message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular blog site edits the uploaded photos like this. It cuts a rectangular area out of them so that the ratio of height to width (i.e. the height / width quotient) can vary from 0.8 to 1.25 inclusively. Besides, at least one side of the cut area should have a size, equal to some power of number 2 (2x for some integer x). If those rules don't indicate the size of the cut are clearly, then the way with which the cut part possesses the largest area is chosen. Of course, both sides of the cut area should be integer. If there are several answers to this problem, you should choose the answer with the maximal height.
Input
The first line contains a pair of integers h and w (1 ≤ h, w ≤ 109) which are the height and width of the uploaded photo in pixels.
Output
Print two integers which are the height and width of the cut area.
Examples
Input
2 1
Output
1 1
Input
2 2
Output
2 2
Input
5 5
Output
5 4
Submitted Solution:
```
from math import log2, ceil, floor
h, w = map(int, input().split())
res = []
for i in range(int(log2(h)) + 1):
temp_h = 2 ** i
if int(ceil(temp_h / 1.25)) <= w:
temp_w = min(w, int(floor(temp_h / .8)))
s = temp_w * temp_h
res.append([s, temp_h, temp_w])
for i in range(int(log2(w)) + 1):
temp_w = 2 ** i
if int(ceil(temp_w * 0.8)) <= h:
temp_h = min(h, int(floor(temp_w * 1.25)))
s = temp_w * temp_h
res.append([s, temp_h, temp_w])
res.sort(reverse=True)
print(*res[0][1:])
``` | instruction | 0 | 25,139 | 23 | 50,278 |
Yes | output | 1 | 25,139 | 23 | 50,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular blog site edits the uploaded photos like this. It cuts a rectangular area out of them so that the ratio of height to width (i.e. the height / width quotient) can vary from 0.8 to 1.25 inclusively. Besides, at least one side of the cut area should have a size, equal to some power of number 2 (2x for some integer x). If those rules don't indicate the size of the cut are clearly, then the way with which the cut part possesses the largest area is chosen. Of course, both sides of the cut area should be integer. If there are several answers to this problem, you should choose the answer with the maximal height.
Input
The first line contains a pair of integers h and w (1 ≤ h, w ≤ 109) which are the height and width of the uploaded photo in pixels.
Output
Print two integers which are the height and width of the cut area.
Examples
Input
2 1
Output
1 1
Input
2 2
Output
2 2
Input
5 5
Output
5 4
Submitted Solution:
```
import math
h,w = map(int,input().split())
if h/w>=0.8 and h/w<=1.25 and ((math.log(h,2)%1==0) or (math.log(w,2)%1==0)):
print(h,w)
else:
w1 = 2**(math.log(w,2)//1)
h1 = min(h,(w1*1.25)//1)
h2 = 2**(math.log(h,2)//1)
w2 = min(w,(h2*1.25)//1)
if (h1/w1>=0.8 and h1/w1<=1.25) and (h2/w2>=0.8 and h2/w2<=1.25):
if h1>=h2 and h1*w1>=h2*w2:
print(int(h1),int(w1))
else:
print(int(h2),int(w2))
elif (h1/w1>=0.8 and h1/w1<=1.25):
print(int(h1),int(w1))
else:
print(int(h2),int(w2))
``` | instruction | 0 | 25,140 | 23 | 50,280 |
Yes | output | 1 | 25,140 | 23 | 50,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular blog site edits the uploaded photos like this. It cuts a rectangular area out of them so that the ratio of height to width (i.e. the height / width quotient) can vary from 0.8 to 1.25 inclusively. Besides, at least one side of the cut area should have a size, equal to some power of number 2 (2x for some integer x). If those rules don't indicate the size of the cut are clearly, then the way with which the cut part possesses the largest area is chosen. Of course, both sides of the cut area should be integer. If there are several answers to this problem, you should choose the answer with the maximal height.
Input
The first line contains a pair of integers h and w (1 ≤ h, w ≤ 109) which are the height and width of the uploaded photo in pixels.
Output
Print two integers which are the height and width of the cut area.
Examples
Input
2 1
Output
1 1
Input
2 2
Output
2 2
Input
5 5
Output
5 4
Submitted Solution:
```
from math import floor
def pic(x,y): #x/y
factors=[2**i for i in range(0,50)]
ans=[]
p=0
i=0
tx,ty=1,1
for i in factors:
if i<=x:
tx=i
else:
break
for i in factors:
if i<=y:
ty=i
else:
break
for ay in range(floor(tx/0.8),floor(tx/1.25)-1,-1):
if ay<=y :
c=tx/ay
if c>= 0.8 and c <= 1.25:
ans.append([tx,ay])
break
for ax in range(floor(ty*1.25),floor(ty*0.8)-1,-1):
if ax<=x :
c=ax/ty
if c>=0.8 and c<=1.25:
ans.append([ax,ty])
break
ans=sorted(ans,key=lambda s:s[0],reverse=True)
maxi=max(ans,key=lambda s:s[0]*s[1])
print(*maxi)
return ""
a,b=map(int,input().strip().split())
print(pic(a,b))
``` | instruction | 0 | 25,141 | 23 | 50,282 |
Yes | output | 1 | 25,141 | 23 | 50,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular blog site edits the uploaded photos like this. It cuts a rectangular area out of them so that the ratio of height to width (i.e. the height / width quotient) can vary from 0.8 to 1.25 inclusively. Besides, at least one side of the cut area should have a size, equal to some power of number 2 (2x for some integer x). If those rules don't indicate the size of the cut are clearly, then the way with which the cut part possesses the largest area is chosen. Of course, both sides of the cut area should be integer. If there are several answers to this problem, you should choose the answer with the maximal height.
Input
The first line contains a pair of integers h and w (1 ≤ h, w ≤ 109) which are the height and width of the uploaded photo in pixels.
Output
Print two integers which are the height and width of the cut area.
Examples
Input
2 1
Output
1 1
Input
2 2
Output
2 2
Input
5 5
Output
5 4
Submitted Solution:
```
from math import ceil,floor
l = []
for i in range(31):
l.append(2**i)
h,w = map(int,input().split())
h1 = 0
w1 = 0
maxi = 0
for i in l:
if i<w:
a = ceil(i*0.8)
b = floor(i*1.25)
if b<=h:
if i*b>=maxi:
maxi = i * b
h1 = b
w1 = i
elif a<=h:
if i*a>=maxi:
maxi = i * a
h1 = a
w1 = i
elif i==w:
a = ceil(i*0.8)
b = floor(i*1.25)
if b<=h:
if i*b>=maxi:
maxi = i * b
h1 = b
w1 = i
elif a<=h:
if i*a>=maxi:
maxi = i * a
h1 = a
w1 = i
h2 = 0
w2 = 0
w,h = h,w
maxi = 0
for i in l:
if i<w:
a = ceil(i/(1.25))
b = floor(i/(0.8))
if b<=h:
if i*b>=maxi:
maxi = i * b
h2 = b
w2 = i
elif a<=h:
if i*a>=maxi:
maxi = i * a
h2 = a
w2 = i
elif i==w:
a = ceil(i/(1.25))
b = floor(i/(0.8))
if b<=h:
if i*b>=maxi:
maxi = i * b
h2 = b
w2 = i
elif a<=h:
if i*a>=maxi:
maxi = i * a
h2 = a
w2 = i
if h1*w1>h2*w2:
print(h1,w1)
elif h1*w1 == h2*w2:
print(max(h1, h2), w1)
else:
print(h2,w2)
``` | instruction | 0 | 25,142 | 23 | 50,284 |
No | output | 1 | 25,142 | 23 | 50,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular blog site edits the uploaded photos like this. It cuts a rectangular area out of them so that the ratio of height to width (i.e. the height / width quotient) can vary from 0.8 to 1.25 inclusively. Besides, at least one side of the cut area should have a size, equal to some power of number 2 (2x for some integer x). If those rules don't indicate the size of the cut are clearly, then the way with which the cut part possesses the largest area is chosen. Of course, both sides of the cut area should be integer. If there are several answers to this problem, you should choose the answer with the maximal height.
