message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543 | instruction | 0 | 43,626 | 22 | 87,252 |
"Correct Solution:
```
n = int(input())
a=[0]*(n+1)
for i in range(2,n+1):
x=i
for j in range(2,n+1):
while x%j==0:
a[j]+=1
x//=j
def num(n):
return len(list(filter(lambda x: x>=n-1,a)))
print(num(75) + num(25) * (num(3)-1) + num(15)*(num(5)-1) + num(5)*(num(5)-1)*(num(3)-2)//2)
``` | output | 1 | 43,626 | 22 | 87,253 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543 | instruction | 0 | 43,627 | 22 | 87,254 |
"Correct Solution:
```
n = int(input())
d = [0]* 101
for i in range(n+1):
i0 = i + 0
for j in range(2, i0+1):
while i % j == 0:
d[j] += 1
i = i //j
d.sort()
d1 = [i for i in d if i >=2]
def num(m): # e の要素のうち m-1 以上のものの個数
return len(list(filter(lambda x: x >= m-1, d1)))
print(num(75) + num(25) * (num(3) - 1) + num(15) * (num(5) - 1) + num(5) * (num(5) - 1) * (num(3) - 2) // 2)
``` | output | 1 | 43,627 | 22 | 87,255 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543 | instruction | 0 | 43,628 | 22 | 87,256 |
"Correct Solution:
```
from collections import defaultdict as ddict
d = ddict(int)
n = int(input())
for i in range(1,n+1):
for p in range(2,int(i**.5)+2):
while i%p == 0:
d[p] += 1
i//=p
if i>1:
d[i] += 1
g = [2,4,14,24,74]
f = [0]*5
for x in d.values():
for i in range(5):
if x >= g[i]:
f[i] += 1
ans = f[1]*(f[1]-1)//2*(f[0]-2) + f[3]*(f[0]-1) + f[2]*(f[1]-1) + f[4]
print(ans)
``` | output | 1 | 43,628 | 22 | 87,257 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543 | instruction | 0 | 43,629 | 22 | 87,258 |
"Correct Solution:
```
c=[0]*101
for i in range(1,int(input())+1):
for j in range(2,i+1):
while i%j==0:
c[j]+=1
i//=j
def f(n):
return sum(i>n-2 for i in c)
print(f(75)+f(25)*(f(3)-1)+f(15)*(f(5)-1)+f(5)*(f(5)-1)*(f(3)-2)//2)
``` | output | 1 | 43,629 | 22 | 87,259 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543 | instruction | 0 | 43,630 | 22 | 87,260 |
"Correct Solution:
```
from collections import defaultdict
d = defaultdict(int)
N = int(input())
for i in range(2, N+1):
for j in range(2, i+1):
while i % j == 0:
d[j] += 1
i = i//j
if i == 1:
break
# print(d)
c74, c24, c14, c4, c2 = 0, 0, 0, 0, 0
for v in d.values():
if v >= 74:
c74 += 1
if v >= 24:
c24 += 1
if v >= 14:
c14 += 1
if v >= 4:
c4 += 1
if v >= 2:
c2 += 1
ans = 0
ans += c74
ans += c24*(c2-1)
ans += c14*(c4-1)
ans += c4*(c4-1)//2*(c2-2)
print(ans)
``` | output | 1 | 43,630 | 22 | 87,261 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543 | instruction | 0 | 43,631 | 22 | 87,262 |
"Correct Solution:
```
N = int(input())
e = [0] * (N+1)
for i in range(2, N+1):
cur = i
for j in range(2, i+1):
while cur % j == 0:
e[j] += 1
cur //= j
def num(m): # eの要素のうち m-1 以上のものの個数
return len(list(filter(lambda x: x >= m-1, e)))
answer = num(75) + num(25) * (num(3)-1) + num(15) * (num(5) - 1)\
+ num(5) * (num(5) - 1) * (num(3) - 2) // 2
print(answer)
``` | output | 1 | 43,631 | 22 | 87,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
N = int(input())
d = {}
for n in range(2,N+1):
i = 2
while i <= n:
while n % i == 0:
n //= i
if i not in d:
d[i] = 0
d[i] += 1
i += 1
ans_d = {3:0,5:0,15:0,25:0,75:0}
for p in d:
for n in ans_d:
if d[p] >= n-1:
ans_d[n] += 1
ans = ans_d[5]*(ans_d[5]-1)*(ans_d[3]-2)//2+ans_d[75]
for n1 in ans_d:
for n2 in ans_d:
if n1 < n2 and n1*n2 == 75:
ans += ans_d[n2]*(ans_d[n1]-1)
print(ans)
``` | instruction | 0 | 43,632 | 22 | 87,264 |
Yes | output | 1 | 43,632 | 22 | 87,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
def f(p,N):
res=0
while N>=p:
N//=p
res+=N
return res
N=int(input())
ps=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]
A=[len([p for p in ps if f(p,N)>=k]) for k in [2,4,14,24,74]]
a,b,c,d,e=A[0],A[1],A[2],A[3],A[4]
ans=0
# 3*5^2
ans+=b*(b-1)//2*(a-2)
# 3*25
ans+=d*(a-1)
# 5*15
ans+=c*(b-1)
# 75
ans+=e
print(ans)
``` | instruction | 0 | 43,633 | 22 | 87,266 |
Yes | output | 1 | 43,633 | 22 | 87,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
import collections
n = int(input())
counts = collections.defaultdict(int)
for x in range(2, n + 1):
for y in range(2, n + 1):
while x % y == 0:
counts[y] += 1
x //= y
def f(a):
return sum(1 for b in counts if counts[b] >= a - 1)
answer = f(75) \
+ f(5) * (f(5) - 1) * (f(3) - 2) // 2 \
+ f(15) * (f(5) - 1) \
+ f(25) * (f(3) - 1)
print(answer)
``` | instruction | 0 | 43,634 | 22 | 87,268 |
Yes | output | 1 | 43,634 | 22 | 87,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
n=int(input())
l={}
if n<=9:
print(0)
else:
for i in range(2,n+1):
x=i
for j in range(2,i+1):
while x%j==0:
if j in l:
l[j]+=1
else:
l[j]=1
x=x//j
c=0
if l[2]>=74:
c+=1
c3=0
c5=0
c25=0
c15=0
for i in l:
if l[i]>=2:
c3+=1
if l[i]>=4:
c5+=1
if l[i]>=14:
c15+=1
if l[i]>=24:
c25+=1
print(c5*(c5-1)*(c3-2)//2+c25*(c3-1)+c15*(c5-1)+c)
``` | instruction | 0 | 43,635 | 22 | 87,270 |
Yes | output | 1 | 43,635 | 22 | 87,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
N = int(input())
M = [0] * (101)
def prime_factorize(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
for i in range(2, N + 1):
for m in prime_factorize(i):
M[m] += 1
print(M[:N+1])
n3, n5, n15, n25, n75 = 0, 0, 0, 0, 0
for i in range(N+1):
if M[i] >= 2: n3 += 1
if M[i] >= 4: n5 += 1
if M[i] >= 14: n15 += 1
if M[i] >= 24: n25 += 1
if M[i] >= 74: n75 += 1
#print(n3, n5, n15, n25, n75)
answer = n75 + n25 * (n3 - 1) + n15 * (n5 - 1) + n5 * (n5 - 1) * (n3 - 2) // 2
