message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19). | instruction | 0 | 80,062 | 22 | 160,124 |
Tags: math
Correct Solution:
```
import sys
def read_in():
stri = sys.stdin.readline()
return stri.strip()
def from_str_to_list(stri):
res_List = []
for i in stri.split():
res_List.append(int(i))
return res_List
def main():
t = int(read_in())
k = 0
for i in range(t):
data = from_str_to_list(read_in())
A = data[0]
B = data[1]
while 10 ** k - 1 <= B:
k += 1
k -= 1
print(A * k)
k = 0
return
main()
``` | output | 1 | 80,062 | 22 | 160,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Jan 25 19:04:48 2020
@author: SnowA
"""
from sys import stdin, stdout
def m(a,b,aLetra,bLetra):
mayor=max(a,b)
if(mayor==a):
orden=(len(aLetra))
else:
orden=(len(bLetra))
contador=a*(orden-1)
if(int("9"*orden)<=b):
contador+=a
return contador
def main():
resp=""
numeros=int(stdin.readline())
for i in range(numeros):
linea = stdin.readline().split()
resp+=str(m(int(linea[0]),int(linea[1]),linea[0],linea[1]))+"\n"
stdout.write(resp)
main()
``` | instruction | 0 | 80,070 | 22 | 160,140 |
No | output | 1 | 80,070 | 22 | 160,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S. Over all pairs of subsets A and B, such that:
* B ⊂ A;
* |B| = |A| - 1;
* greatest common divisor of all elements in A is equal to one;
find the sum of ∑_{x ∈ A}{x} ⋅ ∑_{x ∈ B}{x}, modulo 998 244 353.
Input
The first line contains one integer m (1 ≤ m ≤ 10^5): the number of different values in the multiset S.
Each of the next m lines contains two integers a_i, freq_i (1 ≤ a_i ≤ 10^5, 1 ≤ freq_i ≤ 10^9). Element a_i appears in the multiset S freq_i times. All a_i are different.
Output
Print the required sum, modulo 998 244 353.
Examples
Input
2
1 1
2 1
Output
9
Input
4
1 1
2 1
3 1
6 1
Output
1207
Input
1
1 5
Output
560
Note
A multiset is a set where elements are allowed to coincide. |X| is the cardinality of a set X, the number of elements in it.
A ⊂ B: Set A is a subset of a set B.
In the first example B=\{1\}, A=\{1,2\} and B=\{2\}, A=\{1, 2\} have a product equal to 1⋅3 + 2⋅3=9. Other pairs of A and B don't satisfy the given constraints. | instruction | 0 | 80,143 | 22 | 160,286 |
Tags: combinatorics, math, number theory
Correct Solution:
```
import sys
input=sys.stdin.readline
max_n=10**5+1
mod=998244353
spf=[i for i in range(max_n)]
prime=[True for i in range(max_n)]
mobius=[0 for i in range(max_n)]
prime[0]=prime[1]=False
mobius[1]=1
primes=[]
for i in range(2,max_n):
if(prime[i]):
spf[i]=i
mobius[i]=-1
primes.append(i)
for j in primes:
prod=j*i
if(j>spf[i] or prod>=max_n):
break
spf[prod]=j
prime[prod]=False
if(spf[i]==j):
mobius[prod]=0
else:
mobius[prod]=-mobius[i]
m=int(input())
freq=[0 for i in range(max_n)]
ans=0
for i in range(m):
a,f=map(int,input().split())
freq[a]=f
for i in range(1,max_n):
if(mobius[i]!=0):
cnt=0
val=0
val_square=0
for j in range(i,max_n,i):
cnt+=freq[j]
val=(val+(j*freq[j]))%mod
val_square=(val_square+(j*j*freq[j]))%mod
if(cnt<2):
continue
p=pow(2,cnt-2,mod)
n1=(cnt-1)*p%mod
n2=p
if(cnt>=3):
n2=(n2+(cnt-2)*pow(2,cnt-3,mod))
diff_val=(val*val-val_square)%mod
ans=(ans+mobius[i]*(n1*val_square+n2*diff_val))%mod
print(ans)
``` | output | 1 | 80,143 | 22 | 160,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S. Over all pairs of subsets A and B, such that:
* B ⊂ A;
* |B| = |A| - 1;
* greatest common divisor of all elements in A is equal to one;
find the sum of ∑_{x ∈ A}{x} ⋅ ∑_{x ∈ B}{x}, modulo 998 244 353.
Input
The first line contains one integer m (1 ≤ m ≤ 10^5): the number of different values in the multiset S.
Each of the next m lines contains two integers a_i, freq_i (1 ≤ a_i ≤ 10^5, 1 ≤ freq_i ≤ 10^9). Element a_i appears in the multiset S freq_i times. All a_i are different.
Output
Print the required sum, modulo 998 244 353.
Examples
Input
2
1 1
2 1
Output
9
Input
4
1 1
2 1
3 1
6 1
Output
1207
Input
1
1 5
Output
560
Note
A multiset is a set where elements are allowed to coincide. |X| is the cardinality of a set X, the number of elements in it.
A ⊂ B: Set A is a subset of a set B.
