message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
def getInts(): return list(map(int,input().split()))
#MOD = 10**9 + 7
MOD = 998244353
def ncr(n,r):
num = denom = 1
for i in range(r):
num = (num*(n-i))%MOD
denom = (denom*(i+1))%MOD
return (num * pow(denom,MOD-2,MOD))%MOD
"""
Chains of multiples
Iterate over first of chain
"""
N, K = map(int,input().split())
if N < K:
print(0)
else:
if K == 1:
print(N)
else:
ans = 0
for i in range(1,N+1):
#how many multiples of i are less than or equal to N?
if N//i > K:
ans += ncr(N//i-1,K-1)
ans %= MOD
elif N//i == K:
ans += 1
ans %= MOD
else:
break
print(ans)
``` | instruction | 0 | 100,502 | 22 | 201,004 |
Yes | output | 1 | 100,502 | 22 | 201,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
n,k=map(int,input().split())
from math import factorial
if n<k:
print(0)
else:
ans=0
m=998244353
def ncr(n, r, p):
# initialize numerator
# and denominator
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
if k==1:
print(n)
else:
for i in range(1,n):
if n//i>k:
ans+=ncr(n//i-1,k-1,m)
ans%=m
elif n//i==k:
ans+=1
ans%=m
else:
break
print(ans)
``` | instruction | 0 | 100,504 | 22 | 201,008 |
Yes | output | 1 | 100,504 | 22 | 201,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
#!/usr/bin/env python
# coding:utf-8
# Copyright (C) dirlt
MOD = 998244353
def pow(a, x):
ans = 1
while x > 0:
if x % 2 == 1:
ans = (ans * a) % MOD
a = (a * a) % MOD
x = x // 2
return ans
def run(n, k):
fac = [1] * (n + 1)
for i in range(1, n + 1):
fac[i] = (fac[i - 1] * i) % MOD
ans = 0
for a0 in range(1, n + 1):
nn = n // a0 - 1
if (nn - k + 1) < 0: break
a = fac[nn]
b = fac[k - 1]
c = fac[nn - k + 1]
d = (b * c) % MOD
e = pow(d, MOD - 2)
f = (a * e) % MOD
ans = (ans + f) % MOD
return ans
# this is codeforces main function
def main():
from sys import stdin
def read_int():
return int(stdin.readline())
def read_int_array(sep=None):
return [int(x) for x in stdin.readline().split(sep)]
def read_str_array(sep=None):
return [x.strip() for x in stdin.readline().split(sep)]
import os
if os.path.exists('tmp.in'):
stdin = open('tmp.in')
n, k = read_int_array()
ans = run(n, k)
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 100,505 | 22 | 201,010 |
Yes | output | 1 | 100,505 | 22 | 201,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
from sys import stdin as f
from math import factorial as fac
def binom_coeff(n, k):
return fac(n) / (fac(k) * fac(n - k))
def find_arr_num(n, k, m):
if k == 1:
return n
result, a = 0, 1
while n // a >= k:
count = n // a
result = result + (binom_coeff(count-1, k-1) % m)
a = a + 1
#print(a)
#print(result)
return int(result % m)
m = 998244353
n, k = [int(i) for i in f.readline().strip().split()]
print(find_arr_num(n, k, m))
``` | instruction | 0 | 100,506 | 22 | 201,012 |
No | output | 1 | 100,506 | 22 | 201,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
n,k=map(int,input().split())
from math import factorial
if n<k:
print(0)
else:
ans=0
m=998244353
for i in range(1,n):
if n//i>k:
ans+=factorial(n//i-1)//(factorial(k-1)*factorial(n//i-k))
ans%=m
elif n//i==k:
ans+=1
ans%=m
else:
break
print(ans)
``` | instruction | 0 | 100,507 | 22 | 201,014 |
No | output | 1 | 100,507 | 22 | 201,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
n,k = map(int,input().split())
p = 998244353
def pot(a,d):
res = 1
mno = a
while d > 0:
if d%2 == 1:
res = (res*mno)%p
mno = (mno*mno)%p
d//=2
return res
