message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.
* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.
You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
Constraints
* 0 ≤ K ≤ 50 \times 10^{16}
Input
Input is given from Standard Input in the following format:
K
Output
Print a solution in the following format:
N
a_1 a_2 ... a_N
Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
Examples
Input
0
Output
4
3 3 3 3
Input
1
Output
3
1 0 3
Input
2
Output
2
2 2
Input
3
Output
7
27 0 0 0 0 0 0
Input
1234567894848
Output
10
1000 193 256 777 0 1 1192 1234567891011 48 425
Submitted Solution:
```
K = int(input())
a = K//50
b = K %50
Ans = [100+a-b]*b + [49+a-b]*(50-b)
print(50)
print(*Ans)
``` | instruction | 0 | 48,663 | 5 | 97,326 |
Yes | output | 1 | 48,663 | 5 | 97,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.
* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.
You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
Constraints
* 0 ≤ K ≤ 50 \times 10^{16}
Input
Input is given from Standard Input in the following format:
K
Output
Print a solution in the following format:
N
a_1 a_2 ... a_N
Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
Examples
Input
0
Output
4
3 3 3 3
Input
1
Output
3
1 0 3
Input
2
Output
2
2 2
Input
3
Output
7
27 0 0 0 0 0 0
Input
1234567894848
Output
10
1000 193 256 777 0 1 1192 1234567891011 48 425
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7
def resolve():
k = int(input())
nums = list(range(50))
if k <= 50:
for i in range(k):
for j in range(50):
if j == i:
nums[j] += 50
else:
nums[j] -= 1
print(50)
print(*nums)
else:
q, r = divmod(k, 50)
nums = [num + q for num in nums]
for i in range(r):
for j in range(50):
if j == i:
nums[j] += 50
else:
nums[j] -= 1
print(50)
print(*nums)
if __name__ == '__main__':
resolve()
``` | instruction | 0 | 48,664 | 5 | 97,328 |
Yes | output | 1 | 48,664 | 5 | 97,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.
* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.
You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
Constraints
* 0 ≤ K ≤ 50 \times 10^{16}
Input
Input is given from Standard Input in the following format:
K
Output
Print a solution in the following format:
N
a_1 a_2 ... a_N
Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
Examples
Input
0
Output
4
3 3 3 3
Input
1
Output
3
1 0 3
Input
2
Output
2
2 2
Input
3
Output
7
27 0 0 0 0 0 0
Input
1234567894848
Output
10
1000 193 256 777 0 1 1192 1234567891011 48 425
Submitted Solution:
```
k = int(input())
n = 50
d = k // n
a = [d + i for i in range(n)] # d*n operations
k = k % n
for i in range(k):
a[i] += n
for j in range(n):
if j != i:
a[j] -= 1
print(n)
print(' '.join(map(str, a)))
``` | instruction | 0 | 48,665 | 5 | 97,330 |
Yes | output | 1 | 48,665 | 5 | 97,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.
* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.
You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
Constraints
* 0 ≤ K ≤ 50 \times 10^{16}
Input
Input is given from Standard Input in the following format:
K
Output
Print a solution in the following format:
N
a_1 a_2 ... a_N
Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
Examples
Input
0
Output
4
3 3 3 3
Input
1
Output
3
1 0 3
Input
2
Output
2
2 2
Input
3
Output
7
27 0 0 0 0 0 0
Input
1234567894848
Output
10
1000 193 256 777 0 1 1192 1234567891011 48 425
Submitted Solution:
```
K = int(input())
ans = [K//50 + i for i in range(50)]
if K%50:
for i in range(K%50):
ans[i] += N
print(50)
print(" ".join(ans))
``` | instruction | 0 | 48,666 | 5 | 97,332 |
No | output | 1 | 48,666 | 5 | 97,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.
* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.
You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
Constraints
* 0 ≤ K ≤ 50 \times 10^{16}
Input
Input is given from Standard Input in the following format:
K
Output
Print a solution in the following format:
N
a_1 a_2 ... a_N
Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
Examples
Input
0
Output
4
3 3 3 3
Input
1
Output
3
1 0 3
Input
2
Output
2
2 2
Input
3
Output
7
27 0 0 0 0 0 0
Input
1234567894848
Output
10
1000 193 256 777 0 1 1192 1234567891011 48 425
Submitted Solution:
```
k = int(input())
n = 50
m = n * (k + 1) - 1
series = str(m)
for i in range(50):
series = series + " 0"
print(str(n))
print(series)
``` | instruction | 0 | 48,667 | 5 | 97,334 |
No | output | 1 | 48,667 | 5 | 97,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.
* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.
You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
Constraints
* 0 ≤ K ≤ 50 \times 10^{16}
Input
Input is given from Standard Input in the following format:
K
Output
Print a solution in the following format:
N
a_1 a_2 ... a_N
Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
Examples
Input
0
Output
4
3 3 3 3
Input
1
Output
3
1 0 3
Input
2
Output
2
2 2
Input
3
Output
7
27 0 0 0 0 0 0
Input
1234567894848
Output
10
1000 193 256 777 0 1 1192 1234567891011 48 425
Submitted Solution:
```
K = int(input())
b = int(K % 50)
a = int((K - b) / 50)
list = [0] * 50
j = 0
while (j <= 49 - b) :
list[j] = 49 + a - b
j += 1
while (j >49 - b and j <= 49):
list[j] = 2 * 50 - b
j += 1
list = map(str, list)
print('50')
print(' '.join(list))
``` | instruction | 0 | 48,668 | 5 | 97,336 |
No | output | 1 | 48,668 | 5 | 97,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.
* Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.