print(answer)
``` | instruction | 0 | 43,636 | 22 | 87,272 |
No | output | 1 | 43,636 | 22 | 87,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
N = int(input())
elem = [0 for _ in range(N+1)]
for i in range(1, N+1):
cur = i
for j in range(2, i+1):
while cur % j == 0:
elem[j] += 1
cur //= j
def num(m):
return len(list(filter(lambda x: x >= m-1, elem)))
print(num(75) + num(25)*(num(3)-1) + num(15)*(num(3)-1) + num(5)*(num(5)-1)*(num(3)-2) // 2)
``` | instruction | 0 | 43,637 | 22 | 87,274 |
No | output | 1 | 43,637 | 22 | 87,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
N=int(input())
def factoring(M,n=2):
for i in range(n,int(M**0.5)+1):
if M%i==0:
ret = factoring(M//i,i)
ret[i] = ret.get(i,0)+1
return ret
return {M:1}
p = {}
for i in range(2,N+1):
for k,v in factoring(i).items():
p[k] = p.get(k,0)+v
c = [0]*4
for n in p.values():
if n>=24:c[3]+=1
elif n>=14:c[2]+=1
elif n>=4:c[1]+=1
elif n>=2:c[0]+=1
ans = 0
ans += p[2]>=74
ans += c[3]*(sum(c)-1)
c[2] += c[3]
ans += c[2]*(c[1]+c[2]-1)
c[1] +=c[2]
ans += c[0]*c[1]*(c[1]-1)//2
ans += c[1]*(c[1]-1)*(c[1]-2)//2
print(ans)
``` | instruction | 0 | 43,638 | 22 | 87,276 |
No | output | 1 | 43,638 | 22 | 87,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?
Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.
Constraints
* 1 \leq N \leq 100
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of the Shichi-Go numbers that are divisors of N!.
Examples
Input
9
Output
0
Input
10
Output
1
Input
100
Output
543
Submitted Solution:
```
n = int(input())
def prime_factorize(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
li = []
for i in range(1,n+1):
li += prime_factorize(i)
from collections import Counter
c = Counter(li)
kumi = [0]*101
for i,j in c.items():
kumi[j] += 1
point = 0
#553
point += kumi[2]*(sum(kumi[4:])*(sum(kumi[4:])-1)//2)
#15 5
point += kumi[4]*sum(kumi[14:])
#25 3
point += kumi[3]*sum(kumi[24:])
#75
point += kumi[74]
print(point)
``` | instruction | 0 | 43,639 | 22 | 87,278 |
No | output | 1 | 43,639 | 22 | 87,279 |
Provide a correct Python 3 solution for this coding contest problem.
C: Divisor Game
problem
tsutaj is trying to play about a few games.
In a divisor game, a natural number N of 2 or more is given first, and then the game progresses according to the following procedure.
* Declare one integer from the divisors of N other than N. However, at this time, it is not possible to declare a divisor of an integer that has already been declared.
* Repeat this as long as there is an integer that can be declared, and if there is nothing that can be declared, the game ends.
Find the minimum and maximum number of possible declarations before the end of the game.
Input format
Input is given on one line.
N
Constraint
* 2 \ leq N \ leq 10 ^ {12}
Output format
Output the minimum and maximum number of declarations on one line separated by spaces.
Input example 1
18
Output example 1
twenty five
The following is an example of setting the number of declarations to 2.
* Declare 9.
* Declare 6. (6 is not a divisor of 9, so it can be declared)
If you do this, any integer that is a divisor of 18 and not 18 will be a divisor of the integers you have declared, and the game will end.
Note that you cannot declare anything that is a divisor of an integer that you have already declared. For example, you cannot declare 3 after declaring 9. Because 3 is a divisor of 9.
Input example 2
99
Output example 2
twenty five
Input example 3
10000000019
Output example 3
1 1
The input may not fit in a 32-bit integer type.
Example
Input
18
Output
2 5 | instruction | 0 | 43,770 | 22 | 87,540 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
'''
------------------------
author : iiou16
------------------------
'''
import copy
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
divisors.sort()
return divisors
def main():
N = int(input())
yakusu = make_divisors(N)
max_times = len(yakusu) - 1
min_times = 0
result = []
for i in reversed(yakusu[:-1]):
if i in result:
continue
result.extend(make_divisors(i))
# print(result)
min_times += 1
print(min_times, max_times)
if __name__ == '__main__':
main()
``` | output | 1 | 43,770 | 22 | 87,541 |
Provide a correct Python 3 solution for this coding contest problem.
C: Divisor Game
problem
tsutaj is trying to play about a few games.
In a divisor game, a natural number N of 2 or more is given first, and then the game progresses according to the following procedure.
* Declare one integer from the divisors of N other than N. However, at this time, it is not possible to declare a divisor of an integer that has already been declared.
* Repeat this as long as there is an integer that can be declared, and if there is nothing that can be declared, the game ends.
Find the minimum and maximum number of possible declarations before the end of the game.
Input format
Input is given on one line.
N
Constraint
* 2 \ leq N \ leq 10 ^ {12}
Output format
Output the minimum and maximum number of declarations on one line separated by spaces.
Input example 1
18
Output example 1
twenty five
The following is an example of setting the number of declarations to 2.
* Declare 9.
* Declare 6. (6 is not a divisor of 9, so it can be declared)
If you do this, any integer that is a divisor of 18 and not 18 will be a divisor of the integers you have declared, and the game will end.