silnia = [1] * (n+1)
for i in range(2,n+1):
silnia[i] = (i*silnia[i-1])%p
#dla a_1 = i ile mozliwych (k-1)-tek liczb od 2 do n//a_i
wyn = 0
for a1 in range(1,n):
if n//a1 < k:
break
else:
zb = (n//a1)-1
#wyn += zb po k-1
wyn += silnia[zb] * pot(silnia[zb-k+1], p-2) * pot(silnia[k-1], p-2)
print(wyn%p)
``` | instruction | 0 | 100,508 | 22 | 201,016 |
No | output | 1 | 100,508 | 22 | 201,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,542 | 22 | 201,084 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int
p=input
N=i(p());L=p();R=p()
l=i(L,2);r=i(R,2);a=R
if L[0]<R[0]:a='1'*N
elif L==R:a=L
elif L[-1]=='1'and l+1==r:a=R
elif l//2<r//2:a=R[:-1]+'1'
print(a)
``` | output | 1 | 100,542 | 22 | 201,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,543 | 22 | 201,086 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
n = int(input())
l = input(); L = int(l,2)
r = input(); R = int(r,2)
if (l[0] != r[0]):
print('1'*n)
exit()
if (R - L < 2 or r[-1] == 1):
print(r)
else:
print(r[:-1] + '1')
``` | output | 1 | 100,543 | 22 | 201,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,544 | 22 | 201,088 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
n = int(input())
l = input()
r = input()
if n == 1:
print(r)
elif l[0] == '0':
print('1'*n)
elif r[-1] == '0' and int(l,2)+1 < int(r,2):
print(r[:-1] + "1")
else:
print(r)
``` | output | 1 | 100,544 | 22 | 201,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,545 | 22 | 201,090 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int;p=input
N=i(p());L=p();R=p()
l=i(L,2);r=i(R,2);a=R
if l-l%2<r:a=R[:-1]+'1'
if i(L[-1])and l+1==r:a=R
if L==R:a=L
if L[0]<R[0]:a='1'*N
print(a)
``` | output | 1 | 100,545 | 22 | 201,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,546 | 22 | 201,092 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
n = int(input());l = input();r = input()
if n == 1: print(r)
elif l[0] == '0': print('1'*n)
elif r[-1] == '0' and int(l,2)+1 < int(r,2): print(r[:-1] + "1")
else: print(r)
``` | output | 1 | 100,546 | 22 | 201,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,547 | 22 | 201,094 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve2(l, r):
n = len(l)
if l[0] != r[0]:
return [1]*n
if r[-1] == 1:
return r
x = r.copy()
for j in range(2):
if x == l:
return r
for k in range(n-1,-1,-1):
if x[k] == 1:
x[k] = 0
break
else:
x[k] = 1
x = r.copy()
r[-1] = 1
return r
def solve():
n = int(input())
l = list(map(int,input().strip()))
r = list(map(int,input().strip()))
ans = solve2(l, r)
print(''.join(map(str,ans)))
solve()
``` | output | 1 | 100,547 | 22 | 201,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,548 | 22 | 201,096 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int
p=input
N=i(p());L=p();R=p();a=R
if L[0]<R[0]:a='1'*N
elif L==R:a=L
elif L[-1]=='1'and i(L,2)+1==i(R,2):a=R
elif i(L,2)//2<i(R,2)//2:a=''.join(R[:-1])+'1'
print(a)
``` | output | 1 | 100,548 | 22 | 201,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x ⊕ (x+1) ⊕ ... ⊕ (y-1) ⊕ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l ≤ x ≤ y ≤ r.
Output f(l, r).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the binary representation of r.
The second line contains the binary representation of l — a string of length n consisting of digits 0 and 1 (0 ≤ l < 2^n).
The third line contains the binary representation of r — a string of length n consisting of digits 0 and 1 (0 ≤ r < 2^n).
