message stringlengths 2 59.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?
Input
The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
Output
First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.
Examples
Input
1
Output
5
5 6 7 8 9
Input
5
Output
0
Note
The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
Submitted Solution:
```
def main():
m = int(input())
zc = 0
i = 0
while zc < m:
i += 5
cr = i
while cr % 5 == 0:
zc += 1
cr = int(cr / 5)
if zc == m:
print(5)
ans = []
for k in range(i, i + 5):
ans.append(str(k))
print(' '.join(ans))
else:
print(0)
if __name__ == '__main__':
main()
``` | instruction | 0 | 36,552 | 20 | 73,104 |
Yes | output | 1 | 36,552 | 20 | 73,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?
Input
The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
Output
First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.
Examples
Input
1
Output
5
5 6 7 8 9
Input
5
Output
0
Note
The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
Submitted Solution:
```
m = int(input())
help = 1
data = [5]
while sum(data) <= m:
data.append(data[-1] * 5)
data.pop()
if m == sum(data):
print(0)
else:
print(5)
print(*range(m * 5, m * 5 + 5))
``` | instruction | 0 | 36,553 | 20 | 73,106 |
No | output | 1 | 36,553 | 20 | 73,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?
Input
The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
Output
First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.
Examples
Input
1
Output
5
5 6 7 8 9
Input
5
Output
0
Note
The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
Submitted Solution:
```
def main():
k = int(input())
answer = 0
q = []
for i in range(1, 1000000 + 1):
a = i // 2
b = i // 5
if min(a, b) == k:
q.append(i)
answer += 1
print(answer)
print(*q)
if __name__ == "__main__":
main()
``` | instruction | 0 | 36,554 | 20 | 73,108 |
No | output | 1 | 36,554 | 20 | 73,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?
Input
The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
Output
First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.
Examples
Input
1
Output
5
5 6 7 8 9
Input
5
Output
0
Note
The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
Submitted Solution:
```
#!/usr/bin/env python3
import collections, itertools, fractions, functools, heapq, math, queue, operator
def fac(n):
return functools.reduce(operator.mul, range(1, n+1))
def cnt0(n):
n = int(n)
r = 0
while n%10 == 0:
r += 1
n //= 10
return r
def solve(n):
if n%6 == 5:
print(0)
return
r = 5*(n - n//6)
print(5, n//6)
print(*range(r, r+5))
if __name__ == '__main__':
solve(int(input()))
``` | instruction | 0 | 36,555 | 20 | 73,110 |
No | output | 1 | 36,555 | 20 | 73,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?
Input
The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
Output
First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.
Examples
Input
1
Output
5
5 6 7 8 9
Input
5
Output
0
Note
The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
Submitted Solution:
```
n = m = int(input())
if (n == 5):
print('0')
exit()
k = 1
while k <= m:
k = k*5 + 1
while k:
if (n % 5 == 0):
print('0')
exit()
k //= 5
n -= m // k
m = m % k
s = ' '.join([str(5*n+i) for i in range(5)])
print('5')
print(s)
``` | instruction | 0 | 36,556 | 20 | 73,112 |
No | output | 1 | 36,556 | 20 | 73,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You generate real numbers s1, s2, ..., sn as follows:
* s0 = 0;
* si = si - 1 + ti, where ti is a real number chosen independently uniformly at random between 0 and 1, inclusive.
You are given real numbers x1, x2, ..., xn. You are interested in the probability that si ≤ xi is true for all i simultaneously.
It can be shown that this can be represented as <image>, where P and Q are coprime integers, and <image>. Print the value of P·Q - 1 modulo 998244353.
Input
The first line contains integer n (1 ≤ n ≤ 30).
The next n lines contain real numbers x1, x2, ..., xn, given with at most six digits after the decimal point (0 < xi ≤ n).
Output
Print a single integer, the answer to the problem.
Examples
Input
4
1.00
2
3.000000
4.0
Output
1
Input
1
0.50216
Output
342677322
Input
2
0.5
1.0
Output
623902721
Input
6
0.77
1.234567
2.1
1.890
2.9999
3.77
Output
859831967
Note
In the first example, the sought probability is 1 since the sum of i real numbers which don't exceed 1 doesn't exceed i.
In the second example, the probability is x1 itself.
In the third example, the sought probability is 3 / 8.
Submitted Solution:
```
# incomplete
# http://codeforces.com/contest/913/problem/H
from fractions import Fraction
import sys
import math
sys.exit()
mod = 998244353
def inverse(x, y):
a, ax, ay = x, 1, 0
b, bx, by = y, 0, 1
while b:
q = a // b
a , b = b , a - q * b
ax, bx = bx, ax - q * bx
ay, by = by, ay - q * by
assert a == 1
return ax % y
class poly:
def __init__(self, x = None):
''' DO NOT pass a mutable list here! '''
self.x = x or [0]
def __str__(self):
return ' + '.join(
'{}'.format(v) + (
'' if i == 0
else 'x' if i == 1
else 'x^{i}'.format(i=i)
)
for i, v in enumerate(self.x))
def __add__(a, b):
a = a.x[:]
b = b.x[:]
if len(a) > len(b):
b.extend([0] * (len(a)-len(b)))
elif len(b) > len(a):
a.extend([0] * (len(b)-len(a)))
return poly(list(map(sum, zip(a, b))))
def integrate(self):
return poly([0] + [v/(i+1) for i,v in enumerate(self.x)])
def __mul__(a, b):
''' Polynomial multiplication. '''
if isinstance(b, poly):
ans = [0] * (len(a.x) + len(b.x) - 1)
for i, ai in enumerate(a.x):
for j, bj in enumerate(b.x):
ans[i+j] += ai * bj
return poly(ans) # safe
else: # probably a number
return poly([b * v for v in a.x])
def __lshift__(self, a):
''' Multiply this with x**a. '''
return poly([0]*a + self.x)
def __matmul__(a, b):
''' Function composition. '''
ans = poly() # zero
base = poly([1])
for v in a.x:
ans += base * v
base *= b
return ans
def __neg__(self):
return poly([-v for v in self.x])
def __call__(self, a): # polynomial evaluation
base = 1
ans = 0
for v in self.x:
ans += base * v
base *= a
return ans
class func:
''' Continuous piecewise polynomial. '''
def __init__(self, t = None):
''' Initialize to the function f(x) = max(0, x) '''
self.t = t or [poly(), 0, poly([0, 1])]