Input
The first line contains a pair of integers h and w (1 ≤ h, w ≤ 109) which are the height and width of the uploaded photo in pixels.
Output
Print two integers which are the height and width of the cut area.
Examples
Input
2 1
Output
1 1
Input
2 2
Output
2 2
Input
5 5
Output
5 4
Submitted Solution:
```
from math import ceil,floor
l = []
for i in range(39):
l.append(2**i)
h,w = map(int,input().split())
h1 = 0
w1 = 0
maxi = 0
for i in l:
if i<=w:
a = ceil(i*0.8)
b = floor(i*1.25)
if b<=h:
if i*b>=maxi:
maxi = i * b
h1 = b
w1 = i
elif a<=h:
if i*a>=maxi:
maxi = i * a
h1 = a
w1 = i
h2 = 0
w2 = 0
w,h = h,w
maxi = 0
for i in l:
if i<=w:
a = ceil(i/(1.25))
b = floor(i/(0.8))
if b<=h:
if i*b>=maxi:
maxi = i * b
h2 = b
w2 = i
elif a<=h:
if i*a>=maxi:
maxi = i * a
h2 = a
w2 = i
w2,h2 = h2,w2
if h1*w1>h2*w2:
print(h1,w1)
elif h1*w1 == h2*w2:
if h1>h2:
print(h1,w1)
else:
print(h2,w2)
else:
print(h2,w2)
``` | instruction | 0 | 25,143 | 23 | 50,286 |
No | output | 1 | 25,143 | 23 | 50,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular blog site edits the uploaded photos like this. It cuts a rectangular area out of them so that the ratio of height to width (i.e. the height / width quotient) can vary from 0.8 to 1.25 inclusively. Besides, at least one side of the cut area should have a size, equal to some power of number 2 (2x for some integer x). If those rules don't indicate the size of the cut are clearly, then the way with which the cut part possesses the largest area is chosen. Of course, both sides of the cut area should be integer. If there are several answers to this problem, you should choose the answer with the maximal height.
Input
The first line contains a pair of integers h and w (1 ≤ h, w ≤ 109) which are the height and width of the uploaded photo in pixels.
Output
Print two integers which are the height and width of the cut area.
Examples
Input
2 1
Output
1 1
Input
2 2
Output
2 2
Input
5 5
Output
5 4
Submitted Solution:
```
"""
def cf31B():
#print("Hello World")
from sys import stdin,stdout
inp = list(stdin.readline().strip().split('@'))
flag = True
if len(inp)>=2:
if len(inp[0])==0 or len(inp[-1])==0:
flag = False
if flag:
for i in range(1,len(inp)-1):
if len(inp[i]) < 2:
flag = False
break
else:
flag = False
answer = ""
if flag:
for i , j in enumerate(inp):
if i==0:
answer+=j
elif i==len(inp)-1:
answer+='@'+j
else:
answer+='@'+j[:-1]+','+j[-1]
else:
answer = "No solution"
stdout.write(answer+"\n")
def cf75B():
from sys import stdin, stdout
myname = stdin.readline().strip()
points = {}
for _ in range(int(stdin.readline())):
inp = stdin.readline().strip().split()
a = inp[0]
b = inp[-2][:-2]
p = 0
if inp[1][0] == 'p':
p=15
elif inp[1][0] == 'c':
p=10
else:
p = 5
if a==myname:
if b in points:
points[b] += p
else:
points[b] = p
elif b==myname:
if a in points:
points[a] += p
else:
points[a] = p
else:
if a not in points:
points[a] = 0
if b not in points:
points[b] = 0
revpoints = {}
mylist = []
for key in points:
if points[key] in revpoints:
revpoints[points[key]].append(key)
else:
revpoints[points[key]]=[key,]
for key in revpoints:
revpoints[key].sort()
for key in revpoints:
mylist.append((key,revpoints[key]))
mylist.sort(reverse=True)
for i,j in enumerate(mylist):
for name in j[1]:
stdout.write(name+'\n')
def cf340B():
from sys import stdin, stdout
maxim = -1e9
permutation = []
chosen = [False for x in range(300)]
def search():
if len(permutation)==4:
maxim = max(maxim,calculate_area(permutation))
else:
for i in range(len(colist)):
if chosen[i]:
continue
chosen[i]=True
permutation.append(colist[i])
search()
chosen[i]=False
permutation = permutation[:-1]
def calculate_area(arr):
others = []
leftmost = arr[0]
rightmost = arr[0]
for i in arr:
if i[0]<leftmost[0]:
leftmost = i
elif i[0]>rightmost[0]:
rightmost = i
for i in arr:
if i!=leftmost and i!=rightmost:
others.append(i)
base_length = ((leftmost[0]-rightmost[0])**2 + (leftmost[1]-rightmost[1])**2)**0.5
#print(base_length)
if base_length == 0:
return 0
m = (rightmost[1] - leftmost[1])/(rightmost[0]-leftmost[0])
k = leftmost[1] + (-1)*leftmost[0]*m
#print(m)
#print(k)
t1 = abs(k+m*others[0][0]-others[0][1])/((1+m*m)**0.5)*base_length*0.5
t2 = abs(k+m*others[1][0]-others[1][1])/((1+m*m)**0.5)*base_length*0.5
#print(t1)
#print(t2)
return t1+t2
colist = []
for _ in range(int(stdin.readline())):
x,y = map(int,stdin.readline().split())
colist.append((x,y))
#print(colist)
#ans = calculate_area((0,0),(0,4),(4,0),(4,4))
#print(ans)
search()
stdout.write(str(maxim)+'\n')
def cf29B():
from sys import stdin,stdout
l,d,v,g,r=map(int,stdin.readline().split())
t=0
t+=d/v
cycle=g+r
rem = t
while rem > cycle:
rem -= cycle
if rem >= g:
rem -=g
t+=(r-rem)
t+=(l-d)/v
stdout.write("{0:.6f}\n".format(t))
"""
def cf53B():
from sys import stdin,stdout
from math import floor
h,w = map(int,stdin.readline().split())
maxim = -1e9
ans = None
for i in range(31):
temp = 1<<i
if temp <= h:
for other in range(floor(1/0.8*temp),floor(1/1.25*temp)+1):
if other <=w:
if temp*other > maxim:
maxim = temp*other
ans = (temp,other)
if temp*other == maxim:
if temp >= ans[0]:
ans = (temp,other)
if temp <= w:
for other in range(floor(0.8*temp),floor(1.25*temp)+1):
if other <=h:
if temp*other > maxim:
maxim = temp*other
ans=(other,temp)
if temp*other == maxim:
if other >= ans[0]:
ans = (other,temp)
stdout.write(str(ans[0])+' '+str(ans[1])+'\n')
if __name__=='__main__':
cf53B()
``` | instruction | 0 | 25,144 | 23 | 50,288 |
No | output | 1 | 25,144 | 23 | 50,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular blog site edits the uploaded photos like this. It cuts a rectangular area out of them so that the ratio of height to width (i.e. the height / width quotient) can vary from 0.8 to 1.25 inclusively. Besides, at least one side of the cut area should have a size, equal to some power of number 2 (2x for some integer x). If those rules don't indicate the size of the cut are clearly, then the way with which the cut part possesses the largest area is chosen. Of course, both sides of the cut area should be integer. If there are several answers to this problem, you should choose the answer with the maximal height.