It is guaranteed that l ≤ r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example. | instruction | 0 | 100,549 | 22 | 201,098 |
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int
p=input
N=i(p());L=p();R=p()
l=i(L,2);r=i(R,2);a=R
if l//2<r//2:a=R[:-1]+'1'
if i(L[-1])and l+1==r:a=R
if L==R:a=L
if L[0]<R[0]:a='1'*N
print(a)
``` | output | 1 | 100,549 | 22 | 201,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each. | instruction | 0 | 100,670 | 22 | 201,340 |
Tags: greedy, math
Correct Solution:
```
#dimaag ka bhosda krne wala question..
#dont try to take all cases simultaneously
#better it doing box by box
k,a,b,v=map(int,input().split())
ans=0
while a>0:
ans+=1
kn=min(k,b+1)
a-=v*kn #filling v nuts in a box
b=max((b+1)-k,0) #removing k divisors that is minimum number of divisor in a box
print(ans)
``` | output | 1 | 100,670 | 22 | 201,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each. | instruction | 0 | 100,673 | 22 | 201,346 |
Tags: greedy, math
Correct Solution:
```
import sys
import string
from collections import Counter, defaultdict
from math import fsum, sqrt, gcd, ceil, factorial
from operator import add
inf = float("inf")
# input = sys.stdin.readline
flush = lambda: sys.stdout.flush
comb = lambda x, y: (factorial(x) // factorial(y)) // factorial(x - y)
# inputs
# ip = lambda : input().rstrip()
ip = lambda: input()
ii = lambda: int(input())
r = lambda: map(int, input().split())
rr = lambda: list(r())
k, a, b, v = r()
k-=1
ans = 0
while a > 0 and b > 0:
x = min(b , k)
# print('x ' , x)
a -= (x+1) * v
b -= x
ans += 1
if a > 0:
ans += (a + v- 1)//v
print(ans)
``` | output | 1 | 100,673 | 22 | 201,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each. | instruction | 0 | 100,674 | 22 | 201,348 |
Tags: greedy, math
Correct Solution:
```
k, a, b, v = list(map(int, input().split()))
x = 0
while a > 0:
c = 1
if b > k-1:
c += k-1
b -= k-1
else:
c += b
b = 0
a -= v*c
x += 1
print(str(x))
``` | output | 1 | 100,674 | 22 | 201,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have written on a piece of paper an array of n positive integers a[1], a[2], ..., a[n] and m good pairs of integers (i1, j1), (i2, j2), ..., (im, jm). Each good pair (ik, jk) meets the following conditions: ik + jk is an odd number and 1 ≤ ik < jk ≤ n.
In one operation you can perform a sequence of actions:
* take one of the good pairs (ik, jk) and some integer v (v > 1), which divides both numbers a[ik] and a[jk];
* divide both numbers by v, i. e. perform the assignments: <image> and <image>.
Determine the maximum number of operations you can sequentially perform on the given array. Note that one pair may be used several times in the described operations.
Input
The first line contains two space-separated integers n, m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100).
The second line contains n space-separated integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — the description of the array.
The following m lines contain the description of good pairs. The k-th line contains two space-separated integers ik, jk (1 ≤ ik < jk ≤ n, ik + jk is an odd number).
It is guaranteed that all the good pairs are distinct.
Output
Output the answer for the problem.