In the first example B=\{1\}, A=\{1,2\} and B=\{2\}, A=\{1, 2\} have a product equal to 1⋅3 + 2⋅3=9. Other pairs of A and B don't satisfy the given constraints. | instruction | 0 | 80,144 | 22 | 160,288 |
Tags: combinatorics, math, number theory
Correct Solution:
```
import sys
input=sys.stdin.readline
max_n=10**5+1
mod=998244353
spf=[i for i in range(max_n)]
prime=[True for i in range(max_n)]
mobius=[0 for i in range(max_n)]
prime[0]=prime[1]=False
mobius[1]=1
primes=[]
for i in range(2,max_n):
if(prime[i]):
spf[i]=i
mobius[i]=-1
primes.append(i)
for j in primes:
prod=j*i
if(j>spf[i] or prod>=max_n):
break
spf[prod]=j
prime[prod]=False
if(spf[i]==j):
mobius[prod]=0
else:
mobius[prod]=-mobius[i]
m=int(input())
freq=[0 for i in range(max_n)]
ans=0
for i in range(m):
a,f=map(int,input().split())
freq[a]=f
for i in range(1,max_n):
if(mobius[i]!=0):
cnt=0
val=0
square_val=0
for j in range(i,max_n,i):
cnt+=freq[j]
val=(val+(j*freq[j]))%mod
square_val=(square_val+(j*j*freq[j]))%mod
tmp=0
if(cnt>1):
f1=1
g1=1
if(cnt<3):
f1=mod-2
g1=-1
f2=1
g2=1
if(cnt<2):
f2=mod-2
g2=-1
p1=cnt*pow(2,g1*(cnt-3)*f1,mod)%mod
p2=(cnt-1)*pow(2,g2*(cnt-2)*f2,mod)%mod
tmp=((val*val*p1)%mod + square_val*(p2-p1)%mod)%mod
ans=(ans+mobius[i]*tmp)%mod
print(ans)
``` | output | 1 | 80,144 | 22 | 160,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S. Over all pairs of subsets A and B, such that:
* B ⊂ A;
* |B| = |A| - 1;
* greatest common divisor of all elements in A is equal to one;
find the sum of ∑_{x ∈ A}{x} ⋅ ∑_{x ∈ B}{x}, modulo 998 244 353.
Input
The first line contains one integer m (1 ≤ m ≤ 10^5): the number of different values in the multiset S.
Each of the next m lines contains two integers a_i, freq_i (1 ≤ a_i ≤ 10^5, 1 ≤ freq_i ≤ 10^9). Element a_i appears in the multiset S freq_i times. All a_i are different.
Output
Print the required sum, modulo 998 244 353.
Examples
Input
2
1 1
2 1
Output
9
Input
4
1 1
2 1
3 1
6 1
Output
1207
Input
1
1 5
Output
560
Note
A multiset is a set where elements are allowed to coincide. |X| is the cardinality of a set X, the number of elements in it.
A ⊂ B: Set A is a subset of a set B.
In the first example B=\{1\}, A=\{1,2\} and B=\{2\}, A=\{1, 2\} have a product equal to 1⋅3 + 2⋅3=9. Other pairs of A and B don't satisfy the given constraints.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
"""bassem.benhamed@enet.usf.tn
bassem.benhamed@gmail.com
Objet: GInfo1.IPSAS
EXo: Insertion des coeficiant de A et B
Affichage:
Resolution"""
from math import gcd as pgcd
#Etree des paramètres
n =int(input())
S=[]
for i in range(n):
ligne = input()
S.append((int((ligne.split())[0]),int((ligne.split())[1])))
#Ensemble de partition de S
def partiesliste(seq):
p = []
i, imax = 0, 2**len(seq)-1
while i <= imax:
s = []
j, jmax = 0, len(seq)-1
while j <= jmax:
if (i>>j)&1 == 1:
s.append(seq[j])
j += 1
p.append(s)
i += 1
return p
P = partiesliste(S)
#Ensemble de posibilité de A
P_A =[]
for X in P:
if len(X) >=2:
i=0
repeter = True
while i<(len(X)-1) and repeter == True:
j = i+1
while j < len(X) and (pgcd(X[i][0], X[j][0]) == 1) :
j += 1
#j prend la valeur len(X) si le test se deroule bien et inférieure
#si non. Dans le second cas , on quite la boucle.
if(j < len(X)):
repeter = False
#si nom, le test se poursuit avec l'indice suivant
i+=1
#si le test se déroule bien, i prend la valeur de len(x) - 1
if i == (len(X) - 1 )and repeter == True:
#on ajoute X dans P_A
P_A.append(X)
#Détermination de B pour chaque valeur de A et calcul de la somme
sommes = 0
for X in P_A:
P_X = partiesliste(X)
x=[elt[0] for elt in X]
for Y in P_X:
if len(Y) == len(X) - 1:
y=[elt[0] for elt in Y]
sommes += sum(x) * sum(y)
print(sommes)
``` | instruction | 0 | 80,146 | 22 | 160,292 |
No | output | 1 | 80,146 | 22 | 160,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,320 | 22 | 160,640 |
Tags: combinatorics, math, number theory
Correct Solution:
```
MOD = 10**9+7
def SieveOfEratosthenes(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
l = []
for p in range(2, n):
if prime[p]:
l.append(p)
return l
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,p - 2, p)) % p
primes = SieveOfEratosthenes(31623)
n = int(input())
l = [int(zax) for zax in input().split()]
map = {}
for x in l:
for p in primes:
if p>x:
break
if x%p==0:
check = False
while x%p==0:
if p in map:
map[p]+=1
else:
map[p]=1
x//=p
if x!=1:
if x in map:
map[x]+=1
else:
map[x]=1
count = 1
for x in map:
count *= ncr(n-1+map[x],n-1,MOD)
count %= MOD
count %= MOD
print(count)
``` | output | 1 | 80,320 | 22 | 160,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,321 | 22 | 160,642 |
Tags: combinatorics, math, number theory
Correct Solution:
```
f=[1]*15001
fi=[1]*15001
a=1
b=1
m=10**9+7
from collections import defaultdict
for i in range(1,15001):
a*=i
b*=pow(i,m-2,m)
a%=m
b%=m
f[i]=a
fi[i]=b
d=defaultdict(int)
def factorize(n):
count = 0;
while ((n % 2 > 0) == False):
# equivalent to n = n / 2;
n >>= 1;
count += 1;
if (count > 0):
d[2]+=count
for i in range(3, int(n**0.5) + 1):
count = 0;
while (n % i == 0):
count += 1;
n = int(n / i);
if (count > 0):
d[i]+=count
i += 2;
