Tasks Details
hard
Find a path in given matrix, such that the product of all the numbers on the path has the minimal number of trailing zeros.
Task Score
100%
Correctness
100%
Performance
100%
Matrix A, consisting of N rows and N columns of non-negative integers, is given. Rows are numbered from 0 to N−1 (from top to bottom). Columns are numbered from 0 to N−1 (from left to right). We would like to find a path that starts at the upper-left corner (0, 0) and, moving only right and down, finishes at the bottom-right corner (N−1, N−1). Then, all the numbers on this path will be multiplied together.
Find a path such that the product of all the numbers on the path contain the minimal number of trailing zeros. We assume that 0 has 1 trailing zero.
Write a function:
def solution(A)
that, given matrix A, returns the minimal number of trailing zeros.
Examples:
1. Given matrix A below:
the function should return 1. The optimal path is: (0,0) → (0,1) → (0,2) → (1,2) → (2,2) → (2,3) → (3,3). The product of numbers 2, 10, 1, 4, 2, 1, 1 is 160, which has one trailing zero. There is no path that yields a product with no trailing zeros.
2. Given matrix A below:
the function should return 2. One of the optimal paths is: (0,0) → (1,0) → (1,1) → (1,2) → (2,2) → (3,2) → (3,3). The product of numbers 10, 1, 1, 1, 10, 1, 1 is 100, which has two trailing zeros. There is no path that yields a product with fewer than two trailing zeros.
3. Given matrix A below:
the function should return 1. One of the optimal paths is: (0,0) → (0,1) → (1,1) → (1,2) → (2,2). The product of numbers 10, 10, 0, 10, 10 is 0, which has one trailing zero. There is no path that yields a product with no trailing zeros.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..500];
- each element of matrix A is an integer within the range [0..1,000,000,000].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Python
Time spent on task 3 minutes
Notes
not defined yet
Code: 21:35:53 UTC,
java,
autosave
Code: 21:36:14 UTC,
py,
autosave
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding minimum number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
#elif n[1]<m[1]: return (n)
#elif m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second =a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating 1st line
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:36:24 UTC,
py,
autosave
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding minimum number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second =a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating 1st line
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:36:35 UTC,
py,
autosave
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding minimum number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second =a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating 1st line
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:36:49 UTC,
py,
autosave
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding minimum number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating 1st line
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:37:05 UTC,
py,
autosave
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating 1st line
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:37:35 UTC,
py,
autosave
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating first 1st line(it is different from other lines
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:37:47 UTC,
py,
autosave
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating first 1st line(it is different from other lines
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:38:06 UTC,
py,
autosave
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating first 1st line(it is different from other lines
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
#finding result(last number in array), and comparing it to 1 if there was there was 0 element somewhere
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:38:30 UTC,
py,
autosave
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating first 1st line(it is different from other lines
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
#finding result(zeros in last element of array), and comparing it to 1 if there was there was 0 element somewhere
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Code: 21:38:38 UTC,
py,
verify,
result: Passed
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating first 1st line(it is different from other lines
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
#finding result(zeros in last element of array), and comparing it to 1 if there was there was 0 element somewhere
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Analysis
Code: 21:38:43 UTC,
py,
verify,
result: Passed
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating first 1st line(it is different from other lines
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
#finding result(zeros in last element of array), and comparing it to 1 if there was there was 0 element somewhere
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)
Analysis
Code: 21:38:45 UTC,
py,
final,
score:
100
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162
#minimizing numbers and counting zeros
def change_number(n):
if n==0: return([0,1])
zeros=0
if (n%10)==0:
zeros=zeros_amount(n)
zero=int('1'+zeros*'0')
n=n/zero
if (n%5)==0: n=5
elif (n%2)==0: n=2
else: n=1
return ([n,zeros])
#counting amount of zeros in a number
def zeros_amount(n):
return (len(str(n)) - len(str(n).rstrip('0')))
#finding smaller number with minimum amount of zeroes
def find_min(n,m):
if m[0]==n[0]==0: return ([0,1])
if m[0]==0 or n[1]<m[1]: return(n)
elif n[0]==0 or m[1]<n[1]: return (m)
elif n[0]<m[0]: return (n)
else: return(m)
#multiple two array elements
def mult(a,b):
if a[0]==0 or b[0]==0: return([0,1])
first,second= a[0]*b[0], a[1]+b[1]
if first%10==0:
first=first//10
second+=1
return [first,second]
def solution(A):
N=len(A)
zero_included=0
#creating first 1st line(it is different from other lines
firstline=A[0]
firstline[0]=change_number(firstline[0])
for x in range(1,N): firstline[x] = mult(firstline[x-1], change_number(firstline[x]))
m=1
while m<N:
secondline=A[m]
newline=[]
for x in range(0,N):
secondline[x]=change_number(secondline[x])
if (x==0): newline.append(mult(firstline[0],secondline[0]))
elif (secondline[x][0]==0):
zero_included=1
newline.append([0,1])
else:
newline.append(find_min(mult(firstline[x],secondline[x]),mult(secondline[x],newline[-1])))
firstline=newline
m +=1
#finding result(zeros in last element of array), and comparing it to 1 if there was there was 0 element somewhere
result=firstline[-1][1]
if zero_included!=0: result=min(result,1)
return(result)Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N**2*log(max(max(A))))
expand all
Correctness tests
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
4.
0.036 s
OK
5.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.040 s
OK
4.
0.036 s
OK
5.
0.036 s
OK
6.
0.036 s
OK
7.
0.036 s
OK
8.
0.036 s
OK
9.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
4.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
4.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.040 s
OK
2.
0.036 s
OK
3.
0.040 s
OK
4.
0.036 s
OK
5.
0.036 s
OK
6.
0.036 s
OK
7.
0.036 s
OK
8.
0.036 s
OK
9.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
expand all
Performance tests
1.
0.036 s
OK
1.
0.040 s
OK
2.
0.048 s
OK
3.
0.080 s
OK
1.
0.060 s
OK
big_zero_result
all elements in A are neither dividable by 2 or 5; length of A is close to 500
all elements in A are neither dividable by 2 or 5; length of A is close to 500
✔
OK
1.
0.072 s
OK
2.
0.184 s
OK
3.
0.376 s
OK
4.
0.660 s
OK
5.
1.024 s
OK
big_all_dividable
all elements in A are dividable by either power of 2 or 5; length of A is close to 500
all elements in A are dividable by either power of 2 or 5; length of A is close to 500
✔
OK
1.
0.076 s
OK
2.
0.204 s
OK
3.
0.416 s
OK
4.
0.792 s
OK
5.
1.284 s
OK
1.
1.188 s
OK
2.
1.220 s
OK
3.
1.176 s
OK
4.
1.168 s
OK
5.
1.152 s
OK
1.
1.728 s
OK
2.
0.528 s
OK
1.
0.096 s
OK
2.
0.092 s
OK
3.
0.076 s
OK
4.
0.280 s
OK
5.
0.272 s
OK
6.
0.244 s
OK
7.
0.596 s
OK
8.
0.580 s
OK
9.
0.528 s
OK
10.
1.096 s
OK
11.
1.040 s
OK
12.
0.968 s
OK
13.
1.648 s
OK
14.
1.588 s
OK
15.
1.528 s
OK
1.
1.008 s
OK