message stringlengths 2 59.7k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an equation:
Ax2 + Bx + C = 0.
Your task is to find the number of distinct roots of the equation and print all of them in ascending order.
Input
The first line contains three integer numbers A, B and C ( - 105 ≤ A, B, C ≤ 105). Any coefficient may be equal to 0.
Output
In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.
Examples
Input
1 -5 6
Output
2
2.0000000000
3.0000000000 | instruction | 0 | 85,997 | 20 | 171,994 |
Tags: math
Correct Solution:
```
a,b,c = [int(x) for x in input().split()]
if a == 0 and b == 0 and c == 0:
print(-1)
elif a == 0:
if b == 0:
print(0)
else:
print(1)
print(-c/b)
else:
disc = b**2 - 4*a*c
denom = 2*a
if disc == 0:
print(1)
print(-b/denom)
elif disc < 0:
print(0)
else:
print(2)
if a > 0:
print((-b-disc**0.5)/denom)
print((-b+disc**0.5)/denom)
else:
print((-b+disc**0.5)/denom)
print((-b-disc**0.5)/denom)
``` | output | 1 | 85,997 | 20 | 171,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,054 | 20 | 172,108 |
Tags: implementation
Correct Solution:
```
n = int(input())
for i in range(1,n):
print(*((i*j//n)*10+(i*j%n) for j in range(1,n)))
``` | output | 1 | 86,054 | 20 | 172,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,055 | 20 | 172,110 |
Tags: implementation
Correct Solution:
```
def turn(i, j, k):
p = i * j
ans = ""
while p > 0:
t = p % k
ans += str(t)
p //= k
return ans[::-1]
k = int(input())
for i in range(1, k):
t = []
for j in range(1, k):
t.append(turn(i, j, k))
print(*t, sep=" ")
``` | output | 1 | 86,055 | 20 | 172,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,056 | 20 | 172,112 |
Tags: implementation
Correct Solution:
```
def baseN(num,b,numerals="0123456789abcdefghijklmnopqrstuvwxyz"):
return ((num == 0) and numerals[0]) or (baseN(num // b, b, numerals).lstrip(numerals[0]) + numerals[num % b])
n=int(input())
for i in range(1,n):
for j in range(1,n):
print(baseN(i*j,n),end=" ")
print()
``` | output | 1 | 86,056 | 20 | 172,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,057 | 20 | 172,114 |
Tags: implementation
Correct Solution:
```
from sys import stdout
from sys import stdin
def get():
return stdin.readline().strip()
def getf():
return [int(i) for i in get().split()]
def put(a, end = "\n"):
stdout.write(str(a) + end)
def putf(a, sep = " ", end = "\n"):
stdout.write(sep.join(map(str, a)) + end)
def matrix(n, m, a = 0):
return [[a for i in range(m)]for j in range(n)]
def transf(a, k):
s = ""
while(a >= k):
s += str(a % k)
a //= k
s += str(a)
return s[::-1]
def main():
n = int(get())
t = matrix(n - 1, n - 1)
for i in range(1, n):
for j in range(1, n):
t[i - 1][j - 1] = transf(i*j, n)
for i in t:
putf(i)
main()
``` | output | 1 | 86,057 | 20 | 172,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,058 | 20 | 172,116 |
Tags: implementation
Correct Solution:
```
k=int(input())
a=[[1]+[' '*(len(str(k**2))-1-len(str(i)))+str(i) for i in range(2,k)]]+[[i+1]+[0 for j in range(k-2)] for i in range(1,k-1)]
for i in range(2,k):
for j in range(2,k):
b=i*j
c=''
while b>0:
c+=str(b%k)
b//=k
c=' '*(len(str(k**2))-1-len(c))+c[::-1]
a[i-1][j-1]=c
for i in range(k-1):
print(*a[i])
``` | output | 1 | 86,058 | 20 | 172,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,059 | 20 | 172,118 |
Tags: implementation
Correct Solution:
```
k=int(input())
for i in range(1,k):
z,a=i,[]
for j in range(k-1):
p,s=z,""
while p:
s=str(p%k)+s
p//=k
z+=i
a.append(s)
print(*a)
``` | output | 1 | 86,059 | 20 | 172,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,060 | 20 | 172,120 |
Tags: implementation
Correct Solution:
```
n = int(input())
f = lambda x: x if x < n else f(x // n) * 10 + x % n
for i in range(1, n): print(*[f(i * j) for j in range(1, n)])
``` | output | 1 | 86,060 | 20 | 172,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11 | instruction | 0 | 86,061 | 20 | 172,122 |
Tags: implementation
Correct Solution:
```
def convert_base(number, base):
if base < 2:
return False
remainders = []
while number > 0:
remainders.append(str(number % base))
number //= base
remainders.reverse()
return ''.join(remainders)
n = int(input())
for i in range(1,n):
s = ''
for j in range(1,n):
s += str(convert_base(i * j, n)) + ' '
s = s[:-1]
print(s)
``` | output | 1 | 86,061 | 20 | 172,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
k = int(input())
def convert_to_base_10(a):
res = []
while a:
res.append(a%k)
a//=k
return ''.join(map(str,res[::-1]))
for i in range(1,k):
output = []
for j in range(1,k):
output.append(convert_to_base_10((i*j)))
print(*output)
``` | instruction | 0 | 86,062 | 20 | 172,124 |
Yes | output | 1 | 86,062 | 20 | 172,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
def convert(n,base):
x=""
while n>0:
x+=str(n%base)
n//=base
return x[::-1]
def func(x,n):
ans=[]
ans.append(x)
for i in range(2,n):
ans.append(int(convert(x*i,n)))
return ans
n=int(input())
ans=[]
for i in range(1,n):
if i==1:
x=[j for j in range(1,n)]
ans.append(x)
else:
x=func(i,n)
ans.append(x)
for s in ans:
print(*s)
``` | instruction | 0 | 86,063 | 20 | 172,126 |
Yes | output | 1 | 86,063 | 20 | 172,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
def b(n, k):
v = ''
while n:
n, v = n // k, str(n % k) + v
return v