# if n at the end is a prime number.
if (n > 2):
d[n]+=1
ans=1
n=int(input())
l=list(map(int,input().split()))
for i in l:
factorize(i)
for i in d:
ans*=(f[d[i]+n-1]*fi[n-1]*fi[d[i]])%m
ans%=m
print(ans)
``` | output | 1 | 80,321 | 22 | 160,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,322 | 22 | 160,644 |
Tags: combinatorics, math, number theory
Correct Solution:
```
# Design_by_JOKER
import math
import cmath
Mod = 1000000007
MAX = 33000
n = int(input())
A = list(map(int, input().split()))
B = [0] * MAX
bePrime = [0] * MAX;
primNum = []
C = []
fac = [1]
for j in range(1, MAX):
fac.append(fac[-1] * j % Mod)
def calc(M, N):
return fac[M] * pow(fac[N] * fac[M - N] % Mod, Mod - 2, Mod) % Mod
for j in range(2, MAX):
if bePrime[j] == 0:
primNum.append(j)
i = j
while i < MAX:
bePrime[i] = 1
i = i + j
for x in A:
tmp = x
for j in primNum:
while tmp % j == 0:
tmp /= j
B[j] += 1
if tmp > 1:
C.append(tmp)
ans = 1
for j in range(2, MAX):
if B[j] > 0:
ans = ans * calc(n + B[j] - 1, n - 1) % Mod
l = len(C)
for j in range(0, l):
num = 0;
for k in range(0, l):
if C[k] == C[j]:
num = num + 1
if k > j:
num = 0
break
if num > 0:
ans = ans * calc(n + num - 1, n - 1) % Mod
print(str(ans % Mod))
``` | output | 1 | 80,322 | 22 | 160,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,323 | 22 | 160,646 |
Tags: combinatorics, math, number theory
Correct Solution:
```
# Made By Mostafa_Khaled
bot = True
Mod = 1000000007
MAX = 33000
n = int( input() )
A = list( map( int, input().split() ) )
B = [0] * MAX
bePrime = [0] * MAX;
primNum = []
C = []
fac=[1]
for j in range(1, MAX):
fac.append( fac[-1] * j % Mod )
def calc( M, N ):
return fac[M] * pow( fac[N] * fac[M-N] % Mod, Mod-2,Mod ) % Mod
for j in range(2, MAX):
if bePrime[j] == 0:
primNum.append( j )
i = j
while i < MAX:
bePrime[i] = 1
i = i + j
for x in A:
tmp = x
for j in primNum:
while tmp % j == 0:
tmp /= j
B[j] += 1
if tmp > 1:
C.append( tmp )
ans = 1
for j in range(2,MAX):
if B[j] > 0:
ans = ans * calc( n + B[j] -1 , n - 1 ) % Mod
l = len( C )
for j in range(0, l ):
num= 0;
for k in range(0, l ):
if C[k] == C[j]:
num = num + 1
if k > j:
num = 0
break
if num > 0:
ans = ans * calc( n + num -1, n - 1 ) % Mod
print( str( ans % Mod ) )
# Made By Mostafa_Khaled
``` | output | 1 | 80,323 | 22 | 160,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,324 | 22 | 160,648 |
Tags: combinatorics, math, number theory
Correct Solution:
```
ans={}
#######find prime factors
# Python3 program to print prime factors
# and their powers.
import math
# Function to calculate all the prime
# factors and count of every prime factor
def factorize(n):
count = 0;
# count the number of
# times 2 divides
while ((n % 2 > 0) == False):
# equivalent to n = n / 2;
n >>= 1;
count += 1;
# if 2 divides it
if (count > 0):
if 2 not in ans:
ans[2]=count
else:ans[2]+=count
# check for all the possible
# numbers that can divide it
for i in range(3, int(math.sqrt(n)) + 1):
count = 0;
while (n % i == 0):
count += 1;
n = int(n / i);
if (count > 0):
if i in ans:
ans[i]+=count;
else:ans[i]=count
i += 2;
# if n at the end is a prime number.
if (n > 2):
if n not in ans:
ans[n]=1
else:ans[n]+=1
mod=10**9+7
def fact(n):
f=[1 for x in range(n+1)]
for i in range(1,n+1):
f[i]=(f[i-1]*i)%mod
return f
f=fact(200000)
def ncr(n,r):
result=(f[n]*pow(f[r],mod-2,mod)*pow(f[n-r],mod-2,mod))%mod
return result
n=int(input())
arr=[int(x) for x in input().split()]
for item in arr:
factorize(item)
result=1
for i,val in ans.items():
if val>0:
result=result*ncr(val+n-1,n-1)
result=result%mod
print(result)
``` | output | 1 | 80,324 | 22 | 160,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,325 | 22 | 160,650 |
Tags: combinatorics, math, number theory
Correct Solution:
```
from collections import defaultdict
m = 1000000007
f = [0] * 15001
f[0] = 1
for i in range(1, 15001): f[i] = (f[i - 1] * i) % m
def c(n, k): return (f[n] * pow((f[k] * f[n - k]) % m, m - 2, m)) % m
def prime(n):
m = int(n ** 0.5) + 1
t = [1] * (n + 1)
for i in range(3, m):
if t[i]: t[i * i :: 2 * i] = [0] * ((n - i * i) // (2 * i) + 1)
return [2] + [i for i in range(3, n + 1, 2) if t[i]]
p = prime(31650)
s = defaultdict(int)
def g(n):
for j in p:
while n % j == 0:
n //= j
s[j] += 1
if j * j > n:
s[n] += 1
break
n = int(input()) - 1
a = list(map(int, input().split()))
for i in a: g(i)
if 1 in s: s.pop(1)
d = 1
for k in s.values(): d = (d * c(k + n, n)) % m
print(d)
``` | output | 1 | 80,325 | 22 | 160,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,326 | 22 | 160,652 |
Tags: combinatorics, math, number theory
Correct Solution:
```
Mod = 1000000007
MAX = 33000
n = int( input() )
A = list( map( int, input().split() ) )
B = [0] * MAX
bePrime = [0] * MAX;
primNum = []
C = []
fac=[1]
for j in range(1, MAX):
fac.append( fac[-1] * j % Mod )
def calc( M, N ):
return fac[M] * pow( fac[N] * fac[M-N] % Mod, Mod-2,Mod ) % Mod
for j in range(2, MAX):
if bePrime[j] == 0:
primNum.append( j )
i = j
while i < MAX:
bePrime[i] = 1
i = i + j
for x in A:
tmp = x
for j in primNum:
while tmp % j == 0:
tmp /= j
B[j] += 1
if tmp > 1:
C.append( tmp )
ans = 1
for j in range(2,MAX):
if B[j] > 0:
ans = ans * calc( n + B[j] -1 , n - 1 ) % Mod
l = len( C )
for j in range(0, l ):
num= 0;
for k in range(0, l ):
if C[k] == C[j]:
num = num + 1
if k > j:
num = 0
break
if num > 0:
ans = ans * calc( n + num -1, n - 1 ) % Mod
print( str( ans % Mod ) )
``` | output | 1 | 80,326 | 22 | 160,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci. | instruction | 0 | 80,327 | 22 | 160,654 |
Tags: combinatorics, math, number theory
Correct Solution:
```
import math
import sys
input=sys.stdin.readline
p=(10**9)+7
pri=p
fac=[1 for i in range((10**6)+1)]
for i in range(2,len(fac)):
fac[i]=(fac[i-1]*(i%pri))%pri
def modi(x):
return (pow(x,p-2,p))%p;
def ncr(n,r):
x=(fac[n]*((modi(fac[r])%p)*(modi(fac[n-r])%p))%p)%p
return x;
def prime(x):
ans=[]
while(x%2==0):
x=x//2
ans.append(2)
for i in range(3,int(math.sqrt(x))+1,2):
while(x%i==0):
ans.append(i)
x=x//i
if(x>2):
ans.append(x)
return ans;
n=int(input())
z=list(map(int,input().split()))
ans=[]
for i in range(len(z)):
m=prime(z[i])
ans.extend(m)
ans.sort()
if(ans.count(1)==len(ans)):
print(1)
exit()
cn=[]
count=1
for i in range(1,len(ans)):
if(ans[i]==ans[i-1]):
count+=1
else:
cn.append(count)
count=1
cn.append(count)
al=1
for i in range(len(cn)):
al=al*ncr(n+cn[i]-1,n-1)
al%=pri
print(al)
``` | output | 1 | 80,327 | 22 | 160,655 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
from math import factorial as f
def primes(n):
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
p = primes(31627)
s = [0]*(31623)
s1={}
def factorize(n):
for i in p:
if n<=1:
return 56
while n%i==0:
s[p.index(i)]+=1
n//=i
if n>1:
if n in s1:
s1[n]+=1
else:
s1[n]=1
n = int(input())
for i in map(int,input().split()):
factorize(i)
s = list(filter(lambda a: a != 0, s))
for i in s1.values():
s.append(i)
ans = 1
for i in s:
ans*=f(i+n-1)//(f(n-1)*f(i))
print(int(ans)%1000000007)
``` | instruction | 0 | 80,328 | 22 | 160,656 |
Yes | output | 1 | 80,328 | 22 | 160,657 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from collections import Counter
def factor(x,cou):
while not x%2:
x /= 2
cou[2] += 1
for i in range(3,int(x**0.5)+1,2):
while not x%i:
x //= i
cou[i] += 1
if x != 1:
cou[x] += 1
def main():
mod = 10**9+7
fac = [1]
for ii in range(1,10**5+1):
fac.append((fac[-1]*ii)%mod)
fac_in = [pow(fac[-1],mod-2,mod)]
for ii in range(10**5,0,-1):
fac_in.append((fac_in[-1]*ii)%mod)
fac_in.reverse()
n = int(input())
a = list(map(int,input().split()))
cou = Counter()
for i in a:
factor(i,cou)
ans = 1
for i in cou:
a,b = cou[i]+n-1,n-1
ans = (ans*fac[a]*fac_in[b]*fac_in[a-b])%mod
print(ans)
# Fast IO Region
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")
if __name__ == "__main__":
main()
``` | instruction | 0 | 80,329 | 22 | 160,658 |
Yes | output | 1 | 80,329 | 22 | 160,659 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
Mod = 1000000007
MAX = 33000
n = int( input() )
A = list( map( int, input().split() ) )
B = [0] * MAX
bePrime = [0] * MAX;
primNum = []
C = []
fac=[1]
for j in range(1, MAX):
fac.append( fac[-1] * j % Mod )
def calc( M, N ):
return fac[M] * pow( fac[N] * fac[M-N] % Mod, Mod-2,Mod ) % Mod
for j in range(2, MAX):
if bePrime[j] == 0:
primNum.append( j )
i = j
while i < MAX:
bePrime[i] = 1
i = i + j
for x in A:
tmp = x
for j in primNum:
while tmp % j == 0:
tmp /= j
B[j] += 1
if tmp > 1:
C.append( tmp )
ans = 1
for j in range(2,MAX):
if B[j] > 0:
ans = ans * calc( n + B[j] -1 , n - 1 ) % Mod
l = len( C )
for j in range(0, l ):
num= 0;
for k in range(0, l ):
if C[k] == C[j]:
num = num + 1
if k > j:
num = 0
break
if num > 0:
ans = ans * calc( n + num -1, n - 1 ) % Mod
print( str( ans % Mod ) )
# Made By Mostafa_Khaled
``` | instruction | 0 | 80,330 | 22 | 160,660 |
Yes | output | 1 | 80,330 | 22 | 160,661 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
from math import sqrt, factorial as f
from collections import Counter
from operator import mul
from functools import reduce
def comb(n, m):
o = n - m
if m and o:
if m < o:
m, o = o, m
return reduce(mul, range(m + 1, n), n) // f(o)
return 1
def main():
n = int(input())
aa = list(map(int, input().split()))
if n == 1:
print(1)
return
lim = int(sqrt(max(aa)) // 6) + 12
sieve = [False, True, True] * lim
lim = lim * 3 - 1
for i, s in enumerate(sieve):
if s:
p, pp = i * 2 + 3, (i + 3) * i * 2 + 3
le = (lim - pp) // p + 1
if le > 0:
sieve[pp::p] = [False] * le
else:
break
sieve[0] = sieve[3] = True
primes = [i * 2 + 3 for i, f in enumerate(sieve) if f]
for i, p in enumerate((2, 3, 5, 7)):
primes[i] = p
del sieve
c = Counter()
for x in aa:
for p in primes:
cnt = 0
while not x % p:
x //= p
cnt += 1
if cnt:
c[p] += cnt
if x == 1:
break
if x > 1:
c[x] += 1
x, inf = 1, 1000000007
for p, cnt in c.items():
x = x * comb(cnt + n - 1, n - 1) % inf
print(x)
if __name__ == '__main__':
main()
``` | instruction | 0 | 80,331 | 22 | 160,662 |
Yes | output | 1 | 80,331 | 22 | 160,663 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
MOD = 10**9+7
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,p - 2, p)) % p
def primes(n):
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
p = primes(3170)
s = [0]*10000
def factorize(n):
for i in p:
if n<=1:
return 56
while n%i==0:
s[p.index(i)]+=1