Input
The first line contains a pair of integers h and w (1 ≤ h, w ≤ 109) which are the height and width of the uploaded photo in pixels.
Output
Print two integers which are the height and width of the cut area.
Examples
Input
2 1
Output
1 1
Input
2 2
Output
2 2
Input
5 5
Output
5 4
Submitted Solution:
```
import math
h,w = map(int,input().split())
if h/w>=0.8 and h/w<=1.25 and ((math.log(h,2)%1==0) or (math.log(w,2)%1==0)):
print(h,w)
else:
if h>=w:
w = 2**(math.log(w,2)//1)
h = min(h,(w*1.25)//1)
else:
h = 2**(math.log(h,2)//1)
w = min(h,(w*1.25)//1)
print(int(h),int(w))
``` | instruction | 0 | 25,145 | 23 | 50,290 |
No | output | 1 | 25,145 | 23 | 50,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,156 | 23 | 50,312 |
Tags: constructive algorithms
Correct Solution:
```
import sys
def check(arr):
n = len(arr)
for i in range(n):
for j in range(i + 1, n):
diff = 0
for k in range(n):
if arr[i][k] == arr[j][k]:
diff += 1
else:
diff -= 1
if diff:
return (arr[i], arr[j])
return False
def inv(arr):
return [(not x) for x in arr]
def gen(n):
if n == 1:
return [[True]]
curr = gen(n // 2)
ans = []
ans.extend((x + x for x in curr))
ans.extend((x + inv(x) for x in curr))
return ans
k = int(input())
ans = gen(2 ** k)
for x in ans:
sys.stdout.write("".join(["+" if el else "*" for el in x]))
sys.stdout.write("\n")
``` | output | 1 | 25,156 | 23 | 50,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,157 | 23 | 50,314 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
ans=[[1]]
k=0
while k<n:
k+=1
for i in range(len(ans)):
ans[i].extend(ans[i])
for i in range(len(ans)):
ans.append(ans[i][:])
for i in range(int(len(ans)/2),len(ans)):
for j in range(int(len(ans[i])/2),len(ans[i])):
ans[i][j]*=(-1)
# print(ans)
for i in ans:
for j in i:
if j==-1:
print('*',end='')
else:
print('+',end='')
print()
``` | output | 1 | 25,157 | 23 | 50,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,158 | 23 | 50,316 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
h = [1]
for i in range(n):
if i == 0:
h = [[1,1], [1,-1]]
else:
x = []
for row in h:
p = row.copy()
p.extend(p)
x.append(p)
for row in h:
p = row.copy()
p.extend([-x for x in p])
x.append(p)
# print(x)
h = x.copy()
if n == 0:
print('+')
else:
# print(h)
for row in h:
for i in row:
c = '+' if i == 1 else '*'
print(c,end='')
print('')
``` | output | 1 | 25,158 | 23 | 50,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,159 | 23 | 50,318 |
Tags: constructive algorithms
Correct Solution:
```
from functools import reduce
from operator import *
from math import *
from sys import *
from string import *
from collections import *
setrecursionlimit(10**7)
dX= [-1, 1, 0, 0,-1, 1,-1, 1]
dY= [ 0, 0,-1, 1, 1,-1,-1, 1]
RI=lambda: list(map(int,input().split()))
RS=lambda: input().rstrip().split()
#################################################
def inv(x):
s=""
for i in x:
s+= "+*"[i=='+']
return s
x=['+']
n=RI()[0]
for i in range(n):
t=[]
for j in range(len(x)):
t.append(x[j]+x[j])
t.append(x[j]+inv(x[j]))
x=t
print(*x, sep='\n')
``` | output | 1 | 25,159 | 23 | 50,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,160 | 23 | 50,320 |
Tags: constructive algorithms
Correct Solution:
```
inv=lambda t:''.join('*'if u=='+'else'+'for u in t)
n=int(input())
r=['+']
for _ in range(n):
r=[r[i] * 2 for i in range(len(r))] + [r[i]+inv(r[i]) for i in range(len(r))]
print(*r,sep='\n')
``` | output | 1 | 25,160 | 23 | 50,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,161 | 23 | 50,322 |
Tags: constructive algorithms
Correct Solution:
```
def vec(x):
if x == 0:
return [[1]]
x -= 1
s = vec(x)
y = vec(x)
for i in range(2**x):
for j in range(2**x):
y[i][j] = -y[i][j]
a = [s[i]+y[i] for i in range(2**x)]
b = [s[i]+s[i] for i in range(2**x)]
return a + b
x = int(input())
s = vec(x)
for i in range(2**x):
res = ''
for j in range(2**x):
res += '+' if s[i][j] == 1 else '*'
print(res)
``` | output | 1 | 25,161 | 23 | 50,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,162 | 23 | 50,324 |
Tags: constructive algorithms
Correct Solution:
```
def f(n):
if n == 0:
return ["+"]
else:
pat = f(n-1)
new = [c+c for c in pat]
new.extend(["".join(["+" if d == "*" else "*" for d in c]) + c for c in pat])
return new
n = int(input())
for i in f(n):
print(i)
``` | output | 1 | 25,162 | 23 | 50,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0 | instruction | 0 | 25,163 | 23 | 50,326 |
Tags: constructive algorithms
Correct Solution:
```
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
#from __future__ import print_function, division #while using python2
# from itertools import accumulate
# from collections import defaultdict, Counter
def modinv(n,p):
return pow(n,p-2,p)
def flip(s):
temp = []
for c in s:
if c == '+':
temp.append('*')
else:
temp.append('+')
return "".join(temp)
# def product(strings):
# for i in range(len(strings)):
# for j in range(i+1, len(strings)):
# ans = 0
# s1 = strings[i]
# s2 = strings[j]
# for k in range(len(s1)):
# if s1[k] == s2[k]:
# ans += 1
# else:
# ans -= 1
# print(s1, s2, ans)
def main():
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
k = int(input())