k = int(input())
for i in range(1, k):
print(' '.join(b(i * j, k) for j in range(1, k)))
``` | instruction | 0 | 86,064 | 20 | 172,128 |
Yes | output | 1 | 86,064 | 20 | 172,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
radix = int(input())
for i in range(1, radix):
row = []
for j in range(1, radix):
x = i * j
digits = []
while x != 0:
digits.append(str(x % radix))
x //= radix
row.append(''.join(reversed(digits)))
print(' '.join(row))
``` | instruction | 0 | 86,065 | 20 | 172,130 |
Yes | output | 1 | 86,065 | 20 | 172,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
k=int(input())
for i in range(1,k):
for j in range(1,k):
print(i*j,end=' ')
print()
``` | instruction | 0 | 86,066 | 20 | 172,132 |
No | output | 1 | 86,066 | 20 | 172,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline
n=int(input())
l=1
for i in range(1,n):
for j in range(1,n):
print(j*l,end=' ')
print()
l+=1
``` | instruction | 0 | 86,067 | 20 | 172,134 |
No | output | 1 | 86,067 | 20 | 172,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
n = int(input())
for i in range(1,11):
for j in range(1,n+1):
print(i*j,end=" ")
print()
``` | instruction | 0 | 86,068 | 20 | 172,136 |
No | output | 1 | 86,068 | 20 | 172,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya studies positional notations. He has already learned to add and subtract numbers in the systems of notations with different radices and has moved on to a more complicated action — multiplication. To multiply large numbers one has to learn the multiplication table. Unfortunately, in the second grade students learn only the multiplication table of decimals (and some students even learn it in the first grade). Help Petya make a multiplication table for numbers in the system of notations with the radix k.
Input
The first line contains a single integer k (2 ≤ k ≤ 10) — the radix of the system.
Output
Output the multiplication table for the system of notations with the radix k. The table must contain k - 1 rows and k - 1 columns. The element on the crossing of the i-th row and the j-th column is equal to the product of i and j in the system of notations with the radix k. Each line may have any number of spaces between the numbers (the extra spaces in the samples are put for clarity).
Examples
Input
10
Output
1 2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 9 12 15 18 21 24 27
4 8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Input
3
Output
1 2
2 11
Submitted Solution:
```
n=int(input())
for i in range(1,n):
for j in range(1,n):
print(i*j,end=" ")
print()
``` | instruction | 0 | 86,069 | 20 | 172,138 |
No | output | 1 | 86,069 | 20 | 172,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,153 | 20 | 172,306 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
import sys
input=sys.stdin.readline
s=input().rstrip()
l=list(s.split(" "))
p=1
m=0
for ss in l:
if ss=="+":
p+=1
if ss=="-":
m+=1
n=int(l[-1])
if n*p-m<n or p-m*n>n:
print("Impossible")
exit()
arr_p=[1]*p
arr_m=[1]*m
r=p-m
for i in range(p):
xx=min(n-1,max(n-r,0))
arr_p[i]+=xx
r+=xx
for i in range(m):
xx=min(n-1,max(r-n,0))
arr_m[i]+=xx
r-=xx
ans=[]
pre=1
for i in range(len(l)):
if l[i]=="?":
if pre:
ans.append(str(arr_p.pop()))
else:
ans.append(str(arr_m.pop()))
else:
if l[i]=="+":
pre=1
ans.append(l[i])
elif l[i]=="-":
pre=0
ans.append(l[i])
else:
ans.append(str(l[i]))
print("Possible")
print(" ".join(ans))
``` | output | 1 | 86,153 | 20 | 172,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,154 | 20 | 172,308 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
def binp(a,b):
st=a[0]
fin=a[1]
b=[-b[1],-b[0]]
while True:
s=(st+fin)//2
if s>=b[0]:
if s<=b[1]:
return(s)
else:
fin=s-1
else:
st=s+1
s=input()
for i in range(len(s)):
if s[i]=='=':
y=i
break
y+=1
n=int(s[y:])
now='+'
max_=0
min_=0
ctr=0
pls=0
minus=0
g=[[0,0]]
for i in s:
ctr+=1
if i=='?':
if now=='+':
max_+=n
min_+=1
pls+=1
else:
max_-=1
min_-=n
minus+=1
g.append([min_,max_])
elif i=='-' or i=='+':
now=i
elif i=='=':
break
v=n
if v<min_ or v>max_:
print('Impossible')
else:
print('Possible')
a=[pls*1,pls*v]
b=[minus*(-v)-n,minus*(-1)-n]
#print(a,b)
ctr=0
ctr=binp(a,b)
fir=ctr
j=[0]*pls
if pls!=0:
j=[fir//pls]*pls
for u in range(fir%pls):
j[u]+=1
sec=n-ctr
k=[0]*minus
if minus!=0:
k=[((-sec)//minus)]*minus
#print(k)
#print((-sec)%minus)
for u in range((-sec)%minus):
k[u]+=1
p=0
m=0
now='+'
for i in s:
if i=='?':
if now=='+':
print(j[p],end='')
p+=1
else:
print(k[m],end='')
m+=1
elif i=='-' or i=='+':
now=i
print(i,end='')
else:
print(i,end='')
``` | output | 1 | 86,154 | 20 | 172,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,155 | 20 | 172,310 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
s = ('+ ' + input()).strip().split()
plus = 0
minus = 0
for c in s:
if c == '+':
plus += 1
elif c == '-':
minus += 1
n = int(s[-1])
maxn = plus * n - minus
minn = plus - minus * n
if not (maxn >= n >= minn):
print("Impossible")
exit()
ans = []
for i in range(1, len(s)):
need = min(maxn - n, n - 1)
if s[i-1] == '+':
ans.append(str(n - need))
maxn -= need
elif s[i-1] == '-':
ans.append(str(1 + need))
maxn -= need
else:
ans.append(s[i])
print("Possible")
print(' '.join(ans))
``` | output | 1 | 86,155 | 20 | 172,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,156 | 20 | 172,312 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