# structure of t: [poly, num, poly, num, ..., poly, num, poly]
def __neg__(self):
return func([
-v if i % 2 == 0 # poly
else v # index
for i, v in enumerate(self.t)
])
def __add__(a, b):
a = [-math.inf] + a.t
b = [-math.inf] + b.t
ans = []
while a and b: # are non-empty
if a[-2] < b[-2]:
ans.append(a[-1] + b.pop())
ans.append(b.pop())
elif b[-2] < a[-2]:
ans.append(b[-1] + a.pop())
ans.append(a.pop())
else: # ==
ans.append(a.pop() + b.pop())
a.pop()
ans.append(b.pop())
assert (not a) and (not b) # both must be empty now
assert ans[-1] == -math.inf
ans.pop()
return func(ans[::-1])
def integrate(self):
ans = func([
v.integrate() if i % 2 == 0 # poly
else v # index
for i, v in enumerate(self.t)
])
for i in range(2, len(ans.t), 2):
# fit poly ans.t[i] with ans.t[i-2]
last_val = ans.t[i-2](ans.t[i-1])
this_val = ans.t[i ](ans.t[i-1])
ans.t[i].x[0] += last_val - this_val
assert ans.t[i ](ans.t[i-1]) == ans.t[i-2](ans.t[i-1])
return ans
def __lshift__(self, a):
''' Shift this function left (--) by (a) units. '''
p = poly([a, 1])
return func([
v @ p if i % 2 == 0 # poly
else v - a # index
for i, v in enumerate(self.t)
])
def __str__(self):
return 'Piecewise[{' + \
','.join(
'{{{}, x < {}}}'.format(v, r) for v, r in zip(self.t[:-1:2], self.t[1:-1:2])
) + \
'}, ' + \
str(self.t[-1]) + ']'
x = func()
print(x)
x += x << 1
print(x)
n = int(sys.stdin.readline())
DENOM = 1000000
x = [Fraction(round(float(sys.stdin.readline()) * DENOM), DENOM) for _ in range(n)]
``` | instruction | 0 | 36,605 | 20 | 73,210 |
No | output | 1 | 36,605 | 20 | 73,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let's focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal (∀) and existential (∃). You can read more about them here.
The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
* ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false.
* ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true.
The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
* ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true.
* ∃ x,x>x-1 is read as: there exists a real number x such that x is greater than x-1. This statement is true.
Moreover, these quantifiers can be nested. For example:
* ∀ x,∃ y,x<y is read as: for all real numbers x, there exists a real number y such that x is less than y. This statement is true since for every x, there exists y=x+1.
* ∃ y,∀ x,x<y is read as: there exists a real number y such that for all real numbers x, x is less than y. This statement is false because it claims that there is a maximum real number: a number y larger than every x.
Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
where ∧ denotes logical AND. That is, f(x_1,…, x_n) is true if every inequality x_{j_i}<x_{k_i} holds. Otherwise, if at least one inequality does not hold, then f(x_1,…,x_n) is false.
Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
Note that the order the variables appear in the statement is fixed. For example, if f(x_1,x_2):=(x_1<x_2) then you are not allowed to make x_2 appear first and use the statement ∀ x_2,∃ x_1, x_1<x_2. If you assign Q_1=∃ and Q_2=∀, it will only be interpreted as ∃ x_1,∀ x_2,x_1<x_2.
Input
The first line contains two integers n and m (2≤ n≤ 2⋅ 10^5; 1≤ m≤ 2⋅ 10^5) — the number of variables and the number of inequalities in the formula, respectively.
The next m lines describe the formula. The i-th of these lines contains two integers j_i,k_i (1≤ j_i,k_i≤ n, j_i≠ k_i).
Output
If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
On the next line, output a string of length n, where the i-th character is "A" if Q_i should be a universal quantifier (∀), or "E" if Q_i should be an existential quantifier (∃). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.
Examples
Input
2 1
1 2
Output
1
AE
Input
4 3
1 2
2 3
3 1
Output
-1
Input
3 2
1 3
2 3
Output
2
AAE
Note
For the first test, the statement ∀ x_1, ∃ x_2, x_1<x_2 is true. Answers of "EA" and "AA" give false statements. The answer "EE" gives a true statement, but the number of universal quantifiers in this string is less than in our answer.
For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
For the third test, the statement ∀ x_1, ∀ x_2, ∃ x_3, (x_1<x_3)∧ (x_2<x_3) is true: We can set x_3=max\\{x_1,x_2\}+1. | instruction | 0 | 36,973 | 20 | 73,946 |
Tags: dfs and similar, dp, graphs, math
Correct Solution:
```
import sys;input=sys.stdin.readline
N, M = map(int, input().split())
G1 = [[] for _ in range(N+1)]
G2 = [[] for _ in range(N+1)]
for _ in range(M):
x, y = map(int, input().split())
G1[x].append(y)
G2[y].append(x)
lens = [len(G2[i]) for i in range(N+1)]
stack = [x for x in range(N+1) if lens[x] == 0 and x != 0]
topo = [x for x in range(N+1) if lens[x] == 0 and x != 0]
#print(topo)
while stack:
cur = stack.pop()
for nb in G1[cur]:
lens[nb] -= 1
if lens[nb] == 0:
topo.append(nb)
stack.append(nb)
#print(topo)
if len(topo) != N:
print(-1)
sys.exit(0)
R = [0] * (N+1)
v1 = set()
v2 = set()
c = 0
for s in range(1, N+1):
if s in v1 or s in v2:
R[s] = 'E'
else:
R[s] = 'A'
c += 1
stack = [s]
while stack:
v = stack.pop()
for u in G1[v]:
if u in v1:
continue
v1.add(u)
stack.append(u)
stack = [s]
while stack:
v = stack.pop()
for u in G2[v]:
if u in v2:
continue
v2.add(u)
stack.append(u)
print(c)
print("".join(R[1:]))
``` | output | 1 | 36,973 | 20 | 73,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let's focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal (∀) and existential (∃). You can read more about them here.
The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
* ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false.
* ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true.
The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
* ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true.
* ∃ x,x>x-1 is read as: there exists a real number x such that x is greater than x-1. This statement is true.
Moreover, these quantifiers can be nested. For example:
* ∀ x,∃ y,x<y is read as: for all real numbers x, there exists a real number y such that x is less than y. This statement is true since for every x, there exists y=x+1.