prev = ["++", "+*"]
if k == 0:
print("+")
return
if k == 1:
for p in prev:
print(p)
return
for i in range(2, k+1):
p1 = [x for x in prev]
p2 = [flip(x) for x in prev]
temp = []
for j in range(len(prev)):
temp.append(p1[j] + prev[j])
for j in range(len(prev)):
temp.append(p2[j] + prev[j])
prev = temp
for p in prev:
print(p)
# product(prev)
#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__ == '__main__':
main()
``` | output | 1 | 25,163 | 23 | 50,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
def vec(x):
if x == 0:
return [[1]]
s = vec(x-1)
y = vec(x-1)
for i in range(2**(x-1)):
for j in range(2**(x-1)):
y[i][j] = -y[i][j]
out = [s[i]+y[i] for i in range(2**(x-1))]
out2 = [s[i]+s[i] for i in range(2**(x-1))]
return (out+out2)
x = int(input())
s=vec(x)
for i in range(2**x):
ret = ""
for j in range(2**x):
if s[i][j] == 1:
ret+="+"
else:
ret+="*"
print(ret)
``` | instruction | 0 | 25,164 | 23 | 50,328 |
Yes | output | 1 | 25,164 | 23 | 50,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
k = int(input())
def reflect(matrix):
temp = list(matrix)
for i in range(len(matrix)):
matrix[i] = list(matrix[i])
temp.reverse()
return temp
def invert(matrix):
temp = list(matrix)
for i in range(len(temp)):
temp[i] = list(temp[i])
for i in range(len(temp)):
for j in range(len(temp[i])):
temp[i][j] = not temp[i][j]
return temp
seed = [[True]]
while k > 0:
refl = reflect(seed)
inve = invert(refl)
seedlen = len(seed)
invelen = len(inve)
for i in range(seedlen):
seed[i].extend(seed[i])
for i in range(invelen):
inve[i].extend(refl[i])
seed.extend(inve)
k -= 1
line = ''
for i in range(len(seed)):
for j in range(len(seed[i])):
if seed[i][j] == True:
line += '+'
else:
line += '*'
print(line)
line = ''
``` | instruction | 0 | 25,165 | 23 | 50,330 |
Yes | output | 1 | 25,165 | 23 | 50,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
def minusv(L):
return [-x for x in L]
def adamar(M):
return [L*2 for L in M] + [L + minusv(L) for L in M]
k = int(input())
a = [[1]]
for i in range(k):
a = adamar(a)
for L in a:
print(''.join(['+' if c==1 else '*' for c in L]))
# Made By Mostafa_Khaled
``` | instruction | 0 | 25,166 | 23 | 50,332 |
Yes | output | 1 | 25,166 | 23 | 50,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
import sys
if False:
inp = open('C.txt', 'r')
else:
inp = sys.stdin
def inverse(string):
ans = ''
for char in string:
if char == '*':
ans += '+'
else:
ans += '*'
return ans
def recursion(k):
if k == 0:
return ['+']
ans = recursion(k-1)
ans = ans + ans
for i in range(2**(k-1)):
ans[i] = 2*ans[i]
ans[i + 2**(k-1)] += inverse(ans[i + 2**(k-1)])
return ans
k = int(inp.readline())
for string in recursion(k):
print(string)
``` | instruction | 0 | 25,167 | 23 | 50,334 |
Yes | output | 1 | 25,167 | 23 | 50,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
def flip(str):
ret = ""
for k in str:
if k == "+":
ret += "*"
else:
ret += "+"
return ret
k = int(input())
a2 = ["++","+*"]
a1 = []
# if k == 1:
# print("+")
# print("*")
# exit(0)
while k > 1:
a3 = []
flag = True
for p in range(2):
for i in range(len(a2)):
if flag:
temp = a2[i] + a2[i]
a3.append(temp)
else:
temp = a2[i] + flip(a2[i])
a3.append(temp)
flag = False
a2 = a3
k -= 1
for i in a2:
print(i)
``` | instruction | 0 | 25,168 | 23 | 50,336 |
No | output | 1 | 25,168 | 23 | 50,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
import itertools
n=int(input())
if n==0 :
print('+')
exit()
if n==1 :
print('++')
print('+*')
exit()
for i in itertools.combinations_with_replacement('+*',(2**n)//2) :
if i[0]=='+' :
S=''
for j in range(len(i)) :
S=S+i[j]
S1=''
print(S+S)
for j in range(len(S)) :
if S[j]=='+' :
S1=S1+'*'
else :
S1=S1+'+'
print(S+S1)
``` | instruction | 0 | 25,169 | 23 | 50,338 |
No | output | 1 | 25,169 | 23 | 50,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
import logging
import copy
import math
import sys
#logging.basicConfig(stream=sys.stderr, level=logging.DEBUG)
def testPrevLine(lines, num, curretLine):
for i in range(0, curretLine):
if lines[i] == num or lines[i] & num != 0:
return False
return True
def printLine(m, num):
lineStr = ""
for i in range(m):
if num == 0 or num %2 == 0:
lineStr = "*" + lineStr
else :
lineStr = "+" + lineStr
num = int(num / 2)
print(lineStr)
def solve(n):
m = int(math.pow(2, n))
maxNum = int(math.pow(2, n + 1))
lines = []
for i in range(m):
for newNum in range(maxNum):
if testPrevLine(lines, newNum, i):
lines.append(i)
break
for line, j in enumerate(lines) :
printLine(m, line)
return
def main():
firstLine = input().split()
firstLine = int(firstLine[0])
solve(firstLine)
def log(*message):
logging.debug(message)
if __name__ == "__main__":
main()
``` | instruction | 0 | 25,170 | 23 | 50,340 |
No | output | 1 | 25,170 | 23 | 50,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:
<image>.
Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?
Input
The only line of the input contains a single integer k (0 ≤ k ≤ 9).
Output
Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to - 1, and must be equal to ' + ' if it's equal to + 1. It's guaranteed that the answer always exists.
If there are many correct answers, print any.