n//=i
n1 = int(input())
for i in map(int,input().split()):
factorize(i)
s = list(filter(lambda a: a != 0, s))
ans = 1
for i in s:
ans*=ncr(n1-1+i,n1-1,MOD)
ans%=MOD
print(int(ans)%1000000007)
``` | instruction | 0 | 80,332 | 22 | 160,664 |
No | output | 1 | 80,332 | 22 | 160,665 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
MOD = 10**9+7
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,p - 2, p)) % p
def primes(n):
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
p = primes(33000)
s = [0]*len(p)
def factorize(n):
for i in p:
if n<=1:
return 56
while n%i==0:
s[p.index(i)]+=1
n//=i
n1 = int(input())
for i in map(int,input().split()):
factorize(i)
s = list(filter(lambda a: a != 0, s))
ans = 1
for i in s:
ans*=ncr(n1-1+i,n1-1,MOD)
ans%=MOD
print(int(ans)%MOD)
``` | instruction | 0 | 80,333 | 22 | 160,666 |
No | output | 1 | 80,333 | 22 | 160,667 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
n, m, pr, mn = int(input()), list(map(int, input().split())), 1, []
for i in m:
pr *= i
for i in range(1, int(pr ** 0.5) + 1):
if pr % i == 0:
mn.append(i)
mn.append(pr // i)
if n == 1:
print(1)
else:
print(len(mn) // 2 * n)
``` | instruction | 0 | 80,334 | 22 | 160,668 |
No | output | 1 | 80,334 | 22 | 160,669 |
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 m as a product of integers a1, a2, ... an <image>. Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.
Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo 1000000007 (109 + 7).
Input
The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007 (109 + 7).
Examples
Input
1
15
Output
1
Input
3
1 1 2
Output
3
Input
2
5 7
Output
4
Note
In the second sample, the get a decomposition of number 2, you need any one number out of three to equal 2, and the rest to equal 1.
In the third sample, the possible ways of decomposing into ordered multipliers are [7,5], [5,7], [1,35], [35,1].
A decomposition of positive integer m into n ordered multipliers is a cortege of positive integers b = {b1, b2, ... bn} such that <image>. Two decompositions b and c are considered different, if there exists index i such that bi ≠ ci.
Submitted Solution:
```
def SieveOfEratosthenes(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
l = []
for p in range(2, n):
if prime[p]:
l.append(p)
return l
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
MOD = 10**9+7
primes = SieveOfEratosthenes(10**5)
n = int(input())
l = [int(zax) for zax in input().split()]
map = {}
for x in l:
check = True
for p in primes:
if x%p==0:
check = False
while x%p==0:
if p in map:
map[p]+=1
else:
map[p]=1
x//=p
if check:
if x!=1:
if x in map:
map[x]+=1
else:
map[x]=1
count = 1
for x in map:
count*=ncr(n-1+map[x],n-1,MOD)
count%=MOD
print(count)
``` | instruction | 0 | 80,335 | 22 | 160,670 |
No | output | 1 | 80,335 | 22 | 160,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,831 | 22 | 161,662 |
Tags: math
Correct Solution:
```
def countSetBits(n):
count = 0
while (n):
n &= (n-1)
count+=1
return count
for _ in range(int(input())):
n=int(input())
k=(countSetBits(n))
print(pow(2,k))
``` | output | 1 | 80,831 | 22 | 161,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,832 | 22 | 161,664 |
Tags: math
Correct Solution:
```
from sys import stdin
def readLine():
return stdin.readline()
def readInt():
return int(readLine())
f = lambda a, x: a - (a ^ x) - x
def main():
n = readInt()
for i in range(n):
a = readInt()
bn = bin(a)[2:]
ans = 1
for i in bn:
if i == '1':
ans *= 2
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 80,832 | 22 | 161,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,833 | 22 | 161,666 |
Tags: math
Correct Solution:
```
from functools import reduce
for i in range(int(input())):
#a-x = a xor x
a = bin(int(input()))
r = 1
for i in a[2:]:
r *= 2 if i=='1' else 1
print(r)
``` | output | 1 | 80,833 | 22 | 161,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,834 | 22 | 161,668 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a = int(input())
print(2**bin(a).count('1'))
``` | output | 1 | 80,834 | 22 | 161,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,835 | 22 | 161,670 |
Tags: math
Correct Solution:
```
for i in range(int(input())):
print(2**(bin(int(input())).count('1')))
``` | output | 1 | 80,835 | 22 | 161,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,836 | 22 | 161,672 |
Tags: math
Correct Solution:
```
# Author: πα
n = int(input())
for i in range(n):
b = bin(int(input()))
count = 0
for c in b:
if c == '1':
count += 1
print(pow(2, count))
``` | output | 1 | 80,836 | 22 | 161,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,837 | 22 | 161,674 |
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
ans = 0
for i in range(32):
if n & (1 << i):
ans += 1
ans = 1 << ans
print(ans)
``` | output | 1 | 80,837 | 22 | 161,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2. | instruction | 0 | 80,838 | 22 | 161,676 |
Tags: math
Correct Solution:
```
#codeforces_1064B_live
gi = lambda : list(map(int,input().split()))
print("\n".join(list(map(str,[2**bin(gi()[0])[2:].count("1") for e in range(gi()[0])]))))