* ∃ y,∀ x,x<y is read as: there exists a real number y such that for all real numbers x, x is less than y. This statement is false because it claims that there is a maximum real number: a number y larger than every x.
Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
where ∧ denotes logical AND. That is, f(x_1,…, x_n) is true if every inequality x_{j_i}<x_{k_i} holds. Otherwise, if at least one inequality does not hold, then f(x_1,…,x_n) is false.
Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
Note that the order the variables appear in the statement is fixed. For example, if f(x_1,x_2):=(x_1<x_2) then you are not allowed to make x_2 appear first and use the statement ∀ x_2,∃ x_1, x_1<x_2. If you assign Q_1=∃ and Q_2=∀, it will only be interpreted as ∃ x_1,∀ x_2,x_1<x_2.
Input
The first line contains two integers n and m (2≤ n≤ 2⋅ 10^5; 1≤ m≤ 2⋅ 10^5) — the number of variables and the number of inequalities in the formula, respectively.
The next m lines describe the formula. The i-th of these lines contains two integers j_i,k_i (1≤ j_i,k_i≤ n, j_i≠ k_i).
Output
If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
On the next line, output a string of length n, where the i-th character is "A" if Q_i should be a universal quantifier (∀), or "E" if Q_i should be an existential quantifier (∃). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.
Examples
Input
2 1
1 2
Output
1
AE
Input
4 3
1 2
2 3
3 1
Output
-1
Input
3 2
1 3
2 3
Output
2
AAE
Note
For the first test, the statement ∀ x_1, ∃ x_2, x_1<x_2 is true. Answers of "EA" and "AA" give false statements. The answer "EE" gives a true statement, but the number of universal quantifiers in this string is less than in our answer.
For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
For the third test, the statement ∀ x_1, ∀ x_2, ∃ x_3, (x_1<x_3)∧ (x_2<x_3) is true: We can set x_3=max\\{x_1,x_2\}+1. | instruction | 0 | 36,974 | 20 | 73,948 |
Tags: dfs and similar, dp, graphs, math
Correct Solution:
```
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input().strip()
return(list(s[:len(s)]))
def invr():
return(map(int,input().split()))
def from_file(f):
return f.readline
def build_graph(n, A, reversed=False):
edges = [[] for _ in range(n)]
for i, j in A:
i -= 1
j -= 1
if reversed:
j, i = i, j
edges[i].append(j)
return edges
def fill_min(s, edges, visited_dfs, visited, container):
visited[s] = True
visited_dfs.add(s)
for c in edges[s]:
if c in visited_dfs:
# cycle
return -1
if not visited[c]:
res = fill_min(c, edges, visited_dfs, visited, container)
if res == -1:
return -1
container[s] = min(container[s], container[c])
visited_dfs.remove(s)
return 0
def dfs(s, edges, visited, container):
stack = [s]
colors = {s: 0}
while stack:
v = stack.pop()
if colors[v] == 0:
colors[v] = 1
stack.append(v)
else:
# all children are visited
tmp = [container[c] for c in edges[v]]
if tmp:
container[v] = min(min(tmp), container[v])
colors[v] = 2 # finished
visited[v] = True
for c in edges[v]:
if visited[c]:
continue
if c not in colors:
colors[c] = 0 # white
stack.append(c)
elif colors[c] == 1:
# grey
return -1
return 0
def iterate_topologically(n, edges, container):
visited = [False] * n
for s in range(n):
if not visited[s]:
# visited_dfs = set()
# res = fill_min(s, edges, visited_dfs, visited, container)
res = dfs(s, edges, visited, container)
if res == -1:
return -1
return 0
def solve(n, A):
edges = build_graph(n, A, False)
container_forward = list(range(n))
container_backward = list(range(n))
res = iterate_topologically(n, edges, container_forward)
if res == -1:
return None
edges = build_graph(n, A, True)
iterate_topologically(n, edges, container_backward)
container = [min(i,j) for i,j in zip(container_forward, container_backward)]
res = sum((1 if container[i] == i else 0 for i in range(n)))
s = "".join(["A" if container[i] == i else "E" for i in range(n)])
return res, s
# with open('5.txt') as f:
# input = from_file(f)
n, m = invr()
A = []
for _ in range(m):
i, j = invr()
A.append((i, j))
result = solve(n, A)
if not result:
print (-1)
else:
print(f"{result[0]}")
print(f"{result[1]}")
``` | output | 1 | 36,974 | 20 | 73,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let's focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal (∀) and existential (∃). You can read more about them here.
The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
* ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false.
* ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true.
The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
* ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true.
* ∃ x,x>x-1 is read as: there exists a real number x such that x is greater than x-1. This statement is true.
Moreover, these quantifiers can be nested. For example:
* ∀ x,∃ y,x<y is read as: for all real numbers x, there exists a real number y such that x is less than y. This statement is true since for every x, there exists y=x+1.
* ∃ y,∀ x,x<y is read as: there exists a real number y such that for all real numbers x, x is less than y. This statement is false because it claims that there is a maximum real number: a number y larger than every x.
Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
where ∧ denotes logical AND. That is, f(x_1,…, x_n) is true if every inequality x_{j_i}<x_{k_i} holds. Otherwise, if at least one inequality does not hold, then f(x_1,…,x_n) is false.
Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
Note that the order the variables appear in the statement is fixed. For example, if f(x_1,x_2):=(x_1<x_2) then you are not allowed to make x_2 appear first and use the statement ∀ x_2,∃ x_1, x_1<x_2. If you assign Q_1=∃ and Q_2=∀, it will only be interpreted as ∃ x_1,∀ x_2,x_1<x_2.
Input
The first line contains two integers n and m (2≤ n≤ 2⋅ 10^5; 1≤ m≤ 2⋅ 10^5) — the number of variables and the number of inequalities in the formula, respectively.
The next m lines describe the formula. The i-th of these lines contains two integers j_i,k_i (1≤ j_i,k_i≤ n, j_i≠ k_i).
Output
If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
On the next line, output a string of length n, where the i-th character is "A" if Q_i should be a universal quantifier (∀), or "E" if Q_i should be an existential quantifier (∃). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.
Examples
Input
2 1
1 2
Output
1
AE
Input
4 3
1 2
2 3
3 1
Output
-1
Input
3 2
1 3
2 3
Output
2
AAE
Note
For the first test, the statement ∀ x_1, ∃ x_2, x_1<x_2 is true. Answers of "EA" and "AA" give false statements. The answer "EE" gives a true statement, but the number of universal quantifiers in this string is less than in our answer.