data = input()
data = data.split()
data.reverse()
n = int(data[0])
data.reverse()
pos = 1
neg = 0
for x in data:
if x == '-':
neg += 1
if x == '+':
pos += 1
ans = pos - neg
if ans <= n:
tmp = n - ans
if tmp > pos * (n - 1):
print('Impossible')
exit(0)
else:
print('Possible')
fin = ''
for x in range(len(data)):
if data[x] != '?':
fin += data[x] + ' '
continue
if x == 0 or data[x - 1] == '+':
t = min(n - 1, tmp)
tmp -= t
fin += str(1 + t) + ' '
else:
fin += '1 '
print(fin)
else:
tmp = ans - n
if tmp > neg * (n - 1):
print('Impossible')
exit(0)
else:
print('Possible')
fin = ''
for x in range(len(data)):
if data[x] != '?':
fin += data[x] + ' '
continue
if x != 0 and data[x - 1] == '-':
t = min(n - 1, tmp)
tmp -= t
fin += str(1 + t) + ' '
else:
fin += '1 '
print(fin)
``` | output | 1 | 86,156 | 20 | 172,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,157 | 20 | 172,314 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
def max(a, b):
if(a>=b):
return a
else:
return b
####################################
s=input().split()
length=len(s)
n=int(s[length-1])
plus=1
minus=0
for i in s:
if(i== '+'):
plus+=1
if(i== '-'):
minus+=1
if(plus*n - minus < n or plus - n*minus > n):
print('Impossible')
exit()
else:
print('Possible')
for i in range(0, length-1, 2): #initializing all placeholders with 1
s[i]='1'
# if(minus==0):
# s[0]= repr(n-plus+1)
# else:
# diff=plus-1-minus
# if(diff>0):
# s[0]=repr(n-diff)
# for i in range(2, length-1, 2):
# s[i]='1'
# flag=0
# if(diff<=0):
# s[0]=repr(n)
# for i in range(2, length-1, 2):
# s[i]='1'
# if(flag==0 and s[i-1] == '+'):
# flag=1
# s[i]=repr(1-diff)
res=n-plus+minus
for i in range(0, length-1, 2):
if((i==0 or s[i-1]=='+' ) and res>0):
val=int(s[i])
if(res>n-val):
res-=(n-val)
s[i]=repr(n)
else:
val+=res
s[i]=repr(val)
res=0
elif(s[i-1]=='-' and res<0):
val=int(s[i])
if(res<val-n):
res+=(n-val)
s[i]=repr(n)
else:
val-=res
s[i]=repr(val)
res=0
print (' '.join(s))
``` | output | 1 | 86,157 | 20 | 172,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,158 | 20 | 172,316 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
s=input()
a=s.split()
n=int(a[-1])
pos=1
neg=0
for item in a:
if item=='+':
pos+=1
if item=='-':
neg+=1
if n<pos-neg*n or n>pos*n-neg:
print("Impossible")
else:
print("Possible")
parr=[]
narr=[]
if pos>neg:
for i in range(neg):
narr.append(n)
x=int(n*(neg+1)//pos)
for i in range(pos):
if(i<n*(neg+1)%pos):
parr.append(x+1)
else:
parr.append(x)
else:
for i in range(pos):
parr.append(n)
x=int(n*(pos-1)//neg)
for i in range(neg):
if(i<n*(pos-1)%neg):
narr.append(x+1)
else:
narr.append(x)
sgn=1
for c in s:
if (c=='?'):
if (sgn==1):
print(parr[0],end="")
del parr[0]
else:
print(narr[0],end="")
del narr[0]
continue
elif (c=='+'):
sgn=1
elif c=='-':
sgn=-1
print(c,end="")
``` | output | 1 | 86,158 | 20 | 172,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,159 | 20 | 172,318 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
a = list(input().split())
n = int(a[-1])
R = int(a[-1])
one, two = 0, 0
for i in range(1,len(a),2):
if a[i] == '=':
break
if a[i] == '+':
R -= 1
one += 1
else:
R += 1
two += 1
R -= 1
one += 1
if R >= 0:
if one * (n - 1) >= R:
print('Possible')
for i in range(0, len(a), 2):
if i > 0 and a[i - 1] == '=':
print(a[i])
else:
if i == 0 or a[i - 1] == '+':
print(min(n - 1, R) + 1, a[i + 1], end = ' ')
R -= min(n - 1, R)
else:
print(1, a[i + 1], end = ' ')
else:
print('Impossible')
else:
if two * (1 - n) <= R:
print('Possible')
for i in range(0, len(a), 2):
if i > 0 and a[i - 1] == '=':
print(a[i])
else:
if i > 0 and a[i - 1] == '-':
print(-(max(1 - n, R) - 1), a[i + 1], end = ' ')
R -= max(1 - n, R)
else:
print(1, a[i + 1], end = ' ')
else:
print('Impossible')
``` | output | 1 | 86,159 | 20 | 172,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 86,160 | 20 | 172,320 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip().split()
def smas(x,mas):
for i in range(len(s)):
if s[i]==1 and (i==0 or s[i-1]=="+"):
if x>0:
y=min(n-1,x)
s[i]+=y
x-=y
def smen(x,men):
for i in range(len(s)):
if s[i]==1 and i>0 and s[i-1]=="-":
if x>0:
y=min(n-1,x)
s[i]+=y
x-=y
n=int(s[-1])
ans=0
y=0
if s[0]=="-":
y=1
ind=-1
men=0
mas=0
for i in range(y,len(s)-1,2):
if i!=0 and s[i-1]=="-" :
ans-=1
men+=1
else:
mas+=1
if ind==-1:
ind=i
else:
ans+=1
s[i]=1
l=[-(n*men),-men]
t=True
for i in range(mas,n*mas+1):
if i>=n:
x=n-i
else:
continue
if x>=l[0] and x<=l[1]:
t=False
smas(i-mas,mas)
smen((-x)-men,men)
break
if t:
print("Impossible")
else:
print("Possible")
print(*s)
``` | output | 1 | 86,160 | 20 | 172,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
If you write a positive integer in decimal notation (without leading 0) and look at the digit numbers in order, when the number increases and decreases alternately, the number is "zigza". Let's call it. For example, 2947 is a zigzag number because the digit numbers are in the order of 2 → 9 → 4 → 7 and increase → decrease → increase. In addition, 71946 is a zigzag number because it is in the order of decrease → increase → decrease → increase. On the other hand, 123, 71446, 71442 and 88 are not zigzag numbers. A one-digit positive integer is considered to be a zigzag number.