Examples
Input
2
Output
++**
+*+*
++++
+**+
Note
Consider all scalar products in example:
* Vectors 1 and 2: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
* Vectors 1 and 3: ( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 1 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 3: ( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
* Vectors 2 and 4: ( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
* Vectors 3 and 4: ( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0
Submitted Solution:
```
k = int(input())
answer = [['+', '+'], ['+', '*']]
length = 2
for i in range(k - 1):
new = []
for i in answer:
temp = []
for j in range(length):
if i[j] == '+':
temp += ['+', '+']
else:
temp += ['*', '*']
new.append(temp)
for i in answer:
temp = []
for j in range(length):
if i[j] == '+':
temp += ['+', '*']
else:
temp += ['*', '+']
new.append(temp)
answer = new
length *= 2
print('\n'.join([''.join(i) for i in answer]))
``` | instruction | 0 | 25,171 | 23 | 50,342 |
No | output | 1 | 25,171 | 23 | 50,343 |
Provide a correct Python 3 solution for this coding contest problem.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2 | instruction | 0 | 25,550 | 23 | 51,100 |
"Correct Solution:
```
def gcd(a,b):
while b:a,b=b,a%b
return a
if __name__=="__main__":
a,b=map(int,input().split())
b//=gcd(a,b)
a,c=2,1
while a**2<=b:
if b%a==0:
c*=a
while b%a==0: b//=a
a+=1
print(c if b==1 else c*b)
``` | output | 1 | 25,550 | 23 | 51,101 |
Provide a correct Python 3 solution for this coding contest problem.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2 | instruction | 0 | 25,551 | 23 | 51,102 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#A
def A():
while 1:
n,m = LI()
if n == 0 and m == 0:
break
v = [[] for i in range(n)]
for i in range(m):
a,b = LI()
a -= 1
b -= 1
v[a].append(b)
v[b].append(a)
bfs_map = [1 for i in range(n)]
bfs_map[0] = 0
f = [0 for i in range(n)]
q = deque()
q.append(0)
fl = 1
while q:
if not fl:break
x = q.popleft()
for y in v[x]:
if bfs_map[y]:
bfs_map[y] = 0
f[y] = (1-f[x])
q.append(y)
else:
if f[y] == f[x]:
print(0)
fl = 0
break
if fl:
ans = []
k = sum(f)
if k%2 == 0:
ans.append(k//2)
k = len(f)-sum(f)
if k%2 == 0:
ans.append(k//2)
ans = list(set(ans))
ans.sort()
print(len(ans))
for i in ans:
print(i)
return
#B
def B():
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
def factorize(n):
if n < 4:
return {n:1}
i = 2
d = defaultdict(int)
m = n
while i**2 <= n:
if m%i == 0:
while m%i == 0:
m//=i
d[i] += 1
i += 1
d[m] += 1
return d
p,q = LI()
g = gcd(p,q)
ans = q//g
if ans == 1:
print(1)
else:
d = factorize(ans)
ans = 1
for i in d.keys():
ans *= i
print(ans)
return
#C
def C():
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#I
def I_():
return
#J
def J():
return
#Solve
if __name__ == "__main__":
B()
``` | output | 1 | 25,551 | 23 | 51,103 |
Provide a correct Python 3 solution for this coding contest problem.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2 | instruction | 0 | 25,553 | 23 | 51,106 |
"Correct Solution:
```
def gcd(a, b):
while b: a, b = b, a % b
return a
a,b=map(int,input().split())
b//=gcd(a,b)
a,c=2,1
while a*a<=b:
if b%a==0:
c*=a
while b%a==0:b//=a
a+=1
print([c*b,c][b==1])
``` | output | 1 | 25,553 | 23 | 51,107 |
Provide a correct Python 3 solution for this coding contest problem.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2 | instruction | 0 | 25,556 | 23 | 51,112 |
"Correct Solution:
```
import math
p,q=map(int,input().split())
q//=math.gcd(p,q)
i=2
a=1
while i*i<=q:
if q%i==0:
a*=i
while q%i==0:
q//=i
i+=1
print(a*q)
``` | output | 1 | 25,556 | 23 | 51,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2
Submitted Solution:
```
def gcd(m, n):
r = m % n
return gcd(n, r) if r else n
p, q = map(int, input().split())
q //= gcd(p, q)
x = q; y = 1; k = 2
while k*k <= x:
if x % k == 0:
while x % k == 0: x //= k
y *= k
k += 1
y *= x
print(y)
``` | instruction | 0 | 25,558 | 23 | 51,116 |
Yes | output | 1 | 25,558 | 23 | 51,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2
Submitted Solution:
```
import math
p = int(input())
q = int(input())
print(int(q/(math.gcd(p,q))))
``` | instruction | 0 | 25,559 | 23 | 51,118 |
No | output | 1 | 25,559 | 23 | 51,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2
Submitted Solution:
```
import math
p = int(input())
q = int(input())
print(int(q/math.gcd(p,q)))
``` | instruction | 0 | 25,560 | 23 | 51,120 |
No | output | 1 | 25,560 | 23 | 51,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2
Submitted 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 = 998244353
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():
rr = []
while True:
p,q = LI()
g = fractions.gcd(p,q)
rr.append(max(q//g,2))
break
return '\n'.join(map(str, rr))
print(main())
``` | instruction | 0 | 25,561 | 23 | 51,122 |
No | output | 1 | 25,561 | 23 | 51,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's solve the geometric problem
Mr. A is still solving geometric problems today. It is important to be aware of floating point errors when solving geometric problems.
Floating-point error is the error caused by the rounding that occurs when representing a number in binary finite decimal numbers. For example, 0.1 in decimal is an infinite decimal number of 0.00011001100110011 ... in binary, but an error occurs when rounding this to a finite number of digits.
Positive integers p and q are given in decimal notation. Find the b-ary system (b is an integer greater than or equal to 2) so that the rational number p / q can be expressed as a decimal number with a finite number of digits. If there are more than one, output the smallest one.
Constraints
* 0 <p <q <10 ^ 9
Input Format
Input is given from standard input in the following format.
p q
Output Format
Print the answer in one line.
Sample Input 1
1 2
Sample Output 1
2
1/2 is binary 0.1
Sample Input 2
21 30
Sample Output 2
Ten
21/30 is 0.7 in decimal
Example
Input
1 2
Output
2
Submitted Solution:
```
prime = [2]
def check(x):
for i in prime:
if x % i ==0:
return False
elif x < i * i:
break
return True
def set():
for i in range(3,10**5,2):
if check(i):
prime.append(i)
set()
#print(prime)
p,q = [int(i) for i in input().split(' ')]
for i in prime:
while True:
if p % i ==0 and q % i == 0:
p = p//i
q = q//i
else:
break
ans = 1
for i in prime:
if q % i == 0:
# print(q,i)
q = q // i
ans *= i
while q % i ==0:
q = q // i
ans *= q
print(ans)
``` | instruction | 0 | 25,562 | 23 | 51,124 |
No | output | 1 | 25,562 | 23 | 51,125 |
Provide a correct Python 3 solution for this coding contest problem.