You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
Constraints
* 0 ≤ K ≤ 50 \times 10^{16}
Input
Input is given from Standard Input in the following format:
K
Output
Print a solution in the following format:
N
a_1 a_2 ... a_N
Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
Examples
Input
0
Output
4
3 3 3 3
Input
1
Output
3
1 0 3
Input
2
Output
2
2 2
Input
3
Output
7
27 0 0 0 0 0 0
Input
1234567894848
Output
10
1000 193 256 777 0 1 1192 1234567891011 48 425
Submitted Solution:
```
if __name__ == "__main__":
K = int(input())
N = 50
result = list(range(N))
for x in range(N):
result[x] += K // N
K %= N
for x in range(K):
result[x] += N + 1
for i in range(N):
result[x] -= 1
print (" ".join(map(str, result)))
``` | instruction | 0 | 48,669 | 5 | 97,338 |
No | output | 1 | 48,669 | 5 | 97,339 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028 | instruction | 0 | 48,770 | 5 | 97,540 |
"Correct Solution:
```
from string import digits
from itertools import product
S = input() + "$"
def parse(pri):
code = "+-*"
cur = 0
def number():
nonlocal cur
c = S[cur]
if c == '(':
cur += 1 # '('
v = expr(0)
cur += 1 # ')'
return v
num = 0
while S[cur] in digits:
num = 10*num + int(S[cur])
cur += 1
return num
def expr(level):
nonlocal cur
op = '+'
result = 0
while 1:
if level == 2:
v = number()
else:
v = expr(level+1)
if op == '+':
result += v
elif op == '-':
result -= v
else:
result *= v
c = S[cur]
if code.find(c) == -1 or pri[code.index(c)] != level:
break
op = c
cur += 1
return result
return expr(0)
ans = -10**30
for pri in product(range(3), repeat=3):
ans = max(ans, parse(pri))
print(ans)
``` | output | 1 | 48,770 | 5 | 97,541 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028 | instruction | 0 | 48,771 | 5 | 97,542 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
s = S()
t = []
k = 0
kf = False
for c in s:
if '0' <= c <= '9':
k *= 10
k += int(c)
kf = True
else:
if kf:
t.append(k)
k = 0
t.append(c)
kf = False
if kf:
t.append(k)
def calc(a,b,o):
if o == '+':
return a + b
if o == '-':
return a - b
return a * b
def f(s, ops):
t = s[:]
ff = True
while ff:
ff = False
ti = -1
for i in range(len(t)-1,-1,-1):
if t[i] == '(':
ti = i
break
if ti < 0:
break
tj = -1
for i in range(ti+1,len(t)):
if t[i] == ')':
tj = i + 1
break
t[ti:tj] = [f(t[ti+1:tj-1], ops)]
ff = True
for i in range(3):
nt = []
for c in t:
if isinstance(c, int):
if len(nt) > 1 and ops[nt[-1]] == i:
kt = calc(nt[-2],c,nt[-1])
nt[-2] = kt
del nt[-1]
else:
nt.append(c)
else:
nt.append(c)
t = nt
return t[0]
r = -inf
for a in itertools.product(range(3), repeat=3):
tr = f(t, {'+': a[0], '-': a[1], '*': a[2]})
if r < tr:
r = tr
return r
print(main())
``` | output | 1 | 48,771 | 5 | 97,543 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028 | instruction | 0 | 48,772 | 5 | 97,544 |
"Correct Solution:
```
from itertools import product
S = input()
for x in "()+*-":
S = S.replace(x, " {} ".format(x))
S = S.split()
def parse(pri, cur):
def number():
nonlocal cur
if S[cur].isdigit():
val = int(S[cur])
cur += 1
return val
cur += 1
val = expr(0)
cur += 1
return val
def expr(level):
nonlocal cur
if level == 3:
this = number()
else:
this = expr(level+1)
while cur < len(S) and S[cur] in "+-*" and pri[S[cur]] == level:
op = S[cur]
cur += 1
that = expr(level+1)
if op == "+":
this += that
elif op == "-":
this -= that
else:
this *= that
return this
return expr(0)
pri = {}
ans = -float("inf")
for P in product(range(3), repeat=3):
pri["+"], pri["-"], pri["*"] = P
ans = max(ans, parse(pri, 0))
print(ans)
``` | output | 1 | 48,772 | 5 | 97,545 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028 | instruction | 0 | 48,773 | 5 | 97,546 |
"Correct Solution:
```
INF = 10 ** 20
ops = ["+", "-", "*"]
nums = [str(i) for i in range(0, 10)]
priorities = []
for op1 in ops:
for op2 in ops:
for op3 in ops:
if op1 != op2 and op1 != op3 and op2 != op3:
priorities.append([[op1], [op2], [op3]])
priorities.append([[op1], [op2, op3]])
priorities.append([[op1, op2], [op3]])
priorities.append([[op1, op2, op3]])
def calc(num_lst, op_lst, priority):
for x in range(len(priority)):
while op_lst:
for i, op in enumerate(op_lst):
if op in priority[x]:
if op == "+":
num_lst = num_lst[:i] + [num_lst[i] + num_lst[i + 1]] + num_lst[i + 2:]
elif op == "-":
num_lst = num_lst[:i] + [num_lst[i] - num_lst[i + 1]] + num_lst[i + 2:]
else:
num_lst = num_lst[:i] + [num_lst[i] * num_lst[i + 1]] + num_lst[i + 2:]
op_lst.pop(i)
break
else:
break
return num_lst[0]
def parse(s, priority):