Examples
Input
3 2
8 3 8
1 2
2 3
Output
0
Input
3 2
8 12 8
1 2
2 3
Output
2
Submitted Solution:
```
def primefactorial(n):
primes = []
d = 2
while d*d <= n:
while (n % d) == 0:
primes.append(d)
n /= d
d += 1
if n > 1:
primes.append(n)
return primes
def arrays():
n, m= [int(i) for i in input().split()]
numlist= [int(i) for i in input().split()]
good= []
for i in range(m):
good.append([int(k)-1 for k in input().split()])
#print(good)
mylist= []
for i in numlist:
mylist.append(primefactorial(i))
#print(mylist)
#comparing using good list
ans= 0
for i in range(len(good)):
a= mylist[good[i][0]]
b= mylist[good[i][1]]
for j in a[:]:
if j in b:
a.remove(j)
b.remove(j)
ans+= 1
print(ans)
return
arrays()
``` | instruction | 0 | 100,715 | 22 | 201,430 |
No | output | 1 | 100,715 | 22 | 201,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have written on a piece of paper an array of n positive integers a[1], a[2], ..., a[n] and m good pairs of integers (i1, j1), (i2, j2), ..., (im, jm). Each good pair (ik, jk) meets the following conditions: ik + jk is an odd number and 1 ≤ ik < jk ≤ n.
In one operation you can perform a sequence of actions:
* take one of the good pairs (ik, jk) and some integer v (v > 1), which divides both numbers a[ik] and a[jk];
* divide both numbers by v, i. e. perform the assignments: <image> and <image>.
Determine the maximum number of operations you can sequentially perform on the given array. Note that one pair may be used several times in the described operations.
Input
The first line contains two space-separated integers n, m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100).
The second line contains n space-separated integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — the description of the array.
The following m lines contain the description of good pairs. The k-th line contains two space-separated integers ik, jk (1 ≤ ik < jk ≤ n, ik + jk is an odd number).
It is guaranteed that all the good pairs are distinct.
Output
Output the answer for the problem.
Examples
Input
3 2
8 3 8
1 2
2 3
Output
0
Input
3 2
8 12 8
1 2
2 3
Output
2
Submitted Solution:
```
def primefactorial(n):
primes= []
i= 2
while i**2<= n:
while n%i== 0:
primes.append(i)
n/= i
i+= 1
if n> 1:
primes.append(int(n))
return primes
def arrays():
n, m= [int(i) for i in input().split()]
numlist= [int(i) for i in input().split()]
good= []
for i in range(m):
good.append([int(k)-1 for k in input().split()])
#print(good)
mylist= []
for i in numlist:
mylist.append(primefactorial(i))
#print(mylist)
#comparing using good list
ans= 0
for i in range(len(good)):
a= mylist[good[i][0]]
b= mylist[good[i][1]]
for j in a:
if j in b:
a.remove(j)
b.remove(j)
ans+= 1
for k in b:
if k in a:
a.remove(k)
b.remove(k)
ans+= 1
#print(mylist)
print(ans)
return
arrays()
``` | instruction | 0 | 100,717 | 22 | 201,434 |
No | output | 1 | 100,717 | 22 | 201,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,737 | 22 | 201,474 |
Tags: math
Correct Solution:
```
k, a, b = map(int, input().split())
if a % k:
a += k - a % k
b -= b % k
print(b // k - a // k + 1)
``` | output | 1 | 100,737 | 22 | 201,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,738 | 22 | 201,476 |
Tags: math
Correct Solution:
```
k, a, b = map(int, input().split())
if (a >= 0 and b > 0) or (a < 0 and b <= 0):
print(max(abs(a), abs(b)) // k - ((min(abs(a), abs(b))) - 1) // k)
else:
print(abs(a) // k + abs(b) // k + 1)
``` | output | 1 | 100,738 | 22 | 201,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,739 | 22 | 201,478 |
Tags: math
Correct Solution:
```
(k,a,b)=map(int, input().split())
print(b//k-(a-1)//k)
``` | output | 1 | 100,739 | 22 | 201,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,740 | 22 | 201,480 |
Tags: math
Correct Solution:
```
from math import floor
def seal(a, b):
return (a + b - 1)//b
k,a,b = map(int,input().split())
ans = abs(b)//k + abs(a)//k
if(a < 0 and b >=0):
values = b - a + 1
print(ans + 1)
else:
a,b = abs(a), abs(b)
a,b = min(a,b), max(a,b)
lower = seal(a,k)