Note that you cannot declare anything that is a divisor of an integer that you have already declared. For example, you cannot declare 3 after declaring 9. Because 3 is a divisor of 9.
Input example 2
99
Output example 2
twenty five
Input example 3
10000000019
Output example 3
1 1
The input may not fit in a 32-bit integer type.
Example
Input
18
Output
2 5 | instruction | 0 | 43,771 | 22 | 87,542 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = SR()
return l
mod = 1000000007
#A
"""
n,m = LI()
a = LI()
b = LI()
li = list(set(a)|set(b))
li2 = list(set(a)&set(b))
li.sort()
li2.sort()
print(len(li2),len(li))
for i in li2:
print(i)
for i in li:
print(i)
"""
#B
"""
n = I()
p = LI()
p.append(float("inf"))
ans = ""
i = 0
l = 0
k = [i+1 for i in range(n)]
while i < n:
if p[i] < p[i+1]:
a = p[l:i+1][::-1]
if a != k[l:i+1]:
print(":(")
quit()
l = i+1
for d in range(len(a)):
ans += "("
for d in range(len(a)):
ans += ")"
i += 1
print(ans)
"""
#C
def fact(n):
i = 2
a = n
if n < 4:
return [1,n],[n]
li = [1,n]
while i**2 <= a:
if n%i == 0:
li.append(i)
if i != n//i:
li.append(n//i)
i += 1
li.sort()
i = 2
if len(li) == 2:
return li,[a]
k = []
b = a
while i**2 <= b:
if a%i == 0:
k.append(i)
while a%i == 0:
a//= i
i += 1
if a!=1:
k.append(a)
return li,k
n = I()
l,k = fact(n)
if len(l) == 2:
print(1,1)
else:
print(len(k),len(l)-1)
#D
#E
#F
#G
#H
#I
#J
#K
#L
#M
#N
#O
#P
#Q
#R
#S
#T
``` | output | 1 | 43,771 | 22 | 87,543 |
Provide a correct Python 3 solution for this coding contest problem.
C: Divisor Game
problem
tsutaj is trying to play about a few games.
In a divisor game, a natural number N of 2 or more is given first, and then the game progresses according to the following procedure.
* Declare one integer from the divisors of N other than N. However, at this time, it is not possible to declare a divisor of an integer that has already been declared.
* Repeat this as long as there is an integer that can be declared, and if there is nothing that can be declared, the game ends.
Find the minimum and maximum number of possible declarations before the end of the game.
Input format
Input is given on one line.
N
Constraint
* 2 \ leq N \ leq 10 ^ {12}
Output format
Output the minimum and maximum number of declarations on one line separated by spaces.
Input example 1
18
Output example 1
twenty five
The following is an example of setting the number of declarations to 2.
* Declare 9.
* Declare 6. (6 is not a divisor of 9, so it can be declared)
If you do this, any integer that is a divisor of 18 and not 18 will be a divisor of the integers you have declared, and the game will end.
Note that you cannot declare anything that is a divisor of an integer that you have already declared. For example, you cannot declare 3 after declaring 9. Because 3 is a divisor of 9.
Input example 2
99
Output example 2
twenty five
Input example 3
10000000019
Output example 3
1 1
The input may not fit in a 32-bit integer type.
Example
Input
18
Output
2 5 | instruction | 0 | 43,772 | 22 | 87,544 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(int(1e7))
from collections import deque
def inpl(): return list(map(int, input().split()))
def primes(N):
sieve = [1]*(N+1)
sieve[0], sieve[1] = 0, 0
P = []
for i in range(2, N+1):
if sieve[i]:
P.append(i)
for j in range(2*i, N+1, i):
sieve[j] = 0
return P
N = int(input())
P = primes(10**6 + 10)
factors = [0]*len(P)
for i in range(len(P)):
p = P[i]
while N%p == 0:
factors[i] += 1
N = N//p
factors += [N!=1]
a = sum([f > 0 for f in factors])
b = 1
for i in range(len(P)+1):
b *= factors[i] + 1
b -= 1
print(a, b)
``` | output | 1 | 43,772 | 22 | 87,545 |
Provide a correct Python 3 solution for this coding contest problem.
C: Divisor Game
problem
tsutaj is trying to play about a few games.
In a divisor game, a natural number N of 2 or more is given first, and then the game progresses according to the following procedure.
* Declare one integer from the divisors of N other than N. However, at this time, it is not possible to declare a divisor of an integer that has already been declared.
* Repeat this as long as there is an integer that can be declared, and if there is nothing that can be declared, the game ends.
Find the minimum and maximum number of possible declarations before the end of the game.
Input format
Input is given on one line.
N
Constraint
* 2 \ leq N \ leq 10 ^ {12}
Output format
Output the minimum and maximum number of declarations on one line separated by spaces.
Input example 1
18
Output example 1
twenty five
The following is an example of setting the number of declarations to 2.
* Declare 9.
* Declare 6. (6 is not a divisor of 9, so it can be declared)
If you do this, any integer that is a divisor of 18 and not 18 will be a divisor of the integers you have declared, and the game will end.
Note that you cannot declare anything that is a divisor of an integer that you have already declared. For example, you cannot declare 3 after declaring 9. Because 3 is a divisor of 9.
Input example 2
99
Output example 2
twenty five
Input example 3
10000000019
Output example 3
1 1
The input may not fit in a 32-bit integer type.
Example
Input
18
Output
2 5 | instruction | 0 | 43,773 | 22 | 87,546 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 5 + 10)
def input(): return sys.stdin.readline().strip()
def resolve():
n=int(input())
import math
# cf. https://qiita.com/suecharo/items/14137fb74c26e2388f1f
def make_prime_list_2(num):
if num < 2:
return []
# 0のものは素数じゃないとする
prime_list = [i for i in range(num + 1)]
prime_list[1] = 0 # 1は素数ではない
num_sqrt = math.sqrt(num)
for prime in prime_list:
if prime == 0:
continue
if prime > num_sqrt:
break
for non_prime in range(2 * prime, num, prime):
prime_list[non_prime] = 0
return [prime for prime in prime_list if prime != 0]
def prime_factorization_2(num):
"""numの素因数分解
素因数をkeyに乗数をvalueに格納した辞書型dict_counterを返す"""
if num <= 1:
# 例えば1を食ったときの対処の仕方は問題によって違うと思うのでそのつど考える。
# cf. https://atcoder.jp/contests/abc110/submissions/12688244
return False
else:
num_sqrt = math.floor(math.sqrt(num))
prime_list = make_prime_list_2(num_sqrt)
dict_counter = {} # 何度もこの関数を呼び出して辞書を更新したい時はこれを引数にして
# cf. https://atcoder.jp/contests/arc034/submissions/12251452
for prime in prime_list:
while num % prime == 0:
if prime in dict_counter:
dict_counter[prime] += 1
else:
dict_counter[prime] = 1
num //= prime
if num != 1:
if num in dict_counter:
dict_counter[num] += 1
else:
dict_counter[num] = 1
return dict_counter
d=prime_factorization_2(n)
val=1
for v in d.values():
val*=(v+1)
print(len(d),val-1)
resolve()
``` | output | 1 | 43,773 | 22 | 87,547 |
Provide a correct Python 3 solution for this coding contest problem.