num_lst = []
op_lst = []
p = 0
if s[p] == "(":
p += 1
num, addp = parse(s[p:], priority)
num_lst.append(num)
p += addp
elif s[p] in nums:
acc = ""
while s[p] in nums:
acc += s[p]
p += 1
num_lst.append(int(acc))
while p < len(s) and s[p] != ")":
op_lst.append(s[p])
p += 1
acc = ""
if s[p] == "(":
p += 1
num, addp = parse(s[p:], priority)
num_lst.append(num)
p += addp
else:
#print(s[p])
while s[p] in nums:
acc += s[p]
p += 1
num_lst.append(int(acc))
p += 1
return (calc(num_lst, op_lst, priority), p)
def main():
s = "(" + input() + ")"
ans = -INF
for priority in priorities:
a, _ = parse(s, priority)
ans = max(ans, a)
print(ans)
main()
``` | output | 1 | 48,773 | 5 | 97,547 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028 | instruction | 0 | 48,774 | 5 | 97,548 |
"Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit(10 ** 8)
O = {'+': lambda l, r: l + r,
'-': lambda l, r: l - r,
'*': lambda l, r: l * r}
P = {'+': 0,
'-': 0,
'*': 0}
class Source():
def __init__(self, S, i=0):
self.S = S
self.pos = i
def peek(S):
return S.S[S.pos] if S.pos < len(S.S) else 'a'
def next(S):
S.pos += 1
def expr(S, i):
# print(S.pos)
if i == 0:
left = factor(S)
else:
left = expr(S, i - 1)
while peek(S) in O and P[peek(S)] == i:
ope = peek(S)
next(S)
if i == 0:
right = factor(S)
else:
right = expr(S, i - 1)
left = O[ope](left, right)
# print(left, i)
return left
def factor(S):
if peek(S) == '(':
next(S)
res = expr(S, 2)
next(S)
else:
res = num(S)
return res
def num(S):
sign = 1
if peek(S) == '-':
sign = -1
next(S)
res = 0
while '0' <= peek(S) <= '9':
res = res * 10 + int(peek(S))
next(S)
return sign * res
S = input()
ans = -int(1e19)
for plus in range(3):
P['+'] = plus
for minus in range(3):
P['-'] = minus
for times in range(3):
P['*'] = times
ans = max(ans, expr(Source(S), 2))
print(ans)
``` | output | 1 | 48,774 | 5 | 97,549 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028 | instruction | 0 | 48,775 | 5 | 97,550 |
"Correct Solution:
```
S = input()
y = [0,0,0]
ans = -1e64
def calc(s):
#print(s)
A = []
i = 0
cntr = 0
s2 = ""
for j,c in enumerate(s):
if cntr > 0:
if c == ')':
cntr -= 1
if cntr == 0:
A.append(calc(s2))
continue
if c == '(':
cntr += 1
s2 = s2 + c
else:
if c.isdigit():
i *= 10
i += int(c)
else:
if c == '(':
cntr += 1
if cntr == 1:
s2 = ""
continue
if s[j - 1] != ')':
A.append(i)
i = 0
A.append(c)
A.append(i)
#print(A)
for i in range(3):
j = 0
while j < len(A):
a = A[j]
if a == "+" and y[0] == i:
A[j - 1] += A[j + 1]
del A[j : j + 2]
elif a == "-" and y[1] == i:
A[j - 1] -= A[j + 1]
del A[j : j + 2]
elif a == "*" and y[2] == i:
A[j - 1] *= A[j + 1]
del A[j : j + 2]
else:
j += 1
return A[0]
for i in range(27):
y[0] = i // 9
y[1] = (i % 9) // 3
y[2] = i % 3
ans = max(ans, calc(S))
print(ans)
``` | output | 1 | 48,775 | 5 | 97,551 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028 | instruction | 0 | 48,776 | 5 | 97,552 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
from itertools import product
def parse(S, op):
"""
構文木を返す
Sは普通の記法の式
opは優先順序 通常は(*/: 1, +-: 2)
"""
S = "0+({})".format(S)
V = []
for x in list("()+-*"):
S = S.replace(x, " {} ".format(x))
i = 0
rank = 0
O = []
for s in S.split():
if s == "(":
rank += 1
elif s == ")":
rank -= 1
elif s.isdigit():
V.append(s)
i += 1
else:
V.append(s)
O.append([-rank, op[s], i])
i += 1
G = [[] for _ in range(len(V))]
P = [-1]*len(V)
O = sorted(O)
def get_pair(i):
while P[i] != -1:
i = P[i]
return i
for _, _, i in O:
l, r = get_pair(i-1), get_pair(i+1)
G[i].extend([l, r])
P[l], P[r] = i, i
p = O[-1][2]
return G, V, p
def make(G, V, p):
def call(i):
if V[i].isdigit():
return V[i]
else:
assert len(G[i]) == 2
left = call(G[i][0])
right = call(G[i][1])
return "({}{}{})".format(left, V[i], right)
return call(p)
S = input()
ans = -float("inf")
for a, b, c in product(range(3), repeat=3):
op = {}
op["+"] = a
op["-"] = b
op["*"] = c
ans = max(ans, eval(make(*parse(S, op))))
print(ans)
``` | output | 1 | 48,776 | 5 | 97,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028
Submitted Solution:
```
INF = 10 ** 20
ops = ["+", "-", "*"]
nums = [str(i) for i in range(0, 10)]
priorities = []
for op1 in ops:
for op2 in ops:
for op3 in ops:
if op1 != op2 and op1 != op3 and op2 != op3:
priorities += [[op1], [op2], [op3]]
priorities += [[op1], [op2, op3]]
priorities += [[op1, op2], [op3]]
priorities += [[op1, op2, op3]]
def calc(num_lst, op_lst, priority):
for x in range(len(priority)):
while op_lst:
for i, op in enumerate(op_lst):
if op in priority[x]:
if op == "+":
num_lst = num_lst[:i] + [num_lst[i] + num_lst[i + 1]] + num_lst[i + 2:]
elif op == "-":
num_lst = num_lst[:i] + [num_lst[i] - num_lst[i + 1]] + num_lst[i + 2:]
else:
num_lst = num_lst[:i] + [num_lst[i] * num_lst[i + 1]] + num_lst[i + 2:]
op_lst.pop(i)
break
else:
break
return num_lst[0]
def parse(s, priority):
num_lst = []
op_lst = []
p = 0
if s[p] == "(":
p += 1
num, addp = parse(s[p:], priority)
num_lst.append(num)
p += addp
elif s[p] in nums:
acc = ""
while s[p] in nums:
acc += s[p]
p += 1
num_lst.append(int(acc))
while p < len(s) and s[p] != ")":
op_lst.append(s[p])
p += 1
acc = ""
if s[p] == "(":
p += 1
num, addp = parse(s[p:], priority)
num_lst.append(num)
p += addp
else:
#print(s[p])
while s[p] in nums:
acc += s[p]
p += 1
num_lst.append(int(acc))
p += 1
return (calc(num_lst, op_lst, priority), p)
def main():
s = "(" + input() + ")"
ans = -INF
for priority in priorities:
a, _ = parse(s, priority)
ans = max(ans, a)
print(ans)
main()
``` | instruction | 0 | 48,777 | 5 | 97,554 |
No | output | 1 | 48,777 | 5 | 97,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028
Submitted Solution:
```
INF = 10 ** 20
ops = ["+", "-", "*"]
nums = [str(i) for i in range(0, 10)]
priorities = []
for op1 in ops:
for op2 in ops:
for op3 in ops:
if op1 != op2 and op1 != op3 and op2 != op3:
priorities.append([[op1], [op2], [op3]])
priorities.append([[op1], [op2, op3]])
priorities.append([[op1, op2], [op3]])
priorities.append([[op1, op2, op3]])
def calc(num_lst, op_lst, priority):
for x in range(len(priority)):
while op_lst:
for i, op in enumerate(op_lst):
if op in priority[x]:
if op == "+":
num_lst = num_lst[:i] + [num_lst[i] + num_lst[i + 1]] + num_lst[i + 2:]
elif op == "-":
num_lst = num_lst[:i] + [num_lst[i] - num_lst[i + 1]] + num_lst[i + 2:]
else:
num_lst = num_lst[:i] + [num_lst[i] * num_lst[i + 1]] + num_lst[i + 2:]
op_lst.pop(i)
break
else:
break
return num_lst[0]
def parse(s, priority):
num_lst = []
op_lst = []
p = 0
if s[p] == "(":
p += 1
num, addp = parse(s[p:], priority)
num_lst.append(num)
p += addp
elif s[p] in nums:
acc = ""
while s[p] in nums:
acc += s[p]
p += 1
num_lst.append(int(acc))
while p < len(s) and s[p] != ")":
op_lst.append(s[p])
p += 1
acc = ""
if s[p] == "(":
p += 1
num, addp = parse(s[p:], priority)
num_lst.append(num)
p += addp
else:
#print(s[p])
while s[p] in nums:
acc += s[p]
p += 1
num_lst.append(int(acc))
p += 1
return (calc(num_lst, op_lst, priority), p)
def main():
s = "(" + input() + ")"
ans = -INF
for priority in priorities:
a, _ = parse(s, priority)
print(priority, a)
ans = max(ans, a)
print(ans)
main()
``` | instruction | 0 | 48,778 | 5 | 97,556 |
No | output | 1 | 48,778 | 5 | 97,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Ikta, an elementary school student, received a piece of paper with mathematical formulas from his grandfather. Apparently, the grandfather will give you as much money as the answer to the formula. Ikta has only learned addition, subtraction, and multiplication, so only addition, subtraction, and multiplication are used in mathematical formulas. In normal calculation, multiplication must be calculated before addition and subtraction, but Ikta had a vague understanding of operator precedence, so for the time being, it is convenient to maximize the calculation result of the formula. I decided to consider a good priority.
Given the three binary operators + − × and a formula containing parentheses. Change the precedence of the three operators as you like and answer the calculation result when the formula is maximized.
However, note the following points.
* Operators are always left-associative. (Operators with the same precedence are always calculated from the left side of the formula.)
* Different operators may have the same precedence.
* Do not change the priority while calculating one formula.
Input
The input is given in the following format.
A formula consisting of numbers from 0 to 9 and the operators'+','-','*' and parentheses'(',')'
* To be precise, the input is in the format shown in BNF below.
> <expr> :: = (<expr>) | <number> | <expr> <op> <expr>
> <op> :: = + |-| *
<number> represents a non-negative integer.
Constraints
The input satisfies the following constraints.
* The formula is 200 characters or less.
* No matter what priority is set, it will not overflow as a result of calculation or in the middle of it as a 64-bit integer type.
Output
Output the maximum value obtained from the formula in one line.