Create a program that finds the remainder of the number of zigzags divided by 10000 out of multiples of M between A and B.
input
The input consists of three lines, with one positive integer written on each line.
The integer on the first line represents A, the integer on the second line represents B, and the integer on the third line represents M. These satisfy 1 ≤ A ≤ B ≤ 10500 and 1 ≤ M ≤ 500.
* Note that the values of A and B may not fit in the data types that represent ordinary integers.
output
Output the remainder of the number of zigzag numbers divided by 10000 out of multiples of M between A and B in one line.
Input / output example
Input example 1
100
200
Five
Output example 1
13
In I / O example 1, the number of zigzags that are multiples of 5 from 100 to 200 is 13 of 105, 120, 130, 140, 150, 160, 165, 170, 175, 180, 185, 190, 195. ..
Input example 2
6
1234567
3
Output example 2
246
In I / O example 2, there are 50246 zigzag numbers that are multiples of 3 from 6 to 1234567, so 246, which is the remainder of dividing it by 10000, is output.
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
100
200
5
Output
13
Submitted Solution:
```
def main():
A = int(input())
B = int(input())
M = int(input())
num = []
for x in range(A, B + 1):
if x % M == 0:
num.append(x)
else:
pass
cnt = 0
for x in range(len(num)):
num[x] = list(str(num[x]))
for y in range(len(num[x])):
num[x][y] = int(num[x][y])
if len(num[x]) < 2:
cnt += 1
elif len(num[x]) == 2:
if num[x][0] == num[x][1]:
pass
else:
cnt += 1
elif len(num[x]) > 2:
listcnt = 0
if num[x][0] > num[x][1]:
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
pass
elif y % 2 == 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
pass
if listcnt == len(num[x]) - 2:
cnt += 1
else:
pass
elif num[x][0] < num[x][1]:
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
pass
elif y % 2 == 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
pass
if listcnt == len(num[x]) - 2:
cnt += 1
else:
pass
elif num[x][0] == num[x][1]:
pass
hoge = ""
for y in range(len(num[x])):
num[x][y] = str(num[x][y])
hoge += num[x][y]
num[x] = int(hoge)
print(cnt % 10000)
if __name__ == "__main__":
main()
``` | instruction | 0 | 86,485 | 20 | 172,970 |
No | output | 1 | 86,485 | 20 | 172,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
If you write a positive integer in decimal notation (without leading 0) and look at the digit numbers in order, when the number increases and decreases alternately, the number is "zigza". Let's call it. For example, 2947 is a zigzag number because the digit numbers are in the order of 2 → 9 → 4 → 7 and increase → decrease → increase. In addition, 71946 is a zigzag number because it is in the order of decrease → increase → decrease → increase. On the other hand, 123, 71446, 71442 and 88 are not zigzag numbers. A one-digit positive integer is considered to be a zigzag number.
Create a program that finds the remainder of the number of zigzags divided by 10000 out of multiples of M between A and B.
input
The input consists of three lines, with one positive integer written on each line.
The integer on the first line represents A, the integer on the second line represents B, and the integer on the third line represents M. These satisfy 1 ≤ A ≤ B ≤ 10500 and 1 ≤ M ≤ 500.
* Note that the values of A and B may not fit in the data types that represent ordinary integers.
output
Output the remainder of the number of zigzag numbers divided by 10000 out of multiples of M between A and B in one line.
Input / output example
Input example 1
100
200
Five
Output example 1
13
In I / O example 1, the number of zigzags that are multiples of 5 from 100 to 200 is 13 of 105, 120, 130, 140, 150, 160, 165, 170, 175, 180, 185, 190, 195. ..
Input example 2
6
1234567
3
Output example 2
246
In I / O example 2, there are 50246 zigzag numbers that are multiples of 3 from 6 to 1234567, so 246, which is the remainder of dividing it by 10000, is output.
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
100
200
5
Output
13
Submitted Solution:
```
def main():
A = int(input())
B = int(input())
M = int(input())
num = []
for x in range(A, B + 1):
if x % M == 0:
num.append(x)
else:
pass
cnt = 0
for x in range(len(num)):
num[x] = list(str(num[x]))
c = len(num[x]) #checker
if c == 1:
cnt += 1
elif c == 2:
if num[x][0] == num[x][1]:
pass
else:
cnt += 1
elif c >= 3:
if num[x][0] == num[x][1]:
pass
elif num[x][0] > num[x][1]:
listcnt = 0
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
break
if y % 2 == 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
break
if listcnt == len(num[x]) - 2:
cnt += 1
elif num[x][0] < num[x][1]:
listcnt = 0
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
break
if y % 2 == 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
break
if listcnt == len(num[x]) - 2:
cnt += 1
#hoge = ""
#for y in range(len(num[x])):
# hoge += num[x][y]
#num[x] = int(hoge)
ans = cnt % 10000
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 86,486 | 20 | 172,972 |
No | output | 1 | 86,486 | 20 | 172,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
If you write a positive integer in decimal notation (without leading 0) and look at the digit numbers in order, when the number increases and decreases alternately, the number is "zigza". Let's call it. For example, 2947 is a zigzag number because the digit numbers are in the order of 2 → 9 → 4 → 7 and increase → decrease → increase. In addition, 71946 is a zigzag number because it is in the order of decrease → increase → decrease → increase. On the other hand, 123, 71446, 71442 and 88 are not zigzag numbers. A one-digit positive integer is considered to be a zigzag number.
