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A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def erase_digits(n, remainder, count):
cnt = 0
i = len(n) - 1
s = n
while i >= 0 and cnt < count:
digit = int(s[i])
if digit % 3 == remainder:
cnt += 1
s = s[:i] + s[i + 1 :]
i -= 1
if len(s) == len(n) or cnt < count:
return ""
while len(s) > 1 and s[0] == "0":
s = s[1:]
return s
n = input()
sum = 0
for i in range(len(n)):
digit = int(n[i])
sum = (sum + digit) % 3
if sum == 0:
print(n)
else:
n1 = erase_digits(n, sum, 1)
n2 = erase_digits(n, 3 - sum, 2)
if len(n1) > 0 and len(n1) > len(n2):
print(n1)
elif len(n2) > 0:
print(n2)
else:
print(-1) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN STRING WHILE FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING ASSIGN VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | import sys
s = list(input())
digsum = 0
rem = [(0) for i in range(len(s))]
for i in range(0, len(s)):
dig = ord(s[i]) - ord("0")
digsum += dig
rem[i] = dig % 3
currem = digsum % 3
if currem == 0:
print("".join(s))
sys.exit()
if len(s) == 1:
print(-1)
sys.exit()
inds1 = []
inds2 = []
for i in range(len(rem) - 1, -1, -1):
if rem[i] == 2 and len(inds2) < 2:
inds2.append(i)
if rem[i] == 1 and len(inds1) < 2:
inds1.append(i)
vs = []
if currem == 1:
if len(inds1) >= 1:
vs.append(s[: inds1[0]] + s[inds1[0] + 1 :])
inds2.sort()
if len(inds2) == 2:
vs.append(s[: inds2[0]] + s[inds2[0] + 1 : inds2[1]] + s[inds2[1] + 1 :])
if currem == 2:
if len(inds2) >= 1:
vs.append(s[: inds2[0]] + s[inds2[0] + 1 :])
inds1.sort()
if len(inds1) == 2:
vs.append(s[: inds1[0]] + s[inds1[0] + 1 : inds1[1]] + s[inds1[1] + 1 :])
ans = -1
anslen = 0
for i in range(len(vs)):
ln = len(vs[i])
for j in range(len(vs[i])):
if vs[i][j] != "0" or j == len(vs[i]) - 1:
break
ln -= 1
vs[i] = vs[i][-ln:]
if ln != 0 and (ans == -1 or anslen < ln):
anslen = ln
ans = i
if ans == -1:
print(-1)
else:
print("".join(vs[ans])) | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST IF VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER IF VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR STRING VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | n = input()
if len(n) < 3:
if len(n) == 1:
if int(n) % 3 != 0:
print(-1)
else:
print(n)
if len(n) == 2:
if (int(n[0]) + int(n[1])) % 3 == 0:
print(n)
elif int(n[0]) % 3 == 0:
print(n[0])
elif int(n[1]) % 3 == 0:
print(n[1])
else:
print(-1)
else:
for i in range(len(n)):
if n[i] != "0":
n = n[i:]
break
if n[0] == "0":
n = "0"
dig_sum = sum([int(x) for x in n]) % 3
found = dig_sum == 0
result = "0"
if not found:
for i in range(len(n) - 1, -1, -1):
if (dig_sum - int(n[i])) % 3 == 0:
found = True
tmp = n[:i] + n[i + 1 :]
break
if found:
for i in range(len(tmp)):
if tmp[i] != "0":
tmp = tmp[i:]
break
if tmp[0] == "0":
tmp = "0"
if found:
result = tmp
found_two = False
if len(n) > 2:
for i in range(len(n) - 1, -1, -1):
if int(n[i]) % 3 != 0:
dig_sum -= int(n[i])
tmp = n[:i] + n[i + 1 :]
break
for i in range(len(tmp) - 1, -1, -1):
if (dig_sum - int(tmp[i])) % 3 == 0:
found_two = True
tmp = tmp[:i] + tmp[i + 1 :]
break
if found_two:
for i in range(len(tmp)):
if tmp[i] != "0":
tmp = tmp[i:]
break
if tmp[0] == "0":
tmp = "0"
if not found:
result = tmp
elif found_two and len(result) < len(tmp):
result = tmp
else:
result = n
print(result) | ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR IF VAR NUMBER STRING ASSIGN VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR STRING IF VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR IF VAR NUMBER STRING ASSIGN VAR STRING IF VAR ASSIGN VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR IF VAR NUMBER STRING ASSIGN VAR STRING IF VAR ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | import sys
a = list(map(int, list(input())))
n = len(a)
su = sum(a)
l1 = [i for i in range(n) if a[i] % 3 == 1]
l2 = [i for i in range(n) if a[i] % 3 == 2]
def erase_leading_zeroes(li):
i = 0
while i < n:
if len(li) <= 1:
return
if li[i] == 0:
li.pop(i)
else:
return
if su % 3 == 1:
tmp1 = a[:]
tmp2 = a[:]
if len(l1) >= 1:
tmp1.pop(l1[-1])
erase_leading_zeroes(tmp1)
if len(l2) >= 2:
tmp2.pop(l2[-1])
tmp2.pop(l2[-2])
erase_leading_zeroes(tmp2)
if sum(tmp1) % 3 == 0 or sum(tmp2) % 3 == 0:
if sum(tmp1) % 3 == 0:
a = tmp1
if sum(tmp2) % 3 == 0 and (sum(tmp1) % 3 != 0 or len(tmp2) > len(tmp1)):
a = tmp2
else:
print(-1)
sys.exit()
elif su % 3 == 2:
tmp1 = a[:]
tmp2 = a[:]
if len(l2) >= 1:
tmp1.pop(l2[-1])
erase_leading_zeroes(tmp1)
if len(l1) >= 2:
tmp2.pop(l1[-1])
tmp2.pop(l1[-2])
erase_leading_zeroes(tmp2)
if sum(tmp1) % 3 == 0 or sum(tmp2) % 3 == 0:
if sum(tmp1) % 3 == 0:
a = tmp1
if sum(tmp2) % 3 == 0 and (sum(tmp1) % 3 != 0 or len(tmp2) > len(tmp1)):
a = tmp2
else:
print(-1)
sys.exit()
if len(a) >= 1:
print("".join(map(str, a)))
else:
print(-1) | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def f(n, r, k):
tmp = n
for t in range(k):
for i in range(len(tmp) - 1, -1, -1):
if int(tmp[i]) % 3 == r:
tmp = tmp[:i] + tmp[i + 1 :]
break
for i in range(len(tmp)):
if int(tmp[i]) != 0:
return "".join(tmp[i:])
if len(tmp) == 0 or 0 < int(tmp[len(tmp) - 1]):
return ""
else:
return "0"
n = list(input())
cnt = {i: (0) for i in range(3)}
for d in n:
cnt[int(d) % 3] += 1
r = (cnt[1] + 2 * cnt[2]) % 3
if r == 0:
print("".join(n))
else:
ans = []
if 0 < cnt[r]:
ans.append(f(n, r, 1))
if 1 < cnt[3 - r]:
ans.append(f(n, 3 - r, 2))
if len(ans) == 0 or len("".join(ans)) == 0:
print(-1)
elif len(ans[-1]) <= len(ans[0]):
print(ans[0])
else:
print(ans[-1]) | FUNC_DEF ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER RETURN FUNC_CALL STRING VAR VAR IF FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN STRING RETURN STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR LIST IF NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER IF NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL STRING VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | import sys
n = list(input(""))
l = [(int(c) % 3) for c in n][::-1]
d = sum(l) % 3
if len(n) == 1:
if n == ["3"]:
print(3)
else:
print(-1)
sys.exit(0)
def li(x):
global n
del n[len(n) - 1 - l.index(x)]
del l[l.index(x)]
def cond(x):
x = x.lstrip("0")
if len(x) == 0:
return "0"
return x
if d == 0:
pass
elif d in l:
if l.index(d) == len(l) - 1 and int(n[1]) == 0 and l.count(3 - d) >= 2:
li(3 - d)
li(3 - d)
else:
li(d)
else:
li(3 - d)
li(3 - d)
if len(n) > 0:
q = cond("".join(n))
print(q)
else:
print(-1) | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER IF VAR LIST STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER FUNC_DEF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER RETURN STRING RETURN VAR IF VAR NUMBER IF VAR VAR IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR BIN_OP NUMBER VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | s = input()
l = []
n = len(s)
ans = 0
r1, r2 = [], []
def red(l):
for i in range(len(l)):
if l[i] != 0:
return l[i:]
return l
def pri(l):
print("".join([str(k) for k in l]))
exit()
for i in range(n):
l.append(int(s[i]))
ans += l[-1]
if l[-1] % 3 == 1 and len(r1) < 3:
r1.append(i)
if l[-1] % 3 == 2 and len(r2) < 3:
r2.append(i)
if ans % 3 == 0:
if sum(l) == 0:
if len(l) == 0:
print(-1)
else:
print(0)
exit()
pri(l)
elif ans % 3 == 1:
l1, l2 = [], []
if len(r1) > 0:
t = r1[-1]
l1 = l[:t] + l[t + 1 :]
if len(r2) > 1:
p, q = r2[-1], r2[-2]
l2 = l[:q] + l[q + 1 : p] + l[p + 1 :]
if len(r1) == 0 and len(r2) < 2:
print(-1)
exit()
l1 = red(l1)
l2 = red(l2)
if len(l1) > len(l2):
l = l1
else:
l = l2
if sum(l) == 0:
if len(l) == 0:
print(-1)
else:
print(0)
exit()
pri(l)
else:
l1, l2 = [], []
if len(r2) > 0:
t = r2[-1]
l1 = l[:t] + l[t + 1 :]
if len(r1) > 1:
p, q = r1[-1], r1[-2]
l2 = l[:q] + l[q + 1 : p] + l[p + 1 :]
if len(r2) == 0 and len(r1) < 2:
print(-1)
exit()
l1 = red(l1)
l2 = red(l2)
if len(l1) > len(l2):
l = l1
else:
l = l2
if sum(l) == 0:
if len(l) == 0:
print(-1)
else:
print(0)
exit()
pri(l) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR LIST LIST FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER RETURN VAR VAR RETURN VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR LIST LIST IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR LIST LIST IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | n = input()
a = [0] * len(n)
b1 = [0] * len(n)
b2 = [0] * len(n)
k = 0
c1 = 0
c2 = 0
ans = ""
for i in range(len(n)):
a[i] = int(n[i])
b1[i] = int(n[i])
b2[i] = int(n[i])
k += a[i]
k %= 3
if k != 0:
k1 = k
for i in range(len(b1) - 1, -1, -1):
if b1[i] == 1 or b1[i] == 4 or b1[i] == 7:
del b1[i]
c1 += 1
k1 -= 1
if k1 % 3 == 0:
if len(a) != 0:
for i in range(len(n)):
if len(b1) > 1:
if b1[0] == 0:
c1 += 1
del b1[0]
else:
break
break
if c1 == 0:
k1 = k
for i in range(len(b1) - 1, -1, -1):
if b1[i] == 5 or b1[i] == 2 or b1[i] == 8:
del b1[i]
c1 += 1
k1 -= 2
if k1 % 3 == 0:
if len(b1) != 0:
for i in range(len(n)):
if len(b1) > 1:
if b1[0] == 0:
c1 += 1
del b1[0]
else:
break
break
k2 = k
for i in range(len(b2) - 1, -1, -1):
if b2[i] == 2 or b2[i] == 5 or b2[i] == 8:
del b2[i]
c2 += 1
k2 -= 2
if k2 % 3 == 0:
if len(b2) != 0:
for i in range(len(n)):
if len(b2) > 1:
if b2[0] == 0:
c2 += 1
del b2[0]
else:
break
break
if c2 == 0:
k2 = k
for i in range(len(b2) - 1, -1, -1):
if b2[i] == 1 or b2[i] == 4 or b2[i] == 7:
del b2[i]
c2 += 1
k2 -= 1
if k2 % 3 == 0:
if len(b2) != 0:
for i in range(len(n)):
if len(b2) > 1:
if b2[0] == 0:
c2 += 1
del b2[0]
else:
break
break
if len(b2) >= len(b1):
if k2 % 3 == 0:
a = b2
elif k1 % 3 == 0:
a = b1
else:
a = []
elif k1 % 3 == 0:
a = b1
elif k2 % 3 == 0:
a = b2
else:
a = []
if len(a) == 0:
print("-1")
else:
for i in range(len(a)):
ans += str(a[i])
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR LIST IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR LIST IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | n = input()
n = list(n[::-1])
ans1, ans2, ans3, ans4 = [], [], [], []
s = 0
for i in n:
s += int(i)
if s % 3 == 0:
pass
else:
flag1 = flag2 = flag3 = flag4 = False
if s % 3 == 1:
ans1 = n[:]
for i in n:
if int(i) % 3 == 1:
ans1.remove(i)
flag1 = True
break
ans2 = n[:]
c = 0
for i in n:
if int(i) % 3 == 2:
ans2.remove(i)
c += 1
if c == 2:
flag2 = True
break
else:
ans3 = n[:]
for i in n:
if int(i) % 3 == 2:
ans3.remove(i)
flag3 = True
break
ans4 = n[:]
c = 0
for i in n:
if int(i) % 3 == 1:
ans4.remove(i)
c += 1
if c == 2:
flag4 = True
break
if len(ans1) != 0:
ans1 = "".join(ans1)
if ans1.count("0") == len(ans1):
ans1 = "0"
else:
ans1 = ans1[::-1].lstrip("0")
else:
ans1 = "-1"
if len(ans2) != 0:
ans2 = "".join(ans2)
if ans2.count("0") == len(ans2):
ans2 = "0"
else:
ans2 = ans2[::-1].lstrip("0")
else:
ans2 = "-1"
if len(ans3) != 0:
ans3 = "".join(ans3)
if ans3.count("0") == len(ans3):
ans3 = "0"
else:
ans3 = ans3[::-1].lstrip("0")
else:
ans3 = "-1"
if len(ans4) != 0:
ans4 = "".join(ans4)
if ans4.count("0") == len(ans4):
ans4 = "0"
else:
ans4 = ans4[::-1].lstrip("0")
else:
ans4 = "-1"
n = "".join(n[::-1])
if s % 3 == 0:
print(n)
elif flag1 == True or flag2 == True or flag3 == True or flag4 == True:
if s % 3 == 1:
if flag1 == flag2 == True:
print(max(ans1, ans2, key=len))
elif flag1 == True:
print(ans1)
elif flag2 == True:
print(ans2)
elif flag3 == flag4 == True:
print(max(ans3, ans4, key=len))
elif flag3 == True:
print(ans3)
elif flag4 == True:
print(ans4)
else:
print(-1) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR LIST LIST LIST LIST ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FOR VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR NUMBER STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR NUMBER STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR NUMBER STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR NUMBER STRING ASSIGN VAR STRING ASSIGN VAR FUNC_CALL STRING VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def deleteD(n, num):
rel = list(reversed(s))
for digit in reversed(s):
if int(digit) % 3 == n:
rel.remove(digit)
num -= 1
if num == 0:
if len(rel) == 0:
return []
while rel[-1] == "0" and len(rel) > 1:
rel.pop()
return list(reversed(rel))
return []
s = input()
n = sum([int(c) for c in s]) % 3
if n == 0:
print(s)
else:
if n == 1:
a = deleteD(1, 1)
b = deleteD(2, 2)
else:
a = deleteD(1, 2)
b = deleteD(2, 1)
s = max(a, b, key=len)
if len(s):
print("".join(s))
else:
print(-1) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER IF FUNC_CALL VAR VAR NUMBER RETURN LIST WHILE VAR NUMBER STRING FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR RETURN LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | s = list(input())
def db3(n, sm):
new = n.copy()
sm %= 3
if sm == 0:
return "".join(new)
else:
if sm == 2:
for j in range(1, len(new)):
if int(new[j]) % 3 == 2:
new[j] = ""
return "".join(new)
for j in range(1, len(new)):
if int(new[j]) % 3 == 1:
new[j] = ""
sm -= 1
if sm == 0:
return "".join(new)
else:
for j in range(1, len(new)):
if int(new[j]) % 3 == 1:
new[j] = ""
return "".join(new)
sm = 2
for j in range(1, len(new)):
if int(new[j]) % 3 == 2:
new[j] = ""
sm -= 1
if sm == 0:
return "".join(new)
if "0" in new:
return "0"
else:
return ""
a = 3
i = 0
l = []
Sm = sum([int(x) for x in s])
val = 0
while a and i < len(s):
if s[i] != "0":
l.append(db3(s[i:], Sm - val))
a -= 1
val += int(s[i])
i += 1
ans = max(l, key=len)
if ans:
print(ans)
else:
print(-1) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL STRING VAR IF VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR STRING RETURN FUNC_CALL STRING VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL STRING VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR STRING RETURN FUNC_CALL STRING VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL STRING VAR IF STRING VAR RETURN STRING RETURN STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | import sys
def lencheck(item1, item2):
if len(item1) > len(item2):
return True
def leadingzero(string):
while string[0:1] == "0" and len(string) != 1:
string = string[1:]
return string
data = sys.stdin.read().strip()
sumOfDigits = 0
for i in data:
sumOfDigits += int(i)
data2 = ""
if sumOfDigits % 3 == 0:
data2 = leadingzero(data)
count = 0
record = -1
if sumOfDigits % 3 == 1:
for i in data[0:]:
if int(i) % 3 == 1:
record = count
count += 1
if record != -1:
if lencheck(leadingzero(data[0:record] + data[record + 1 :]), data2):
data2 = leadingzero(data[0:record] + data[record + 1 :])
if len(data2) < len(data) - 1:
for i in range(0, len(data)):
if int(data[i]) % 3 == 2:
for j in range(i + 1, len(data)):
if int(data[j]) % 3 == 2 and lencheck(
leadingzero(data[0:i] + data[i + 1 : j] + data[j + 1 :]), data2
):
data2 = leadingzero(data[0:i] + data[i + 1 : j] + data[j + 1 :])
record = j
if (
len(data) > 2
and int(data[2:]) % 3 == 0
and lencheck(leadingzero(data[2:]), data2)
):
data2 = leadingzero(data[2:])
elif (
len(data) > 1
and int(data[1:]) % 3 == 0
and lencheck(leadingzero(data[1:]), data2)
):
data2 = leadingzero(data[1:])
if sumOfDigits % 3 == 2:
for i in data[0:]:
if int(i) % 3 == 2:
record = count
count += 1
if record != -1 and lencheck(
leadingzero(data[0:record] + data[record + 1 :]), data2
):
data2 = leadingzero(data[0:record] + data[record + 1 :])
if len(data2) < len(data) - 1:
for i in range(0, len(data)):
if int(data[i]) % 3 == 1:
for j in range(i + 1, len(data)):
if int(data[j]) % 3 == 1 and lencheck(
leadingzero(data[0:i] + data[i + 1 : j] + data[j + 1 :]), data2
):
data2 = leadingzero(data[0:i] + data[i + 1 : j] + data[j + 1 :])
record = j
if (
len(data) > 2
and int(data[2:]) % 3 == 0
and lencheck(leadingzero(data[2:]), data2)
):
data2 = leadingzero(data[2:])
elif (
len(data) > 1
and int(data[1:]) % 3 == 0
and lencheck(leadingzero(data[1:]), data2)
):
data2 = leadingzero(data[1:])
if data2 == "":
print(-1)
if data2 != "":
print(data2) | IMPORT FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN NUMBER FUNC_DEF WHILE VAR NUMBER NUMBER STRING FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FOR VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FOR VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR STRING EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def checkBeauty(digi):
total = 0
for i in range(0, len(digi)):
total += int(digi[i])
return total % 3
def modThree(digi):
one = []
two = []
for i in range(0, len(digi)):
if int(digi[i]) % 3 == 1:
one.insert(0, i)
elif int(digi[i]) % 3 == 2:
two.insert(0, i)
modResult = []
modResult.append(one)
modResult.append(two)
return modResult
def checkHead(digi):
if len(digi) == 0:
return -1
while len(digi) > 1:
if int(digi[0]) == 0:
digi.pop(0)
else:
break
return "".join(digi)
def checkOneOrTwo(digiOne, digiTwo):
resultOne = checkHead(digiOne)
resultTwo = checkHead(digiTwo)
finalResult = -1
if resultOne == -1 and resultTwo != -1:
finalResult = resultTwo
elif resultOne != -1 and resultTwo == -1:
finalResult = resultOne
elif resultOne != -1 and resultTwo != -1:
finalResult = resultTwo if len(resultTwo) > len(resultOne) else resultOne
return finalResult
def beauty(numDel):
delOne = []
delTwo = []
if numDel == 0:
print(checkHead(x))
return
elif numDel == 1:
if len(modResult[0]) != 0:
delOne = x[:]
delOne.pop(modResult[0][0])
if len(delOne) != 0 and int(delOne[0]) != 0:
print(checkHead(delOne))
return
if len(modResult[1]) <= 1:
print(checkHead(delOne))
return
elif len(modResult[1]) > 1:
delTwo = x[:]
delTwo.pop(modResult[1][0])
delTwo.pop(modResult[1][1])
if len(delTwo) != 0 and int(delTwo[0]) != 0:
print(checkHead(delTwo))
return
else:
print(checkOneOrTwo(delOne, delTwo))
return
else:
if len(modResult[1]) != 0:
delTwo = x[:]
delTwo.pop(modResult[1][0])
if len(delTwo) != 0 and int(delTwo[0]) != 0:
print(checkHead(delTwo))
return
if len(modResult[0]) <= 1:
print(checkHead(delTwo))
return
elif len(modResult[0]) > 1:
delOne = x[:]
delOne.pop(modResult[0][0])
delOne.pop(modResult[0][1])
if len(delOne) != 0 and int(delOne[0]) != 0:
print(checkHead(delOne))
return
else:
print(checkOneOrTwo(delOne, delTwo))
return
x = list(input())
modResult = modThree(x)
numToDel = checkBeauty(x)
beauty(numToDel) | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER WHILE FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER RETURN FUNC_CALL STRING VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN IF VAR NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN IF FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN IF FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | a = input()
if len(a) == 1:
if int(a) % 3 == 0:
print(a)
else:
print(-1)
exit(0)
one = []
two = []
sum = 0
zs, zf = 0, 0
for i in range(len(a)):
q = int(a[i])
sum += q
if q == 0:
if zs == 0:
zs = i
elif zs != 0 and zf == 0:
zf = i
if q % 3 == 1:
if len(one) <= 4:
one.append(i)
elif q % 3 == 2:
if len(two) <= 4:
two.append(i)
if zf == 0 and zs != 0:
zf = len(a) - 1
sol = []
s = ""
if sum % 3 == 1:
if len(one) > 0:
s = a[: one[-1]] + a[one[-1] + 1 :]
sol.append(s)
if len(a) > 1 and len(two) > 1:
s = a[: two[-2]] + a[two[-2] + 1 : two[-1]] + a[two[-1] + 1 :]
sol.append(s)
if sum % 3 == 2:
if len(two) > 0:
s = a[: two[-1]] + a[two[-1] + 1 :]
sol.append(s)
if len(a) > 1 and len(one) > 1:
s = a[: one[-2]] + a[one[-2] + 1 : one[-1]] + a[one[-1] + 1 :]
sol.append(s)
if sum % 3 == 0:
print(a)
else:
mini = 0
max = 0
for i in range(len(sol)):
if sol[i] == "":
sol[i] = "-1"
elif sol[i][0] == "0":
sol[i] = sol[i][zf - zs :]
if len(sol[i]) > 0:
if sol[i][0] == "0":
f = -1
for j in range(len(sol[i])):
if sol[i][j] != "0":
f = j
break
if f != -1:
sol[i] = sol[i][f:]
else:
sol[i] = "0"
if len(sol[i]) == 0:
sol[i] = "0"
if len(sol[i]) > max:
max = len(sol[i])
mini = i
print(sol[mini]) | ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR STRING IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR STRING IF VAR VAR NUMBER STRING ASSIGN VAR VAR VAR VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR STRING ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR STRING IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR STRING IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | s = input().strip()
m = {(0): 0, (1): 0, (2): 0}
bol = [0] * len(s)
two = []
one = []
i = 0
bc = 0
for a in s:
m[int(a) % 3] = m[int(a) % 3] + 1
if int(a) % 3 == 0:
bol[i] = 1
bc += 1
elif int(a) % 3 == 1:
one.append(i)
elif int(a) % 3 == 2:
two.append(i)
i += 1
out = "" + s
i = 0
for a in s:
if a != "0":
break
else:
out = "" + out[i + 1 :]
i += 1
if (m[1] + 2 * m[2]) % 3 == 1:
tm = ""
tp = ""
if m[1] >= 1:
tm += s
tm = tm[0 : one[len(one) - 1]] + tm[one[len(one) - 1] + 1 :]
i = 0
for a in tm:
if a != "0":
break
else:
tm = "" + tm[1:]
i += 1
if m[2] > 1:
tp += s
tp = tp[: two[len(two) - 1]] + tp[two[len(two) - 1] + 1 :]
tp = tp[: two[len(two) - 2]] + tp[two[len(two) - 2] + 1 :]
i = 0
for a in tp:
if a != "0":
break
else:
tp = "" + tp[1:]
i += 1
if len(tp) > len(tm):
out = tp
else:
out = tm
elif (m[1] + 2 * m[2]) % 3 == 2:
tm = ""
tp = ""
if m[2] >= 1:
tm += s
tm = tm[: two[len(two) - 1]] + tm[two[len(two) - 1] + 1 :]
i = 0
for a in tm:
if a != "0":
break
else:
tm = "" + tm[1:]
i += 1
if m[1] > 1:
tp += s
for z in range(len(one) - 1, len(one) - 3, -1):
tp = tp[: one[z]] + tp[one[z] + 1 :]
i = 0
for a in tp:
if a != "0":
break
else:
tp = "" + tp[1:]
i += 1
if len(tp) > len(tm):
out = tp
else:
out = tm
if out == "" and m[0] > 0:
out = "0"
elif out == "" and m[0] == 0:
out = "-1"
print(out) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP STRING VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR BIN_OP STRING VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER NUMBER NUMBER ASSIGN VAR STRING ASSIGN VAR STRING IF VAR NUMBER NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR BIN_OP STRING VAR NUMBER VAR NUMBER IF VAR NUMBER NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR BIN_OP STRING VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER NUMBER NUMBER ASSIGN VAR STRING ASSIGN VAR STRING IF VAR NUMBER NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR BIN_OP STRING VAR NUMBER VAR NUMBER IF VAR NUMBER NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR BIN_OP STRING VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR STRING VAR NUMBER NUMBER ASSIGN VAR STRING IF VAR STRING VAR NUMBER NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | s = input()
def rm(s, m):
if s == None or len(s) == 1:
return None
i = len(s) - 1
while i >= 0:
if int(s[i]) % 3 == m:
break
i -= 1
if i == -1:
return None
elif i == 0:
k = i + 1
while k < len(s) and s[k] == "0":
k += 1
if k == len(s):
return "0"
else:
return s[k:]
elif i == len(s) - 1:
return s[:i]
else:
return s[:i] + s[i + 1 :]
def ans(s):
s_sum = 0
i = 0
while i < len(s):
s_sum += int(s[i])
s_sum = s_sum % 3
i += 1
if s_sum == 0:
return s
elif s_sum == 1:
s1 = rm(s, 1)
s2 = rm(rm(s, 2), 2)
if s1 == None and s2 == None:
return -1
elif s1 == None:
return s2
elif s2 == None:
return s1
elif len(s1) > len(s2):
return s1
else:
return s2
elif s_sum == 2:
s1 = rm(s, 2)
s2 = rm(rm(s, 1), 1)
if s1 == None and s2 == None:
return -1
elif s1 == None:
return s2
elif s2 == None:
return s1
elif len(s1) > len(s2):
return s1
else:
return s2
t = ans(s)
if t == None:
print(-1)
else:
print(t) | ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR NONE FUNC_CALL VAR VAR NUMBER RETURN NONE ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER IF VAR NUMBER RETURN NONE IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER IF VAR FUNC_CALL VAR VAR RETURN STRING RETURN VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR RETURN BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER RETURN VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR NONE VAR NONE RETURN NUMBER IF VAR NONE RETURN VAR IF VAR NONE RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR RETURN VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR NONE VAR NONE RETURN NUMBER IF VAR NONE RETURN VAR IF VAR NONE RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NONE EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | n = list(input())