Examples
Input
4 5 2
0 1 2 1
0 2 1 2
1 2 1 1
1 3 1 3
2 3 2 1
Output
6
Input
Output | instruction | 0 | 25,569 | 23 | 51,138 |
"Correct Solution:
```
V,E,F=map(int,input().split())
EDGE=dict()
EDGE2=[[] for i in range(V)]
for i in range(E):
x,to,c,d=map(int,input().split())
EDGE[(x,to)]=[c,d]
EDGE2[x].append(to)
BACK=[-1]*V
P=[0]*V
# 一回ベルマンフォードをしてポテンシャルを求める.
ANS=[float("inf")]*V
ANS[0]=0
for rep in range(V):
for x,to in EDGE:
c,d=EDGE[(x,to)]
if c>0 and ANS[to]>ANS[x]+d:
ANS[to]=ANS[x]+d
BACK[to]=x
P=[ANS[i] for i in range(V)]
LA=ANS[V-1]
NOW=V-1
while NOW!=0:
fr=BACK[NOW]
EDGE[(fr,NOW)][0]-=1
if (NOW,fr) in EDGE:
EDGE[(NOW,fr)][0]+=1
else:
EDGE[(NOW,fr)]=[1,-EDGE[fr,NOW][1]]
EDGE2[NOW].append(fr)
NOW=fr
for x,to in EDGE:
EDGE[(x,to)][1]+=P[x]-P[to]
# あとはダイクストラを使える.
import heapq
for i in range(F-1):
ANS=[float("inf")]*V
Q=[(0,0)]
ANS[0]=0
while Q:
time,ind=heapq.heappop(Q)
if time>ANS[ind]:
continue
for to in EDGE2[ind]:
c,d=EDGE[(ind,to)]
if c>0 and ANS[to]>ANS[ind]+d:
ANS[to]=ANS[ind]+d
BACK[to]=ind
heapq.heappush(Q,(ANS[to],to))
LA+=ANS[V-1]+P[V-1]-P[0]
if LA==float("inf"):
break
P=[P[i]-ANS[i] for i in range(V)]
NOW=V-1
while NOW!=0:
fr=BACK[NOW]
EDGE[(fr,NOW)][0]-=1
if (NOW,fr) in EDGE:
EDGE[(NOW,fr)][0]+=1
else:
EDGE[(NOW,fr)]=[1,-EDGE[fr,NOW][1]]
EDGE2[NOW].append(fr)
NOW=fr
for x,to in EDGE:
EDGE[(x,to)][1]+=ANS[to]-ANS[x]
if LA==float("inf"):
print(-1)
else:
print(LA)
``` | output | 1 | 25,569 | 23 | 51,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,685 | 23 | 51,370 |
Tags: implementation, math
Correct Solution:
```
#code
p=10**9+7
h,w = map(int,input().split())
r = list(map(int,input().split()))
c = list(map(int,input().split()))
grid=[]
f=0
x=0
for i in range(h):
l = list('1'*r[i]+'2'+'0'*(w-r[i]-1))
grid.append(l)
#print(grid)
for i in range(w):
if c[i]==0 and grid[0][i]=='1':
f=1
break
elif c[i]==0:
grid[0][i]='2'
pass
else:# grid[r[c[i]-1]-1][i]=='1' or grid[r[c[i]-1]-1][i-1 if i!=0 else i]=='0':
for j in range(0,c[i]):
if j<c[i]:
if r[j]!=i:
if grid[j][i]=='2':
f=1
break
grid[j][i]='1'
else:
f=1
break
if c[i]!=h:
if grid[c[i]][i]=='1':
f=1
break
grid[c[i]][i]='2'
for i in range(h):
q=e=0
for j in range(w):
if grid[i][j]=='0':
q+=1
x+=q
#print(grid)
#print(x,pow(2,x,p))
if f:
print(0)
else:
print(pow(2,x,p))
``` | output | 1 | 25,685 | 23 | 51,371 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,686 | 23 | 51,372 |
Tags: implementation, math
Correct Solution:
```
def powe(a,b,mod):
res = 1
a = a%mod
while b>0:
if(b&1==1):
res = (res*a)%mod
b=b>>1
a = (a*a)%mod
return res
h,w = map(int,input().split())
ri = list(map(int,input().split()))
ci = list(map(int,input().split()))
const = [[0 for _ in range(w)] for _ in range(h)]
flag=True
for i in range(h):
for j in range(ri[i]+1):
if j<ri[i]:
const[i][j]=1
else:
if j<w:
const[i][j]=-1
for i in range(w):
for j in range(ci[i]+1):
if j<ci[i]:
if const[j][i]!=-1:
const[j][i]=1
else:
flag=False
break
else:
if j<h:
if const[j][i]==1:
flag=False
break
const[j][i]=-1
if flag==False:
break
count_0 = 0
if flag==False:
print(0)
else:
for i in range(1,h):
for j in range(1,w):
if const[i][j]==0:
count_0+=1
print(powe(2,count_0,1000000007))
``` | output | 1 | 25,686 | 23 | 51,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,687 | 23 | 51,374 |
Tags: implementation, math
Correct Solution:
```
#import collections
# import random
# import math
#import itertools
#import math
#mport math
from collections import defaultdict
# import itertools
# from sys import stdin, stdout
#import math
import sys
# import operator
# from decimal import Decimal
# sys.setrecursionlimit(10**6)
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def li(): return [int(i) for i in input().split()]
def lli(rows): return [li() for _ in range(rows)]
def si(): return input()
def ii(): return int(input())
def ins(): return input().split()
def solve():
mod = 10**9+7
h,w = LI(); r = LI(); c= LI()
grid = [[0]*w for _ in range(h)]
for i in range(h):
if r[i]>0:
for j in range(r[i]):
grid[i][j] = 1
if j+1<w:
grid[i][j+1] =-1
else:
grid[i][0] = -1
for i in range(w):
if c[i]>0:
for j in range(c[i]):
if grid[j][i] == -1:
print(0)
return
grid[j][i] = 1
if j+1<h:
if grid[j+1][i] ==1:
print(0)
return
grid[j+1][i] =-1
else:
if grid[0][i] == 1:
print(0)
return
grid[0][i] = -1
ans = 1
for i in range(h):
for j in range(w):
if grid[i][j] == 0:
ans = (ans*2)%mod
print(ans)
return
def main():
solve()
# for _ in range(II()):
# sys.stdout.write(str(solve()) + "\n")
# z += str(ans) + '\n'
# print(len(ans), ' '.join(map(str, ans)), sep='\n')
# stdout.write(z)
# for interactive problems
# print("? {} {}".format(l,m), flush=True)
# or print this after each print statement
# sys.stdout.flush()
if __name__ == "__main__":
main()
``` | output | 1 | 25,687 | 23 | 51,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,688 | 23 | 51,376 |
Tags: implementation, math
Correct Solution:
```
h,w=map(int,input().split())
ar=list(map(int,input().split()))
br=list(map(int,input().split()))
flag=True
count=0
for i in range(h):
for j in range(w):
c_1=''
c_2=''
if(ar[i]>j):
c_1='B'
elif(ar[i]<j):
c_1='E'
elif(ar[i]==j):
c_1='W'
if(br[j]>i):
c_2='B'
elif(br[j]<i):
c_2='E'
elif(br[j]==i):
c_2='W'
if(c_1==c_2 and c_1=='E'):
count+=1
elif(c_1!=c_2 and c_1!='E' and c_2!='E'):
flag=False
break
if(flag):
print((2**count)%(10**9+7))
else:
print(0)
``` | output | 1 | 25,688 | 23 | 51,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,689 | 23 | 51,378 |
Tags: implementation, math
Correct Solution:
```
a,b = map(int, input().split())
h = map(int, input().split())
v = map(int, input().split())
room = [-1 for j in range(a*b)]
for ind, i in enumerate(h):
for j in range(i):
room[b*ind + j] = 1
if i != b:
room[b*ind + i] = 0
#print(room)
for ind, i in enumerate(v):
for j in range(i):
if room[b*j + ind] == 0:
print(0)
quit()
room[b*j + ind] = 1
if i != a:
if room[b*i + ind] == 1:
print(0)
quit()
room[b*i + ind] = 0
#print(room)