Create a program that finds the remainder of the number of zigzags divided by 10000 out of multiples of M between A and B.
input
The input consists of three lines, with one positive integer written on each line.
The integer on the first line represents A, the integer on the second line represents B, and the integer on the third line represents M. These satisfy 1 ≤ A ≤ B ≤ 10500 and 1 ≤ M ≤ 500.
* Note that the values of A and B may not fit in the data types that represent ordinary integers.
output
Output the remainder of the number of zigzag numbers divided by 10000 out of multiples of M between A and B in one line.
Input / output example
Input example 1
100
200
Five
Output example 1
13
In I / O example 1, the number of zigzags that are multiples of 5 from 100 to 200 is 13 of 105, 120, 130, 140, 150, 160, 165, 170, 175, 180, 185, 190, 195. ..
Input example 2
6
1234567
3
Output example 2
246
In I / O example 2, there are 50246 zigzag numbers that are multiples of 3 from 6 to 1234567, so 246, which is the remainder of dividing it by 10000, is output.
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
100
200
5
Output
13
Submitted Solution:
```
def main():
A = int(input())
B = int(input())
M = int(input())
num = []
for x in range(A, B + 1):
if x % M == 0:
num.append(x)
else:
pass
cnt = 0
for x in range(len(num)):
num[x] = list(str(num[x]))
c = len(num[x]) #checker
if c == 1:
cnt += 1
elif c == 2:
if num[x][0] == num[x][1]:
pass
else:
cnt += 1
elif c >= 3:
if num[x][0] == num[x][1]:
pass
elif num[x][0] > num[x][1]:
listcnt = 0
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
break
if y % 2 == 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
break
if listcnt == len(num[x]) - 2:
cnt += 1
elif num[x][0] < num[x][1]:
listcnt = 0
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
break
if y % 2 == 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
break
if listcnt == len(num[x]) - 2:
cnt += 1
hoge = ""
for y in range(len(num[x])):
hoge += num[x][y]
num[x] = int(hoge)
ans = cnt % 10000
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 86,487 | 20 | 172,974 |
No | output | 1 | 86,487 | 20 | 172,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
If you write a positive integer in decimal notation (without leading 0) and look at the digit numbers in order, when the number increases and decreases alternately, the number is "zigza". Let's call it. For example, 2947 is a zigzag number because the digit numbers are in the order of 2 → 9 → 4 → 7 and increase → decrease → increase. In addition, 71946 is a zigzag number because it is in the order of decrease → increase → decrease → increase. On the other hand, 123, 71446, 71442 and 88 are not zigzag numbers. A one-digit positive integer is considered to be a zigzag number.
Create a program that finds the remainder of the number of zigzags divided by 10000 out of multiples of M between A and B.
input
The input consists of three lines, with one positive integer written on each line.
The integer on the first line represents A, the integer on the second line represents B, and the integer on the third line represents M. These satisfy 1 ≤ A ≤ B ≤ 10500 and 1 ≤ M ≤ 500.
* Note that the values of A and B may not fit in the data types that represent ordinary integers.
output
Output the remainder of the number of zigzag numbers divided by 10000 out of multiples of M between A and B in one line.
Input / output example
Input example 1
100
200
Five
Output example 1
13
In I / O example 1, the number of zigzags that are multiples of 5 from 100 to 200 is 13 of 105, 120, 130, 140, 150, 160, 165, 170, 175, 180, 185, 190, 195. ..
Input example 2
6
1234567
3
Output example 2
246
In I / O example 2, there are 50246 zigzag numbers that are multiples of 3 from 6 to 1234567, so 246, which is the remainder of dividing it by 10000, is output.
