev[0]
prev=l[i]
else:
if l[i][0]>prev[0]:
prev=l[i]
if i==n-1:
sum=sum+prev[0]
return sum
t=int(input())
for _ in range(t):
n=int(input())
l=list(set(map(mapf,filter(fil,input().split()))))
l.sort(key=sortf,reverse=True)
n=len(l)
if l==[]:
print(0)
else:
print(func(l,n))
``` | output | 1 | 96,242 | 22 | 192,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,243 | 22 | 192,486 |
Tags: greedy, number theory
Correct Solution:
```
from heapq import *
t=int(input())
for _ in range(t):
n=int(input())
it=list(map(int,input().split()))
it=[-i for i in it if i%2==0]
heapify(it)
tot=0
tot=0
ss=set()
while it:
no=it.pop()
if no in ss:
continue
ss.add(no)
if no%4==0:
heappush(it,no//2)
tot+=1
print(tot)
``` | output | 1 | 96,243 | 22 | 192,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd. | instruction | 0 | 96,244 | 22 | 192,488 |
Tags: greedy, number theory
Correct Solution:
```
if __name__ == "__main__":
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
aset = set(a)
alist = list(aset)
# print(alist)
distances = {}
alist.sort()
ans = 0
for i in range(len(alist)):
num = alist[i]
distance = 0
num2 = num
found = False
while num2 % 2 == 0:
distance += 1
num2 = num2 // 2
if num2 in distances:
found = True
distance += distances[num2]
break
distances[num] = distance
if found:
distances.pop(num2)
ans = sum(distances.values())
print(ans)
``` | output | 1 | 96,244 | 22 | 192,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
MOD = 1000000007
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
for _ in range(ii()):
n=ii()
s=set()
c=0
l=sorted(il(),reverse=True)
for i in l:
while i%2==0:
if not i in s:
c+=1
s.add(i)
i//=2
print(c)
``` | instruction | 0 | 96,245 | 22 | 192,490 |
Yes | output | 1 | 96,245 | 22 | 192,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
for _ in range(int(input())):
input()
a = sorted(map(int, input().split()), reverse=True)
seen = set()
ret = 0
for val in a:
if val & 1 or val in seen:
continue
while val & 1 == 0:
seen.add(val)
val >>= 1
ret += 1
print(ret)
``` | instruction | 0 | 96,246 | 22 | 192,492 |
Yes | output | 1 | 96,246 | 22 | 192,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The sum of n for all test cases in the input doesn't exceed 2⋅10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] — all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
t = int(input(''))
for v in range(t):
n = input('')
a = list(map(int,input('').split(' ')))
d = {}
for nu in a:
if(nu % 2 == 0):
d[nu] = True
dd = {}
ans = 0
for k in d:
t = k
while(t % 2 == 0):
if(t not in dd):
ans += 1
dd[t] = True
else:
break
t = t//2
print(ans)
``` | instruction | 0 | 96,247 | 22 | 192,494 |
Yes | output | 1 | 96,247 | 22 | 192,495 |
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