Examples
Input
3-2*3
Output
3
Input
(5-3*4)*(0-2+1)
Output
21
Input
1-2+3-4+5-6*0
Output
3
Input
(1989967-3*1-211+4487)
Output
8511076028
Submitted Solution:
```
from string import digits
from itertools import product
S = input() + "$"
def parse(pri):
code = "+-*"
cur = 0
def number():
nonlocal cur
c = S[cur]
if c == '(':
cur += 1 # '('
v = expr(0)
cur += 1 # ')'
return v
num = 0
while S[cur] in digits:
num = 10*num + int(S[cur])
cur += 1
return num
def expr(level):
nonlocal cur
op = '+'
result = 0
while 1:
if level == 2:
v = number()
else:
v = expr(level+1)
if op == '+':
result += v
elif op == '-':
result -= v
else:
result *= v
c = S[cur]
if code.find(c) == -1 or pri[code.index(c)] != level:
break
op = c
cur += 1
return result
return expr(0)
ans = 0
for pri in product(range(3), repeat=3):
ans = max(ans, parse(pri))
print(ans)
``` | instruction | 0 | 48,779 | 5 | 97,558 |
No | output | 1 | 48,779 | 5 | 97,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
from math import *
from collections import *
from functools import *
from bisect import *
from itertools import *
from heapq import *
inf = float('inf')
ninf = -float('inf')
ip = input
alphal = "abcdefghijklmnopqrstuvwxyz"
alphau = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
mod = (10 ** 9) + 7
def ipl():
return list(map(int, ip().split()))
def ipn():
return int(ip())
def ipf():
return float(ip())
def ptrres(a):
for i in a:
print(i, end=" ")
print()
def solve():
n, a, b = ipl()
if a == 1:
if (n-1) % b == 0:
print("Yes")
else:
print("No")
return
x = 1
while x <= n:
if n % b == x % b:
print("YES")
return
x *= a
print("NO")
t = ipn()
for _ in range(t):
solve()
``` | instruction | 0 | 49,098 | 5 | 98,196 |
Yes | output | 1 | 49,098 | 5 | 98,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
def solve():
n, a, b = map(int, input().split())
if a == 1:
if (n - 1) % b == 0:
return 'YES'
return 'NO'
poww = 1
while poww <= n:
if poww % b == n % b:
return 'YES'
poww *= a
return 'NO'
t = int(input())
for _ in range(t):
ans = solve()
print(ans)
``` | instruction | 0 | 49,099 | 5 | 98,198 |
Yes | output | 1 | 49,099 | 5 | 98,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
# STEPS FOR B- Plus and Multiply
# write any random term of infinite set and simplify it you will get a
# general term i.e. a**(max power of a possible) + b*(function of a).
# so to achieve n if we can go from n to 1 backwards then n is in the set.
# so we can subtract b till it gets 0 and then divide by a till it becomes 1.But
# this will give us TLE.
# so we will first subtract by a then check if its divisible by b. if not then
# subtract by a**2/a**3/a**4,...,etc till it does not get divisible by b.
# as we are subtracting by exponetial elements it will take much less time.
# but for a==1 we do it seperately as in that case remainder=1(always)
# but the problem when a==1 and n%b==1 is that if a=b=n=1 then n%b==0
# so it will print NO but if we see n=1 is always possible.Thus (n-1)%b==0
# is a valid option.
def void():
n,a,b=map(int,input().split())
if a==1:
if (n-1)%b==0:
print("Yes")
else:
print("No")
else:
i=1
valid=False
while i<=n:
if (n-i)%b==0:
valid=True
break
i*=a
if valid:
print("Yes")
else:
print("No")
for i in range(int(input())):
void()
``` | instruction | 0 | 49,100 | 5 | 98,200 |
Yes | output | 1 | 49,100 | 5 | 98,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
import os,sys,io
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def readint():
return int(input().decode())
t = readint()
for _ in range(t):
n,a,b = [int(x) for x in input().decode().split(' ')]
if a == 1:
if (n-1)%b:
sys.stdout.write('No\n')
else:
sys.stdout.write('Yes\n')
continue
subtractfrom = 1
ans = False
while subtractfrom <= n:
diff = n - subtractfrom
if diff%b == 0:
ans = True
break
subtractfrom *= a
if ans:
sys.stdout.write('Yes\n')
else:
sys.stdout.write('No\n')
``` | instruction | 0 | 49,101 | 5 | 98,202 |
Yes | output | 1 | 49,101 | 5 | 98,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
for _ in range(int(input())):
n,a,b = [int(x) for x in input().split()]
if a == 1:
if n%b == 1:
print("YES")
else:
print("NO")
else:
x = 1
ch = True
while (x<=n):
if (n-x)%b == 0:
ch = False; break
x*=a
if ch:
print("NO")
else:
print("YEs")
``` | instruction | 0 | 49,102 | 5 | 98,204 |
No | output | 1 | 49,102 | 5 | 98,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
import sys
from sys import stdout
input = lambda: sys.stdin.readline().strip()
P = lambda: list(map(int, input().split()))
from math import factorial as f, gcd
from collections import deque, defaultdict as dd, Counter as C
from heapq import heapify, heappop, heappush, heappushpop, heapreplace, merge
from random import randint, choice, sample
import time
mod = 10**9+7
a = ord('a')
start = time.time()
def fast_exp(x, exp):
ans = 1
base = x
while exp:
if exp & 1:
ans *= base
base *= base
base %= mod
ans %= mod
exp >>= 1
return ans
def countBits(n):
count = 0
while n:
count += n & 1
n >>= 1
return count
def submasks(n):
#this is cool