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
100
200
5
Output
13
Submitted Solution:
```
def main():
A = int(input())
B = int(input())
M = int(input())
num = []
for x in range(A, B + 1):
if x % M == 0:
num.append(x)
else:
pass
cnt = 0
for x in range(len(num)):
num[x] = list(str(num[x]))
c = len(num[x]) #checker
if c == 1:
cnt += 1
elif c == 2:
if num[x][0] == num[x][1]:
pass
else:
cnt += 1
elif c >= 3:
listcnt = 0
if num[x][0] == num[x][1]:
pass
elif num[x][0] > num[x][1]:
#listcnt = 0
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
break
if y % 2 == 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
break
if listcnt == len(num[x]) - 2:
cnt += 1
elif num[x][0] < num[x][1]:
#listcnt = 0
for y in range(1, len(num[x]) - 1):
if y % 2 != 0:
if num[x][y] > num[x][y + 1]:
listcnt += 1
else:
break
if y % 2 == 0:
if num[x][y] < num[x][y + 1]:
listcnt += 1
else:
break
if listcnt == len(num[x]) - 2:
cnt += 1
#hoge = ""
#for y in range(len(num[x])):
# hoge += num[x][y]
#num[x] = int(hoge)
ans = cnt % 10000
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 86,488 | 20 | 172,976 |
No | output | 1 | 86,488 | 20 | 172,977 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,509 | 20 | 173,018 |
"Correct Solution:
```
def solve(n):
ans = 0
mp = {}
while (n >= 10):
s = str(n)
next = -1
for i in range(1, len(s)):
next = max(next, int(s[0:i])*int(s[i:]))
if next in mp:
ans = -1
break
mp[next] = True
n = next
ans += 1
return ans
def main():
q = int(input())
for i in range(q):
n = int(input())
print(solve(n))
main()
``` | output | 1 | 86,509 | 20 | 173,019 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,510 | 20 | 173,020 |
"Correct Solution:
```
from math import log10
for i in range(int(input())):
n = int(input())
j = 0
while n >= 10:
n = max((n // (10**k)) * (n % (10**k)) for k in range(1, int(log10(n)) + 1))
j += 1
print(j)
``` | output | 1 | 86,510 | 20 | 173,021 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,511 | 20 | 173,022 |
"Correct Solution:
```
#!/usr/bin/env python3
q = int(input())
for _ in range(q):
n = input()
for k in range(len(n)**2):
if len(n) == 1:
print(k)
break
n = str(max(int(n[:i]) * int(n[i:]) for i in range(1, len(n))))
``` | output | 1 | 86,511 | 20 | 173,023 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,512 | 20 | 173,024 |
"Correct Solution:
```
# your code goes here
#times volume24 2424
Q=int(input())
for i in range(Q):
N=str(input())
c=0
while len(N)>1 and c<15:
M=0
for j in range(1,len(N)):
# print(N[:j])
l=int(N[:j])
r=int(N[j:])
l*=r
if M<l:
M=l
# print(M)
N=str(M)
c+=1
# print(N)
if c>=15:
d=[int(N)]
while len(N)>1 and c>=0 and c<50:
M=0
# print(d)
for j in range(1,len(N)):
# print(N[:j])
l=int(N[:j])
r=int(N[j:])
l*=r
if M<l:
M=l
# print(M)
k=0
while k<len(d) and M>d[k]:
k+=1
# print(k)
if k>=len(d):
k=-1
elif M==d[k]:
c=-1
d.insert(k,M)
N=str(M)
c+=1
# print(k)
print(c)
``` | output | 1 | 86,512 | 20 | 173,025 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,513 | 20 | 173,026 |
"Correct Solution:
```
"かけざん"
"最大のものを取得して、一桁になるまでに操作を行う回数を答える"
def kakezan(n):
ret = 0
str_n = str(n)
digit_amount = len(str_n)
for i in range(digit_amount-1):
# print(str_n[:i+1])
# print(str_n[i+1:])
# print("")
ret = max(ret, int(str_n[:i+1])*int(str_n[i+1:]))
return ret
Q = int(input()) # 入力される整数の個数
N = [int(input()) for i in range(Q)]
for n in N:
cnt = 0
while n >= 10:
n = kakezan((n))
cnt += 1
print(cnt)
``` | output | 1 | 86,513 | 20 | 173,027 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,514 | 20 | 173,028 |
"Correct Solution:
```
def calc(n):
if n>=10 and n<100:
return (n//10)*(n%10)
elif n>=100 and n<1000:
r1=(n//10)*(n%10)
r2=(n//100)*(n%100)
return max(r1,r2)
elif n>=1000 and n<10000:
r1=(n//10)*(n%10)
r2=(n//100)*(n%100)
r3=(n//1000)*(n%1000)
return max(r1,r2,r3)
elif n>=10000 and n<100000:
r1=(n//10)*(n%10)
r2=(n//100)*(n%100)
r3=(n//1000)*(n%1000)
r4=(n//10000)*(n%10000)
return max(r1,r2,r3,r4)
elif n>=100000 and n<1000000:
r1=(n//10)*(n%10)
r2=(n//100)*(n%100)
r3=(n//1000)*(n%1000)
r4=(n//10000)*(n%10000)
r5=(n//100000)*(n%100000)
return max(r1,r2,r3,r4,r5)
else:
r1=(n//10)*(n%10)
r2=(n//100)*(n%100)
r3=(n//1000)*(n%1000)
r4=(n//10000)*(n%10000)
r5=(n//100000)*(n%100000)
r6=(n//1000000)*(n%1000000)
return max(r1,r2,r3,r4,r5,r6)
n=int(input())
for i in range(n):
cnt=0
a=int(input())
while a>=10:
a=calc(a)
cnt+=1
print(cnt)
``` | output | 1 | 86,514 | 20 | 173,029 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,515 | 20 | 173,030 |
"Correct Solution:
```
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')
#A
def A():
while 1:
n = II()
ans = 0
if n == 0:
break
i = 1
while i < n / 2:
tmp = 0
l = i
while tmp < n:
tmp += l
l += 1
if tmp == n:
ans += 1
i += 1
print(ans)
return
#B
def B():
def f(x):
tmp = 0
for i in range(len(x)-1):
x1 = int(x[:i+1])
x2 = int(x[i+1:])
tmp = max(tmp, x1 * x2)
return tmp
q = II()
for i in range(q):
ans = 0
n = II()
while n > 9:
ans += 1
n = f(str(n))
print(ans)
return
#C
def C():
return
#D
def D():
return
#E
def E():
R, C, K = LI()
n = II()
rc = LIR_(n)
fldr = [0 for _ in range(R)]
fldc = [0 for _ in range(C)]
for r, c in rc:
fldr[r] += 1
fldc[c] += 1
fldcs = fldc[::1]
fldcs.sort()
fldrs = fldr[::1]
fldrs.sort()
ans = 0
dr = defaultdict(int)
dc = defaultdict(int)
for k in range(R):
dr[fldr[k]] += 1
for k in range(C):
dc[fldc[k]] += 1
for k in range(K+1):
ans += dr[K - k] * dc[k]
for r, c in rc:
a = fldr[r] + fldc[c]
if a == K:
ans -= 1
elif a == K + 1:
ans += 1
print(ans)
return
#F
def F():
n, m = LI()
uvl = LIR_(m)
dist = [[inf] * n for _ in range(n)]
dist0 = defaultdict(int)
full = []
d = 0
for u, v, l in uvl:
if u == 0:
dist0[v] = l + 1
full.append(v)
d += 1
elif v == 0:
dist0[u] = l + 1
d += 1
full.append(u)
else:
l += 1
dist[u][v] = l
dist[v][u] = l
for i in range(n):
dist[i][i] = 0
for k in range(n):
for i in range(n):