upper = int(b//k)
print(upper - lower + 1)
# if(a % k == 0 and b % k == 0 and k != 1):
# ans += 1
# print(ans)
``` | output | 1 | 100,740 | 22 | 201,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,741 | 22 | 201,482 |
Tags: math
Correct Solution:
```
k, a, b = [int(i) for i in input().split()]
count = 0
if(a%k!=0):
a = (a + k) - (a%k)
if(b%k!=0):
b = b - (b%k)
if(b<a):
print(0)
else:
print(1 + (b-a)//k)
``` | output | 1 | 100,741 | 22 | 201,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,742 | 22 | 201,484 |
Tags: math
Correct Solution:
```
k,a,b=map(int,input().split())
m=b//k
if a>0:
x=(a//k)
m-=x
if a%k==0:
m+=1
else:
m+=(abs(a)//k)
m+=1
print(m)
``` | output | 1 | 100,742 | 22 | 201,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,743 | 22 | 201,486 |
Tags: math
Correct Solution:
```
n,l,h=input().split()
n=int(n)
l=int(l)
h=int(h)
if l>0 and h>0:
print(h//n-(l-1)//n)
else:
print(h//n+abs(l)//n+1)
``` | output | 1 | 100,743 | 22 | 201,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5 | instruction | 0 | 100,744 | 22 | 201,488 |
Tags: math
Correct Solution:
```
import math
def main():
k, a, b = map(int, input().split())
low = (a+k-1)//k
big = b//k
ans = big - low+1
# if a<0 and b>0:
# ans += 1
print(int(ans))
if __name__ == '__main__':
main()
``` | output | 1 | 100,744 | 22 | 201,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
# your code goes here
k, a, b = map(int, input().split())
#//=floor division
floor_a = a//k
floor_b = b//k
ans = floor_b - floor_a
if a%k==0:
ans += 1
print(ans)
``` | instruction | 0 | 100,745 | 22 | 201,490 |
Yes | output | 1 | 100,745 | 22 | 201,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
k,a,b=map(int,input().split())
if a<=0 and b>=0:
res=abs(a)//k+b//k+1
print(res)
else:
a,b=abs(a),abs(b)
a,b=max(a,b),min(a,b)-1
res=a//k-(b//k)
print(res)
``` | instruction | 0 | 100,746 | 22 | 201,492 |
Yes | output | 1 | 100,746 | 22 | 201,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
k,a,b = list(map(int,input().split()))
if a%k!=0:
n1 = a+k-(a%k)
else:
n1 = a
n2 = b-b%k
n = ((n2-n1)//k)+1
print(n)
``` | instruction | 0 | 100,747 | 22 | 201,494 |
Yes | output | 1 | 100,747 | 22 | 201,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
k, l, r = map(int, input().split())
p = l // k
if(l % k > 0):
p += 1
l = p * k
o = (r - l) // k + 1
if(l > r):
o = 0
print(o)
``` | instruction | 0 | 100,748 | 22 | 201,496 |
Yes | output | 1 | 100,748 | 22 | 201,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
k, a, b = map(int, input().split())
counter = 0
if (a == b):
if (abs(a) % k == 0 or a == 0):
counter = 0
elif ((b > 0 and a >= 0) or (a < 0 and b <= 0)):
n1 = abs(b)//k
n2 = abs(a)//k
counter = max(n1-n2+1, n2-n1+1)
elif(b > 0 and a < 0):
n1 = b//k
n2 = abs(a)//k
counter = n1+n2+1
print(counter)
``` | instruction | 0 | 100,749 | 22 | 201,498 |
No | output | 1 | 100,749 | 22 | 201,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
from math import ceil
def delimost(k, a, b):
if a == b == 0:
return 1
return ceil((b + 1 - a) / k)
K, A, B = [int(j) for j in input().split()]
print(delimost(K, A, B))
``` | instruction | 0 | 100,750 | 22 | 201,500 |
No | output | 1 | 100,750 | 22 | 201,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
import sys
c = 0
k,a,b = map(int,sys.stdin.readline().split())
if(k==1):
print(b-a+1)
elif(k==2):
print(((b-a)/2)+1)
else:
for i in range(a,b+1):
if(i%k==0):
c =c+1
print(c)
``` | instruction | 0 | 100,751 | 22 | 201,502 |
No | output | 1 | 100,751 | 22 | 201,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Examples
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
Submitted Solution:
```
#!/bin/python3
import math
k, a, b = [int(i) for i in input().split()]
if k > 2:
print(sum([1 for x in range(a,b+1) if x%k==0]))
elif k == 1:
print(b-a+1)
elif k == 2:
print(math.ceil((b-a+1)/2))
``` | instruction | 0 | 100,752 | 22 | 201,504 |
No | output | 1 | 100,752 | 22 | 201,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is an integer N.