For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
For the third test, the statement ∀ x_1, ∀ x_2, ∃ x_3, (x_1<x_3)∧ (x_2<x_3) is true: We can set x_3=max\\{x_1,x_2\}+1.
Submitted Solution:
```
import sys;input=sys.stdin.readline
N, M = map(int, input().split())
G1 = [[] for _ in range(N+1)]
G2 = [[] for _ in range(N+1)]
for _ in range(M):
x, y = map(int, input().split())
G1[x].append(y)
G2[y].append(x)
vs = set()
for s in range(1, N+1):
stack = [s]
if s in vs:
continue
vs.add(s)
ps = set([s])
while stack:
v = stack.pop()
for u in G1[v]:
if u in ps:
print(-1)
sys.exit()
if u in vs:
continue
ps.add(u)
vs.add(u)
stack.append(u)
R = [0] * (N+1)
v1 = set()
v2 = set()
c = 0
for s in range(1, N+1):
if s in v1 or s in v2:
R[s] = 'E'
else:
R[s] = 'A'
c += 1
stack = [s]
while stack:
v = stack.pop()
for u in G1[v]:
if u in v1:
continue
v1.add(u)
stack.append(u)
stack = [s]
while stack:
v = stack.pop()
for u in G2[v]:
if u in v2:
continue
v2.add(u)
stack.append(u)
print(c)
print("".join(R[1:]))
``` | instruction | 0 | 36,975 | 20 | 73,950 |
No | output | 1 | 36,975 | 20 | 73,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let's focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal (∀) and existential (∃). You can read more about them here.
The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
* ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false.
* ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true.
The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
* ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true.
* ∃ x,x>x-1 is read as: there exists a real number x such that x is greater than x-1. This statement is true.
Moreover, these quantifiers can be nested. For example:
* ∀ x,∃ y,x<y is read as: for all real numbers x, there exists a real number y such that x is less than y. This statement is true since for every x, there exists y=x+1.
* ∃ y,∀ x,x<y is read as: there exists a real number y such that for all real numbers x, x is less than y. This statement is false because it claims that there is a maximum real number: a number y larger than every x.
Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
where ∧ denotes logical AND. That is, f(x_1,…, x_n) is true if every inequality x_{j_i}<x_{k_i} holds. Otherwise, if at least one inequality does not hold, then f(x_1,…,x_n) is false.
Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
Note that the order the variables appear in the statement is fixed. For example, if f(x_1,x_2):=(x_1<x_2) then you are not allowed to make x_2 appear first and use the statement ∀ x_2,∃ x_1, x_1<x_2. If you assign Q_1=∃ and Q_2=∀, it will only be interpreted as ∃ x_1,∀ x_2,x_1<x_2.
Input
The first line contains two integers n and m (2≤ n≤ 2⋅ 10^5; 1≤ m≤ 2⋅ 10^5) — the number of variables and the number of inequalities in the formula, respectively.
The next m lines describe the formula. The i-th of these lines contains two integers j_i,k_i (1≤ j_i,k_i≤ n, j_i≠ k_i).
Output
If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
On the next line, output a string of length n, where the i-th character is "A" if Q_i should be a universal quantifier (∀), or "E" if Q_i should be an existential quantifier (∃). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.
Examples
Input
2 1
1 2
Output
1
AE
Input
4 3
1 2
2 3
3 1
Output
-1
Input
3 2
1 3
2 3
Output
2
AAE
Note
For the first test, the statement ∀ x_1, ∃ x_2, x_1<x_2 is true. Answers of "EA" and "AA" give false statements. The answer "EE" gives a true statement, but the number of universal quantifiers in this string is less than in our answer.
For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
For the third test, the statement ∀ x_1, ∀ x_2, ∃ x_3, (x_1<x_3)∧ (x_2<x_3) is true: We can set x_3=max\\{x_1,x_2\}+1.
Submitted Solution:
```
import sys;input=sys.stdin.readline
N, M = map(int, input().split())
G1 = [[] for _ in range(N+1)]
G2 = [[] for _ in range(N+1)]
for _ in range(M):
x, y = map(int, input().split())
G1[x].append(y)
G2[y].append(x)
vs = set()
for s in range(1, N+1):
stack = [s]
vs.add(s)
ps = set([s])
while stack:
v = stack.pop()
for u in G1[v]:
if u in ps:
print(-1)
sys.exit()
if u in vs:
continue
ps.add(u)
vs.add(u)
stack.append(u)
R = [0] * (N+1)
v1 = set()
v2 = set()
c = 0
for s in range(1, N+1):
if s in v1 or s in v2:
R[s] = 'E'
else:
R[s] = 'A'
c += 1
stack = [s]
while stack:
v = stack.pop()
for u in G1[v]:
if u in v1:
continue
v1.add(u)
stack.append(u)
stack = [s]
while stack:
v = stack.pop()
for u in G2[v]:
if u in v2:
continue
v2.add(u)
stack.append(u)
print(c)
print("".join(R[1:]))
``` | instruction | 0 | 36,976 | 20 | 73,952 |
No | output | 1 | 36,976 | 20 | 73,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let's focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal (∀) and existential (∃). You can read more about them here.
The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
* ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false.
* ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true.
The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
* ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true.
* ∃ x,x>x-1 is read as: there exists a real number x such that x is greater than x-1. This statement is true.
Moreover, these quantifiers can be nested. For example:
* ∀ x,∃ y,x<y is read as: for all real numbers x, there exists a real number y such that x is less than y. This statement is true since for every x, there exists y=x+1.
* ∃ y,∀ x,x<y is read as: there exists a real number y such that for all real numbers x, x is less than y. This statement is false because it claims that there is a maximum real number: a number y larger than every x.
Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
where ∧ denotes logical AND. That is, f(x_1,…, x_n) is true if every inequality x_{j_i}<x_{k_i} holds. Otherwise, if at least one inequality does not hold, then f(x_1,…,x_n) is false.
Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
Note that the order the variables appear in the statement is fixed. For example, if f(x_1,x_2):=(x_1<x_2) then you are not allowed to make x_2 appear first and use the statement ∀ x_2,∃ x_1, x_1<x_2. If you assign Q_1=∃ and Q_2=∀, it will only be interpreted as ∃ x_1,∀ x_2,x_1<x_2.
Input
The first line contains two integers n and m (2≤ n≤ 2⋅ 10^5; 1≤ m≤ 2⋅ 10^5) — the number of variables and the number of inequalities in the formula, respectively.