C: Divisor Game
problem
tsutaj is trying to play about a few games.
In a divisor game, a natural number N of 2 or more is given first, and then the game progresses according to the following procedure.
* Declare one integer from the divisors of N other than N. However, at this time, it is not possible to declare a divisor of an integer that has already been declared.
* Repeat this as long as there is an integer that can be declared, and if there is nothing that can be declared, the game ends.
Find the minimum and maximum number of possible declarations before the end of the game.
Input format
Input is given on one line.
N
Constraint
* 2 \ leq N \ leq 10 ^ {12}
Output format
Output the minimum and maximum number of declarations on one line separated by spaces.
Input example 1
18
Output example 1
twenty five
The following is an example of setting the number of declarations to 2.
* Declare 9.
* Declare 6. (6 is not a divisor of 9, so it can be declared)
If you do this, any integer that is a divisor of 18 and not 18 will be a divisor of the integers you have declared, and the game will end.
Note that you cannot declare anything that is a divisor of an integer that you have already declared. For example, you cannot declare 3 after declaring 9. Because 3 is a divisor of 9.
Input example 2
99
Output example 2
twenty five
Input example 3
10000000019
Output example 3
1 1
The input may not fit in a 32-bit integer type.
Example
Input
18
Output
2 5 | instruction | 0 | 43,774 | 22 | 87,548 |
"Correct Solution:
```
import math
n = int(input())
p = list()
m = int(math.sqrt(n)+1)
for i in range(2,m):
if n%i == 0:
c = 0
while n%i == 0:
c += 1
n //= i
p.append(c)
if n != 1:
p.append(1)
maxi = 1
for x in p:
maxi *= (x+1)
print(len(p),maxi-1)
``` | output | 1 | 43,774 | 22 | 87,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56 | instruction | 0 | 44,305 | 22 | 88,610 |
Tags: brute force, constructive algorithms, math, number theory
Correct Solution:
```
x=int(input())
s=set()
s.add(x)
s.add(x+1)
s.add(x*(x+1))
if len(s)!=3:
print(-1)
else:
print(*list(s))
``` | output | 1 | 44,305 | 22 | 88,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56 | instruction | 0 | 44,306 | 22 | 88,612 |
Tags: brute force, constructive algorithms, math, number theory
Correct Solution:
```
n =int(input())
a = n
b = n+1
c = n*(n+1)
if(2*a*b*c==n*(a*b+a*c+b*c) and a!=b and b!=c and a!=c):
print(n,n+1,n*(n+1))
else:
print(-1)
``` | output | 1 | 44,306 | 22 | 88,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56 | instruction | 0 | 44,307 | 22 | 88,614 |
Tags: brute force, constructive algorithms, math, number theory
Correct Solution:
```
n = int(input())
if n == 1:
print(-1)
elif n == 2:
print(2, 3, 6)
else:
for i in range(1, n + 1):
if n % i == 0:
print(n, n + 1, n * n + n)
exit()
``` | output | 1 | 44,307 | 22 | 88,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56 | instruction | 0 | 44,308 | 22 | 88,616 |
Tags: brute force, constructive algorithms, math, number theory
Correct Solution:
```
n=int(input())
if n==1:
print(-1)
else:
print(n,end=" ")
print(n+1,end=" ")
print(n*(n+1))
``` | output | 1 | 44,308 | 22 | 88,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56 | instruction | 0 | 44,309 | 22 | 88,618 |
Tags: brute force, constructive algorithms, math, number theory
Correct Solution:
```
n = int(input())
if n == 1:
print(-1)
else:
print(n, n + 1, (n + 1) * n)
``` | output | 1 | 44,309 | 22 | 88,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56 | instruction | 0 | 44,310 | 22 | 88,620 |
Tags: brute force, constructive algorithms, math, number theory
Correct Solution:
```
n = int(input())
if n == 1:
print('-1')
else:
print(n, n + 1, n * n + n)
``` | output | 1 | 44,310 | 22 | 88,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56 | instruction | 0 | 44,311 | 22 | 88,622 |
Tags: brute force, constructive algorithms, math, number theory
Correct Solution:
```
n = int(input())
print('%d %d %d' % (n, n + 1, n * (n + 1)) if n != 1 else -1)
``` | output | 1 | 44,311 | 22 | 88,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56
Submitted Solution:
```
# import itertools
# import bisect
# import math
from collections import defaultdict, Counter
import os
import sys
from io import BytesIO, IOBase
# sys.setrecursionlimit(10 ** 5)
ii = lambda: int(input())
lmii = lambda: list(map(int, input().split()))
slmii = lambda: sorted(map(int, input().split()))
li = lambda: list(input())
mii = lambda: map(int, input().split())
msi = lambda: map(str, input().split())
def gcd(a, b):
if b == 0: return a
return gcd(b, a % b)
def lcm(a, b): return (a * b) // gcd(a, b)
def main():
# for _ in " " * int(input()):
n=ii()
print(*[n,n+1,n*(n+1)] if n-1 else [-1])
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 44,316 | 22 | 88,632 |
Yes | output | 1 | 44,316 | 22 | 88,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56
Submitted Solution:
```
n = int(input())
dividers = 0
for i in range(2, n):
if ((n + 1) % i == 0):
dividers += 1
if (dividers == 0):
print(-1)
else:
print(n, n + 1, (n + 1) * n)
``` | instruction | 0 | 44,318 | 22 | 88,636 |
No | output | 1 | 44,318 | 22 | 88,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
Input
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
Input
3
Output
2 7 42
Input
7
Output
7 8 56
Submitted Solution:
```
import math
n = int(input())
ok = False
if n == 1:
print(-1)
else:
for i in range(2, n+1):
for j in range(n+1, 2 + (2*n)):
lcm = math.gcd(math.gcd(i, j), n)*(i*j*n)/(math.gcd(i, j)*math.gcd(i, n)*math.gcd(j, n))
# print(i, j, lcm)
if (1/i) + (1/j) + (1/lcm) == (2/n):
print(i, j, int(lcm))
ok = True
break
if ok:
break
if not ok:
print(-1)
``` | instruction | 0 | 44,320 | 22 | 88,640 |
No | output | 1 | 44,320 | 22 | 88,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n, we denote the integer obtained by reversing the decimal notation of n (without leading zeroes) by rev(n). For example, rev(123) = 321 and rev(4000) = 4.