Takahashi chooses an integer a from the positive integers not greater than N with equal probability.
Find the probability that a is odd.
Constraints
* 1 \leq N \leq 100
Input
Input is given from Standard Input in the following format:
N
Output
Print the probability that a is odd. Your output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-6}.
Examples
Input
4
Output
0.5000000000
Input
5
Output
0.6000000000
Input
1
Output
1.0000000000
Submitted Solution:
```
def prime_divisor(n):
prime_divisor_set = {1}
end = int(n ** 0.5) + 1
for i in range(2, end):
if not (n % i):
prime_divisor_set.add(i)
while (n % i) == 0:
n //= i
return prime_divisor_set
a, b = map(int, input().split())
s1, s2 = prime_divisor(a), prime_divisor(b)
ans = len(s1 & s2)
print(ans)
``` | instruction | 0 | 100,966 | 22 | 201,932 |
No | output | 1 | 100,966 | 22 | 201,933 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,156 | 22 | 204,312 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
colors = ''
used = [0]*11
count = 0
for ai in a:
for i in range(11):
if ai%primes[i] is 0:
if used[i] is 0:
count += 1
used[i] = count
colors += str(used[i]) + ' '
break
print("{0}\n{1}".format(count, colors))
``` | output | 1 | 102,156 | 22 | 204,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,157 | 22 | 204,314 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
import math
t = int(input())
def check(c, a, n):
for i in range(n):
for j in range(n):
if (c[i] == c[j] and math.gcd(a[i], a[j]) == 1):
return False
return True
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
c = [0 for i in range(n)]
freq = [[0, i] for i in range(101)]
need = max(0, n-11)
for j in range(2, 101):
for i in range(n):
if a[i] % j == 0:
freq[j][0] += 1
cur = 1
if need > 0:
for frequency, num in freq:
if num <= 1:
continue
did = False
for i in range(n):
if a[i] % num == 0 and c[i] == 0:
c[i] = cur
did = True
if did:
cur += 1
if all([i != 0 for i in c]):
break
for i in range(n):
if c[i] == 0:
c[i] = cur
cur += 1
print(max(c))
print(*c)
``` | output | 1 | 102,157 | 22 | 204,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,158 | 22 | 204,316 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
testcases=int(input())
for j in range(testcases):
n=int(input())
comps=list(map(int,input().split()))
primes=[2,3,5,7,11,13,17,19,23,29,31]
dict1={}
for s in range(len(comps)):
for k in primes:
if comps[s]%k==0:
if k in dict1:
dict1[k].append(s)
else:
dict1[k]=[s]
break
#
counter=1
for b in dict1:
for l in dict1[b]:
comps[l]=str(counter)
counter+=1
print(counter-1)
print(" ".join(comps))
``` | output | 1 | 102,158 | 22 | 204,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,159 | 22 | 204,318 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
from math import sqrt
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
dd = dict()
for j in range(n):
i = a[j]
# firstfactor = 2
for firstfactor in range(2, int(sqrt(i)) + 1):
if i % firstfactor == 0:
if firstfactor in dd:
dd[firstfactor].append(j)
else:
dd[firstfactor] = [j]
break
# print(dd)
ks = sorted(dd.keys())
ans = [0] * n
print(len(ks))
for i in range(len(ks)):
for x in dd[ks[i]]:
ans[x] = i+1
print(*ans)