leng = 0
mp = [0, 0, 0]
for x in n:
leng += 1
mp[int(x) % 3] += 1
tot = (mp[1] + 2 * mp[2]) % 3
if tot == 0:
print("".join(n))
exit()
if mp[tot] == 0:
do = tot ^ 3
cnt = 2
elif mp[tot] == 1 and int(n[0]) % 3 == tot and n[1:3] == ["0", "0"]:
do = tot ^ 3
cnt = 2
if mp[do] == 0:
do = tot
cnt = 1
else:
do = tot
cnt = 1
index = leng - 1
if cnt >= leng:
print(-1)
exit()
for x in range(cnt):
while int(n[index]) % 3 != do:
index -= 1
n[index] = ""
index -= 1
index = 0
ans = "".join(n)
while ans[index] == "0" and index < leng - cnt - 1:
index += 1
print(ans[index:]) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER FOR VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER VAR NUMBER NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER LIST STRING STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR WHILE BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR WHILE VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | s = input()
rem = 0
rems = [0] * len(s)
one = []
two = []
for i in range(len(s)):
rems[i] = int(s[i]) % 3
if rems[i] == 1:
one.append(i)
elif rems[i] == 2:
two.append(i)
rem += int(s[i])
rem %= 3
include = [1] * len(s)
include1 = [1] * len(s)
f = 0
if rem == 1:
if len(one):
include[one[len(one) - 1]] = 0
if len(two) >= 2:
include1[two[len(two) - 1]] = 0
include1[two[len(two) - 2]] = 0
elif rem == 2:
one, two = two, one
if len(one):
include[one[len(one) - 1]] = 0
if len(two) >= 2:
include1[two[len(two) - 1]] = 0
include1[two[len(two) - 2]] = 0
ans = []
ans1 = []
for i in range(len(s)):
if include[i]:
ans.append(int(s[i]))
if include1[i]:
ans1.append(int(s[i]))
fans = []
fans1 = []
for i in ans:
if f == 0 and i == 0:
continue
if i:
f = 1
fans.append(i)
elif f:
fans.append(i)
ans.sort()
if len(ans):
if ans[0] == ans[len(ans) - 1] == 0:
fans.append(0)
f = 0
for i in ans1:
if f == 0 and i == 0:
continue
if i:
f = 1
fans1.append(i)
elif f:
fans1.append(i)
ans1.sort()
if len(ans1):
if ans1[0] == ans1[len(ans1) - 1] == 0:
fans1.append(0)
f = 0
if (len(fans) == 0 or sum(fans) % 3) and (len(fans1) == 0 or sum(fans1) % 3):
print(-1)
else:
if len(fans) < len(fans1):
fans, fans1 = fans1, fans
if sum(fans) % 3 == 0:
for i in fans:
print(i, end="")
else:
for i in fans1:
print(i, end="") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def find1right(s, c, orig):
r = s.rfind(c)
if r != -1:
res = orig[:r] + orig[r + 1 :]
res_strip = res.lstrip("0")
ans = len(orig) - len(res_strip), res_strip
if "0" in res and res_strip == "":
ans = len(s) - 1, "0"
return ans
return len(orig), ""
def find2right(s, c, orig):
r = s.rfind(c)
if r != -1:
rr = s.rfind(c, 0, r)
if rr != -1:
cur = orig[:rr] + orig[rr + 1 : r] + orig[r + 1 :]
cur_strip = cur.lstrip("0")
ans = len(orig) - len(cur_strip), cur_strip
if "0" in cur and cur_strip == "":
ans = len(s) - 1, "0"
return ans
return len(s), ""
s = input()
n = len(s)
rest = [(int(c) % 3) for c in s]
rest_str = "".join(str(i) for i in rest)
if len(s) == 1:
if int(s) % 3 == 0:
print(s)
else:
print(-1)
exit()
if s[1:] == "0" * (n - 1):
print(0)
exit()
q = sum(rest) % 3
if q == 0:
ans = s
elif q == 1:
ans = n, ""
ans = min(ans, find1right(rest_str, "1", s))
ans = min(ans, find2right(rest_str, "2", s))
ans = ans[1]
else:
ans = n, ""
ans = min(ans, find1right(rest_str, "2", s))
ans = min(ans, find2right(rest_str, "1", s))
ans = ans[1]
print(ans if ans != "" else -1) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF STRING VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING RETURN VAR RETURN FUNC_CALL VAR VAR STRING FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF STRING VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING RETURN VAR RETURN FUNC_CALL VAR VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER BIN_OP STRING BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR STRING VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR STRING VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR STRING VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR STRING VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR STRING VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def clean_trailing_zeros(a):
if not a:
return a
i = len(a) - 1
while i >= 1 and a[i] == 0:
i -= 1
return a[: i + 1]
def remove_digits(a, i1, i2):
if i1:
i1 = i1[0]
result1 = clean_trailing_zeros(a[:i1] + a[i1 + 1 :])
else:
result1 = []
if len(i2) >= 2:
i21, i22 = i2[:2]
result2 = clean_trailing_zeros(a[:i21] + a[i21 + 1 : i22] + a[i22 + 1 :])
else:
result2 = []
if len(result1) > len(result2):
return result1
else:
return result2
def main():
a = map(int, input())
a = list(a)[::-1]
if len(a) == 1:
a = a[0]
if a % 3 == 0:
return a
else:
return -1
a_mod_3 = [(x % 3) for x in a]
sum_mod_3 = sum(a) % 3
if sum_mod_3 == 0:
return "".join(map(str, a[::-1]))
if sum_mod_3 == 1:
i1 = [i for i, x in enumerate(a_mod_3) if x == 1]
i2 = [i for i, x in enumerate(a_mod_3) if x == 2]
else:
i1 = [i for i, x in enumerate(a_mod_3) if x == 2]
i2 = [i for i, x in enumerate(a_mod_3) if x == 1]
result = remove_digits(a, i1, i2)
if result:
return "".join(map(str, result[::-1]))
else:
return -1
print(main()) | FUNC_DEF IF VAR RETURN VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER VAR VAR NUMBER VAR NUMBER RETURN VAR BIN_OP VAR NUMBER FUNC_DEF IF VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN VAR RETURN NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL STRING FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR RETURN FUNC_CALL STRING FUNC_CALL VAR VAR VAR NUMBER RETURN NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def remove_1_remainder(num, num_sum):
dig_idx = len(num) - 1
to_remove = []
found = True
while num_sum % 3 != 0:
if dig_idx < 0 or len(num) - len(to_remove) == 1:
found = False
break
curr_num = int(num[dig_idx])
if curr_num % 3 == 1:
to_remove.append(dig_idx)
num_sum -= curr_num
dig_idx -= 1
return to_remove, found
def remove_2_remainder(num, num_sum):
dig_idx = len(num) - 1
to_remove = []
found = True
while num_sum % 3 != 0:
if dig_idx < 0 or len(num) - len(to_remove) == 1:
found = False
break
curr_num = int(num[dig_idx])
if curr_num % 3 == 2:
to_remove.append(dig_idx)
num_sum -= curr_num
dig_idx -= 1
return to_remove, found
def rebuild_string(num, indices):
new_str = ""
indices = set(indices)
for idx in range(len(num)):
if idx in indices:
continue
new_str += num[idx]
while len(new_str) != 0 and new_str[0] == "0":
if new_str == "0":
return new_str
new_str = new_str[1:]
return new_str
num = input()
num_sum = sum(int(p) for p in num)
ops = 0
dig_idx = len(num) - 1
remainder = num_sum % 3
to_print = None
if remainder == 0:
to_print = num
elif len(num) == 1:
to_print = -1
elif remainder == 1:
idx_removed, is_valid = remove_1_remainder(num, num_sum)
idx_removed_2, is_valid_2 = remove_2_remainder(num, num_sum)
if not is_valid and not is_valid_2:
to_print = -1
elif is_valid and is_valid_2:
str_1, str_2 = rebuild_string(num, idx_removed), rebuild_string(
num, idx_removed_2
)
to_print = str_1 if len(str_1) > len(str_2) else str_2
elif is_valid:
to_print = rebuild_string(num, idx_removed)
else:
to_print = rebuild_string(num, idx_removed_2)
elif remainder == 2:
idx_removed, is_valid = remove_1_remainder(num, num_sum)
idx_removed_2, is_valid_2 = remove_2_remainder(num, num_sum)
if not is_valid and not is_valid_2:
to_print = -1
elif is_valid and is_valid_2:
str_1, str_2 = rebuild_string(num, idx_removed), rebuild_string(
num, idx_removed_2
)
to_print = str_1 if len(str_1) > len(str_2) else str_2
elif is_valid:
to_print = rebuild_string(num, idx_removed)
else:
to_print = rebuild_string(num, idx_removed_2)
if (
to_print is None
or isinstance(to_print, str)
and len(to_print) > 1
and to_print[0] == "0"
):
raise Exception()
print(to_print) | FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER NUMBER IF VAR NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER NUMBER IF VAR NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR WHILE FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING IF VAR STRING RETURN VAR ASSIGN VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NONE IF VAR NUMBER ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NONE FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING FUNC_CALL VAR EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | num = list(map(int, list(input())))
num2 = sum(num) % 3
j = 0
for i in range(1, len(num)):
if num[i] % 3 == num2:
j = i
num3 = 3 - num2
z = []
for i in range(1, len(num)):
if num[i] % 3 == num3:
z += [i]
num4 = num[0] % 3
if num2 == 0:
for i in range(len(num)):
print(num[i], end="")
elif j != 0:
for i in range(len(num)):
if i != j:
print(num[i], end="")
elif num4 == num2 and len(num) != 1 and (len(z) < 2 or num[1] != 0):
i2 = len(num)
for i in range(1, len(num)):
if num[i] != 0:
i2 = i
break
for i in range(i2, len(num)):
print(num[i], end="")
if i2 == len(num):
print(0)
elif len(z) >= 2:
for i in range(len(num)):
if i != z[0] and i != z[1]:
print(num[i], end="")
elif num4 == 0:
print(-1)
elif num4 == num3 and len(z) == 0:
print(-1)
elif len(num) < 3:
print(-1)
else:
i2 = len(num)
for i in range(1, len(num)):
if num[i] != 0:
i2 = i
break
if i2 == z[0] and z[0] != len(num) - 1:
for i in range(i2 + 1, len(num)):
i2 = i
if num[i] != 0:
i2 = i
break
if i2 == len(num):
print(0)
else:
for i in range(i2, len(num)):
print(num[i], end="")
elif i2 == z[0] and z[0] == len(num) - 1:
print(0)
else:
for i in range(i2, len(num)):
if i != z[0]:
print(num[i], end="") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR LIST VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING IF VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING IF VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING IF VAR VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def countremove(n, mod):
res = 0
for i in range(len(n)):
if mod[len(n) - 1 - i] == 3:
continue
if n[i] == "0":
res += 1
else:
break
return res
def markremove(n):
mod = [(int(x) % 3) for x in n][::-1]
excess = sum(mod) % 3
ones = mod.count(1)
twos = mod.count(2)
if excess == 1:
remove1, remove2 = 104206969, 104206969
if ones > 0:
mod[mod.index(1)] = 3
remove1 = countremove(n, mod) + 1
mod[mod.index(3)] = 1
if twos > 1:
mod[mod.index(2)] = 3
mod[mod.index(2)] = 3
remove2 = countremove(n, mod) + 2
mod[mod.index(3)] = 2
mod[mod.index(3)] = 2
if remove1 < remove2:
mod[mod.index(1)] = 3
else:
mod[mod.index(2)] = 3
mod[mod.index(2)] = 3
elif excess == 2:
remove1, remove2 = 104206969, 104206969
if twos > 0:
mod[mod.index(2)] = 3
remove1 = countremove(n, mod) + 1
mod[mod.index(3)] = 2
if ones > 1:
mod[mod.index(1)] = 3
mod[mod.index(1)] = 3
remove2 = countremove(n, mod) + 2
mod[mod.index(3)] = 1
mod[mod.index(3)] = 1
if remove1 < remove2:
mod[mod.index(2)] = 3
else:
mod[mod.index(1)] = 3
mod[mod.index(1)] = 3
return mod
def buildans(n, mod):
excess = sum(mod) % 3
ans = ""
for i in range(len(n) - 1, -1, -1):
if mod[i] != 3:
ans += n[len(n) - 1 - i]
remove0s = 0
for i in range(len(ans)):
remove0s = i
if ans[i] != "0":
break
if remove0s > 0:
ans = ans[remove0s:]
if excess != 0 or ans == "":
ans = "-1"
return ans
n = input()
mod = markremove(n)
print(buildans(n, mod)) | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER RETURN VAR FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER VAR STRING ASSIGN VAR STRING RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def func(t):
an = ""
l = len(t)
for i in range(l):
if t[i] != "0":
break
for j in range(i, l):
an += t[j]
return an
for nitish in range(1):
s = input()
ss = 0
cnt = 10**20
cnt1 = 10**20
n = len(s)
for i in range(n):
ss += int(s[i])
if ss % 3 == 0:
print(s)
elif ss % 3 == 1:
ans = []
for i in range(n - 1, -1, -1):
if int(s[i]) % 3 == 1:
ans.append(i)
break
ans1 = []
for i in range(n - 1, -1, -1):
if int(s[i]) % 3 == 2:
ans1.append(i)
if len(ans1) == 2:
break
if len(ans) == 0 and len(ans1) < 2:
print(-1)
else:
cnt = 10**20
cnt1 = 10**20
t = ""
tt = ""
if len(ans) != 0:
t = ""
cnt = 1