``` | output | 1 | 80,838 | 22 | 161,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,981 | 22 | 161,962 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
a.sort()
i,b=0,0
while i < (n-1):
if a[i+1]-a[i]==1:
b+=1
i+=2
else:
i+=1
c,d=0,0
for i in range(n):
if a[i]%2==0:
c+=1
d=n-c
if (c%2==0 and d%2==0) or (c%2!=0 and d%2!=0 and b>0 ):
print("YES")
else:
print("NO")
``` | output | 1 | 80,981 | 22 | 161,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,982 | 22 | 161,964 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
def game():
odd = []
even = []
for i in arr:
if i%2==0:
odd.append(i)
else:
even.append(i)
if (len(odd)%2)==0 and (len(even)%2==0):
print('YES')
return
else:
for i in odd:
for j in even:
if abs(i-j)==1:
print('YES')
return
print('NO')
t = int(input())
for _ in range(t):
n = int(input())
*arr ,= map(int,input().split())
arr.sort()
game()
``` | output | 1 | 80,982 | 22 | 161,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,983 | 22 | 161,966 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
def main():
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
T = int(input())
for t in range(T):
n = int(input())
a = tuple([int(x) for x in input().strip().split()])
even = [aa for aa in a if aa % 2 == 0]
odd = [aa for aa in a if aa % 2]
even_n = len(even)
odd_n = len(odd)
if even_n == 0 or odd_n == 0 or (even_n % 2 == 0 or odd_n % 2 == 0):
print('YES')
continue
f = False
for aa in range(1, 101):
if aa in a:
if f:
print('YES')
break
else:
f = True
else:
f = False
else:
print('NO')
if __name__ == '__main__':
main()
``` | output | 1 | 80,983 | 22 | 161,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,984 | 22 | 161,968 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
import math
#------------------------------warmup----------------------------
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")
#-------------------game starts now----------------------------------------------------
for ik in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
even=[]
odd=[]
for i in range(n):
if l[i]%2==0:
even.append(l[i])
else:
odd.append(l[i])
if len(even)%2==0 and len(odd)%2==0:
print("YES")
else:
f=0
for i in even:
if i-1 in odd or i+1 in odd:
f=1
break
if f==1:
print("YES")
else:
print("NO")
``` | output | 1 | 80,984 | 22 | 161,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,985 | 22 | 161,970 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
import math
for case in range(int(input())):
n = int(input())
a = input().split(' ')
numeven = 0
for i in range(n):
a[i] = int(a[i])
if a[i] % 2 == 0:
numeven = numeven + 1
numodd = n - numeven
ans = -1
if numeven % 2 != 0:
if numodd %2 != 0:
for i in a:
if i+1 in a or i-1 in a:
ans = 1
break
else:
if numodd %2 == 0:
ans = 1
if ans == 1:
print("YES")
else:
print("NO")
``` | output | 1 | 80,985 | 22 | 161,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,986 | 22 | 161,972 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
from collections import Counter
import sys
sys.setrecursionlimit(10 ** 6)
mod = 1000000007
inf = int(1e18)
dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]
def inverse(a):
return pow(a, mod - 2, mod)
def usearch(x, a):
lft = 0
rgt = len(a) + 1
while rgt - lft > 1:
mid = (rgt + lft) // 2
if a[mid] <= x:
lft = mid
else:
rgt = mid
return lft
def solve():
n = int(input())
a = sorted(map(int, input().split()))
even = 0
odd = 0
for x in a:
if x % 2:
odd += 1
else:
even += 1
if odd % 2 == 0 and even % 2 == 0:
print("YES")
return
for i in range(n-1):
if a[i+1] - a[i] == 1:
print("YES")
return
print("NO")
def main():
n = int(input())
for i in range(n):
solve()
main()
``` | output | 1 | 80,986 | 22 | 161,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,987 | 22 | 161,974 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = sorted(map(int, input().split()))
matched = 0
evens = 0
odds = 0
for i in range(n):
if a[i] % 2 == 0:
evens += 1
else:
odds += 1
i = 1
while i < n:
if a[i] - a[i-1] == 1:
matched += 1
i += 1
i += 1
evens -= matched
odds -= matched
if evens % 2 == 0 and odds % 2 == 0:
print("YES")
elif matched != 0 and evens % 2 == 1 and odds % 2 == 1:
print("YES")
else:
print("NO")
``` | output | 1 | 80,987 | 22 | 161,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable. | instruction | 0 | 80,988 | 22 | 161,976 |
Tags: constructive algorithms, graph matchings, greedy, sortings
Correct Solution:
```
for _ in range(int(input())):
input()
l = list(map(int, input().split()))
n = [i for i in l if i%2]
if len(n)%2==0:
print('YES')
else:
l.sort()
for i in range(len(l)-1):
if l[i+1]-l[i]==1:
print('YES')
break
else:
print('NO')
``` | output | 1 | 80,988 | 22 | 161,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable.
Submitted Solution:
```
#https://codeforces.com/contest/1360/problem/C
for i in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
l.sort()
even=0
odd=0
for i in l:
if i%2==0:
even+=1
else:
odd+=1
if even%2==0 and odd%2==0:
print('YES')
continue
f=0
for i in range(1,n):
if l[i]-l[i-1]==1:
f=1
break
if f:
print('YES')
else:
print('NO')
``` | instruction | 0 | 80,990 | 22 | 161,980 |
Yes | output | 1 | 80,990 | 22 | 161,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable.