The next m lines describe the formula. The i-th of these lines contains two integers j_i,k_i (1≤ j_i,k_i≤ n, j_i≠ k_i).
Output
If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
On the next line, output a string of length n, where the i-th character is "A" if Q_i should be a universal quantifier (∀), or "E" if Q_i should be an existential quantifier (∃). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.
Examples
Input
2 1
1 2
Output
1
AE
Input
4 3
1 2
2 3
3 1
Output
-1
Input
3 2
1 3
2 3
Output
2
AAE
Note
For the first test, the statement ∀ x_1, ∃ x_2, x_1<x_2 is true. Answers of "EA" and "AA" give false statements. The answer "EE" gives a true statement, but the number of universal quantifiers in this string is less than in our answer.
For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
For the third test, the statement ∀ x_1, ∀ x_2, ∃ x_3, (x_1<x_3)∧ (x_2<x_3) is true: We can set x_3=max\\{x_1,x_2\}+1.
Submitted Solution:
```
import sys;input=sys.stdin.readline
N, M = map(int, input().split())
G1 = [[] for _ in range(N+1)]
G2 = [[] for _ in range(N+1)]
for _ in range(M):
x, y = map(int, input().split())
G1[x].append(y)
G2[y].append(x)
vs = set()
for s in range(1, N+1):
stack = [s]
vs.add(s)
ps = set([s])
while stack:
v = stack.pop()
for u in G1[v]:
if u in ps:
print(-1)
sys.exit()
if u in vs:
continue
ps.add(u)
vs.add(u)
stack.append(u)
R = [0] * (N+1)
c = 0
for s in range(1, N+1):
if R[s] != 0:
continue
R[s] = 'A'
c += 1
stack = [s]
while stack:
v = stack.pop()
for u in G1[v]:
if R[u] != 0:
continue
R[u] = 'E'
stack.append(u)
stack = [s]
while stack:
v = stack.pop()
for u in G2[v]:
if R[u] != 0:
continue
R[u] = 'E'
stack.append(u)
print(c)
print("".join(R[1:]))
``` | instruction | 0 | 36,977 | 20 | 73,954 |
No | output | 1 | 36,977 | 20 | 73,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let's focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal (∀) and existential (∃). You can read more about them here.
The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
* ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false.
* ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true.
The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
* ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true.
* ∃ x,x>x-1 is read as: there exists a real number x such that x is greater than x-1. This statement is true.
Moreover, these quantifiers can be nested. For example:
* ∀ x,∃ y,x<y is read as: for all real numbers x, there exists a real number y such that x is less than y. This statement is true since for every x, there exists y=x+1.
* ∃ y,∀ x,x<y is read as: there exists a real number y such that for all real numbers x, x is less than y. This statement is false because it claims that there is a maximum real number: a number y larger than every x.
Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
where ∧ denotes logical AND. That is, f(x_1,…, x_n) is true if every inequality x_{j_i}<x_{k_i} holds. Otherwise, if at least one inequality does not hold, then f(x_1,…,x_n) is false.
Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
Note that the order the variables appear in the statement is fixed. For example, if f(x_1,x_2):=(x_1<x_2) then you are not allowed to make x_2 appear first and use the statement ∀ x_2,∃ x_1, x_1<x_2. If you assign Q_1=∃ and Q_2=∀, it will only be interpreted as ∃ x_1,∀ x_2,x_1<x_2.
Input
The first line contains two integers n and m (2≤ n≤ 2⋅ 10^5; 1≤ m≤ 2⋅ 10^5) — the number of variables and the number of inequalities in the formula, respectively.
The next m lines describe the formula. The i-th of these lines contains two integers j_i,k_i (1≤ j_i,k_i≤ n, j_i≠ k_i).
Output
If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
On the next line, output a string of length n, where the i-th character is "A" if Q_i should be a universal quantifier (∀), or "E" if Q_i should be an existential quantifier (∃). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.
Examples
Input
2 1
1 2
Output
1
AE
Input
4 3
1 2
2 3
3 1
Output
-1
Input
3 2
1 3
2 3
Output
2
AAE
Note
For the first test, the statement ∀ x_1, ∃ x_2, x_1<x_2 is true. Answers of "EA" and "AA" give false statements. The answer "EE" gives a true statement, but the number of universal quantifiers in this string is less than in our answer.
For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
For the third test, the statement ∀ x_1, ∀ x_2, ∃ x_3, (x_1<x_3)∧ (x_2<x_3) is true: We can set x_3=max\\{x_1,x_2\}+1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input().strip()
return(list(s[:len(s)]))
def invr():
return(map(int,input().split()))
def from_file(f):
return f.readline
def build_graph(n, A, reversed=False):
edges = [[] for _ in range(n)]
for i, j in A:
i -= 1
j -= 1
if reversed:
j, i = i, j
edges[i].append(j)
return edges
def fill_min(s, edges, visited_dfs, visited, container):
visited[s] = True
visited_dfs[s] = True
for c in edges[s]:
if visited_dfs[c]:
# cycle
return -1
res = fill_min(c, edges, visited_dfs, visited, container)
if res == -1:
return -1
container[s] = min(container[s], container[c])
return 0
def iterate_topologically(n, edges, container):
visited = [False] * n
for s in range(n):
if not visited[s]:
visited_dfs = [False] * n
res = fill_min(s, edges, visited_dfs, visited, container)
if res == -1:
return -1
return 0
def solve(n, m, A):
edges = build_graph(n, A)
container = list(range(n))
res = iterate_topologically(n, edges, container)
if res == -1:
return None
edges = build_graph(n, A, True)
iterate_topologically(n, edges, container)
res = sum((1 if container[i] == i else 0 for i in range(n)))