for i in range(n):
if i != ans[0]:
t += s[i]
for i in t:
if i == "0":
cnt += 1
else:
break
if len(ans1) == 2:
cnt1 = 2
tt = ""
for i in range(n):
if i not in ans1:
tt += s[i]
for i in tt:
if i == "0":
cnt1 += 1
else:
break
if t != "" and cnt <= cnt1:
an = func(t)
if an != "":
print(func(t))
else:
print(-1)
elif tt != "":
an = func(tt)
if an != "":
print(func(tt))
else:
print(-1)
else:
print(-1)
else:
ans = []
for i in range(n - 1, -1, -1):
if int(s[i]) % 3 == 2:
ans.append(i)
break
ans1 = []
for i in range(n - 1, -1, -1):
if int(s[i]) % 3 == 1:
ans1.append(i)
if len(ans1) == 2:
break
if len(ans) == 0 and len(ans1) < 2:
print(-1)
else:
cnt = 10**20
cnt2 = 10**20
t = ""
tt = ""
if len(ans) != 0:
t = ""
cnt = 1
for i in range(n):
if i != ans[0]:
t += s[i]
for i in t:
if i == "0":
cnt += 1
else:
break
if len(ans1) == 2:
cnt1 = 2
tt = ""
for i in range(n):
if i not in ans1:
tt += s[i]
for i in tt:
if i == "0":
cnt1 += 1
else:
break
if t != "" and cnt <= cnt1:
an = func(t)
if an != "":
print(func(t))
else:
print(-1)
elif tt != "":
an = func(tt)
if an != "":
print(func(tt))
else:
print(-1)
else:
print(-1) | FUNC_DEF ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR VAR FOR VAR VAR IF VAR STRING VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR STRING ASSIGN VAR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR VAR FOR VAR VAR IF VAR STRING VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR STRING ASSIGN VAR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | import sys
from itertools import compress
s = input()
n = len(s)
mod = [0] * n
for i, x in enumerate(map(int, s)):
mod[i] = x % 3
total_mod = sum(mod) % 3
def remove_zeros(a):
for i in range(n):
if not a[i]:
continue
if s[i] == "0":
a[i] = 0
else:
return
if total_mod == 0:
a = [1] * n
remove_zeros(a)
ans = "".join(compress(s, a))
if ans:
print(ans)
else:
print(0 if "0" in s else -1)
else:
ans1, ans2 = "", ""
for i in range(n - 1, -1, -1):
if mod[i] == total_mod:
a = [1] * n
a[i] = 0
remove_zeros(a)
ans1 = "".join(compress(s, a))
break
rem = 2
a = [1] * n
for i in range(n - 1, -1, -1):
if mod[i] == 3 - total_mod:
a[i] = 0
rem -= 1
if rem == 0:
remove_zeros(a)
ans2 = "".join(compress(s, a))
break
ans = ans1 if len(ans1) > len(ans2) else ans2
if ans:
print(ans)
else:
print(0 if "0" in s else -1) | IMPORT ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR VAR IF VAR VAR IF VAR VAR STRING ASSIGN VAR VAR NUMBER RETURN IF VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING VAR NUMBER NUMBER ASSIGN VAR VAR STRING STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING VAR NUMBER NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | n = list(map(int, list(input())))
mds = []
for i in n:
mds.append(i % 3)
mod = sum(mds) % 3
try:
if mod == 1:
i = len(mds) - 1
nn = n[:]
while i >= 0:
if mds[i] == 1:
del nn[i]
break
i -= 1
i = len(mds) - 1
count = []
nnn = n[:]
while i >= 0:
if mds[i] == 2:
count.append(i)
if len(count) == 2:
del nnn[count[0]]
del nnn[count[1]]
break
i -= 1
if len(nn) != 1 and nn[0] == 0:
while nn[0] == 0:
del nn[0]
if len(nnn) != 1 and nnn[0] == 0:
while nnn[0] == 0:
del nnn[0]
if (
len(nn) < len(n)
and len(nn) > len(nnn)
or len(nnn) == len(n)
and len(nn) < len(n)
):
n = nn[:]
elif len(nnn) < len(n):
n = nnn[:]
else:
n = [-1]
if mod == 2:
i = len(mds) - 1
nn = n[:]
while i >= 0:
if mds[i] == 2:
del nn[i]
break
i -= 1
i = len(mds) - 1
count = []
nnn = n[:]
while i >= 0:
if mds[i] == 1:
count.append(i)
if len(count) == 2:
del nnn[count[0]]
del nnn[count[1]]
break
i -= 1
if len(nn) != 1 and nn[0] == 0:
while nn[0] == 0:
del nn[0]
if len(nnn) != 1 and nnn[0] == 0:
while nnn[0] == 0:
del nnn[0]
if (
len(nn) < len(n)
and len(nn) > len(nnn)
or len(nnn) == len(n)
and len(nn) < len(n)
):
n = nn[:]
elif len(nnn) < len(n):
n = nnn[:]
else:
n = [-1]
if len(n) != 1 and n[0] == 0:
while n[0] == 0:
del n[0]
if len(n) == 0:
n = [-1]
except:
if 0 in n:
n = [0]
else:
n = [-1]
print("".join(map(str, n))) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER IF VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR VAR WHILE VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER WHILE VAR NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER WHILE VAR NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER IF VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR VAR WHILE VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER WHILE VAR NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER WHILE VAR NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER WHILE VAR NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST NUMBER IF NUMBER VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def find(s):
if len(s) == 1:
if int(s) % 3 == 0:
return s
return -1
digit = [[] for i in range(3)]
zeros = []
for i in range(len(s)):
digit[int(s[i]) % 3] += [i]
if s[i] == "0":
zeros += [i]
total = (len(digit[1]) * 1 + len(digit[2]) * 2) % 3
if total == 0:
return s
if total == 1:
s1 = ""
if len(digit[1]) >= 1:
p1 = digit[1][-1]
s1 = s[:p1] + s[p1 + 1 :]
last_pos = len(s1)
for i in range(len(s1)):
if s1[i] != "0":
last_pos = i
break
if last_pos < len(s1):
s1 = s1[last_pos:]
elif len(s1) > 0:
if s1[-1] == "0":
s1 = "0"
s2 = ""
if len(digit[2]) >= 2:
p1 = digit[2][-2]
p2 = digit[2][-1]
s2 = s[:p1] + s[p1 + 1 : p2] + s[p2 + 1 :]
last_pos = len(s2)
for i in range(len(s2)):
if s2[i] != "0":
last_pos = i
break
if last_pos < len(s2):
s2 = s2[last_pos:]
elif len(s2) > 0:
if s2[-1] == "0":
s2 = "0"
if s1 == "" and s2 == "":
return -1
elif s1 == "":
return s2
elif s2 == "":
return s1
else:
if len(s1) > len(s2):
return s1
return s2
if total == 2:
digit[1], digit[2] = digit[2], digit[1]
s1 = ""
if len(digit[1]) >= 1:
p1 = digit[1][-1]
s1 = s[:p1] + s[p1 + 1 :]
last_pos = len(s1)
for i in range(len(s1)):
if s1[i] != "0":
last_pos = i
break
if last_pos < len(s1):
s1 = s1[last_pos:]
elif len(s1) > 0:
if s1[-1] == "0":
s1 = "0"
s2 = ""
if len(digit[2]) >= 2:
p1 = digit[2][-2]
p2 = digit[2][-1]
s2 = s[:p1] + s[p1 + 1 : p2] + s[p2 + 1 :]
last_pos = len(s2)
for i in range(len(s2)):
if s2[i] != "0":
last_pos = i
break
if last_pos < len(s2):
s2 = s2[last_pos:]
elif len(s2) > 0:
if s2[-1] == "0":
s2 = "0"
if s1 == "" and s2 == "":
return -1
elif s1 == "":
return s2
elif s2 == "":
return s1
else:
if len(s1) > len(s2):
return s1
return s2
print(find(input())) | FUNC_DEF IF FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN VAR RETURN NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER LIST VAR IF VAR VAR STRING VAR LIST VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR NUMBER RETURN VAR IF VAR NUMBER ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR STRING IF VAR STRING VAR STRING RETURN NUMBER IF VAR STRING RETURN VAR IF VAR STRING RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR RETURN VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR STRING IF VAR STRING VAR STRING RETURN NUMBER IF VAR STRING RETURN VAR IF VAR STRING RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | sn = input().strip()
n = len(sn)
def rem3(sn):
return sum(map(int, sn)) % 3
def buildn(sn, com, start=0):
return "".join(sn[start + i] for i in com)
def subminbeau2(sn, start=0):
n = len(sn)
r3 = rem3(sn[start:])
sn3 = "".join(str(int(c) % 3) for c in sn)
nums3 = set(sn3[1 + start :])
if r3 == 0:
return 0, sn[start:]
elif str(r3) in nums3:
p1 = sn3.find(str(r3), start + 1)
return 1, sn[start:p1] + sn[p1 + 1 :]
elif sn3[1 + start :].count(str(3 - r3)) >= 2:
p1 = sn3.find(str(3 - r3), start + 1)
p2 = sn3.find(str(3 - r3), p1 + 1)
return 2, sn[start:p1] + sn[p1 + 1 : p2] + sn[p2 + 1 :]
else:
return -1, None
def minbeau(sn):
n = len(sn)
nnzs = [i for i, c in enumerate(sn) if c != "0"]
res = n, ""
for i in nnzs[:3]:
td, d3n = subminbeau2(sn, i)
if td >= 0:
res = min(res, (i + td, d3n))
if res[0] < n:
return res[1]
elif "0" in sn:
return "0"
else:
return "-1"
print(minbeau(sn)) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER FUNC_DEF NUMBER RETURN FUNC_CALL STRING VAR BIN_OP VAR VAR VAR VAR FUNC_DEF NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR IF VAR NUMBER RETURN NUMBER VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN NUMBER BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP NUMBER VAR FUNC_CALL VAR BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER RETURN NUMBER BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER RETURN NUMBER NONE FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR STRING ASSIGN VAR VAR STRING FOR VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR RETURN VAR NUMBER IF STRING VAR RETURN STRING RETURN STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def RemoveFirstZero(str):
count = 0
for i in range(len(str)):
if str[i] != "0":
count = i
break
return str[count:]
def RemainderOf3(str):
sum = 0
for i in range(len(str)):
sum += int(str[i])
return sum % 3
input1 = input()
setofans = []
t = 0
p = True
q = True
max = 0
if RemainderOf3(input1) == 0:
setofans.append(input1)
elif RemainderOf3(input1) == 1:
for i in range(len(input1) - 1, -1, -1):
if int(input1[i]) % 3 == 1:
if p:
setofans.append(input1[:i] + input1[i + 1 :])
p = False
else:
continue
if int(input1[i]) % 3 == 2:
if q:
max += 1
if max == 1:
t = i
elif max == 2:
setofans.append(input1[:i] + input1[i + 1 : t] + input1[t + 1 :])
q = False
else:
continue
else:
continue
else:
for i in range(len(input1) - 1, -1, -1):
if int(input1[i]) % 3 == 1:
if q:
max += 1
if max == 1:
t = i
elif max == 2:
setofans.append(input1[:i] + input1[i + 1 : t] + input1[t + 1 :])
q = False
else:
continue
else:
continue
if int(input1[i]) % 3 == 2:
if p:
setofans.append(input1[:i] + input1[i + 1 :])
p = False
else:
continue
ans = list(map(RemoveFirstZero, setofans))
max = len(ans[0])
a = ans[0]
for i in ans:
if max < len(i):
max = len(i)
a = i
if a != "":
if a[0] != "0":
print(a)
else:
print("0")
else:
print(-1) | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER IF VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER IF VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER IF VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER IF VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR IF VAR STRING IF VAR NUMBER STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | s = input()
L = len(s)
S = 0
zero = []
one = []
two = []
for i in range(L):
x = int(s[i])
S = (S + x) % 3
if x % 3 == 1:
one.append(i)
elif x % 3 == 2:
two.append(i)
elif x == 0:
zero.append(i)
if S == 0:
print(s)
elif S == 1:
if len(one) >= 1 and (one[-1] > 0 or len(zero) == 0 or zero[0] > 1) and L > 1:
print(s[: one[-1]] + s[one[-1] + 1 :])
elif len(two) >= 2 and two[-2] > 0 and L > 2:
print(s[: two[-2]] + s[two[-2] + 1 : two[-1]] + s[two[-1] + 1 :])
elif len(one) == 1 and L > 1:
print(int(s[1:]))
elif len(two) == 2 and L > 2:
print(int(s[1 : two[-1]] + s[two[-1] + 1 :]))
else:
print(-1)
elif len(two) >= 1 and (two[-1] > 0 or len(zero) == 0 or zero[0] > 1) and L > 1:
print(s[: two[-1]] + s[two[-1] + 1 :])
elif len(one) >= 2 and one[-2] > 0 and L > 2:
print(s[: one[-2]] + s[one[-2] + 1 : one[-1]] + s[one[-1] + 1 :])
elif len(two) == 1 and L > 1:
print(int(s[1:]))
elif len(one) == 2 and L > 2:
print(int(s[1 : one[-1]] + s[one[-1] + 1 :]))
else:
print(-1) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def erase(s, n, m):
p = "147" if n == 1 else "258"
if sum(s.count(c) for c in p) < m:
return []
t = list(reversed(s))
for c in p:
while t.count(c) > 0 and m > 0:
t.remove(c)
m -= 1
while len(t) > 1 and t[-1] == "0":
t.pop()
return list(reversed(t))
def solve(s):
n = sum(int(c) for c in s) % 3
if n == 0:
a = b = s
if n == 1:
a = erase(s, 1, 1)
b = erase(s, 2, 2)
if n == 2:
a = erase(s, 2, 1)
b = erase(s, 1, 2)
s = max(a, b, key=len)
return "-1" if s == [] else s
def divide_by_three(s):
return solve(s)
s = list(input())
result = divide_by_three(s)