``` | output | 1 | 102,159 | 22 | 204,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,160 | 22 | 204,320 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
import math
for h in range(int(input())):
n = int(input())
arr = list(map(int, input().strip().split()))
primes = [2,3,5,7,11,13,17,19,23,29,31,37]
ans = [0 for i in range(n)]
col = 1
dicti = {}
for i in range(n):
for j in primes:
if arr[i]%j == 0:
if j in dicti:
ans[i] = dicti[j]
else:
ans[i] = col
dicti[j] = col
col += 1
break
print(max(ans))
print(*ans)
``` | output | 1 | 102,160 | 22 | 204,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,161 | 22 | 204,322 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
primes=[2,3,5,7,11,13,17,19,23,29,31]
t=int(input())
for i in range(t):
n=int(input())
ls=[int(a) for a in input().split()]
an=[]
for j in range(n):
an.append(0)
ctr=1
for p in range(len(primes)):
nx=False
for j in range(n):
if ls[j]%primes[p]==0 and an[j]==0:
an[j]=ctr
nx=True
if nx==True:
ctr+=1
s=''
for a in an:
s+=str(a)+' '
print(ctr-1)
print(s)
``` | output | 1 | 102,161 | 22 | 204,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,162 | 22 | 204,324 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
k=1
ans=[0]*n
for i in range(2,1001):
f=0
for j in range(n):
if l[j]%i==0 and ans[j]==0:
ans[j]=k
f=1
if f==1:
k+=1
print(k-1)
print(*ans)
``` | output | 1 | 102,162 | 22 | 204,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement. | instruction | 0 | 102,163 | 22 | 204,326 |
Tags: brute force, constructive algorithms, greedy, math, number theory
Correct Solution:
```
c = [0] * 1001
col = 1
for i in range(2, 1001):
if c[i] == 0:
for j in range(i * 2, 1001, i):
if c[j] == 0:
c[j] = col
col += 1
def tc():
n = int(input())
a = [int(x) for x in input().split()]
if n <= 11:
print(n)
print(' '.join(map(str, range(1, n + 1))))
return
ans = [c[x] for x in a]
key = {k: v for k, v in zip(set(ans), range(1, 12))}
ans = [key[x] for x in ans]
print(len(key))
print(' '.join(map(str, ans)))
################################
T = int(input())
for _ in range(T):
tc()
``` | output | 1 | 102,163 | 22 | 204,327 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
arr=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
for t in range(input()):
n=input()
l=map(int,raw_input().split())
ans=[0]*n
d=Counter()
for i in range(n):
for j in range(11):
if l[i]%arr[j]==0:
d[j]=1
ans[i]=j+1
break
d1=Counter()
c=0
for i in range(11):
d1[i]=c
if not d[i]:
c+=1
for i in range(n):
ans[i]-=d1[ans[i]-1]
pr_num(max(ans))
pr_arr(ans)
``` | instruction | 0 | 102,164 | 22 | 204,328 |
Yes | output | 1 | 102,164 | 22 | 204,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement.
Submitted Solution:
```
def answer(n,A):
dp=[1]*32
dp[0]=dp[1]=0
for i in range(2,32):
if dp[i]==1:
p=2*i
while p<=31:
if dp[p]==1:
dp[p]=0
p+=i
count=1
res=[0]*n
for i in range(2,32):
if dp[i]==1:
flag=0
for j in range(n):
if res[j]==0 and A[j]%i==0:
flag=1
res[j]=count
if flag==1:
count+=1
return count,res
t=int(input())
for i in range(t):
n=int(input())
arr=list(map(int,input().split()))
a,b=answer(n,arr)
print(a-1)
print(*b)
``` | instruction | 0 | 102,165 | 22 | 204,330 |
Yes | output | 1 | 102,165 | 22 | 204,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement.