for j in range(n):
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
fulls = itertools.combinations(range(d), 2)
ans = inf
for a, b in fulls:
a, b = full[a], full[b]
tmp = dist0[a] + dist0[b] + dist[a][b]
ans = min(ans, tmp)
print(-1 if ans == inf else ans)
return
#G
def G():
s = S()
s = s[::-1]
M_count = s.count("M")
M_lis = [0] * M_count
M_count //= 2
tmp = 0
k = 0
for i in range(len(s)):
if s[i] == "+":
tmp += 1
elif s[i] == "-":
tmp -= 1
else:
M_lis[k] = tmp
k += 1
M_lis.sort()
print(-sum(M_lis[:M_count]) + sum(M_lis[M_count:]))
return
#H
def H():
def f(t):
return a * t + b * math.sin(c * t * math.pi)
a, b, c = LI()
ok = 99 // a + 1
ng = 0
while 1:
t = (ok + ng) / 2
ft = f(t)
if abs(ft - 100) <= 10 ** (-6):
print(t)
return
if ft > 100:
ok = t
if ft < 100:
ng = t
return
#Solve
if __name__ == '__main__':
B()
``` | output | 1 | 86,515 | 20 | 173,031 |
Provide a correct Python 3 solution for this coding contest problem.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1 | instruction | 0 | 86,516 | 20 | 173,032 |
"Correct Solution:
```
from functools import reduce
g=lambda x,y:x*y
cv=lambda s:str(max([reduce(g,list(map(int,[s[:i],s[i:]]))) for i in range(1,len(s))]))
for _ in range(int(input())):
c=0
n=input()
s=set()
while 1:
if len(n)==1:break
if n in s:
c=-1
break
s.add(n)
n=cv(n)
c+=1
print(c)
``` | output | 1 | 86,516 | 20 | 173,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
def split(x):
assert(isinstance(x, int))
sx = str(x)
l = len(sx)
candidates = []
for idx in range(1, l):
first = sx[idx:]
candidates.append(int(sx[:idx]) * int(sx[idx:]))
if len(candidates) == 0:
return None
else:
return max(candidates)
def problem1():
Q = input()
answers = []
for _ in range(int(Q)):
N = int(input())
x = N
cnt = 0
while True:
next_x = split(x)
if next_x is None:
break
elif x == next_x:
cnt = -1
break
else:
cnt += 1
x = next_x
answers.append(str(cnt))
print('\n'.join(answers))
if __name__ == '__main__':
problem1()
``` | instruction | 0 | 86,517 | 20 | 173,034 |
Yes | output | 1 | 86,517 | 20 | 173,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
def solve(dic, cnt, n):
if n < 10:
return cnt
if n in dic:
return -1
dic[n] = True
s = str(n)
max_score = 0
for i in range(1, len(s)):
a, b = int(s[:i]), int(s[i:])
max_score = max(max_score, a * b)
return solve(dic, cnt + 1, max_score)
q = int(input())
for _ in range(q):
dic = dict()
print(solve(dic, 0, int(input())))
``` | instruction | 0 | 86,518 | 20 | 173,036 |
Yes | output | 1 | 86,518 | 20 | 173,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
def solve(N, cnt):
if len(str(N)) == 1:
return cnt
sN = str(N)
maxNum = 0
for i in range(1, len(sN)):
left = int(sN[0:i])
right = int(sN[i:])
maxNum = max(left * right, maxNum)
ret = solve(maxNum, cnt + 1)
return ret
Q = int(input())
for i in range(Q):
N = int(input())
ans = solve(N, 0)
print(ans)
``` | instruction | 0 | 86,519 | 20 | 173,038 |
Yes | output | 1 | 86,519 | 20 | 173,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
import sys
from collections import defaultdict, Counter, deque
from itertools import accumulate, permutations, combinations
from operator import itemgetter
from bisect import bisect_left, bisect_right, bisect
from heapq import heappop, heappush
from fractions import gcd
from math import ceil, floor, sqrt, cos, sin, pi
from copy import deepcopy
def main():
Q = int(input())
for _ in range(Q):
n = input()
count = 0
while len(n) > 1:
if count == 1000000:
count = -1
break
tmp = 0
for i in range(1,len(n)):
tmp = max(tmp, int(n[:i])*int(n[i:]))
count += 1
n = str(tmp)
print(count)
if __name__ == '__main__':
main()
``` | instruction | 0 | 86,520 | 20 | 173,040 |
Yes | output | 1 | 86,520 | 20 | 173,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
from math import log10
for i in range(int(input())):
n = int(input())
j = 0
while n >= 10:
n = max((n / (10**k)) * (n % (10**k))
for k in range(1, int(log10(n)) + 1))
j += 1
print(j)
``` | instruction | 0 | 86,521 | 20 | 173,042 |
No | output | 1 | 86,521 | 20 | 173,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
for i in range(int(input())):
n = int(input())
j = 0
while n >= 10:
L = [(n // (10**k)) * (n % (10**k))
for k in range(1, int(log10(n)) + 1)]
n = max(L)
j += 1
print(j)
``` | instruction | 0 | 86,522 | 20 | 173,044 |
No | output | 1 | 86,522 | 20 | 173,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
# your code goes here
#times volume24 2424
Q=int(input())
for i in range(Q):
N=str(input())
c=0
while len(N)>1 and c<15:
M=0
for j in range(1,len(N)):
# print(N[:j])
l=int(N[:j])
r=int(N[j:])
l*=r
if M<l:
M=l
# print(M)
N=str(M)
c+=1
# print(N)
if c>=15:
d=[int(N)]
while len(N)>1 and c>=0 and c<50:
M=0
for j in range(1,len(N)):
# print(N[:j])
l=int(N[:j])
r=int(N[j:])
l*=r
if M<l:
M=l
# print(M)
k=0
while k<len(d) and M<d[k]:
k+=1
# print(d)
if k>=len(d):
k=-1
elif M==d[k]:
c=-1
d.insert(k,M)
N=str(N)
# print(N)
print(c)
``` | instruction | 0 | 86,523 | 20 | 173,046 |
No | output | 1 | 86,523 | 20 | 173,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro is an elementary school student who has just learned multiplication. Somehow, he likes multiplication, so when he sees numbers, he wants to multiply. He seems to like to do the following for integers greater than or equal to 0. (Processing flow)
* Procedure 1. If a certain integer n greater than or equal to 0 is a single digit in decimal notation, the process ends there. Otherwise go to step 2
* Step 2. When an integer n of 10 or more is displayed in decimal, it is possible to break it into two numbers by inserting a break between some digits (for example, 2012-> 20, 12). For possible cutting methods like this, multiply the two obtained numbers and set the largest one as the next n, and return to step 1. (For details, see "Supplementary information on step 2" below.)