Submitted Solution:
```
def solve(n, a):
count = {}
count[0] = []
count[1] = []
for i in a: count[i % 2].append(i)
if not (len(count[0]) % 2):
return 1
found = 0
for i in count[1]:
if i-1 in count[0] or i+1 in count[0]:
found = 1
break
return found
def main():
t = int(input())
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
ans = solve(n, a)
if ans:
print('YES')
else:
print('NO')
if __name__ == "__main__":
main()
``` | instruction | 0 | 80,991 | 22 | 161,982 |
Yes | output | 1 | 80,991 | 22 | 161,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort()
odd = 0
even = 0
for i in a:
if i % 2 == 0:
even += 1
else:odd += 1
odd = odd % 2
even = even % 2
dif = 0
j = 0
while j < n - 1:
if abs(a[j] - a[j + 1]) == 1:
dif += 1
j += 1
j += 1
dif = dif % 2
if odd + even == 1:
print("NO")
continue
if odd + even == 2 and dif == 0:
print("NO")
continue
print("YES")
``` | instruction | 0 | 80,993 | 22 | 161,986 |
No | output | 1 | 80,993 | 22 | 161,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable.
Submitted Solution:
```
n= int(input())
for i in range(n):
x=int(input())
l=list(map(int,input().split()))
c=0
d=0
e=0
for i in range(x-1):
if l[i]%2==0:
c+=1
if l[i]%2!=0:
d+=1
if abs(l[i]-l[i+1])==1:
e+=1
if l[-1]%2==0:
c+=1
else:
d+=1
f=0
if c%2!=d%2:
print("NO")
elif c%2==0 and d%2==0:
print("YES")
elif c%2!=0 and d%2!=0:
if e>0:
print("YES")
else:
print("NO")
else:
print("NO")
``` | instruction | 0 | 80,994 | 22 | 161,988 |
No | output | 1 | 80,994 | 22 | 161,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable.
Submitted Solution:
```
def main():
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
T = int(input())
for t in range(T):
n = int(input())
a = sorted([int(x) for x in input().strip().split()])
even = [aa for aa in a if aa % 2 == 0]
odd = [aa for aa in a if aa % 2]
even_n = len(even)
odd_n = len(odd)
if even_n == 0 or odd_n == 0:
print('YES')
continue
even_i = 0
odd_i = 0
even_ = even[0]
odd_ = odd[0]
while True:
if abs(even_ - odd_) == 1:
print('YES')
break
if even_ < odd_:
if even_i + 1 == even_n:
print('NO')
break
even_i = even_i+1
even_ = even[even_i]
else:
if odd_i + 1 == odd_n:
print('NO')
break
odd_i = odd_i+1
odd_ = odd[odd_i]
else:
if abs(even_ - odd_) == 1:
print('YES')
else:
print('NO')
if __name__ == '__main__':
main()
``` | instruction | 0 | 80,995 | 22 | 161,990 |
No | output | 1 | 80,995 | 22 | 161,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x-y|=1. For example, in each of the pairs (2, 6), (4, 3), (11, 7), the numbers are similar to each other, and in the pairs (1, 4), (3, 12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a = [11, 14, 16, 12], there is a partition into pairs (11, 12) and (14, 16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2 ≤ n ≤ 50) — length of array a.
The second line contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each test case print:
* YES if the such a partition exists,
* NO otherwise.
The letters in the words YES and NO can be displayed in any case.
Example
Input
7
4
11 14 16 12
2
1 8
4
1 1 1 1
4
1 2 5 6
2
12 13
6
1 6 3 10 5 8
6
1 12 3 10 5 8
Output
YES
NO
YES
YES
YES
YES
NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable.
Submitted Solution:
```
from sys import stdin, stdout
t = int(input())
for _ in range(t):
n = int(stdin.readline())
nums = list(map(int, stdin.readline().strip().split()))
nums.sort()
d = dict()
for i in nums:
d[i] = 0
temp = list(d.keys())
for k in temp:
if k-1 in d and k in d:
del d[k]
del d[k-1]
elif k+1 in d and k in d:
del d[k]
del d[k+1]
newl = [i for i in nums if i in d]
even, odd = 0, 0
for k in newl:
if k%2==0: even += 1
else: odd += 1
if even%2==0 and odd%2==0:
print('YES')
else:
print('NO')
``` | instruction | 0 | 80,996 | 22 | 161,992 |
No | output | 1 | 80,996 | 22 | 161,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks n integers a1, a2, ..., an are good.
Now she is interested in good sequences. A sequence x1, x2, ..., xk is called good if it satisfies the following three conditions:
* The sequence is strictly increasing, i.e. xi < xi + 1 for each i (1 ≤ i ≤ k - 1).
* No two adjacent elements are coprime, i.e. gcd(xi, xi + 1) > 1 for each i (1 ≤ i ≤ k - 1) (where gcd(p, q) denotes the greatest common divisor of the integers p and q).
* All elements of the sequence are good integers.
Find the length of the longest good sequence.
Input
The input consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 105) — the number of good integers. The second line contains a single-space separated list of good integers a1, a2, ..., an in strictly increasing order (1 ≤ ai ≤ 105; ai < ai + 1).
Output
Print a single integer — the length of the longest good sequence.
Examples
Input
5
2 3 4 6 9
Output
4
Input
9
1 2 3 5 6 7 8 9 10
Output
4
Note
In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4. | instruction | 0 | 81,087 | 22 | 162,174 |
Tags: dp, number theory
Correct Solution:
```
primes=[]
for i in range(2,10**5):
x=0
for j in range(2,int(i**0.5)+1):
if i%j==0:
x=1
break
if x==0:
primes.append(i)
set_p=set(primes)
n=int(input())
a=list(map(int,input().split()))
#a=[100000 for i in range(1,n+1)]
a=a[::-1]
ans=0
store={}
c=0
for i in a:
c=0
b=set()
if i==1:
ans=max(ans,1)
#print(store)
x=i
for i in primes:
if x==1:
break
if x in set_p:
b.add(x)
break
if x%i==0:
b.add(i)
while x%i==0:
x=x//i
for i in b:
if i in store:
c=max(c,store[i]+1)
else:
c=max(c,1)
for i in b:
store[i]=c
if c>ans:
ans=c
print(ans)
``` | output | 1 | 81,087 | 22 | 162,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. F has n positive integers, a_1, a_2, …, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 1.5 ⋅ 10^7).