print(pow(2, room.count(-1), 1000000007))
``` | output | 1 | 25,689 | 23 | 51,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,690 | 23 | 51,380 |
Tags: implementation, math
Correct Solution:
```
h, w = map(int, input().split())
*r, = map(int, input().split())
*c, = map(int, input().split())
s = h*w
minc = [0]*w
for i in range(0, h):
s -= r[i]
if r[i] != w:
s -= 1
if r[i] < w and c[r[i]] >= i+1:
print(0)
exit()
for j in range(w):
if j < r[i] and minc[j] != -1:
minc[j] += 1
if j < r[i] and minc[j] != -1 and c[j] < minc[j]:
print(0)
exit()
if j >= r[i]:
minc[j] = -1
if j > r[i] and (c[j] >= i+1 or c[j]-i == 0):
s -= 1
print(pow(2, s, 10**9+7))
``` | output | 1 | 25,690 | 23 | 51,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,691 | 23 | 51,382 |
Tags: implementation, math
Correct Solution:
```
mod = 1000000007
def fexp(x, y):
ans = 1
while y > 0:
if y % 2 == 1:
ans = ans * x % mod
x = x * x % mod
y //= 2
return ans
h, w = map(int, input().split(' '))
r = list(map(int, input().split(' ')))
c = list(map(int, input().split(' ')))
fixed = [[0 for i in range(w)] for i in range(h)]
ok = True
for i in range(h):
for j in range(r[i]):
fixed[i][j] = 1
if r[i] < w:
fixed[i][r[i]] = -1
for i in range(w):
for j in range(c[i]):
if fixed[j][i] == -1:
ok = False
fixed[j][i] = 1
if c[i] < h:
if fixed[c[i]][i] == 1:
ok = False
fixed[c[i]][i] = -1
if not ok:
print(0)
else:
count = 0
for i in range(h):
for j in range(w):
if fixed[i][j] == 0:
count += 1
print(fexp(2, count))
``` | output | 1 | 25,691 | 23 | 51,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7). | instruction | 0 | 25,692 | 23 | 51,384 |
Tags: implementation, math
Correct Solution:
```
R=lambda:map(int,input().split())
h,w=map(range,R())
r=*R(),0
c=*R(),0
print((pow(2,sum(i>c[j]and j>r[i]for i in h
for j in w),10**9+7),0)[any(i<c[r[i]]for
i in h)|any(i<r[c[i]]for i in w)])
``` | output | 1 | 25,692 | 23 | 51,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
n,m=map(int,input().split())
r=list(map(int,input().split()))
c=list(map(int,input().split()))
g=[]
for i in range(n):
g.append([1]*r[i])
g[-1].extend([0]*(m-r[i]))
for i in range(m):
for j in range(c[i]):
g[j][i]=1
for i in range(n):
for j in range(m):
if g[i][j]!=1:
break
else:
if r[i]!=j+1:
print(0)
exit()
continue
if r[i]!=j:
print(0)
exit()
for i in range(m):
for j in range(n):
if g[j][i]!=1:
break
else:
if c[i]!=j+1:
print(0)
exit()
continue
if c[i]!=j:
print(0)
exit()
x=0
for i in range(n):
for j in range(m):
if g[i][j]==0:
if i>c[j] and j>r[i]:
x+=1
print(pow(2,x,7+10**9))
``` | instruction | 0 | 25,693 | 23 | 51,386 |
Yes | output | 1 | 25,693 | 23 | 51,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
def main():
h, w = map(int, input().split())
rs = list(map(int, input().split()))
cs = list(map(int, input().split()))
data = []
for i in range(h):
row = []
for j in range(w):
row.append([0, False])
data.append(row)
good_r = []
for i in range(h):
for j in range(rs[i]):
if data[i][j] != [0, True]:
data[i][j] = [1, True]
else:
return 0
if rs[i] != w:
if data[i][rs[i]] != [1, True]:
data[i][rs[i]] = [0, True]
else:
return 0
good_r.append(rs[i])
else:
good_r.append(-1)
good_c = []
for i in range(w):
for j in range(cs[i]):
if data[j][i] != [0, True]:
data[j][i] = [1, True]
else:
return 0
if cs[i] != h:
if data[cs[i]][i] != [1, True]:
data[cs[i]][i] = [0, True]
else:
return 0
good_c.append(cs[i])
else:
good_c.append(-1)
#print('\n'.join(' '.join(['1' if x[0] else '0' for x in a]) for a in data))
k = 1
for i in range(h):
for j in range(w):
if not data[i][j][1]:
if 0 <= good_r[i] and 0 <= good_c[j]:
k *= 2
k = k % 1000000007
return k
print(main())
``` | instruction | 0 | 25,694 | 23 | 51,388 |
Yes | output | 1 | 25,694 | 23 | 51,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
h, w = map(int, input().split())
r = list(map(int, input().split()))
c = list(map(int, input().split()))
grid = [[0 for i in range(w)] for i in range(h)]
flag = 1
for i in range(w):
for j in range(c[i]):
grid[j][i] = 1
if c[i] < h:
grid[c[i]][i] = 2
for i in range(h):
for j in range(r[i]):
if grid[i][j] == 2:
flag = 0
grid[i][j] = 1
if r[i] < w:
if grid[i][r[i]] == 1:
flag = 0
grid[i][r[i]] = 2
c = 0
if flag:
for i in grid:
c+=i.count(0)
print(pow(2, c, 1000000007))
else:
print(0)
``` | instruction | 0 | 25,695 | 23 | 51,390 |
Yes | output | 1 | 25,695 | 23 | 51,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
from collections import Counter
if __name__ == '__main__':
l = [int(i) for i in input().split(" ")]
h = l[0]
w = l[1]
matrix = [[0 for j in range(w)] for i in range(h)]
r = [int(i) for i in input().split(" ")]
c = [int(i) for i in input().split(" ")]
for i in range(h):
if r[i] == 0:
matrix[i][0] = -1
else:
for j in range(r[i]):
matrix[i][j] = 1
if r[i] < w:
matrix[i][r[i]] = -1
flag = 0
for j in range(w):
if c[j] == 0:
if matrix[0][j] == 1:
print(0)
flag = 1
break
else:
matrix[0][j] = -1
else:
for i in range(c[j]):
if matrix[i][j] == -1:
print(0)
flag = 1
break
else:
matrix[i][j] = 1
if flag == 1:
break
if c[j] < h:
if matrix[c[j]][j] == 1:
print(0)
flag = 1
break
matrix[c[j]][j] = -1
if flag == 1:
break
if flag == 0:
c = 0
for i in matrix:
c += i.count(0)
print((2**c)%1000000007)
``` | instruction | 0 | 25,696 | 23 | 51,392 |
Yes | output | 1 | 25,696 | 23 | 51,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
h, w = tuple(map(int, input().split(" ")))
#print(h, w)
r = list(map(int, input().split(" ")))
c = list(map(int, input().split(" ")))
fill = 0
for i in range(1, h):
for j in range(1, w):
if i > c[j] and j > r[i]:
fill += 1
if fill == 0:
print(0)
else:
print((2 ** fill) % 1000000007)
``` | instruction | 0 | 25,697 | 23 | 51,394 |
No | output | 1 | 25,697 | 23 | 51,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
#CodeForces Problem Set 1228B
import sys
def possibilities(height, width, r, c):
#board = height*[[0]*width] creates height no of copies of a list, any change in any list is reflected to all lists.