#https://cp-algorithms.com/algebra/all-submasks.html
org = n
while n:
yield n
n = (n-1) & org
def d2k(d, k):
mapping = "0123456789ABCDEF"
res = []
while d != 0:
res.append(d%k)
d //= k
return res[::-1]
def solve():
n, a, b = P()
if a == 1:
print("Yes" if (n-1) % b == 0 else "No")
return
if (n-1) % b == 0:
print('Yes')
return
test = d2k(n, a)
start = 0
# print(test)
while True:
l = len(test)
passdown = 0
for i in range(start, l):
x = test[i]
if i == l-1:
test[i] = x + passdown
print(test)
if (x + passdown) % b == 0:
print('Yes')
return
else:
if start + 1 == l:
print('No')
return
else:
test[start+1] += a
test[start] = 0
start += 1
elif i == start:
# print('run')
passdown = (x-1) * a
test[i] = 1
# print(start, test)
else:
x += passdown
passdown = (x % b) * a
test[i] = x-(x%b)
# print(start)
# print(test)
# print(test)
print('----')
# arr = [0] * 101
# arr[1] = 1
# for i in range(2, 101):
# if (i%a==0 and arr[i//a]) or (i >= b and arr[i-b]):
# arr[i] = 1
# for i, x in enumerate(arr):
# if x:
# print(i)
tc = int(input())
for t in range(1, tc+1):
solve()
# solve()
# print(time.time()-start)
``` | instruction | 0 | 49,103 | 5 | 98,206 |
No | output | 1 | 49,103 | 5 | 98,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
from sys import stdin
input=stdin.readline
rn=lambda:int(input())
rns=lambda:map(int,input().split())
rl=lambda:list(map(int,input().split()))
rs=lambda:input().strip()
YN=lambda x:print('YES') if x else print('NO')
mod=10**9+7
for _ in range(rn()):
n,a,b=rns()
if a==1:
YN((n-1)%b==0 or n==1)
else:
root=a
ans = False
while a<n:
ans = ans or (n-a)%b==0
a*=root
YN(ans or n==1)
``` | instruction | 0 | 49,104 | 5 | 98,208 |
No | output | 1 | 49,104 | 5 | 98,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
from itertools import *
from heapq import *
from bisect import *
from io import BytesIO, IOBase
from typing import overload
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M = 998244353
EPS = 1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def divisors(n):
div=set()
for i in range(1,int(sqrt(n))+1):
if(n%i==0):
div.add(i)
div.add(n//i)
return div
for _ in range(Int()):
n,a,b = value()
ans = "NO"
if(a == 1):
if(n%b == 1): ans = "YES"
else:
x = 1
while(x < n):
if((n - x)%b == 0):
ans = "YES"
break
x *= a
if(n == 1): ans = 'YES'
print(ans)
# ans = set()
# qu = [1]
# for i in range(100):
# x = qu.pop(0)
# ans.add(x)
# qu.append(min(x*a,x + b))
# qu.append(max(x*a,x + b))
# for i in sorted(ans): print(i)
``` | instruction | 0 | 49,105 | 5 | 98,210 |
No | output | 1 | 49,105 | 5 | 98,211 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,513 | 5 | 99,026 |
"Correct Solution:
```
# AOJ 0046 Differential
# Python3 2018.6.14 bal4u
data = []
while True:
try:
data.append(float(input()))
except EOFError:
break
print(max(data)-min(data))
``` | output | 1 | 49,513 | 5 | 99,027 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,514 | 5 | 99,028 |
"Correct Solution:
```
maxv, minv = 0, 10**6
while True:
try:
height = float(input())
except:
break
maxv = height if maxv < height else maxv
minv = height if minv > height else minv
print(maxv-minv)
``` | output | 1 | 49,514 | 5 | 99,029 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,515 | 5 | 99,030 |
"Correct Solution:
```
try:
ls = []
while True:
ls.append(float(input()))
except:
print(max(ls) - min(ls))
``` | output | 1 | 49,515 | 5 | 99,031 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,516 | 5 | 99,032 |
"Correct Solution:
```
d = []
while True:
try:
d.append(float(input()))
except EOFError:
break
print(max(d)-min(d))
``` | output | 1 | 49,516 | 5 | 99,033 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,517 | 5 | 99,034 |
"Correct Solution:
```
import sys
mi=10**7
ma=-1
for l in sys.stdin:
mi=min(mi,float(l))
ma=max(ma,float(l))
print(ma-mi)
``` | output | 1 | 49,517 | 5 | 99,035 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,518 | 5 | 99,036 |
"Correct Solution:
```
x = []
while 1:
try:
h = float(input())
x.append(h)
except EOFError:
break
print(max(x)-min(x))
``` | output | 1 | 49,518 | 5 | 99,037 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,519 | 5 | 99,038 |
"Correct Solution:
```
def multi_input():
while True:
try:
yield float(input())
except EOFError:
break
a = list(multi_input())
print(max(a)-min(a))
``` | output | 1 | 49,519 | 5 | 99,039 |
Provide a correct Python 3 solution for this coding contest problem.
There is data that records the altitude of mountains that have been climbed so far. Create a program that reads this data and outputs the elevation difference between the highest and lowest mountains.
Input
The input is given in the following format:
Mountain height
...
...
The height of the mountain is given over multiple lines. All values entered are real numbers greater than or equal to 0 and less than or equal to 1,000,000. The number of mountain heights entered is 50 or less.
Output
The elevation difference between the highest mountain and the lowest mountain is output as a real number. The output may contain an error of 0.01 or less.