print(*result, sep="") | FUNC_DEF ASSIGN VAR VAR NUMBER STRING STRING IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR WHILE FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING EXPR FUNC_CALL VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR LIST STRING VAR FUNC_DEF RETURN FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR STRING |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def n_by_deleting_r(n, r):
if all(map(lambda x: int(x) % 3 != r, n)):
return ""
for idx in range(len(n) - 1, -1, -1):
if r == int(n[idx]) % 3:
return n[:idx] + n[idx + 1 :]
def n_by_two_digit_remove(n, r):
if sum(map(lambda x: int(x) % 3 == r, n)) < 2:
return ""
L = -1
R = -1
for i in range(len(n) - 1, -1, -1):
if int(n[i]) % 3 == r:
if R == -1:
R = i
else:
L = i
break
return n[:L] + n[L + 1 : R] + n[R + 1 :]
def rem_prefix0(s: str):
for i in range(len(s)):
if s[i] != "0":
return s[i:]
return "0"
def solve(n):
s = sum(map(int, list(n)))
r = s % 3
if r == 0:
return n
sol1 = n_by_deleting_r(n, r)
sol2 = n_by_two_digit_remove(n, 3 - r)
if sol1 == "" and sol2 == "":
return "-1"
if sol1 == "":
return rem_prefix0(sol2)
if sol2 == "":
return rem_prefix0(sol1)
if len(rem_prefix0(sol1)) > len(rem_prefix0(sol2)):
return rem_prefix0(sol1)
else:
return rem_prefix0(sol2)
def main():
print(solve(input()))
main() | FUNC_DEF IF FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR RETURN STRING FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER RETURN BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FUNC_DEF IF FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER RETURN STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR RETURN BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_DEF VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING RETURN VAR VAR RETURN STRING FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR IF VAR STRING VAR STRING RETURN STRING IF VAR STRING RETURN FUNC_CALL VAR VAR IF VAR STRING RETURN FUNC_CALL VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | digits = [int(d) for d in input()]
remainder = sum(digits) % 3
def clip(digits):
digits_copy = digits.copy()
while len(digits_copy) > 0:
if digits_copy[0] == 0 and len(digits_copy) > 1:
del digits_copy[0]
else:
break
if len(digits_copy) != 0:
return "".join([str(c) for c in digits_copy])
ans = None
if remainder == 0:
ans = clip(digits)
elif remainder == 1:
length = len(digits)
broke = 0
digits_del1 = digits.copy()
digits_del2 = digits.copy()
length1, length2 = None, None
for i in range(length):
if digits_del1[length - 1 - i] % 3 == 1:
del digits_del1[length - 1 - i]
broke += 1
break
if broke != 0:
ans = clip(digits_del1)
length1 = None if ans is None else len(ans)
count = 0
mod2_index = []
for i in range(length):
if digits[length - 1 - i] % 3 == 2:
count += 1
mod2_index.append(length - 1 - i)
if count == 2:
break
if count == 2:
del digits_del2[mod2_index[0]]
del digits_del2[mod2_index[1]]
ans2 = clip(digits_del2)
length2 = None if ans2 is None else len(ans2)
if length1 is None:
if length2 is not None:
ans = ans2
elif length2 > length1:
ans = ans2
elif remainder == 2:
length = len(digits)
broke = 0
digits_del1 = digits.copy()
digits_del2 = digits.copy()
length1, length2 = None, None
for i in range(length):
if digits_del1[length - 1 - i] % 3 == 2:
del digits_del1[length - 1 - i]
broke += 1
break
if broke != 0:
ans = clip(digits_del1)
length1 = None if ans is None else len(ans)
count = 0
mod2_index = []
for i in range(length):
if digits[length - 1 - i] % 3 == 1:
count += 1
mod2_index.append(length - 1 - i)
if count == 2:
break
if count == 2:
del digits_del2[mod2_index[0]]
del digits_del2[mod2_index[1]]
ans2 = clip(digits_del2)
length2 = None if ans2 is None else len(ans2)
if length1 is None:
if length2 is not None:
ans = ans2
elif length2 > length1:
ans = ans2
if ans is None:
print(-1)
else:
print(ans) | ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NONE IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NONE NONE FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NONE NONE FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NONE NONE FUNC_CALL VAR VAR IF VAR NONE IF VAR NONE ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NONE NONE FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NONE NONE FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NONE NONE FUNC_CALL VAR VAR IF VAR NONE IF VAR NONE ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NONE EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | n = input()
nums = [(int(i) % 3) for i in n]
ones = nums.count(1)
twos = nums.count(2)
dis = (ones - twos) % 3
def solve():
global n
global nums
global dis
if dis == 0:
return n
t = rfind(dis)
if t > 0 or t == 0 and len(n) > 1 and nums[t + 1] != 0:
return n[:t] + n[t + 1 :]
g, c = findtwo(3 - dis)
if g > 0:
return n[:g] + n[g + 1 : c] + n[c + 1 :]
if nums[0] == dis:
return tr(n[1:])
return tr(n[:g] + n[g + 1 : c] + n[c + 1 :])
def rfind(t):
c = -1
for i in range(len(n)):
if nums[i] == t:
c = i
return c
def findtwo(tr):
c = -1
g = -1
for i in range(len(n)):
if nums[i] == tr:
c = g
g = i
return c, g
def tr(s):
if len(s) == 0:
return "-1"
i = 0
while i < len(s) - 1 and s[i] == "0":
i += 1
return s[i:]
print(solve()) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_DEF IF VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER RETURN BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP NUMBER VAR IF VAR NUMBER RETURN BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR RETURN FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR STRING VAR NUMBER RETURN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | m = list(input())
n = list(map(int, m))
s = sum(n)
l = len(n)
if s % 3 == 0:
print("".join(m))
else:
r = s % 3
for j in range(1, len(n)):
if n[j] % 3 == r:
n.pop(j)
break
if len(n) == l:
flag = 0
if n[0] % 3 == r:
if len(n) > 1:
if n[1] != 0:
n.pop(0)
flag = 1
if flag == 0:
num = []
count = 0
for j in range(1, len(n)):
if n[j] % 3 != 0 and n[j] % 3 != r:
count += 1
num.append(j)
if count == 2:
break
if count == 2:
n.pop(num[1])
n.pop(num[0])
elif count == 1:
n.pop(num[0])
n.pop(0)
else:
n.pop(0)
if len(n) == 0:
print(-1)
else:
j = 0
flag = 0
while flag == 0 and len(n) > 1:
flag = 1
if n[j] == 0:
n.pop(j)
flag = 0
for j in range(len(n)):
print(n[j], end="")
print("") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR STRING |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | def f(x, y):
global n
if d[x] and (d[x][-1] != 0 or n[1] != "0"):
n = n[: d[x][-1]] + n[d[x][-1] + 1 :]
else:
i = 0
if d[x]:
for i in range(1, len(n)):
if n[i] == "0":
continue
else:
break
if i == 2:
n = n[2:]
elif len(d[y]) >= 2:
a = d[y][-2]
b = d[y][-1]
n = n[:a] + n[a + 1 : b] + n[b + 1 :]
ridofzero()
else:
n = n[i:]
def ridofzero():
global n
for i in range(len(n)):
if n[i] == "0":
continue
else:
break
if len(n) > 0:
n = n[i:]
n = str(input())
ch = False
if n[0] == "0":
ch = True
ridofzero()
d = {(0): [], (1): [], (2): []}
s = 0
for i in range(len(n)):
x = int(n[i])
s += x
d[x % 3].append(i)
if s < 3:
if d[0] or ch:
print(0)
else:
print(-1)
elif len(n) == 1:
if not s % 3:
print(n)
else:
print(-1)
else:
x = s % 3
if x == 0:
pass
elif x == 1:
f(1, 2)
else:
f(2, 1)
if len(n) == 0:
print(-1)
else:
print(n) | FUNC_DEF IF VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER STRING ASSIGN VAR BIN_OP VAR VAR VAR NUMBER VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR DICT NUMBER NUMBER NUMBER LIST LIST LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | s = input()
n = len(s)
dp, leadingzero, camefrom, erasepos = [], 0, [], []
for i in range(3):
dp.append([int(1000000000.0)] * (n + 1))
camefrom.append([int(-1)] * (n + 1))
dp[0][n] = 0
for i in range(n - 1, -1, -1):
for j in range(3):
dp[j][i] = min(
dp[j][i], min(1 + dp[j][i + 1], dp[(int(j) - int(s[i])) % 3][i + 1])
)
if dp[j][i] >= int(1000000000.0):
continue
if (
1 + dp[j][i + 1] == dp[j][i]
and 1 + dp[j][i + 1] != dp[(int(j) - int(s[i])) % 3][i + 1]
):
camefrom[j][i] = j
else:
camefrom[j][i] = (int(j) - int(s[i])) % 3
ans = []
parity, initialparity = 0, 0
for i in range(n - 1):
previdx = camefrom[initialparity][i]
if dp[initialparity][i] == 1 + dp[previdx][i + 1]:
initialparity = previdx
continue
initialparity = previdx
ans.append(s[i])
parity += int(s[i]) % 3
if (parity + int(s[n - 1])) % 3 == 0:
ans.append(s[n - 1])
for i in range(len(ans)):
if ans[i] != "0":
break
ans1, ans2, leading1 = ans, [s[0]], i
parity, initialparity = int(s[0]) % 3, -int(s[0]) % 3
for i in range(1, n - 1):
previdx = camefrom[initialparity][i]
if previdx == -1:
break
if dp[initialparity][i] == 1 + dp[previdx][i + 1]:
initialparity = previdx
continue
initialparity = previdx
ans2.append(s[i])
parity += int(s[i]) % 3
if n > 1 and (parity + int(s[n - 1])) % 3 == 0:
ans2.append(s[n - 1])
if len(ans1) - leading1 < len(ans2) and parity % 3 == 0:
print("".join(ans2))
elif len(ans1) != 0:
print("".join(ans1[leading1:]))
else:
print(-1) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR LIST NUMBER LIST LIST FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP LIST FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP LIST FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR NUMBER IF BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR VAR LIST VAR NUMBER VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER IF VAR VAR VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR NUMBER |
A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input
The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).
Output
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.
Examples
Input
1033
Output
33
Input
10
Output
0
Input
11
Output
-1
Note
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | a = [int(x) for x in input()]
n = len(a)
rem = sum(a) % 3
a1 = a[:]
r1 = rem
i = n - 1
while r1 > 0 and i >= 0:
if a1[i] % 3 == 2:
if r1 == 1:
a1[i] = -1
r1 = 2
else:
a1[i] = -1
r1 = 0
i -= 1
cb = n + 1
ab = []
if r1 == 0:
c1 = 0
i = 0
while i < n and (a1[i] == -1 or a1[i] == 0):
c1 += 1
i += 1
while i < n:
c1 += 1 if a1[i] == -1 else 0
i += 1
cb = c1
ab = a1
a2 = a[:]
r2 = rem
i = n - 1
while r2 > 0 and i >= 0:
if a2[i] % 3 == 1:
a2[i] = -1
r2 -= 1
i -= 1
if r2 == 0:
c1 = 0
i = 0
while i < n and (a2[i] == -1 or a2[i] == 0):
c1 += 1
i += 1
while i < n:
c1 += 1 if a2[i] == -1 else 0
i += 1
if c1 < cb:
cb = c1
ab = a2
a3 = a[:]
r3 = rem
i = n - 1
while r3 > 0 and i >= 0:
d1 = 1
d2 = 2
if a3[i] % 3 == 1 and d1 > 0:
a3[i] = -1
r3 -= 1
if a3[i] % 3 == 2 and d2 > 0:
if r3 == 1:
a3[i] = -1
r3 = 2
else:
a3[i] = -1
r3 = 0
i -= 1
if r3 == 0:
c1 = 0
i = 0
while i < n and (a3[i] == -1 or a3[i] == 0):
c1 += 1
i += 1
while i < n:
c1 += 1 if a3[i] == -1 else 0
i += 1
if c1 < cb:
cb = c1
ab = a3
if cb == n + 1:
print(-1)
else:
i = 0
while i < n and (ab[i] == 0 or ab[i] == -1):
i += 1
if i == n:
print(0 if sum(ab) != -n else -1)
else:
while i < n:
if ab[i] != -1:
print(ab[i], end="")
i += 1 | ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER WHILE VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER WHILE VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER WHILE VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER NUMBER WHILE VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING VAR NUMBER |
It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.
We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (x_{i}, y_{i}). Both Adil and Bera can carry only one bottle at once each.
For both Adil and Bera the process looks as follows: Choose to stop or to continue to collect bottles. If the choice was to continue then choose some bottle and walk towards it. Pick this bottle and walk to the recycling bin. Go to step 1.
Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.
They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.
-----Input-----
First line of the input contains six integers a_{x}, a_{y}, b_{x}, b_{y}, t_{x} and t_{y} (0 ≤ a_{x}, a_{y}, b_{x}, b_{y}, t_{x}, t_{y} ≤ 10^9) — initial positions of Adil, Bera and recycling bin respectively.
The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.