s = "".join(["A" if container[i] == i else "E" for i in range(n)])
return res, s
# with open('5.txt') as f:
# input = from_file(f)
n, m = invr()
A = []
for _ in range(m):
i, j = invr()
A.append((i, j))
result = solve(n, m, A)
if not result:
print (-1)
else:
print(f"{result[0]}")
print(f"{result[1]}")
``` | instruction | 0 | 36,978 | 20 | 73,956 |
No | output | 1 | 36,978 | 20 | 73,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,243 | 20 | 74,486 |
Tags: math
Correct Solution:
```
n,t=map(int,input().split())
if n==1 and t==10:print('-1')
elif t==10: print(10**(n-1))
elif t!=10: print(t*10**(n-1))
``` | output | 1 | 37,243 | 20 | 74,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,244 | 20 | 74,488 |
Tags: math
Correct Solution:
```
nt = input().strip().split(' ')
n = int(nt[0])
t = int(nt[1])
ans = int('9'*n) - int('9'*n)%t
if ans<=0:
print(-1)
else:
print(ans)
``` | output | 1 | 37,244 | 20 | 74,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,245 | 20 | 74,490 |
Tags: math
Correct Solution:
```
import math as mt
digits = lambda x: 1 + mt.floor(mt.log(x, 10))
n, t = map(int, input().split())
if digits(t) > n:
print(-1)
else:
print(str(t) * n if t < 10 else 10 ** (n-1))
``` | output | 1 | 37,245 | 20 | 74,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,246 | 20 | 74,492 |
Tags: math
Correct Solution:
```
#!/usr/bin/env python
n, t = map(int, input().split())
for i in range(10**(n - 1), 10**n):
if not i % t:
print(i)
quit()
print(-1)
``` | output | 1 | 37,246 | 20 | 74,493 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,247 | 20 | 74,494 |
Tags: math
Correct Solution:
```
A,B=list(map(int,input().split()))
if(A==1 and B==10):
print(-1)
else:
Aux=10**(A-1)
Cont=Aux%B
print((B-Cont)+Aux)
``` | output | 1 | 37,247 | 20 | 74,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,248 | 20 | 74,496 |
Tags: math
Correct Solution:
```
n,m = list(map(int,input().split()))
k,l,t=10**(n-1),10**(n),0
for i in range(k,l):
if i%m==0:
print(i)
t=t+1
break
if t==0:
print("-1")
``` | output | 1 | 37,248 | 20 | 74,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,249 | 20 | 74,498 |
Tags: math
Correct Solution:
```
n,t=map(int,input().split())
if t==10 and n==1:
print('-1')
else:
if t==10:
n-=1
a=str(t)
for i in range(n-1):
a+=str(0)
print(a)
``` | output | 1 | 37,249 | 20 | 74,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712 | instruction | 0 | 37,250 | 20 | 74,500 |
Tags: math
Correct Solution:
```
from sys import stdin
n,m=map(int,stdin.readline().strip().split())
if(m!=10):
x=m*pow(10,n-1)
else:
x=m*pow(10,n-2)
if x%m!=0:
print(-1)
else:
print(x)
``` | output | 1 | 37,250 | 20 | 74,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
(n,t)=map(int,input().split())
p=pow(10,n-1)
for i in range(1,pow(10,n)):
if p%t==0:
print(p)
break
elif(p<pow(10,n)-1):
p=p+1
else:
print("-1")
``` | instruction | 0 | 37,251 | 20 | 74,502 |
Yes | output | 1 | 37,251 | 20 | 74,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
num = tuple(map(int, input().split()))
n,t = num
if t==10:
if n==1:
print(-1)
else:
for i in range(1,n):
print(1, end="")
print(0, end="")
else:
for i in range(0,n):
print(t, end="")
``` | instruction | 0 | 37,252 | 20 | 74,504 |
Yes | output | 1 | 37,252 | 20 | 74,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
def main():
n, t = list(map(int, input().split()))
if t > 9 and n == 1:
print(-1)
elif t < 10:
print (t*(10**(n-1)))
else:
print (t*(10**(n-2)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 37,253 | 20 | 74,506 |
Yes | output | 1 | 37,253 | 20 | 74,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
import math
def main():
n,t=map(int,input().split(" "))
k=10**(n-1)
m=0
while len(str(k))<=n:
if k%t==0:
print(k)
m+=1
break
k+=1
if m!=1:
print(-1)
main()
``` | instruction | 0 | 37,254 | 20 | 74,508 |
Yes | output | 1 | 37,254 | 20 | 74,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
import random
n,m = [int(x) for x in input().split()]
z = []
cnt = 0
for i in range(n-1):
z.append(random.randint(1,9))
x = ''
for i in z:
x = x+str(i)
for i in range(0,9):
if int((x+str(i)))%m == 0:
print(x+str(i))
cnt = 1
break
if cnt == 0:
print('-1')
``` | instruction | 0 | 37,255 | 20 | 74,510 |
No | output | 1 | 37,255 | 20 | 74,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
n = list(map(int, input().split()))
if n[0] == 1 and n[1] == 10:
print(-1)
else:
if n[1] == 10:
x = str((n[1]) * (n[0] - 1)) + '0'
print(x)
else:
y = str(n[1]) * n[0]
print(y)
``` | instruction | 0 | 37,256 | 20 | 74,512 |
No | output | 1 | 37,256 | 20 | 74,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
n,t= map(int,input().split())
maxx=10**n - 1
result=t
while maxx>result:
result*=t
if result>maxx:
print(int(result/t))
if maxx<t:
print(-1)
``` | instruction | 0 | 37,257 | 20 | 74,514 |
No | output | 1 | 37,257 | 20 | 74,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
3 2
Output
712
Submitted Solution:
```
n,t=map(int,input().split())
a=10**(n-1)
b=10**n-1
flag=False
#print(a,b)
for i in range(a,b):
if(i%t==0):
flag=True
break
if(flag):
print(i)
else:
print(-1)
``` | instruction | 0 | 37,258 | 20 | 74,516 |
No | output | 1 | 37,258 | 20 | 74,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000 | instruction | 0 | 37,405 | 20 | 74,810 |
Tags: brute force, math, number theory
Correct Solution:
```
n, k = [int(x) for x in input().strip().split()]
cn2 = n
cn5 = n
cnt2 = 0
cnt5 = 0
while cn2 % 2 == 0:
if cnt2 == k:
break
cn2 /= 2
cnt2 += 1
while cn5 % 5 == 0:
if cnt5 == k:
break
cn5 /= 5
cnt5 += 1
ans = n * (2 ** (k - cnt2)) * (5 ** (k - cnt5))