Example
Input
3776.0
1819.0
645.2
2004.1
1208.6
Output
3130.8 | instruction | 0 | 49,520 | 5 | 99,040 |
"Correct Solution:
```
h0 = float(input())
min = h0
max = h0
while True:
try:
h = float(input())
except:
break
if h<min:
min = h
if max<h:
max = h
print(f'{max-min:.2f}')
``` | output | 1 | 49,520 | 5 | 99,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
Input
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
Examples
Input
3
1 2 3
Output
1.000000000000000
Input
4
1 2 3 4
Output
2.000000000000000
Input
10
1 10 2 9 3 8 4 7 5 6
Output
4.500000000000000
Note
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
Submitted Solution:
```
def max_sum(nums, shift):
res = 0
res_m = 0
cur_sum = 0
cur_m_sum = 0
for i in range(len(nums)):
cur_sum += (nums[i] + shift)
cur_m_sum += (nums[i] + shift)
res = max(res, cur_sum)
cur_sum = max(0, cur_sum)
res_m = min(res_m, cur_m_sum)
cur_m_sum = min(0, cur_m_sum)
return res, -res_m
def weaks(nums, shift):
return max_sum(nums, shift)
def main():
int(input())
nums = list(map(int, input().split()))
l = -10000
r = 10000
ans = max(weaks(nums, 0))
w1 = 1
w2 = -1
PREC = 10**-6
while abs(w1 - w2) >= PREC and abs(w1 - w2) > PREC * max(w1, w2):
m = (r + l)/2
print (w1,w2,r,l,m)
w1, w2 = weaks(nums, m)
# print(w1, w2)
if w1 > w2:
r = m
else:
l = m
print ((w1 + w2) / 2)
# print (weaks([1,2,3],-2500.0))
main()
``` | instruction | 0 | 50,061 | 5 | 100,122 |
No | output | 1 | 50,061 | 5 | 100,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
n=int(input())
d=dict()
for i in range(1,int(n**0.5)+1):
cnt=0
while n%(i+1)==0:
n//=i+1
cnt+=1
d[i+1]=cnt
if n>1:
if n in d:
d[n]+=1
else:
d[n]=1
ans=0
d=list(d.items())
for i,r in d:
for j in range(1,44):
if j*(j+1)//2>r:
ans+=j-1
break
print(ans)
``` | instruction | 0 | 50,273 | 5 | 100,546 |
Yes | output | 1 | 50,273 | 5 | 100,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
import bisect as bi
n=int(input())
ans=0
summ=[(i+2)*(i+1)//2 for i in range(100000)]
#summ[i]:=1~iの和
for divide in range(2,int(n**0.5)+1):
if n==1:break
cnt=0
while n%divide==0:
cnt+=1
n//=divide
if cnt==0:continue
ans+=bi.bisect_right(summ,cnt)
print(ans+(n!=1))
``` | instruction | 0 | 50,274 | 5 | 100,548 |
Yes | output | 1 | 50,274 | 5 | 100,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
n = int(input())
s = []
for i in range(2, int(n**0.5)+1):
t = 0
while n%i == 0:
n //= i
t += 1
s.append(t)
if n == 1: break
if n > 1: s.append(1)
ans = 0
for i in s:
for j in range(1, i+1):
if i >= j:
i -= j
ans += 1
else:
break
print(ans)
``` | instruction | 0 | 50,275 | 5 | 100,550 |
Yes | output | 1 | 50,275 | 5 | 100,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
from math import sqrt
n=int(input())
t=0
i=2
while i*i<=n:
if n%i==0:
c=0
while n%i==0:
n//=i
c+=1
t+=(int(sqrt(8*c+1))-1)//2
i+=1
if n>1:
t+=1
print(t)
``` | instruction | 0 | 50,276 | 5 | 100,552 |
Yes | output | 1 | 50,276 | 5 | 100,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
N = int(input())
def furui(n):
if n < 2:
return []
plist = [1 for i in range(n)]
plist[0] = plist[1] = 0
for i in range(2, n):
if plist[i]:
for j in range(i*2, n, i):
plist[j] = 0
ret = []
for i in range(n):
if plist[i] != 0:
ret.append(i)
return ret
primes = furui(1000000)
count = 0
if N == 1:
print(0)
else:
for p in primes:
e = 1
while N % (p ** e) == 0:
z = p ** e
N //= z
count += 1
# print(z, N)
e += 1
if count:
print(count)
else:
print(1) # prime
``` | instruction | 0 | 50,277 | 5 | 100,554 |
No | output | 1 | 50,277 | 5 | 100,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
n = int(input())
ans = 0
for i in range(2,int(n**0.5)+1):
count = 1
loop = 1
while n % i == 0:
n = int(n/i)
count -= 1
if count == 0:
ans += 1
loop += 1
count = loop
if ans == 0 and n > 1:
ans = 1
print(ans)
``` | instruction | 0 | 50,278 | 5 | 100,556 |
No | output | 1 | 50,278 | 5 | 100,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
if n == 1:
print(0)
else:
ans = 0
N = n ** (0.5)
for i in range(2,int(N)+1):
if n%i == 0:
ans += 1
n = n//i
if ans == 0:
print(1)
else:
print(ans)
``` | instruction | 0 | 50,279 | 5 | 100,558 |
No | output | 1 | 50,279 | 5 | 100,559 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is participating in a programming contest, AXC001. He has just submitted his code to Problem A.
The problem has N test cases, all of which must be passed to get an AC verdict.
Takahashi's submission has passed M cases out of the N test cases.
Determine whether Takahashi's submission gets an AC.
Constraints
* 1 \leq N \leq 100
* 0 \leq M \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
Output
If Takahashi's submission gets an AC, print `Yes`; otherwise, print `No`.
Examples
Input
3 3
Output
Yes
Input
3 2
Output
No
Input
1 1
Output
Yes | instruction | 0 | 50,281 | 5 | 100,562 |
"Correct Solution:
```
N,M=map(int,input().split())
if N!=M:
print("No")
else:
print("Yes")
``` | output | 1 | 50,281 | 5 | 100,563 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is participating in a programming contest, AXC001. He has just submitted his code to Problem A.
The problem has N test cases, all of which must be passed to get an AC verdict.
Takahashi's submission has passed M cases out of the N test cases.
Determine whether Takahashi's submission gets an AC.
Constraints
* 1 \leq N \leq 100
* 0 \leq M \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
Output
If Takahashi's submission gets an AC, print `Yes`; otherwise, print `No`.