Then follow n lines, each of them contains two integers x_{i} and y_{i} (0 ≤ x_{i}, y_{i} ≤ 10^9) — position of the i-th bottle.
It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.
-----Output-----
Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10^{ - 6}.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if $\frac{|a - b|}{\operatorname{max}(1, b)} \leq 10^{-6}$.
-----Examples-----
Input
3 1 1 2 0 0
3
1 1
2 1
2 3
Output
11.084259940083
Input
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3
Output
33.121375178000
-----Note-----
Consider the first sample.
Adil will use the following path: $(3,1) \rightarrow(2,1) \rightarrow(0,0) \rightarrow(1,1) \rightarrow(0,0)$.
Bera will use the following path: $(1,2) \rightarrow(2,3) \rightarrow(0,0)$.
Adil's path will be $1 + \sqrt{5} + \sqrt{2} + \sqrt{2}$ units long, while Bera's path will be $\sqrt{2} + \sqrt{13}$ units long. | line = list(map(int, input().split(" ")))
A, B, T = zip(line[::2], line[1::2])
n = int(input())
bottles = [tuple(map(int, input().split(" "))) for _ in range(n)]
def dist(x, y):
return abs(complex(*x) - complex(*y))
def keyFun(x, a):
return dist(T, bottles[x]) - dist(a, bottles[x])
if n == 1:
ret = min(
dist(bottles[0], A) + dist(bottles[0], T),
dist(bottles[0], B) + dist(bottles[0], T),
)
print(format(ret, ".100f"))
exit()
ret = float("inf")
allSum = sum(map(lambda x: dist(x, T) * 2, bottles))
for a, b in ((A, B), (B, A)):
best1 = max(range(n), key=lambda x: keyFun(x, a))
best2 = max((i for i in range(n) if i != best1), key=lambda x: keyFun(x, b))
ret = min(
ret,
allSum
+ dist(a, bottles[best1])
- dist(T, bottles[best1])
+ min(0, dist(b, bottles[best2]) - dist(T, bottles[best2])),
)
print(format(ret, ".100f")) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING VAR FUNC_CALL VAR VAR FUNC_DEF RETURN FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR FOR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR STRING |
It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.
We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (x_{i}, y_{i}). Both Adil and Bera can carry only one bottle at once each.
For both Adil and Bera the process looks as follows: Choose to stop or to continue to collect bottles. If the choice was to continue then choose some bottle and walk towards it. Pick this bottle and walk to the recycling bin. Go to step 1.
Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.
They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.
-----Input-----
First line of the input contains six integers a_{x}, a_{y}, b_{x}, b_{y}, t_{x} and t_{y} (0 ≤ a_{x}, a_{y}, b_{x}, b_{y}, t_{x}, t_{y} ≤ 10^9) — initial positions of Adil, Bera and recycling bin respectively.
The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.
Then follow n lines, each of them contains two integers x_{i} and y_{i} (0 ≤ x_{i}, y_{i} ≤ 10^9) — position of the i-th bottle.
It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.
-----Output-----
Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10^{ - 6}.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if $\frac{|a - b|}{\operatorname{max}(1, b)} \leq 10^{-6}$.
-----Examples-----
Input
3 1 1 2 0 0
3
1 1
2 1
2 3
Output
11.084259940083
Input
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3
Output
33.121375178000
-----Note-----
Consider the first sample.
Adil will use the following path: $(3,1) \rightarrow(2,1) \rightarrow(0,0) \rightarrow(1,1) \rightarrow(0,0)$.
Bera will use the following path: $(1,2) \rightarrow(2,3) \rightarrow(0,0)$.
Adil's path will be $1 + \sqrt{5} + \sqrt{2} + \sqrt{2}$ units long, while Bera's path will be $\sqrt{2} + \sqrt{13}$ units long. | ax, ay, bx, by, tx, ty = list(map(int, input().split()))
od = []
co = []
objects = int(input())
for i in range(objects):
x, y = list(map(int, input().split()))
od.append(((tx - x) ** 2 + (ty - y) ** 2) ** 0.5)
co.append([x, y])
asaved = [0, 0]
asaved2 = [0, 0]
ma = -(10**25)
for i in range(objects):
x = co[i][0]
y = co[i][1]
saved = od[i] - ((ax - x) ** 2 + (ay - y) ** 2) ** 0.5
ma = max(ma, saved)
if saved > asaved[0]:
asaved2[0] = asaved[0]
asaved2[1] = asaved[1]
asaved[0] = saved
asaved[1] = i
elif saved > asaved2[0]:
asaved2[0] = saved
asaved2[1] = i
bsaved = [0, 0]
bsaved2 = [0, 0]
mb = -(10**25)
for i in range(objects):
x = co[i][0]
y = co[i][1]
saved = od[i] - ((bx - x) ** 2 + (by - y) ** 2) ** 0.5
mb = max(mb, saved)
if saved > bsaved[0]:
bsaved2[1] = bsaved[1]
bsaved2[0] = bsaved[0]
bsaved[0] = saved
bsaved[1] = i
elif saved > bsaved2[0]:
bsaved2[0] = saved
bsaved2[1] = i
tot = 2 * sum(od)
if mb < 0 and ma < 0:
tot -= max(mb, ma)
elif bsaved[1] == asaved[1]:
tot -= max(asaved[0] + bsaved2[0], asaved2[0] + bsaved[0])
else:
tot -= asaved[0] + bsaved[0]
print(tot) | ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR LIST VAR VAR ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.
We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (x_{i}, y_{i}). Both Adil and Bera can carry only one bottle at once each.
For both Adil and Bera the process looks as follows: Choose to stop or to continue to collect bottles. If the choice was to continue then choose some bottle and walk towards it. Pick this bottle and walk to the recycling bin. Go to step 1.
Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.
They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.
-----Input-----
First line of the input contains six integers a_{x}, a_{y}, b_{x}, b_{y}, t_{x} and t_{y} (0 ≤ a_{x}, a_{y}, b_{x}, b_{y}, t_{x}, t_{y} ≤ 10^9) — initial positions of Adil, Bera and recycling bin respectively.
The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.
Then follow n lines, each of them contains two integers x_{i} and y_{i} (0 ≤ x_{i}, y_{i} ≤ 10^9) — position of the i-th bottle.
It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.
-----Output-----
Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10^{ - 6}.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if $\frac{|a - b|}{\operatorname{max}(1, b)} \leq 10^{-6}$.
-----Examples-----
Input
3 1 1 2 0 0
3
1 1
2 1
2 3
Output
11.084259940083
Input
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3
Output
33.121375178000
-----Note-----
Consider the first sample.
Adil will use the following path: $(3,1) \rightarrow(2,1) \rightarrow(0,0) \rightarrow(1,1) \rightarrow(0,0)$.
Bera will use the following path: $(1,2) \rightarrow(2,3) \rightarrow(0,0)$.
Adil's path will be $1 + \sqrt{5} + \sqrt{2} + \sqrt{2}$ units long, while Bera's path will be $\sqrt{2} + \sqrt{13}$ units long. | import sys
input = sys.stdin.readline
def dist(x1, y1, x2, y2):
return (abs(x1 - x2) ** 2 + abs(y1 - y2) ** 2) ** 0.5
ax, ay, bx, by, tx, ty = map(int, input().split())
n = int(input())
xy = [list(map(int, input().split())) for i in range(n)]
s = 0
for i in range(n):
x, y = xy[i]
s += dist(tx, ty, x, y)
s *= 2
ans = 10**18
ad = [
[dist(ax, ay, xy[i][0], xy[i][1]) - dist(tx, ty, xy[i][0], xy[i][1]), i]
for i in range(n)
]
bd = [
[dist(bx, by, xy[i][0], xy[i][1]) - dist(tx, ty, xy[i][0], xy[i][1]), i]
for i in range(n)
]
ad.sort()
bd.sort()
if ad[0][0] > 0 or bd[0][0] > 0 or n == 1:
md = min(ad[0][0], bd[0][0])
ans = min(ans, s + md)
elif ad[0][1] == bd[0][1]:
md = min(ad[0][0] + bd[1][0], ad[1][0] + bd[0][0], ad[0][0], bd[0][0])
ans = min(ans, s + md)
else:
md = min(ad[0][0] + bd[0][0], ad[0][0], bd[0][0])
ans = min(ans, s + md)
print(ans) | IMPORT ASSIGN VAR VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR LIST BIN_OP FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST BIN_OP FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.
We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (x_{i}, y_{i}). Both Adil and Bera can carry only one bottle at once each.
For both Adil and Bera the process looks as follows: Choose to stop or to continue to collect bottles. If the choice was to continue then choose some bottle and walk towards it. Pick this bottle and walk to the recycling bin. Go to step 1.
Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.
They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.
-----Input-----
First line of the input contains six integers a_{x}, a_{y}, b_{x}, b_{y}, t_{x} and t_{y} (0 ≤ a_{x}, a_{y}, b_{x}, b_{y}, t_{x}, t_{y} ≤ 10^9) — initial positions of Adil, Bera and recycling bin respectively.
The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.
Then follow n lines, each of them contains two integers x_{i} and y_{i} (0 ≤ x_{i}, y_{i} ≤ 10^9) — position of the i-th bottle.
It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.
-----Output-----
Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10^{ - 6}.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if $\frac{|a - b|}{\operatorname{max}(1, b)} \leq 10^{-6}$.
-----Examples-----
Input
3 1 1 2 0 0
3
1 1
2 1
2 3
Output
11.084259940083
Input
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3
Output
33.121375178000
-----Note-----
Consider the first sample.
Adil will use the following path: $(3,1) \rightarrow(2,1) \rightarrow(0,0) \rightarrow(1,1) \rightarrow(0,0)$.
Bera will use the following path: $(1,2) \rightarrow(2,3) \rightarrow(0,0)$.
Adil's path will be $1 + \sqrt{5} + \sqrt{2} + \sqrt{2}$ units long, while Bera's path will be $\sqrt{2} + \sqrt{13}$ units long. | import sys
def calc_dist(x, y, x2, y2):
return ((x - x2) ** 2 + (y - y2) ** 2) ** 0.5
inp = sys.stdin.read().splitlines()
line0 = inp[0].split()
Ax, Ay, Bx, By, Tx, Ty = (
int(line0[0]),
int(line0[1]),
int(line0[2]),
int(line0[3]),
int(line0[4]),
int(line0[5]),
)
totaldist = 0
Alst = []
Blst = []
for i in range(2, int(inp[1]) + 2):
cood = inp[i].split()
x = int(cood[0])
y = int(cood[1])
Tdist = calc_dist(x, y, Tx, Ty)
totaldist += Tdist
Adist = calc_dist(x, y, Ax, Ay)
Bdist = calc_dist(x, y, Bx, By)
Asave = Tdist - Adist
Bsave = Tdist - Bdist
Alst.append(Asave)
Blst.append(Bsave)
Amax = max(Alst)
Bmax = max(Blst)
maxsave = max([Amax + Bmax, Amax, Bmax])
if maxsave == Amax + Bmax:
Aindex = Alst.index(Amax)
Bindex = Blst.index(Bmax)
if Aindex == Bindex:
Alst[Aindex] = -1000000000.0
Amax2 = max(Alst)
Blst[Bindex] = -1000000000.0
Bmax2 = max(Blst)
maxsave = max(Amax2 + Bmax, Amax + Bmax2, Amax, Bmax)
final = totaldist * 2 - maxsave
print(final) | IMPORT FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR LIST BIN_OP VAR VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.
We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (x_{i}, y_{i}). Both Adil and Bera can carry only one bottle at once each.
For both Adil and Bera the process looks as follows: Choose to stop or to continue to collect bottles. If the choice was to continue then choose some bottle and walk towards it. Pick this bottle and walk to the recycling bin. Go to step 1.
Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.
They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.
-----Input-----
First line of the input contains six integers a_{x}, a_{y}, b_{x}, b_{y}, t_{x} and t_{y} (0 ≤ a_{x}, a_{y}, b_{x}, b_{y}, t_{x}, t_{y} ≤ 10^9) — initial positions of Adil, Bera and recycling bin respectively.
The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.
Then follow n lines, each of them contains two integers x_{i} and y_{i} (0 ≤ x_{i}, y_{i} ≤ 10^9) — position of the i-th bottle.
It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.
-----Output-----
Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10^{ - 6}.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if $\frac{|a - b|}{\operatorname{max}(1, b)} \leq 10^{-6}$.
-----Examples-----
Input
3 1 1 2 0 0
3
1 1
2 1
2 3
Output
11.084259940083
Input
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3
Output
33.121375178000
-----Note-----
Consider the first sample.
Adil will use the following path: $(3,1) \rightarrow(2,1) \rightarrow(0,0) \rightarrow(1,1) \rightarrow(0,0)$.
Bera will use the following path: $(1,2) \rightarrow(2,3) \rightarrow(0,0)$.
Adil's path will be $1 + \sqrt{5} + \sqrt{2} + \sqrt{2}$ units long, while Bera's path will be $\sqrt{2} + \sqrt{13}$ units long. | read = lambda: list(map(int, input().split()))
def dis(x1, y1, x2, y2):
return ((x1 - x2) ** 2 + (y1 - y2) ** 2) ** 0.5
ax, ay, bx, by, tx, ty = read()
n = int(input())
a, b = [], []
sum = 0
for i in range(n):
x, y = read()
dist = dis(tx, ty, x, y)
a.append((dis(ax, ay, x, y) - dist, i))
b.append((dis(bx, by, x, y) - dist, i))
sum += dist * 2
a.sort()
b.sort()
if n > 1 and a[0][1] == b[0][1]:
ans = min(a[0][0], b[0][0], a[0][0] + b[1][0], a[1][0] + b[0][0])
else:
ans = min(a[0][0], b[0][0])
if n > 1:
ans = min(a[0][0] + b[0][0], ans)
print(ans + sum) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR LIST LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.
We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (x_{i}, y_{i}). Both Adil and Bera can carry only one bottle at once each.