Taro seems to like this process, but I can't predict how many times step 2 should be repeated, and I think it may have to be done infinitely. So I asked Taro's older brother and college student how many times this step 2 should be done for an integer n greater than or equal to 0.
Your job is to give Q integers greater than or equal to 0 N1 .. NQ, so find out how many steps 2 will be performed on each integer before the end of the process. If you need an infinite number of steps, output -1.
Supplement on step 2
You should also take into account those that have a 0 at the beginning of the digit as a result of the isolation.
For example, when n = 1024, 1 * 024, 10 * 24, and 102 * 4 are calculated as 24,240,408, respectively, so 408 is selected and this is the next n.
Constraints
> 1 ≤ Q ≤ 100
> 0 ≤ Ni ≤ 106
>
Input
> Q
> N1
> N2
> ...
> NQ
>
* Q represents the number of integers greater than or equal to 0 given
* Ni is an integer greater than or equal to 0 that Taro is interested in, and represents the i-th one.
Output
Output Q integers separated by line breaks
> R1
> R2
> ..
> RQ
>
* Ri represents the number of times step 2 is executed before the processing is completed for Ni.
* Ri is -1 if step 2 needs to be performed infinitely for Ni
Examples
Input
3
9
99
123
Output
0
2
3
Input
2
999999
1000000
Output
12
1
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=0;b=[]
while n//10:
a+=1
b+=[n]+b
x=1;s=m=0
while n:
s+=x*(n%10)
n//=10
x*=10
m=max(s*n,m)
n=m
if b[-1]==n:print(-1);break
else:print(a)
``` | instruction | 0 | 86,524 | 20 | 173,048 |
No | output | 1 | 86,524 | 20 | 173,049 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,653 | 20 | 175,306 |
Tags: bitmasks, brute force, math
Correct Solution:
```
n,p=map(int,input().split())
#print("{0:b}".format(n).count('1'))
t=0
while (("{0:b}".format(n).count('1'))>t or n<t) and n>=0:
t+=1
n-=p
if n<0:
print(-1)
else:
print(t)
``` | output | 1 | 87,653 | 20 | 175,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,654 | 20 | 175,308 |
Tags: bitmasks, brute force, math
Correct Solution:
```
n, p = map(int, input().split())
n -= p
f = False
for i in range(100):
s = bin(n)
#print(n, s, i + 1)
if len(s) - 2 >= i + 1 >= s.count('1') > 0 and n > 0:
k = i + 1
f = True
break
else:
n -= p
if f:
print(k)
else:
print(-1)
``` | output | 1 | 87,654 | 20 | 175,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,655 | 20 | 175,310 |
Tags: bitmasks, brute force, math
Correct Solution:
```
n, p = map(int, input().split())
i = 0
while True:
if n - p * i < 0:
print(-1)
break
if bin(n - p * i).count('1') <= i and i <= n - p * i:
print(i)
break
i += 1
``` | output | 1 | 87,655 | 20 | 175,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,656 | 20 | 175,312 |
Tags: bitmasks, brute force, math
Correct Solution:
```
n, p = map(int, input().split())
power = []
power.append(1)
for i in range(1,31):
power.append(power[len(power)-1]*2)
ans = -1
i = 1
while i < 31:
y = n - i*p
if y <= 0 or y < i:
break
a = []
while y:
a.append(y%2)
y = y//2
count = 0
for x in a:
if x:
count += 1
if count <= i:
ans = i
break
i += 1
print(ans)
``` | output | 1 | 87,656 | 20 | 175,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
For example, some -9-binary ("minus nine" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
And if p=-9 we can represent 7 as one number (2^4-9).
Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
Input
The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
Output
If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
Examples
Input
24 0
Output
2
Input
24 1
Output
3
Input
24 -1
Output
4
Input
4 -7
Output
2
Input
1 1
Output
-1
Note
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible. | instruction | 0 | 87,657 | 20 | 175,314 |
Tags: bitmasks, brute force, math
Correct Solution:
```
from math import log
def col(n):
rez = 0
while n > 0:
rez += n%2
n //= 2
return rez
def f(n, p):
for i in range(1, int(log(n, 2) + 20)):
if col(n-p*i) == i:
return i
if col(n-p*i) < i and n-p*i >= i:
return i
return -1
n, p = map(int, input().split())
print(f(n, p))
``` | output | 1 | 87,657 | 20 | 175,315 |
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