You are given a positive integer D. How many positive integers N satisfy rev(N) = N + D?
Constraints
* D is an integer.
* 1 ≤ D < 10^9
Input
Input is given from Standard Input in the following format:
D
Output
Print the number of the positive integers N such that rev(N) = N + D.
Examples
Input
63
Output
2
Input
75
Output
0
Input
864197532
Output
1920
Submitted Solution:
```
"""
https://atcoder.jp/contests/arc075/tasks/arc075_d
桁数を決め打ちする？
abcd
-dcba
ABCD
999a+90b-90c-999d = ABCD
みたいな感じになる
これをいい感じに満たせばおk
aとd,bとcをまとめる
-9～9の19通り
19**4 = 2476099通り
aとdの分配の仕方の通り数をあらかじめ数えておけばおｋ
ただし、a != 0
差は、9999109999
9999019999
差があるとDを超える場合は探索を打ち切ればおｋ
"""
import sys
def dfs(dig,nd,nsum):
P = 10**((dig-1)-nd) - 10**nd #この桁の寄与の基準
maxc = 10**((dig-1)-nd)
if nd == dig//2:
if nsum == D:
if dig % 2 == 0:
return 1
else:
return 10
return 0
elif nd != 0:
ret = 0
for i in range(-9,10):
if nsum+P*i < D-maxc or nsum+P*i > D+maxc:
continue
ret += dfs(dig,nd+1,nsum+P*i) * sums[i]
#print (ret,"y")
return ret
else:
ret = 0
for i in range(-9,10):
if nsum+P*i < D-maxc or nsum+P*i > D+maxc:
continue
tmp = dfs(dig,nd+1,nsum+P*i) * sums_z[i]
#if tmp > 0:
# print (dig,nd+1,i,"is 1",tmp)
ret += tmp
return ret
D = int(input())
ans = 0
sums = [0] * 30
for i in range(-9,10):
for x in range(-9,1):
for y in range(0,10):
if x+y == i:
sums[i] += 1
print (sums , file=sys.stderr)
sums_z = [0] * 30
for i in range(-9,10):
for x in range(-9,0):
for y in range(1,10):
if x+y == i:
sums_z[i] += 1
print (sums_z , file=sys.stderr)
for dig in range(1,20):
now = dfs(dig,0,0)
#print (dig,now)
ans += now
print (ans)
``` | instruction | 0 | 44,535 | 22 | 89,070 |
Yes | output | 1 | 44,535 | 22 | 89,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,709 | 22 | 89,418 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
l=list(map(int,input().split()))
r,c=l[0],l[1]
if(r==1 and c==1):
print(0)
elif r<=c:
rs=[i for i in range(1,r+1)]
cs=[i for i in range(r+1,r+c+1)]
k=[[0 for i in range(c)] for j in range(r)]
#print(rs,cs)
for i in range(r):
for j in range(c):
k[i][j]=rs[i]*cs[j]
for i in range(r):
for j in range(c):
print(k[i][j], end=" ")
print()
else:
cs=[i for i in range(1,c+1)]
rs=[i for i in range(c+1,r+c+1)]
k=[[0 for i in range(c)] for j in range(r)]
for i in range(r):
for j in range(c):
k[i][j]=rs[i]*cs[j]
for i in range(r):
for j in range(c):
print(k[i][j], end=" ")
print()
``` | output | 1 | 44,709 | 22 | 89,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,710 | 22 | 89,420 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def power_set(L):
"""
L is a list.
The function returns the power set, but as a list of lists.
"""
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
#the function could stop here closing with
#return powerset
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
def sortByFirstAndSecond(A):
A = sorted(A,key = lambda x:x[0])
A = sorted(A,key = lambda x:x[1])
return list(A)
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = 1
# t = int(input())
for _ in range(t):
r,c = li()
if r==1 and c==1:
print(0)
continue
if c==1:
for i in range(2,r+2):
print(i)
continue
l = [[0 for i in range(c)] for j in range(r)]
for i in range(r):
for j in range(c):
l[i][j] = (i+1)*(j+r+1)
for i in l:
print(*i)
``` | output | 1 | 44,710 | 22 | 89,421 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,711 | 22 | 89,422 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
n, m = map(int, input().split())
if n == 1 and m == 1:
print(0)
else:
ans = []
if m == 1:
ans = [[q] for q in range(2, n+2)]
else:
ans = [[q for q in range(2, m+2)]]
for q in range(2, n+1):
ans.append([])
for q1 in range(m):
ans[-1].append(ans[0][q1]*(m+q))
for q in ans:
print(*q)
``` | output | 1 | 44,711 | 22 | 89,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,712 | 22 | 89,424 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
h,m=list(map(int,input().split()))
a=[[0]*m for i in range(h)]
if h==min(h,m):
for i in range(h):
a[i][0]=(i+1)*2
for i in range(1,m):
if 2*i<=a[h-1][0]:
a[0][i]=2*i+1
else:
a[0][i]=a[0][i-1]+1
else:
for i in range(m):
a[0][i]=(i+1)*2
for i in range(1,h):
if 2*i<=a[0][m-1]:
a[i][0]=2*i+1
else:
a[i][0]=a[i-1][0]+1
for i in range(1,h):
for j in range(1,m):
a[i][j]=a[i][0]*a[0][j]
if h==1 and m==1:
print(0)
else:
for i in range(h):
print(*a[i])
``` | output | 1 | 44,712 | 22 | 89,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,713 | 22 | 89,426 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
import bisect
from math import gcd,ceil
from heapq import heapify,heappop,heappush
from io import BytesIO, IOBase
def main():
r,c = map(int,input().split())
n = r+c
nums = [i for i in range(c+1,r+c+1)][::-1]
# print(nums)
if r<=1 and c<=1:
print(0)
return
elif r==1:
for i in range(1,r+c):
print(i+1,end=' ')
print()
else:
a = [[0]*c for i in range(r)]
for i in range(r):
for j in range(c):
if j==0:
a[i][j] = nums[i]
else:
a[i][j] = (j+1)*a[i][0]
print(a[i][j],end=' ')
print()
# print(a)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = lambda s: self.buffer.write(s.encode()) if self.writable else None
def read(self):
if self.buffer.tell():
return self.buffer.read().decode("ascii")
return os.read(self._fd, os.fstat(self._fd).st_size).decode("ascii")
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().decode("ascii")
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
def print(*args, sep=" ", end="\n", file=sys.stdout, flush=False):
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(end)