Examples
Input
3 3
Output
Yes
Input
3 2
Output
No
Input
1 1
Output
Yes | instruction | 0 | 50,287 | 5 | 100,574 |
"Correct Solution:
```
n,m = map(int, input().split())
print("Yes") if n-m==0 else print("No")
``` | output | 1 | 50,287 | 5 | 100,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0
Submitted Solution:
```
N = int(input())
V = map(int, input().split())
C = map(int, input().split())
print(sum([v - c for v, c in zip(V, C) if v > c]))
``` | instruction | 0 | 50,321 | 5 | 100,642 |
Yes | output | 1 | 50,321 | 5 | 100,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0
Submitted Solution:
```
n, *d = map(int, open(0).read().split())
print(sum([max(d[i] - d[i+n], 0) for i in range(n)]))
``` | instruction | 0 | 50,322 | 5 | 100,644 |
Yes | output | 1 | 50,322 | 5 | 100,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0
Submitted Solution:
```
N=int(input())
V=[int(x) for x in input().split()]
C=[int(x) for x in input().split()]
Y=0
for x in range(N):
Y+=max(0,V[x]-C[x])
print(Y)
``` | instruction | 0 | 50,323 | 5 | 100,646 |
Yes | output | 1 | 50,323 | 5 | 100,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0
Submitted Solution:
```
n = int(input())
v = list(map(int,input().split()))
c = list(map(int,input().split()))
print(sum([max(0,vv-cc) for vv, cc in zip(v,c)]))
``` | instruction | 0 | 50,324 | 5 | 100,648 |
Yes | output | 1 | 50,324 | 5 | 100,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0
Submitted Solution:
```
import sys
# import bisect
# import math
# import itertools
import numpy as np
# import collections
"""Template"""
class IP:
"""
入力を取得するクラス
"""
def __init__(self):
self.input = sys.stdin.readline
def I(self):
"""
1文字の取得に使います
:return: int
"""
return int(self.input())
def S(self):
"""
1文字の取得(str
:return: str
"""
return self.input()
def IL(self):
"""
1行を空白で区切りリストにします(int
:return: リスト
"""
return list(map(int, self.input().split()))
def SL(self):
"""
1行の文字列を空白区切りでリストにします
:return: リスト
"""
return list(map(str, self.input().split()))
def ILS(self, n):
"""
1列丸々取得します(int
:param n: 行数
:return: リスト
"""
return [int(self.input()) for _ in range(n)]
def SLS(self, n):
"""
1列丸々取得します（str
:param n: 行数
:return: リスト
"""
return [self.input() for _ in range(n)]
def SILS(self, n):
"""
Some Int LineS
横に複数、縦にも複数
:param n: 行数
:return: list
"""
return [self.IL() for _ in range(n)]
def SSLS(self, n):
"""
Some String LineS
:param n: 行数
:return: list
"""
return [self.SL() for _ in range(n)]
class Idea:
def __init__(self):
pass
def HF(self, p):
"""
Half enumeration
半分全列挙です
pの要素の和の組み合わせを作ります。
ソート、重複削除行います
:param p: list : 元となるリスト
:return: list : 組み合わせられた和のリスト
"""
return sorted(set(p[i] + p[j] for i in range(len(p)) for j in range(i, len(p))))
def Bfs2(self, a):
"""
bit_full_search2
bit全探索の改良版
全探索させたら2進数のリストと10進数のリストを返す
:return: list２つ : １個目 2進数(16桁) 2個目 10進数
"""
# 参考
# https://blog.rossywhite.com/2018/08/06/bit-search/
# https://atcoder.jp/contests/abc105/submissions/4088632
value = []
for i in range(1 << len(a)):
output = []
for j in range(len(a)):
if self.bit_o(i, j):
"""右からj+1番目のiが1かどうか判定"""
# output.append(a[j])
output.append(a[j])
value.append([format(i, 'b').zfill(16), sum(output)])
value.sort(key=lambda x: x[1])
bin = [value[k][0] for k in range(len(value))]
val = [value[k][1] for k in range(len(value))]
return bin, val
def S(self, s, r=0, m=-1):
"""
ソート関係行います。色々な設定あります。
:param s: 元となるリスト
:param r: reversするかどうか 0=False 1=True
:param m: （2次元配列）何番目のインデックスのソートなのか
:return: None
"""
r = bool(r)
if m == -1:
s.sort(reverse=r)
else:
s.sort(reverse=r, key=lambda x: x[m])
def bit_n(self, a, b):
"""
bit探索で使います。0以上のときにTrue出します
自然数だからn
:param a: int
:param b: int
:return: bool
"""
return bool((a >> b & 1) > 0)
def bit_o(self, a, b):
"""
bit探索で使います。1のときにTrue出すよ
oneで1
:param a: int
:param b: int
:return: bool
"""
return bool(((a >> b) & 1) == 1)
def ceil(self, x, y):
"""
Round up
小数点切り上げ割り算
:param x: int
:param y: int
:return: int
"""
return -(-x // y)
def ave(self, a):
"""
平均を求めます
:param a: list
:return: int
"""
return sum(a) / len(a)
def gcd(self, x, y):
if y == 0:
return x
else:
return self.gcd(y, x % y)
"""ここからメインコード"""
def main():
# 1文字に省略
r, e, p = range, enumerate, print
ip = IP()
id = Idea()
mod = 10 ** 9 + 7
"""この下から書いてね"""
n = ip.I()
v = ip.IL()
c = ip.IL()
ans = 0
for i in r(2 ** n):
x = []
y = []
for j in r(n):
if (i >> j) & 1:
x.append(v[j])
y.append(c[j])
x = np.array(x)
y = np.array(y)
ans = max(sum(x) - sum(y), ans)
p(ans)
main()
``` | instruction | 0 | 50,325 | 5 | 100,650 |
No | output | 1 | 50,325 | 5 | 100,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0
Submitted Solution:
```
n=int(input())
a=[int(input()) for _ range(n)]
b=[int(input()) for _ range(n)]
ans=0
for i in range(n):
if a[i]>b[i]:
ans+=a[i]
print(ans)
``` | instruction | 0 | 50,326 | 5 | 100,652 |
No | output | 1 | 50,326 | 5 | 100,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0
Submitted Solution:
```
N = int(input())
X = list(map(int, input().split())
Y = list(map(int, input().split())
ans = 0
for i in range(N):
ans += max(X[i] - Y[i]), 0)
print(ans)
``` | instruction | 0 | 50,327 | 5 | 100,654 |
No | output | 1 | 50,327 | 5 | 100,655 |
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