Output
Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print «-1» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1. | instruction | 0 | 81,721 | 22 | 163,442 |
Tags: number theory
Correct Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
# t = int(input())
n = int(input())
l = list(map(int,input().split()))
m = max(l) + 1
prime = [0]*m
cmd = [0]*(m)
def sieve():
for i in range(2,m):
if prime[i] == 0:
for j in range(2*i,m,i):
prime[j] = i
for i in range(2,m):
if prime[i] == 0:
prime[i] = i
g = 0
for i in range(n):
g = gcd(l[i],g)
sieve()
ans = -1
for i in range(n):
ele = l[i]//g
while ele>1:
div = prime[ele]
cmd[div]+=1
while ele%div == 0:
ele//=div
ans = max(ans,cmd[div])
if ans == -1:
print(-1)
exit()
print(n-ans)
``` | output | 1 | 81,721 | 22 | 163,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. F has n positive integers, a_1, a_2, …, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 1.5 ⋅ 10^7).
Output
Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print «-1» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1. | instruction | 0 | 81,722 | 22 | 163,444 |
Tags: number theory
Correct Solution:
```
from math import gcd
R = lambda:map(int, input().split())
n = int(input())
l = list(R())
m = max(l)+1
prime = [0]*(m)
commondivisor = [0]*(m)
def seive():
for i in range(2,m):
if prime[i]==0:
for j in range(i*2,m,i):
prime[j] = i
for i in range(2,m):
if not prime[i]:
prime[i] = i
gc = l[0]
for i in range(1,n):
gc = gcd(gc,l[i])
seive()
mi = -1
for i in range(n):
ele = l[i]//gc
while ele > 1:
div = prime[ele]
commondivisor[div]+=1
while ele%div == 0:
ele//=div
mi = max(mi, commondivisor[div])
if mi == -1:
print(-1)
else:
print(n-mi)
``` | output | 1 | 81,722 | 22 | 163,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. F has n positive integers, a_1, a_2, …, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 1.5 ⋅ 10^7).
Output
Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print «-1» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1. | instruction | 0 | 81,723 | 22 | 163,446 |
Tags: number theory
Correct Solution:
```
from math import gcd
def sieve():
global prime
prime=[0]*(m)
for i in range(2,m):
if prime[i] ==0:
for j in range(i,m,i):
prime[j] =i
n=int(input())
arr=list(map(int,input().split()))
m=max(arr) +1
prime=[0]*(m)
sieve()
gd=0
commondivision=[0]*(m)
for i in range(n):
gd=gcd(gd,arr[i])
ans=-1
for i in range(n):
curr=arr[i] //gd
while curr >1:
div=prime[curr]
while curr %div ==0:
curr //=div
commondivision[div] +=1
ans=max(ans,commondivision[div])
if ans==-1:
print(-1)
else:
print(n-ans)
``` | output | 1 | 81,723 | 22 | 163,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. F has n positive integers, a_1, a_2, …, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 1.5 ⋅ 10^7).
Output
Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print «-1» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1. | instruction | 0 | 81,724 | 22 | 163,448 |
Tags: number theory
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from math import gcd
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
A = list(map(int, readline().split()))
M = max(A) + 1
## 1- index
MinFactor = [0] * M
MinFactor[1] = 1
commondivisor = [0] * M
def seive():
for i in range(2, M):
if not MinFactor[i]:
for j in range(i * 2, M, i):
if not MinFactor[j]: #���ꂠ�邱�Ƃōŏ��̈��q�ƂȂ�
MinFactor[j] = i
for i in range(2, M):
if not MinFactor[i]:
MinFactor[i] = i
gc = A[0]
for i in range(1, N):
gc = gcd(gc, A[i])
seive()
mi = -1
for i in range(N):
a = A[i] // gc
while a > 1:
div = MinFactor[a]
commondivisor[div] += 1
while a % div == 0:
a //= div
mi = max(mi, commondivisor[div])
if mi == -1:
print(-1)
else:
print(N - mi)
``` | output | 1 | 81,724 | 22 | 163,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. F has n positive integers, a_1, a_2, …, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 1.5 ⋅ 10^7).
Output
Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print «-1» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1. | instruction | 0 | 81,725 | 22 | 163,450 |
Tags: number theory
Correct Solution:
```
from math import gcd
n = int(input())
l = list(map(int,input().split()))
m = max(l)+1
prime = [0]*(m)
commondivisor = [0]*(m)
def seive():
for i in range(2,m):
if prime[i] == 0:
for j in range(i*2,m,i):
prime[j] = i
for i in range(2,m):
if not prime[i]:
prime[i] = i
gc = l[0]
for i in range(1,n):
gc = gcd(gc,l[i])
seive()
mi = -1
for i in range(n):
ele = l[i]//gc
while ele > 1:
div = prime[ele]
commondivisor[div]+=1
while ele%div == 0:
ele //= div
mi = max(mi,commondivisor[div])
if mi == -1:
print(-1)
else:
print(n-mi)
``` | output | 1 | 81,725 | 22 | 163,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. F has n positive integers, a_1, a_2, …, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 1.5 ⋅ 10^7).
Output
Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print «-1» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1. | instruction | 0 | 81,726 | 22 | 163,452 |
Tags: number theory
Correct Solution:
```
n = int(input())
m=list(map(int,input().split()))
gc=m[0]
import math
for i in range(1,n):
gc=math.gcd(gc,m[i])
k=max(m)
prime=[0]*(k+1)
pfac=[0]*(k+1)
for i in range(2,k+1):
if prime[i]==0:
for j in range(i*i,k+1,i):
prime[j]=i
for i in range(1,k+1):
if not prime[i]:
prime[i]=i
for i in range(n):
v=m[i]//gc
while v>1:
fac=prime[v]
pfac[fac]+=1
while v%fac==0:
v//=fac
ma=0
ma=max(pfac)
if ma:
print(n-ma)
else:
print(-1)
``` | output | 1 | 81,726 | 22 | 163,453 |
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