if flush:
file.flush()
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
sys.setrecursionlimit(10000)
if __name__ == "__main__":
main()
``` | output | 1 | 44,713 | 22 | 89,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,714 | 22 | 89,428 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
#設定
import sys
input = sys.stdin.buffer.readline
#ライブラリインポート
from collections import defaultdict
#入力受け取り
def getlist():
return list(map(int, input().split()))
#処理内容
def main():
R, C = getlist()
if R == 1 and C == 1:
print(0)
return
if C == 1:
for i in range(R):
print(2 + i)
return
L = [[0] * C for i in range(R)]
s = R + 1
for i in range(C):
L[0][i] = s + i
for j in range(C):
s = L[0][j]
for i in range(R):
L[i][j] = s * (i + 1)
for i in range(R):
print(" ".join(list(map(str, L[i]))))
if __name__ == '__main__':
main()
``` | output | 1 | 44,714 | 22 | 89,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,715 | 22 | 89,430 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
n,m=map(int,input().split())
if(n==m==1):
print(0)
elif(m==1):
for j in range(2,n+2):
print(j)
else:
for i in range(1,n+1):
for j in range(m):
print((n+1)*i+j*i,end=' ')
print()
``` | output | 1 | 44,715 | 22 | 89,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices. | instruction | 0 | 44,716 | 22 | 89,432 |
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
def gcd(a, b):
while b:
a, b = b, a % b
return a
def construct_matrix(columns, rows):
# print(columns, rows)
for r in rows:
for c in columns:
print(r*c//gcd(c,r), end=' ')
print()
def main(r,c):
if r==c==1:
print(0)
return
a=range(1,c+r+1)
rows,columns=(a[:r],a[r:]) if r<=c else (a[c:],a[:c])
construct_matrix(columns, rows)
if __name__ == "__main__":
r,c=map(int, input().split())
main(r,c)
``` | output | 1 | 44,716 | 22 | 89,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
origr, origc = map(int, input().split())
r, c = sorted([origr, origc])
if r == c == 1:
print(0)
else:
rows = range(1, r + 1)
cols = range(r + 1, c + r + 1)
if origr <= origc:
for i in rows:
print(*[i * j for j in cols])
else:
for j in cols:
print(*[i * j for i in rows])
``` | instruction | 0 | 44,717 | 22 | 89,434 |
Yes | output | 1 | 44,717 | 22 | 89,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
r, c = map(int, input().split())
if r+c == 2:
print(0)
elif c==1:
for i in range(2, r+2):
print(i)
else:
for i in range(1, r+1):
for j in range(r+1, r+c+1):
print(i*j, end=' ')
print()
``` | instruction | 0 | 44,718 | 22 | 89,436 |
Yes | output | 1 | 44,718 | 22 | 89,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
from sys import stdin, stdout
import math
from bisect import bisect_left, bisect_right
input = lambda: stdin.readline().strip()
print = lambda s: stdout.write(s)
def lcm(a, b):
return (a*b)//(math.gcd(a, b))
r, c = map(int, input().split())
b = []
for i in range(1, r+c+1):
b.append(i)
mat = []
for i in range(r):
mat.append([])
for j in range(c):
mat[-1].append(0)
for i in range(r):
for j in range(c):
mat[i][j] = lcm(b[i], b[r+j])
if r==1 and c==1:
print('0\n')
elif c==1:
for i in range(r):
print(str(2+i)+'\n')
else:
for i in range(r):
for j in range(c):
print(str(mat[i][j])+' ')
print('\n')
``` | instruction | 0 | 44,719 | 22 | 89,438 |
Yes | output | 1 | 44,719 | 22 | 89,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
from queue import Queue
import itertools
import bisect
import heapq
#sys.setrecursionlimit(100000)
#^^^TAKE CARE FOR MEMORY LIMIT^^^
def main():
pass
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")
def binary(n):
return (bin(n).replace("0b", ""))
def decimal(s):
return (int(s, 2))
def pow2(n):
p = 0
while (n > 1):
n //= 2
p += 1
return (p)
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
l.append(i)
n = n / i
if n > 2:
l.append(int(n))
return (l)
def isPrime(n):
if (n == 1):
return (False)
else:
root = int(n ** 0.5)
root += 1
for i in range(2, root):
if (n % i == 0):
return (False)
return (True)
def maxPrimeFactors(n):
maxPrime = -1
while n % 2 == 0:
maxPrime = 2
n >>= 1
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
maxPrime = i
n = n / i
if n > 2:
maxPrime = n
return int(maxPrime)
def countcon(s, i):
c = 0
ch = s[i]
for i in range(i, len(s)):
if (s[i] == ch):
c += 1
else:
break
return (c)
def lis(arr):
n = len(arr)
lis = [1] * n
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j] and lis[i] < lis[j] + 1:
lis[i] = lis[j] + 1
maximum = 0
for i in range(n):
maximum = max(maximum, lis[i])
return maximum
def isSubSequence(str1, str2):
m = len(str1)
n = len(str2)
j = 0
i = 0
while j < m and i < n:
if str1[j] == str2[i]:
j = j + 1
i = i + 1
return j == m
def maxfac(n):
root = int(n ** 0.5)
for i in range(2, root + 1):
if (n % i == 0):
return (n // i)
return (n)
def p2(n):
c=0
while(n%2==0):
n//=2
c+=1
return c
def seive(n):
primes=[True]*(n+1)
primes[1]=primes[0]=False
for i in range(2,n+1):
if(primes[i]):
for j in range(i+i,n+1,i):
primes[j]=False
p=[]
for i in range(0,n+1):
if(primes[i]):
p.append(i)
return(p)
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def denofactinverse(n,m):
fac=1
for i in range(1,n+1):
fac=(fac*i)%m
return (pow(fac,m-2,m))
def numofact(n,m):
fac=1
for i in range(1,n+1):
fac=(fac*i)%m
return(fac)
def sod(n):
s=0
while(n>0):
s+=n%10
n//=10
return s
n,m=map(int,input().split())
if(n==1 and m==1):
print(0)
else:
if(n==1):
for i in range(0,m):
print(i+2,end=" ")
elif(m==1):
for i in range(0,n):
print(i+2)
else:
ans=[[] for i in range(0,n)]
for i in range(0,m):
ans[0].append(i+2)