Submitted Solution:
```
t=int(input())
sosu=[2,3,5,7,11,13,17,19,23,29,31]
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
ans=[]
ans2=0
for i in range(n):
for j in range(11):
if a[i]%sosu[j]==0:
ans.append(j+1)
ans2=max(j+1,ans2)
break
used=[1]*ans2
for x in ans:
used[x-1]=0
for i in range(ans2-1):
used[i+1]+=used[i]
for i in range(n):
ans[i]-=used[ans[i]-1]
print(ans2-used[-1])
print(*ans)
``` | instruction | 0 | 102,166 | 22 | 204,332 |
Yes | output | 1 | 102,166 | 22 | 204,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement.
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline
first11Primes=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
def getSmallestPrimeFactor(x):
for p in first11Primes:
if x%p==0:
return p
t=int(input())
for _ in range(t):
n=int(input())
a=[int(x) for x in input().split()]
ans=[-1 for __ in range(n)]
mapp=dict() #{prime:colour}
j=1
for i in range(n):
p=getSmallestPrimeFactor(a[i])
if p not in mapp.keys():
mapp[p]=j
j+=1
ans[i]=mapp[p]
print(j-1)
print(' '.join([str(x) for x in ans]))
``` | instruction | 0 | 102,167 | 22 | 204,334 |
Yes | output | 1 | 102,167 | 22 | 204,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement.
Submitted Solution:
```
stat=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547]
from collections import defaultdict
for _ in range(int(input())):
index=defaultdict(lambda:0)
N=int(input())
L=list(map(int,input().split()))
Color=1
for i in stat:
FLAG=0
for j in range(N):
if index[j]==0 and L[j]%i==0:
index[j]=Color
FLAG=1
if FLAG==1:
Color+=1
print(Color-1)
for i in range(N):
print(index[i],end=" ")
# print(index[i],end=" ")
print()
``` | instruction | 0 | 102,168 | 22 | 204,336 |
Yes | output | 1 | 102,168 | 22 | 204,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren't: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a_1,a_2,…,a_n.
She wants to choose an integer m ≤ 11 and color each element one of m colors from 1 to m so that:
* for each color from 1 to m there is at least one element of this color;
* each element is colored and colored exactly one color;
* the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. \gcd(a_i, a_j)>1 for each pair i, j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all a_i ≤ 1000 then she can always solve the task by choosing some m ≤ 11.
Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a_1,a_2,…,a_n (4 ≤ a_i ≤ 1000).
It is guaranteed that the sum of n over all test cases doesn't exceed 10^4.
Output
For each test case print 2 lines. The first line should contain a single integer m (1 ≤ m ≤ 11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ m), where c_i is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
Input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
Output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, \gcd(6,10)=2, \gcd(6,15)=3 and \gcd(10,15)=5. Therefore, it's valid to color all elements the same color. Note that there are other colorings which satisfy Alice's requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice's requirement.
Submitted Solution:
```
from math import ceil
def is_prime(x):
if x==2:
return True
for i in range(2,ceil(x**0.5)+1):
if x%i==0:
return False
return True
for t in range(int(input())):
n = int(input())
factors = {}
ans = [0]*n
arr = list(map(int,input().split()))
for i in range(n):
for j in range(2,ceil(arr[i]**0.5)+1):
if is_prime(j):
if arr[i]%j == 0:
try:
if type(factors[j])==list:
factors[j].append(i)
except:
factors[j] = [i]
# print(factors)
m = 1
done = {}
for i in range(n):
done[i] = False
for i in factors.keys():
for j in factors[i]:
if not done[j]:
ans[j] = m
done[j] = True
m+=1
print(m-1)
print(*ans)
``` | instruction | 0 | 102,169 | 22 | 204,338 |
No | output | 1 | 102,169 | 22 | 204,339 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.