For both Adil and Bera the process looks as follows: Choose to stop or to continue to collect bottles. If the choice was to continue then choose some bottle and walk towards it. Pick this bottle and walk to the recycling bin. Go to step 1.
Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.
They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.
-----Input-----
First line of the input contains six integers a_{x}, a_{y}, b_{x}, b_{y}, t_{x} and t_{y} (0 ≤ a_{x}, a_{y}, b_{x}, b_{y}, t_{x}, t_{y} ≤ 10^9) — initial positions of Adil, Bera and recycling bin respectively.
The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.
Then follow n lines, each of them contains two integers x_{i} and y_{i} (0 ≤ x_{i}, y_{i} ≤ 10^9) — position of the i-th bottle.
It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.
-----Output-----
Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10^{ - 6}.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if $\frac{|a - b|}{\operatorname{max}(1, b)} \leq 10^{-6}$.
-----Examples-----
Input
3 1 1 2 0 0
3
1 1
2 1
2 3
Output
11.084259940083
Input
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3
Output
33.121375178000
-----Note-----
Consider the first sample.
Adil will use the following path: $(3,1) \rightarrow(2,1) \rightarrow(0,0) \rightarrow(1,1) \rightarrow(0,0)$.
Bera will use the following path: $(1,2) \rightarrow(2,3) \rightarrow(0,0)$.
Adil's path will be $1 + \sqrt{5} + \sqrt{2} + \sqrt{2}$ units long, while Bera's path will be $\sqrt{2} + \sqrt{13}$ units long. | ax, ay, bx, by, tx, ty = list(map(int, input().split()))
n = int(input())
a, b = [], []
res = 0
for i in range(n):
x, y = list(map(int, input().split()))
lt = ((tx - x) * (tx - x) + (ty - y) * (ty - y)) ** 0.5
la = ((ax - x) * (ax - x) + (ay - y) * (ay - y)) ** 0.5
lb = ((bx - x) * (bx - x) + (by - y) * (by - y)) ** 0.5
a += [(la - lt, i)]
b += [(lb - lt, i)]
res += lt
a.sort()
b.sort()
res *= 2
if a[0][1] == b[0][1] and n > 1:
res += min(a[0][0], b[0][0], a[0][0] + b[1][0], a[1][0] + b[0][0])
elif a[0][1] == b[0][1]:
res += min(a[0][0], b[0][0])
else:
res += min(a[0][0], b[0][0], a[0][0] + b[0][0])
print(res) | ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR LIST LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR NUMBER VAR LIST BIN_OP VAR VAR VAR VAR LIST BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
lis = [0, -1, -1]
sol = [0]
r = 0
for i in range(len(s)):
r = (r + int(s[i])) % 3
if lis[r] != -1:
sol.append(max(sol[-1], 1 + sol[lis[r]]))
else:
sol.append(sol[-1])
lis[r] = i + 1
print(sol[-1]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER BIN_OP NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
n = len(s)
ans = 0
i = 0
while i < n:
if int(s[i]) % 3 == 0:
ans += 1
i += 1
continue
if i != n - 1 and int(s[i]) % 3 == 1 and int(s[i + 1]) % 3 == 2:
ans += 1
i += 2
continue
if i != n - 1 and int(s[i]) % 3 == 2 and int(s[i + 1]) % 3 == 1:
ans += 1
i += 2
continue
if i < n - 2 and int(s[i]) % 3 == int(s[i + 1]) % 3 == int(s[i + 2]) % 3 == 1:
ans += 1
i += 3
continue
if i < n - 2 and int(s[i]) % 3 == int(s[i + 1]) % 3 == int(s[i + 2]) % 3 == 2:
ans += 1
i += 3
continue
i += 1
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = list(map(int, list(input())))
dp = [0] * len(s)
t = [-1, -1, -1]
k = 0
m = 0
for i in range(len(s)):
k += s[i]
r = -1
if t[k % 3] != -1:
r = dp[t[k % 3]]
elif k % 3 == 0:
r = 0
dp[i] = max(m, r + 1)
t[k % 3] = i
m = max(m, dp[i])
print(dp[-1]) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | MOD = 1000000007
MOD2 = 998244353
ii = lambda: int(input())
si = lambda: input()
dgl = lambda: list(map(int, input()))
f = lambda: map(int, input().split())
il = lambda: list(map(int, input().split()))
ls = lambda: list(input())
let = "@abcdefghijklmnopqrstuvwxyz"
s = si()
n = len(s)
l = [0] * n
rem = [-1] * 3
rem[0], x = 0, 0
for i in range(n):
x = (x + int(s[i])) % 3
if rem[x] != -1:
l[i] = max(l[i - 1], l[rem[x]] + 1)
else:
l[i] = l[i - 1]
rem[x] = i
print(l[n - 1]) | ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | dp = dict()
dp[0] = 0
dp[1] = -1000000000.0
dp[2] = -1000000000.0
s = input()
n = len(s)
sum = 0
ans = 0
for i in range(n):
v = ord(s[i]) - ord("0")
sum += v
sum %= 3
dp[sum] = max(dp[0], dp[1], dp[2], dp[sum] + 1)
ans = max(ans, dp[sum])
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
n = len(s)
ss = 0
f = [1] * n
rem = []
for i in range(n):
rem.append(int(s[i]) % 3)
ans = rem.count(0)
i = 0
while i < n:
if rem[i] == 1:
if i + 1 < n and rem[i + 1] == 2:
ans += 1
i += 2
elif i + 1 < n and rem[i + 1] == 1 and i + 2 < n and rem[i + 2] == 1:
ans += 1
i += 3
else:
i += 1
elif rem[i] == 2:
if i + 1 < n and rem[i + 1] == 1:
ans += 1
i += 2
elif i + 1 < n and rem[i + 1] == 2 and i + 2 < n and rem[i + 2] == 2:
ans += 1
i += 3
else:
i += 1
else:
i += 1
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input().strip()
i = 0
ans = 0
while i < len(s):
rem_digits = len(s) - i
if rem_digits >= 3:
if int(s[i]) % 3 == 0:
i += 1
elif int(s[i : i + 2]) % 3 == 0:
i += 2
elif int(s[i + 1]) % 3 == 0:
i += 2
else:
i += 3
ans += 1
elif rem_digits == 2:
if int(s[i]) % 3 == 0:
i += 1
ans += 1
elif int(s[i : i + 2]) % 3 == 0:
i += 2
ans += 1
else:
i += 1
else:
if int(s[i]) % 3 == 0:
ans += 1
i += 1
print(ans) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
y = []
for i in s:
y.append(int(i))
s = 0
dp = []
count = 0
for i in y:
if i % 3 != 0:
j = 0
dp.append(i)
while j < len(dp) - 1:
dp[j] = dp[j] + i
if dp[j] % 3 == 0:
dp = []
count += 1
break
j += 1
else:
count += 1
dp = []
print(count) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR IF BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR LIST VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | a = list(map(int, input()))
for i in range(len(a)):
a[i] = a[i] % 3
s = 0
ans = 0
for i in range(len(a)):
s += a[i]
if a[i] == 0:
s = 0
ans += 1
elif s % 3 == 0:
s = 0
ans += 1
elif s != a[i] and abs(a[i] - a[i - 1]) == 1:
s = 0
ans += 1
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
n = len(s)
c = [0] * n
r = [0] * n
if s[0] in ["3", "6", "9"]:
c[0] = 1
else:
c[0] = 0
r[0] = int(s[0]) % 3
j = 1
while j < n:
if int(s[j]) % 3 == 0:
c[j] = c[j - 1] + 1
elif r[j - 1] == 4:
r[j] = 0
c[j] = c[j - 1] + 1
elif int(s[j]) % 3 == r[j - 1]:
r[j] = 4
c[j] = c[j - 1]
elif int(s[j]) % 3 != r[j - 1] and r[j - 1] != 0:
r[j] = 0
c[j] = c[j - 1] + 1
else:
r[j] = int(s[j]) % 3
c[j] = c[j - 1]
j = j + 1
print(c[-1]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR IF VAR NUMBER LIST STRING STRING STRING ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | c = 0
f = [0, -1, -1]
z = [0]
for e, i in enumerate(input()):
c = (c + int(i)) % 3
z.append(z[-1])
if f[c] != -1:
z[-1] = max(z[-1], z[f[c]] + 1)
f[c] = e + 1
print(z[-1]) | ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
count, ans, tmp = 0, 0, 0
for i in s:
if int(i) % 3 == 0 or (tmp + int(i)) % 3 == 0 or count == 2:
tmp = 0
ans += 1
count = 0
else:
tmp += int(i)
count += 1
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | l = [*map(lambda x: int(x) % 3, input())]
dp = [0, -float("inf"), -float("inf")]
for d in l:
new_dp = [-float("inf")] * 3
new_dp[d] = max(0, max(dp) + (d == 0))
new_dp[0] = max(new_dp[0], dp[-d % 3] + 1)
for r in (1, 2):
new_dp[r] = max(new_dp[r], dp[(r - d) % 3])
dp = new_dp
print(max(dp)) | ASSIGN VAR LIST FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR ASSIGN VAR LIST NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR STRING FOR VAR VAR ASSIGN VAR BIN_OP LIST FUNC_CALL VAR STRING NUMBER ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER FOR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
k = 0
ans = 0
vis = {0}
for i, c in enumerate(s):
k += int(c)
k %= 3
if k in vis:
ans += 1
k = 0
vis = {0}
else:
vis.add(k)
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
n = 0
ans = 0
for k in s:
t = int(k)
n += t
if n % 3 == 0 or t % 3 == 0 or n % 100 % 3 == 0:
ans += 1
n = 0
else:
n = n * 10
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = str(input())
prev_sum = 0
ans = 0
f_el = 0
need = True
for el in s:
el = int(el)
if el % 3 == 0:
ans += 1
prev_sum = 0
need = True
continue
if need:
f_el = el
need = False
prev_sum = el
continue
prev_sum += el
if prev_sum % 3 == 0 or (prev_sum - f_el) % 3 == 0:
ans += 1
prev_sum = 0
need = True
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input().strip()
have_one = 0
have_two = 0
res = 0
for c in s:
x = int(c)
if x % 3 == 1:
have_one += 1
elif x % 3 == 2:
have_two += 1
else:
res += 1
have_one = 0
have_two = 0
if have_one == 3 or have_two == 3 or have_one > 0 and have_two > 0:
res += 1
have_one = 0
have_two = 0
print(res) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
n = len(s)
r = [0] * n
for i in range(n):
r[i] = int(s[i]) % 3
i = 0
c = 0
while i < n:
if i + 2 < n and (
r[i] == r[i + 1] == r[i + 2] == 1 or r[i] == r[i + 1] == r[i + 2] == 2
):
c += 1
i += 3
elif i + 1 < n and (r[i] == 2 and r[i + 1] == 1 or r[i] == 1 and r[i + 1] == 2):
c += 1
i += 2
elif r[i] == 0:
c += 1
i += 1
else:
i += 1
print(c) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | import sys
sys.setrecursionlimit(10000)
t = 1
for _ in range(t):
s = input()
no = ""
count = 0
for i in s:
if int(i) % 3 == 0:
count += 1
no = ""
else:
no += i
if int(no) % 3 == 0:
count += 1
no = ""
elif int(no[-2:]) % 3 == 0:
count += 1
no = ""
print(count) | IMPORT EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR STRING VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR STRING IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
ans = 0
r = 0
r1 = 0
r2 = 0
for i in s:
n = ord(i) - 48
if n % 3 == 0:
ans += 1
r = r1 = r2 = 0
else:
r = r + n % 3
if r % 3 == 0:
ans += 1
r = r1 = r2 = 0
elif r % 3 == 1:
if r1 == 1:
ans += 1
r = r1 = r2 = 0
else:
r1 = 1
elif r2 == 1:
ans += 1
r = r1 = r2 = 0
else:
r2 = 1
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
count = 0
i = 0
while i < len(s):
if int(s[i]) % 3 == 0:
count += 1
i += 1
elif i < len(s) - 1 and (int(s[i : i + 2]) % 3 == 0 or int(s[i + 1]) % 3 == 0):
count += 1
i += 2
elif i < len(s) - 2 and (
int(s[i + 1 : i + 3]) % 3 == 0 or int(s[i : i + 3]) % 3 == 0 or s[i + 2] == "0"
):
count += 1
i += 3
else:
i += 1
print(count) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | n = input()
ans = 0
r, c = 0, 0
for i in n:
r += int(i)
c += 1
if int(i) % 3 == 0 or r % 3 == 0 or c == 3:
ans += 1
r, c = 0, 0
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | a = input()
su = 0
ans = 0
b = ""
for i in range(len(a)):
if int(a[i]) % 3 == 0:
ans += 1
b += "9"
else:
b += str(int(a[i]) % 3)
for i in range(len(b) - 2):
if int(b[i]) + int(b[i + 1]) == 3:
ans += 1
b = b[:i] + "99" + b[i + 2 :]
if (
int(b[i]) + int(b[i + 1]) + int(b[i + 2]) == 3
or int(b[i]) + int(b[i + 1]) + int(b[i + 2]) == 6
):
ans += 1
b = b[:i] + "999" + b[i + 3 :]
if len(a) >= 2 and int(b[-1]) + int(b[-2]) == 3:
ans += 1
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR STRING VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR STRING VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR STRING VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = str(input())
c = 0
cont = 0
su = 0
for i in range(len(s)):
k = int(s[i]) % 3
if k == 0:
c += 1
su = 0
cont = 0
else:
if k == 1:
su += 1
else:
su += 2
cont += 1
if su % 3 == 0 or cont == 3:
c += 1
cont = 0
su = 0
print(c) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | arr = input()
count = 0
idex = -1
for i in range(len(arr)):
for j in range(i, idex, -1):
next = int(arr[j : i + 1])
if next % 3 == 0:
idex = i
count += 1
break
print(count) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s, k, i = "".join([str(int(i) % 3) for i in input()]) + " ", 0, 0
while i < len(s):
if s[i] == "0":