mul=ans[0][-1]
for r in range(1,n):
tbm=mul+r
for c in range(0,m):
ans[r].append(ans[0][c]*tbm)
for i in ans:
print(*i)
``` | instruction | 0 | 44,720 | 22 | 89,440 |
Yes | output | 1 | 44,720 | 22 | 89,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
read=input().split()
n=int(read[0])
m=int(read[1])
x=m+n
if(n==1 or m==1):
print(0)
else:
for i in range(0,n):
for j in range (0,m):
print(x*(j+1),end=" ")
print('')
x=x-1
``` | instruction | 0 | 44,721 | 22 | 89,442 |
No | output | 1 | 44,721 | 22 | 89,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
r, c = [int(p) for p in input().split()]
if r == 1 and c == 1:
print(0)
exit()
if 2*(r+1) <= r+c:
print(0)
else:
matrix = []
for i in range(1, r+1):
row = []
for j in range(1, c+1):
row.append(i*(r+j))
matrix.append(row)
for i in matrix:
print(*i)
``` | instruction | 0 | 44,722 | 22 | 89,444 |
No | output | 1 | 44,722 | 22 | 89,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
r, c = [int(p) for p in input().split()]
if r == 1 and c == 1:
print(0)
exit()
if 2*(r+1) <= r+c:
print(0)
else:
matrix = []
for i in range(1, r+1):
row = []
for j in range(1, c+1):
row.append(i*(r+j))
matrix.append(row)
print(matrix)
``` | instruction | 0 | 44,723 | 22 | 89,446 |
No | output | 1 | 44,723 | 22 | 89,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column.
We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct.
The magnitude of a matrix equals to the maximum of b_k.
For example, suppose we have the following matrix:
\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix}
We construct the array b:
1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1;
2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4;
3. b_3 is the greatest common divisor of 2 and 4, that is 2;
4. b_4 is the greatest common divisor of 9 and 144, that is 9;
5. b_5 is the greatest common divisor of 7 and 84, that is 7.
So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.
For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0.
Input
The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.
Output
If there is no solution, output a single integer 0.
Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.
Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).
Examples
Input
2 2
Output
4 12
2 9
Input
1 1
Output
0
Note
In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.
In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.
Submitted Solution:
```
import sys
from collections import Counter
import math
def array_int():
return [int(i) for i in sys.stdin.readline().split()]
def vary(arrber_of_variables):
if arrber_of_variables==1:
return int(sys.stdin.readline())
if arrber_of_variables>=2:
return map(int,sys.stdin.readline().split())
def makedict(var):
return dict(Counter(var))
# i am noob wanted to be better and trying hard for that
def printDivisors(n):
divisors=[]
# Note that this loop runs till square root
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
# If divisors are equal, print only one
if (n//i == i) :
divisors.append(i)
else :
# Otherwise print both
divisors.extend((i,n//i))
i = i + 1
return divisors
def countTotalBits(num):
binary = bin(num)[2:]
return(len(binary))
def isPrime(n):
# Corner cases
if (n <= 1) :
return False
if (n <= 3) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
mod=10**9+7
# def ncr(n,r):
# if n<r:
# return 0
# if n==r:
# return 1
# numer=fact[n]
# # print(numer)
# denm=(fact[n-r]*fact[r])
# # print(denm)
# return numer*pow(denm,mod-2,mod)
# def dfs(node):
# global graph,m,cats,count,visited,val
# # print(val)
# visited[node]=1
# if cats[node]==1:
# val+=1
# # print(val)
# for i in graph[node]:
# if visited[i]==0:
# z=dfs(i)
# # print(z,i)
# count+=z
# val-=1
# return 0
# else:
# return 1
# fact=[1]*(1001)
# c=1
# mod=10**9+7
# for i in range(1,1001):
# print(fact)
def comp(x):
# fact[i]=(fact[i-1]*i)%mod
return x[1]
def SieveOfEratosthenes(n):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
# Print all prime numbers
for p in range(2, n+1):
if prime[p]:
primes.append(p)
primes=[]
# SieveOfEratosthenes(2*(10**6))
def lcm(a,b):
return (a*b)//math.gcd(a,b)
def primeFactors(n):
factors=[]
# Print the number of two's that divide n
while n % 2 == 0:
factors.append(2)
n = n // 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
factors.append(i)
n = n // i
# Condition if n is a prime
# number greater than 2
if n > 2:
factors.append(n)
return factors
def transpose(l1, l2):
# iterate over list l1 to the length of an item
for i in range(len(l1[0])):
# print(i)
row =[]
for item in l1:
# appending to new list with values and index positions
# i contains index position and item contains values
row.append(item[i])
l2.append(row)
return l2
mod=10**9+7
testCases=1
# testCases=vary(1)
for _ in range(testCases):
r,c=vary(2)
if r==1 and c==1:
print(0)
else:
if r>=c:
pt=c+1
arr=[]
for i in range(r):
arr.append([pt*j for j in range(1,c+1)])
pt+=1
for i in arr:
print(*i)
else:
pt=c+1
arr=[]
for i in range(r):
arr.append([pt*j for j in range(1,r+1)])
pt+=1
l2=[]
for i in transpose(arr,l2):
print(*i)
``` | instruction | 0 | 44,724 | 22 | 89,448 |
No | output | 1 | 44,724 | 22 | 89,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes investigating different math objects, for example, magic squares. But Sasha understands that magic squares have already been studied by hundreds of people, so he sees no sense of studying them further. Instead, he invented his own type of square — a prime square.