board = []
for i in range(height):
board.append([0]*width)
filled = 0
for i in range(height):
for j in range(r[i]):
filled += 1
board[i][j] = 1
if r[i] < width:
filled += 1
board[i][r[i]] = -1
for i in range(width):
for j in range(c[i]):
if board[j][i] == 0:
filled += 1
board[j][i] = 1
if c[i] < height and board[c[i]][i] == 0:
filled += 1
flexible = height*width - filled
ans = 0
if flexible:
ans = 2
flexible -= 1
for i in range(flexible):
ans = (ans*2)%1000000007
print(ans)
height, width = map(int, sys.stdin.readline().split())
r = list(map(int, sys.stdin.readline().strip().split()))
c = list(map(int, sys.stdin.readline().strip().split()))
possibilities(height, width, r, c)
``` | instruction | 0 | 25,698 | 23 | 51,396 |
No | output | 1 | 25,698 | 23 | 51,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
[h,w]=[int(k) for k in input().split()]
R=[int(k) for k in input().split()]
C=[int(k) for k in input().split()]
c=0
for i in range(h):
for j in range(w):
if i>C[j] and j>R[i]:
c=c+1
if c==0:
print(c)
else:
ans= (2**c)%(1000000007)
print(ans)
``` | instruction | 0 | 25,699 | 23 | 51,398 |
No | output | 1 | 25,699 | 23 | 51,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose there is a h × w grid consisting of empty or full cells. Let's make some definitions:
* r_{i} is the number of consecutive full cells connected to the left side in the i-th row (1 ≤ i ≤ h). In particular, r_i=0 if the leftmost cell of the i-th row is empty.
* c_{j} is the number of consecutive full cells connected to the top end in the j-th column (1 ≤ j ≤ w). In particular, c_j=0 if the topmost cell of the j-th column is empty.
In other words, the i-th row starts exactly with r_i full cells. Similarly, the j-th column starts exactly with c_j full cells.
<image> These are the r and c values of some 3 × 4 grid. Black cells are full and white cells are empty.
You have values of r and c. Initially, all cells are empty. Find the number of ways to fill grid cells to satisfy values of r and c. Since the answer can be very large, find the answer modulo 1000000007 (10^{9} + 7). In other words, find the remainder after division of the answer by 1000000007 (10^{9} + 7).
Input
The first line contains two integers h and w (1 ≤ h, w ≤ 10^{3}) — the height and width of the grid.
The second line contains h integers r_{1}, r_{2}, …, r_{h} (0 ≤ r_{i} ≤ w) — the values of r.
The third line contains w integers c_{1}, c_{2}, …, c_{w} (0 ≤ c_{j} ≤ h) — the values of c.
Output
Print the answer modulo 1000000007 (10^{9} + 7).
Examples
Input
3 4
0 3 1
0 2 3 0
Output
2
Input
1 1
0
1
Output
0
Input
19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
Output
797922655
Note
In the first example, this is the other possible case.
<image>
In the second example, it's impossible to make a grid to satisfy such r, c values.
In the third example, make sure to print answer modulo (10^9 + 7).
Submitted Solution:
```
s=input("")
h,w=(int(x) for x in s.split())
l1=[[2 for i in range(w)]for j in range(h)]
s=input("")
r=[int(x) for x in s.split()]
s=input("")
c=[int(x) for x in s.split()]
for i in range (0,h):
a=r[i]
for u in range(0,a):
l1[i][u]=1
if a<w: l1[i][a]=0
for i in range (0,w):
b=c[i]
for u in range(0,b):
l1[u][i]=1
if b<h: l1[b][i]=0
ans=0
for i in range (0,h):
ans+=l1[i].count(2)
print (int(2**ans))
``` | instruction | 0 | 25,700 | 23 | 51,400 |
No | output | 1 | 25,700 | 23 | 51,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At a geometry lesson Bob learnt that a triangle is called right-angled if it is nondegenerate and one of its angles is right. Bob decided to draw such a triangle immediately: on a sheet of paper he drew three points with integer coordinates, and joined them with segments of straight lines, then he showed the triangle to Peter. Peter said that Bob's triangle is not right-angled, but is almost right-angled: the triangle itself is not right-angled, but it is possible to move one of the points exactly by distance 1 so, that all the coordinates remain integer, and the triangle become right-angled. Bob asks you to help him and find out if Peter tricks him. By the given coordinates of the triangle you should find out if it is right-angled, almost right-angled, or neither of these.
Input
The first input line contains 6 space-separated integers x1, y1, x2, y2, x3, y3 — coordinates of the triangle's vertices. All the coordinates are integer and don't exceed 100 in absolute value. It's guaranteed that the triangle is nondegenerate, i.e. its total area is not zero.
Output
If the given triangle is right-angled, output RIGHT, if it is almost right-angled, output ALMOST, and if it is neither of these, output NEITHER.
Examples
Input
0 0 2 0 0 1
Output
RIGHT
Input
2 3 4 5 6 6
Output
NEITHER
Input
-1 0 2 0 0 1
Output
ALMOST | instruction | 0 | 25,872 | 23 | 51,744 |
Tags: brute force, geometry
Correct Solution:
```
def checkTri(points):
d1 = (points[0] - points[2]) ** 2 + (points[1] - points[3]) ** 2
d2 = (points[0] - points[4]) ** 2 + (points[1] - points[5]) ** 2
d3 = (points[2] - points[4]) ** 2 + (points[3] - points[5]) ** 2
if(d1 ==0 or d2 ==0 or d3 ==0):
return 0
if(d3-d2==d1 or d2-d3==d1 or d2+d3==d1 ):
return 1
else:
return 0
coord=list(map(int, input().split()))
check=checkTri(coord)
if(check==1):
print("RIGHT")
else:
i=0
almostCheck = 0
while(i<6):
almostCoord = coord
almostCoord[i]=almostCoord[i]+1
almostCheck=checkTri(almostCoord)
if(almostCheck==1):
print("ALMOST")
break
almostCoord[i] = almostCoord[i] - 2
almostCheck = checkTri(almostCoord)
if (almostCheck == 1):
print("ALMOST")
break
almostCoord[i] = almostCoord[i] + 1
i=i+1
if(almostCheck != 1):
print("NEITHER")
``` | output | 1 | 25,872 | 23 | 51,745 |
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