print(ans)
``` | output | 1 | 37,405 | 20 | 74,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000 | instruction | 0 | 37,408 | 20 | 74,816 |
Tags: brute force, math, number theory
Correct Solution:
```
s=input().split()
n=int(s[0])
k=int(s[1])
def f(a, b):
while b:
a%=b
a,b=b,a
return a
print(int(n*(10**k)/f(n,10**k)))
``` | output | 1 | 37,408 | 20 | 74,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
Submitted Solution:
```
n, k = [int(x) for x in input().split()]
a = n
count2 = 0
count5 = 0
while a % 2 == 0 and count2 < k:
a = a // 2
count2 += 1
a = n
while a % 5 == 0 and count5 < k:
a = a // 5
count5 += 1
print(n * (2 ** (k - count2)) * (5 ** (k - count5)))
``` | instruction | 0 | 37,411 | 20 | 74,822 |
Yes | output | 1 | 37,411 | 20 | 74,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
Submitted Solution:
```
n, k = [int(i) for i in input().split()]
i = 0
while n % 2 == 0 and i != k:
n //= 2
i += 1
i = 0
while n % 5 == 0 and i != k:
n //= 5
i += 1
print(n * (10 ** k))
``` | instruction | 0 | 37,413 | 20 | 74,826 |
Yes | output | 1 | 37,413 | 20 | 74,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
Submitted Solution:
```
n, k = map(int, input().split())
new_n = n
cnt_2 = 0
while new_n % 2 == 0:
cnt_2 += 1
new_n //=2
cnt_5 = 0
while new_n % 5 == 0:
cnt_5 += 1
new_n //= 5
n *= 5 ** max(0, k - cnt_5)
n *= 2 ** max(0, k - cnt_2)
print(n)
``` | instruction | 0 | 37,414 | 20 | 74,828 |
Yes | output | 1 | 37,414 | 20 | 74,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
Submitted Solution:
```
n,k=map(int,input().split())
curr=k
while 1:
if n%10==5:
n*=2
n//=10
elif (n%10)%2==0 and n%10!=0:
n*=5
n//=10
elif n%10==0:
n//=10
else:
break
print(str(n)+("0"*k))
``` | instruction | 0 | 37,415 | 20 | 74,830 |
No | output | 1 | 37,415 | 20 | 74,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
Submitted Solution:
```
n, k = map(int, input().split())
ans = 1
f = 0
ans *= n
for i in range(k, -1, -1):
if f == 3:
break
if f != 1:
if n % 2 != 0:
ans *= 2 ** i
if f == 0:
f = 1
else:
f = 3
else:
n //= 2
if f != 2:
if n % 5 != 0:
ans *= 5 ** i
if f == 0:
f = 2
else:
f = 3
else:
n //= 5
print(ans)
``` | instruction | 0 | 37,416 | 20 | 74,832 |
No | output | 1 | 37,416 | 20 | 74,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
Submitted Solution:
```
n, k = map(int, input().split())
m = n
num2 = 0
num5 = 0
while m % 2 == 0:
m //= 2
num2 += 1
while m % 5 == 0:
m //= 5
num5 += 1
if k > min(num2, num5):
k -= min(num2, num5)
num2 -= min(num2, num5)
num5 -= min(num2, num5)
while num2 > 0:
n //= 2
num2 -= 1
while num5 > 0:
n //= 5
num5 -= 1
print(n * (10 ** k))
else:
print(n)
``` | instruction | 0 | 37,417 | 20 | 74,834 |
No | output | 1 | 37,417 | 20 | 74,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.
For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the k-rounding of n.
Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).
Output
Print the k-rounding of n.
Examples
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000
Submitted Solution:
```
n,k=[int(x) for x in input().split()]
if k==0:
print(n)
else:
ans=-1
i=1
while i*i<=n:
if (i*(10**k))%n==0:
ans=i*(10**k)
break
i+=1
if ans==-1:
print(n*(10**k))
else:
print(ans)
``` | instruction | 0 | 37,418 | 20 | 74,836 |
No | output | 1 | 37,418 | 20 | 74,837 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,652 | 20 | 75,304 |
"Correct Solution:
```
import re
import sys
s = sys.stdin.readline()
while s:
for i in range(10):
tmp = s.replace("X",str(i))
a,b,c = map(int,re.split("[+=]",tmp))
if a+b-c == 0:
if len(str(a)+str(b)+str(c))+2 == len(tmp.rstrip()):
print(i)
break
else:
continue
else:
print("NA")
s = sys.stdin.readline()
``` | output | 1 | 37,652 | 20 | 75,305 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,653 | 20 | 75,306 |
"Correct Solution:
```
def get_input():
while True:
try:
yield ''.join(input())
except EOFError:
break
N = list(get_input())
for l in range(len(N)):
s1,t = N[l].split('+')
s2,s3 = t.split('=')
ans = -1
if (len(s1) > 1 and s1[0] == "X") or (len(s2) > 1 and s2[0] == "X") or (len(s3) > 1 and s3[0] == "X"):
for i in range(1,10):
n1 = int(s1.replace("X",str(i)))
n2 = int(s2.replace("X",str(i)))
n3 = int(s3.replace("X",str(i)))
if n1 + n2 == n3:
ans = i
break
else:
for i in range(10):
n1 = int(s1.replace("X",str(i)))
n2 = int(s2.replace("X",str(i)))
n3 = int(s3.replace("X",str(i)))
if n1 + n2 == n3:
ans = i
break
if ans >= 0:
print(ans)
else:
print("NA")
``` | output | 1 | 37,653 | 20 | 75,307 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,654 | 20 | 75,308 |
"Correct Solution:
```
if __name__ == '__main__':
while True:
try:
a,z = input().split("=")
x,y = a.split("+")
start = 0
if (len(x) > 1 and x[0] == "X") or (len(y) > 1 and y[0] == "X") or (len(z) > 1 and z[0] == "X"):
start = 1
# x + y = zの形式に変換
ans = False
for i in range(start,10):
x1 = x.replace("X",str(i))
y1 = y.replace("X",str(i))
z1 = z.replace("X",str(i))
if int(x1) + int(y1) == int(z1):
print(i)
ans = True
break
if ans == False:
print("NA")
except EOFError:
break
``` | output | 1 | 37,654 | 20 | 75,309 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,655 | 20 | 75,310 |
"Correct Solution:
```
# Edit: 2014/09/28
# Lang: Python3
# Time: 0.xxs
#
if __name__ == "__main__":
while True:
try:
st0 = input()
# print(st0)
st0 = st0.strip("\n").replace("+", ",").replace("=", ",")
# print(st0)
#strA, strB, strC = str.split(",")
#print(strA, strB, strC)
for i in range(10):
#print(i,str(i))
a, b, c = st0.replace("X", str(i)).split(",")
if a[0] == "0" and len(a) > 1:
continue
if b[0] == "0" and len(b) > 1:
continue
if c[0] == "0" and len(c) > 1:
continue
#print(strI)
#pass
if int(a) + int(b) == int(c):
print(i)
break
else:
print("NA")
except:
break
``` | output | 1 | 37,655 | 20 | 75,311 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,656 | 20 | 75,312 |
"Correct Solution:
```
import re
while True:
try:
s = input()
except EOFError:
break
for x in '0123456789':
t = s.replace('X', x)
a, b, c = re.split(r'[+=]', t)
if len(a) > 1 and a[0] == '0': continue
if len(b) > 1 and b[0] == '0': continue
if len(c) > 1 and c[0] == '0': continue
a, b, c = map(int, (a, b, c))
if a + b != c: continue
print(x)
break
else:
print('NA')
``` | output | 1 | 37,656 | 20 | 75,313 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,657 | 20 | 75,314 |
"Correct Solution:
```
import sys
t='X'
for e in sys.stdin:
for i in'0123456789'[(e[0]==t)*(e[1]!='+')or('+X'in e)*('+X='not in e)or('=X'in e)*(e[-1]==t):]:
l,r=e.replace(t,i).split('=')
if sum(map(int,l.split('+')))==int(r):print(i);break
else:print('NA')
``` | output | 1 | 37,657 | 20 | 75,315 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,658 | 20 | 75,316 |
"Correct Solution:
```
import sys
while True:
try:
s=input()
ans=s.split("=")[1]
a,b=s.split("=")[0].split("+")
for X in range(10):
_a=a.replace("X",str(X))
if str(int(_a))!=_a:
continue
_b=b.replace("X",str(X))
if str(int(_b))!=_b:
continue
_ans=ans.replace("X",str(X))
if str(int(_ans))!=_ans:
continue
if int(_a)+int(_b)==int(_ans):
print(X)
break
else:
print("NA")
except EOFError:
sys.exit(0)
``` | output | 1 | 37,658 | 20 | 75,317 |
Provide a correct Python 3 solution for this coding contest problem.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1 | instruction | 0 | 37,659 | 20 | 75,318 |
"Correct Solution:
```
import sys
t='X'
for e in sys.stdin:
for i in'0123456789'[(e[0]==t)*(e[1]!='+')or('+X'in e)*('+X='not in e)or(e[-2:]=='=X'):]:
l,r=e.replace(t,i).split('=')
if sum(map(int,l.split('+')))==int(r):print(i);break
else:print('NA')
``` | output | 1 | 37,659 | 20 | 75,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1
Submitted Solution:
```
while True:
try:
s = input()
except:
break
t, sc = s.split("=")
sa, sb = t.split("+")
if (len(sa) > 1 and sa[0] == 'X') \
or (len(sb) > 1 and sb[0] == 'X') \
or (len(sc) > 1 and sc[0] == 'X'):
start = 1
else:
start = 0
f = 0
for i in range(start,10):
na = int( sa.replace("X", str(i)) )
nb = int( sb.replace("X", str(i)) )
nc = int( sc.replace("X", str(i)) )
if na + nb == nc:
f += 1
break
if f > 0:
print(i)
else:
print("NA")
``` | instruction | 0 | 37,660 | 20 | 75,320 |
Yes | output | 1 | 37,660 | 20 | 75,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1
Submitted Solution:
```
import re
while 1:
try:f=input()
except:break
s=any([len(x)>1 and x[0]=='X' for x in re.split('[+=]',f)])
f=f.replace('=','==')
for i in '0123456789'[s:]:
if eval(f.replace('X',i)):print(i);break
else:print('NA')
``` | instruction | 0 | 37,661 | 20 | 75,322 |
Yes | output | 1 | 37,661 | 20 | 75,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1
Submitted Solution:
```
import sys
import re
def line():return sys.stdin.readline().strip()
while True:
try:
e = re.findall(r"[0-9X]+",line())
ans = -1
for i in range(10):
a,b,c = [_.replace("X",str(i)) for _ in e]
if i == 0 \
and (\
(a[0] == "0" and len(a) != 1)\
or (b[0] == "0" and len(b) != 1)\
or (c[0] == "0" and len(c) != 1)\
):
continue
if int(a) + int(b) == int(c) :
ans = i
break
if ans == -1:
print("NA")
else:
print(ans)
except:
break
``` | instruction | 0 | 37,662 | 20 | 75,324 |
Yes | output | 1 | 37,662 | 20 | 75,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1
Submitted Solution:
```
import sys
t='X'
for e in sys.stdin:
for i in'0123456789'[(e[0]==t and'+'!=e[1]) or('+X='not in e and'+X'in e )or(e[-1]==t and'=X'in e):]:
l,r=e.replace(t,i).split('=')
if sum(map(int,l.split('+')))==int(r):print(i);break
else:print('NA')
``` | instruction | 0 | 37,663 | 20 | 75,326 |
Yes | output | 1 | 37,663 | 20 | 75,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The problem of hiding a part of a formula and searching for the hidden number is called verbal arithmetic. This time, I'm dealing with an expression in which some numbers in the expression are hidden by X. Enter the following formula and create a program that outputs the result.
Formula
* A simple one-line addition expression in the form of "number string + number string = number string".
* A "number sequence" is a sequence of numbers 0-9 and the letter X.
* It is assumed that the leftmost number in the "number string" with two or more digits is not 0.
* X must be at least one in the entire formula.
result
* The answer to the verbal arithmetic. It is one of 0 to 9 with a value of X such that the formula holds. Suppose there are no more than one answer.
* If there is no answer, the result should be “NA”.
Input
Given multiple datasets. For each dataset, an addition expression containing one or more Xs (a string of up to 126 characters without spaces) is given on one line. The number of datasets does not exceed 150.
Output
For each data set, output the result of the verbal arithmetic on one line. Print the numbers 0-9 or NA.
Example
Input
123+4X6=X79
12X+4X6=X79
XX22+89=X2XX
Output
5
NA
1
Submitted Solution:
```
while True:
try:
l = input().rstrip()
[a,b] = l.split("+")
[c,d] = b.split("=")
for i in range(10):
A = int(a.replace("X",str(i)))
C = int(c.replace("X",str(i)))
D = int(d.replace("X",str(i)))
if A+C == D:
print(i)
break
else:
print('NA')
except:
break
``` | instruction | 0 | 37,664 | 20 | 75,328 |
No | output | 1 | 37,664 | 20 | 75,329 |
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