A square of size n × n is called prime if the following three conditions are held simultaneously:
* all numbers on the square are non-negative integers not exceeding 10^5;
* there are no prime numbers in the square;
* sums of integers in each row and each column are prime numbers.
Sasha has an integer n. He asks you to find any prime square of size n × n. Sasha is absolutely sure such squares exist, so just help him!
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases.
Each of the next t lines contains a single integer n (2 ≤ n ≤ 100) — the required size of a square.
Output
For each test case print n lines, each containing n integers — the prime square you built. If there are multiple answers, print any.
Example
Input
2
4
2
Output
4 6 8 1
4 9 9 9
4 10 10 65
1 4 4 4
1 1
1 1 | instruction | 0 | 44,804 | 22 | 89,608 |
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
for i in range(n-1):
l=n*[0]
l[i]=1
l[i+1]=1
print(*l, sep=" ")
l=n*[0]
l[0]=1
l[n-1]=1
print(*l, sep=" ")
``` | output | 1 | 44,804 | 22 | 89,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes investigating different math objects, for example, magic squares. But Sasha understands that magic squares have already been studied by hundreds of people, so he sees no sense of studying them further. Instead, he invented his own type of square — a prime square.
A square of size n × n is called prime if the following three conditions are held simultaneously:
* all numbers on the square are non-negative integers not exceeding 10^5;
* there are no prime numbers in the square;
* sums of integers in each row and each column are prime numbers.
Sasha has an integer n. He asks you to find any prime square of size n × n. Sasha is absolutely sure such squares exist, so just help him!
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases.
Each of the next t lines contains a single integer n (2 ≤ n ≤ 100) — the required size of a square.
Output
For each test case print n lines, each containing n integers — the prime square you built. If there are multiple answers, print any.
Example
Input
2
4
2
Output
4 6 8 1
4 9 9 9
4 10 10 65
1 4 4 4
1 1
1 1 | instruction | 0 | 44,805 | 22 | 89,610 |
Tags: constructive algorithms, math
Correct Solution:
```
from math import *
sInt = lambda: int(input())
mInt = lambda: map(int, input().split())
lInt = lambda: list(map(int, input().split()))
t = sInt()
for _ in range(t):
s = sInt()
c = 0
for i in range(s-1):
a = [0 for j in range(s)]
a[c] = 1
a[c+1] = 1
c += 1
print(*a,sep = ' ')
a = [0 for i in range(s)]
a[0] = 1
a[-1] = 1
print(*a,sep=' ')
``` | output | 1 | 44,805 | 22 | 89,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes investigating different math objects, for example, magic squares. But Sasha understands that magic squares have already been studied by hundreds of people, so he sees no sense of studying them further. Instead, he invented his own type of square — a prime square.
A square of size n × n is called prime if the following three conditions are held simultaneously:
* all numbers on the square are non-negative integers not exceeding 10^5;
* there are no prime numbers in the square;
* sums of integers in each row and each column are prime numbers.
Sasha has an integer n. He asks you to find any prime square of size n × n. Sasha is absolutely sure such squares exist, so just help him!
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases.
Each of the next t lines contains a single integer n (2 ≤ n ≤ 100) — the required size of a square.
Output
For each test case print n lines, each containing n integers — the prime square you built. If there are multiple answers, print any.
Example
Input
2
4
2
Output
4 6 8 1
4 9 9 9
4 10 10 65
1 4 4 4
1 1
1 1 | instruction | 0 | 44,806 | 22 | 89,612 |
Tags: constructive algorithms, math
Correct Solution:
```
n = 3000
primes = set(range(2, n + 1))
for i in range(2, int(n ** 0.5 + 1)):
if i not in primes:
i += 1
else:
remove = range(i * 2, n + 1, i)
primes.difference_update(remove)
prime = list(primes)
q = int(input())
tmp = 0
for _ in range(q):
n = int(input())
for p in prime:
x = (p - 1) % (n - 1)
y = (p - 1) // (n - 1)
if x == 0 and y not in primes:
tmp = y
break
ans = [[tmp] * n for _ in range(n)]
for i in range(n):
ans[i][i] = 1
for an in ans:
print(" ".join(map(str, an)))
``` | output | 1 | 44,806 | 22 | 89,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes investigating different math objects, for example, magic squares. But Sasha understands that magic squares have already been studied by hundreds of people, so he sees no sense of studying them further. Instead, he invented his own type of square — a prime square.
A square of size n × n is called prime if the following three conditions are held simultaneously:
* all numbers on the square are non-negative integers not exceeding 10^5;
* there are no prime numbers in the square;
* sums of integers in each row and each column are prime numbers.
Sasha has an integer n. He asks you to find any prime square of size n × n. Sasha is absolutely sure such squares exist, so just help him!
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases.
Each of the next t lines contains a single integer n (2 ≤ n ≤ 100) — the required size of a square.
Output
For each test case print n lines, each containing n integers — the prime square you built. If there are multiple answers, print any.
Example
Input
2
4
2
Output
4 6 8 1
4 9 9 9
4 10 10 65
1 4 4 4
1 1
1 1 | instruction | 0 | 44,807 | 22 | 89,614 |
Tags: constructive algorithms, math
Correct Solution:
```
from sys import stdin
class Input:
def __init__(self):
self.it = iter(stdin.readlines())
def line(self):
return next(self.it).strip()
def array(self, sep = ' ', cast = int):
return list(map(cast, self.line().split(sep = sep)))
def testcases(unknown = False):
inpt = Input()
def testcases_decorator(func):
cases, = [float('Inf')] if unknown else inpt.array()
while cases > 0:
try:
func(inpt)
cases -= 1
except StopIteration:
break
return testcases_decorator
'''
@thnkndblv
'''
@testcases()
def solve(inpt):
n, = inpt.array()
two = [[1, 1], [1, 1]]
three = [[1, 1, 1], [1, 1, 1], [1, 1, 1]]
a = [[0] * n for i in range(n)]
while n:
if n == 3:
n -= 3
a[n][n:n+3] = [1, 1, 1]
a[n+1][n:n+3] = [1, 1, 1]
a[n+2][n:n+3] = [1, 1, 1]
else:
n -= 2
a[n][n:n+2] = [1, 1]
a[n+1][n:n+2] = [1, 1]
for x in a:
print(*x)
``` | output | 1 | 44,807 | 22 | 89,615 |
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