k += 1
i += 1
elif s[i : i + 2] in "12.21":
k += 1
i += 2
elif s[i : i + 3] in "111.222":
k += 1
i += 3
else:
i += 1
print(k) | ASSIGN VAR VAR VAR BIN_OP FUNC_CALL STRING FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR STRING NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
ans = 0
temp = ""
for i in s:
if int(i) == 0:
ans += 1
temp = ""
else:
temp += i
k = int(temp)
if k % 3 == 0:
ans += 1
temp = ""
else:
su = 0
for j in temp:
su += int(j)
for j in temp:
su -= int(j)
if su % 3 == 0 and su != 0:
ans += 1
temp = ""
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR STRING VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR FOR VAR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = " " + input()
dp = [0] * len(s)
for i in range(1, len(s)):
sum = 0
for j in range(i, 0, -1):
sum += int(s[j])
if sum % 3 == 0:
dp[i] = max(dp[j - 1] + 1, 1)
break
dp[i] = max(dp[i], dp[i - 1])
print(dp[-1]) | ASSIGN VAR BIN_OP STRING FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | def solve(digits):
solutions = {}
for index, digit in enumerate(digits):
if digit % 3 == 0:
solutions[index] = 1 + solutions.get(index - 1, 0)
else:
cand1 = solutions.get(index - 1, 0)
cand2 = 0
for start_index in range(index - 1, -1, -1):
if sum(digits[start_index : index + 1]) % 3 == 0:
cand2 = solutions.get(start_index - 1, 0) + 1
break
solutions[index] = max(cand1, cand2)
return solutions[len(digits) - 1]
number = int(input())
digits = [int(x) for x in str(number)]
print(solve(digits)) | FUNC_DEF ASSIGN VAR DICT FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s, r, fin = input(), 0, [0, -1, -1]
n = len(s)
z = [(0) for _ in range(n + 1)]
for i, c in enumerate(s):
r = (r + int(c)) % 3
z[i + 1] = z[i]
if fin[r] != -1:
z[i + 1] = max(z[i + 1], z[fin[r]] + 1)
fin[r] = i + 1
print(z[n]) | ASSIGN VAR VAR VAR FUNC_CALL VAR NUMBER LIST NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = list(map(lambda o: int(o), list(input())))
s.reverse()
sm = c = cc = 0
for i in s:
cc += 1
sm += i
if not i % 3 or not sm % 3 or cc == 3:
c += 1
sm = 0
cc = 0
print(c) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER FOR VAR VAR VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
z = 0
t = ""
for i in range(len(s)):
if s[i] == "0" or int(s[i]) % 3 == 0:
z += 1
t += "0"
else:
t += s[i]
a = t.split("0")
num = 0
for i in range(len(a)):
mod = -1
q = 0
for j in range(len(a[i])):
k = int(a[i][j])
if mod == -1:
mod = k % 3
q += 1
else:
mod += k % 3
q += 1
if mod % 3 == 0:
num += 1
mod = -1
q = 0
elif q == 3:
num += 1
mod = -1
q = 0
print(z + num) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR STRING VAR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
l = [0] * 200005
s = "#" + s
for i in range(1, len(s)):
l[i] = l[i - 1]
if i == 1:
if int(s[i]) % 3 == 0:
l[i] = 1
elif i == 2:
if int(s[i]) % 3 == 0:
l[i] = max(l[i], l[i - 1] + 1)
if int(s[i - 1 : i + 1]) % 3 == 0:
l[i] = max(l[i], l[i - 2] + 1)
else:
if int(s[i]) % 3 == 0:
l[i] = max(l[i], l[i - 1] + 1)
if int(s[i - 1 : i + 1]) % 3 == 0:
l[i] = max(l[i], l[i - 2] + 1)
if int(s[i - 2 : i + 1]) % 3 == 0:
l[i] = max(l[i], l[i - 3] + 1)
print(l[len(s) - 1]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP STRING VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER IF VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | c = 0
r = 0
b = 0
for i in input():
n = int(i) % 3
if n == 0 or (n + r) % 3 == 0 or b == 2:
c += 1
r = 0
b = 0
else:
r += n
b += 1
print(c) | ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
i = 0
res = 0
while i < len(s):
while i < len(s) and s[i] == "0":
i += 1
res += 1
contains = [0, 0, 0]
j = i
for j in range(i, len(s)):
val = int(s[j]) % 3
contains[val] += 1
if (
contains[0]
or contains[1]
and contains[2]
or contains[1] == 3
or contains[2] == 3
):
res += 1
break
i = j + 1
print(res) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | def solve(digits):
solutions = {}
fin = {(0): 0}
digit_sum = 0
for index, digit in enumerate(digits):
digit_sum += digit
solutions[index] = solutions.get(index - 1, 0)
if digit_sum % 3 in fin:
solutions[index] = max(
solutions[index], solutions.get(fin[digit_sum % 3], 0) + 1
)
fin[digit_sum % 3] = index
return solutions[len(digits) - 1]
number = int(input())
digits = [int(x) for x in str(number)]
print(solve(digits)) | FUNC_DEF ASSIGN VAR DICT ASSIGN VAR DICT NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR RETURN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | import sys
def main():
s = sys.stdin.readline().strip()
N = len(s)
dp = [0] * N
dp[0] = 1 if int(s[0]) % 3 == 0 else 0
if len(s) == 1:
print(dp[0])
return
dp[1] = dp[0]
if int(s[1]) % 3 == 0:
dp[1] = dp[0] + 1
if (int(s[1]) + int(s[0])) % 3 == 0:
dp[1] = max(dp[1], 1)
if len(s) == 2:
print(dp[1])
return
dp[2] = dp[1]
if int(s[2]) % 3 == 0:
dp[2] = 1 + dp[1]
if (int(s[2]) + int(s[1])) % 3 == 0:
dp[2] = max(dp[2], dp[0] + 1)
if (int(s[2]) + int(s[1]) + int(s[0])) % 3 == 0:
dp[2] = max(dp[2], 1)
for i in range(3, N):
ith = int(s[i])
prev = int(s[i - 1])
prever = int(s[i - 2])
dp[i] = dp[i - 1]
if ith % 3 == 0:
dp[i] = dp[i - 1] + 1
if (prev + ith) % 3 == 0:
dp[i] = max(dp[i], dp[i - 2] + 1)
if (prever + prev + ith) % 3 == 0:
dp[i] = max(dp[i], dp[i - 3] + 1)
print(dp[N - 1])
main() | IMPORT FUNC_DEF ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER RETURN ASSIGN VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER BIN_OP VAR NUMBER NUMBER IF BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER RETURN ASSIGN VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER BIN_OP NUMBER VAR NUMBER IF BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER IF BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
l = []
c = 0
for i in range(len(s)):
if int(s[i]) % 3 == 0:
c += 1
l = []
else:
l.append(int(s[i]))
for j in range(len(l) - 1):
l[j] += int(s[i])
if l[j] % 3 == 0:
c += 1
l = []
break
print(c) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | a = input()
l = len(a)
count = 0
num = ""
for i in range(l):
b = int(a[i])
if b == 0:
count += 1
num = ""
elif b % 3 == 0:
count += 1
num = ""
else:
num += str(b)
if int(num) % 100 % 3 == 0 or int(num) % 3 == 0:
count += 1
num = ""
print(count) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR STRING IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR STRING VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
n = len(s)
ans = 0
c = 0
l = []
for i in range(n):
a = int(s[i]) % 3
if a == 0:
ans += 1
c = 0
l = []
elif c == 0:
l.append(int(s[i]))
c += 1
elif c == 1:
if (a + l[0]) % 3 == 0:
ans += 1
c = 0
l = []
else:
c += 1
else:
ans += 1
c = 0
l = []
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | a = str(input())
a = "x" + a
n = len(a)
f = [(0) for i in range(0, 10**6)]
k = 0
for i in range(1, n):
if int(a[i]) % 3 == 0:
f[i] = f[i - 1] + 1
k = i
else:
for j in range(i - 1, k, -1):
if int(a[j : i + 1]) % 3 == 0:
f[i] = f[k] + 1
k = i
else:
f[i] = f[k]
maxf = 0
for i in range(0, n):
maxf = max(maxf, f[i])
print(maxf) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP STRING VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
f = [0] * len(s)
cnt = 0
for i in range(len(s)):
a = int(s[i])
if a % 3 == 0:
cnt += 1
f[i] = 1
for i in range(len(s) - 1):
if f[i] == 0 and f[i + 1] == 0:
a = int(s[i]) * 10 + int(s[i + 1])
if a % 3 == 0:
cnt += 1
f[i] = f[i + 1] = 1
elif i < len(s) - 2 and f[i] == 0 and f[i + 1] == 0 and f[i + 2] == 0:
a = int(s[i]) * 100 + int(s[i + 1]) * 10 + int(s[i + 2])
if a % 3 == 0:
cnt += 1
f[i] = f[i + 1] = f[i + 2] = 1
for i in range(len(s) - 2):
if f[i] == 0 and f[i + 1] == 0 and f[i + 2] == 0:
a = int(s[i]) * 100 + int(s[i + 1]) * 10 + int(s[i + 2])
if a % 3 == 0:
cnt += 1
f[i] = f[i + 1] = f[i + 2] = 1
for i in range(len(s) - 3):
if f[i] == 0 and f[i + 1] == 0 and f[i + 2] == 0 and f[i + 3] == 0:
a = int(s[i]) * 1000 + int(s[i + 1]) * 100 + int(s[i + 2]) * 10 + int(s[i + 3])
if a % 3 == 0:
cnt += 1
print(cnt) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
ost = []
for i in map(int, s):
ost += [i % 3]
now = 0
last = -1
answer = 0
for i in ost:
if i == 0:
now = []
answer += 1
elif now == 2:
now = 0
answer += 1
elif now == 1:
if last != i:
answer += 1
now = 0
else:
now = 2
else:
now = 1
last = i
print(answer) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR VAR VAR LIST BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR LIST VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
s = "#" + s
res = 0
n = len(s)
dp = [0] * n
l = 0
for i in range(1, n):
j = int(s[i : i + 1])
if j == 0 or j % 3 == 0:
dp[i] = dp[i - 1] + 1
l = i
continue
dp[i] = dp[i - 1]
t = j
for k in range(i - 1, l, -1):
t += int(s[k : k + 1])
if t % 3 == 0:
dp[i] += 1
l = i
print(dp[n - 1]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP STRING VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | a = [int(i) for i in list(input())]
current = 0
l = len(a)
for i in range(1, l):
a[i] += a[i - 1]
a = [0] + a
ans = 0
for i in range(l + 1):
for j in range(current, i):
if (a[i] - a[j]) % 3 == 0:
ans += 1
current = i
break
print(ans) | ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | n = input()
buf = 0
c = 0
m = 0
for i in range(len(n)):
buf += int(n[i])
m += 1
if int(n[i]) % 3 == 0 or buf % 3 == 0 or m >= 3:
buf = 0
c += 1
m = 0
print(c) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | lst = map(int, input())
res, prev, length = 0, [], 0
for i, x in enumerate(lst):
if x % 3 == 0:
res += 1
prev, length = [], 0
else:
k, item = 0, x % 3
for j in range(length - 1, -1, -1):
item = (item + prev[j]) % 3
if item == 0:
res, k, prev, length = res + 1, 1, [], 0
break
if k == 0:
prev.append(x % 3)
length += 1
print(res) | ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER LIST NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR LIST NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER LIST NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | def main():
a = [(int(i) % 3) for i in input()]
d = []
k = 0
ans = 0
summ = 0
for i in a:
if i == 0:
ans += 1
k = 0
summ = 0
d = []
else:
d += [i]
k += 1
summ += i
if k != 1:
if summ % 3 == 0:
ans += 1
summ = 0
k = 0
d = []
elif d[0] == summ % 3:
ans += 1
summ = 0
k = 0
d = []
print(ans)
main() | FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST VAR LIST VAR VAR NUMBER VAR VAR IF VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | u = "0" + input()
n = len(u) - 1
d = [0] * (n + 1)
def div3(k):
if int(k) % 3 == 0:
return 1
return 0
if div3(u[1]):
d[1] = 1
else:
d[1] = 0
for i in range(2, n + 1):
a1 = d[i - 1] + div3(u[i])
a2 = d[i - 2] + div3(u[i] + u[i - 1])
if i > 2:
a3 = d[i - 3] + div3(u[i] + u[i - 1] + u[i - 2])
else:
a3 = 0
d[i] = max(a1, a2, a3)
print(d[-1]) | ASSIGN VAR BIN_OP STRING FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN NUMBER RETURN NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | n = list(input())
cntr = 0
marks = []
for i in range(len(n)):
marks.append(0)
for i in range(len(n)):
if int(n[i]) % 3 == 0:
cntr += 1
marks[i] = 1
elif marks[i] == 0:
marks[i] = 1
tmpcnt = int(n[i])
mrk = -1
for j in range(2):
if i + j + 1 < len(n):
if int(n[i + j + 1]) % 3 != 0:
if (tmpcnt + int(n[i + j + 1])) % 3 == 0:
cntr += 1
mrk = i + j + 1
break
else:
tmpcnt += int(n[i + j + 1])
else:
break
if mrk != -1:
for k in range(i + 1, mrk + 1):
marks[k] = 1
print(cntr) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER NUMBER IF BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = str(input())
n = len(s)
l = [-1] * 3
a = []
pre = [0]
dp = [int(0)] * n
for i in range(len(s)):
a.append(int(s[i]))
pre.append((pre[i] + int(s[i])) % 3)
dp[n - 1] = a[n - 1] % 3 == 0
l[pre[n - 1]] = n - 1
l[pre[n]] = n
for i in range(n - 2, -1, -1):
dp[i] = dp[i + 1] + int(a[i] % 3 == 0)
if l[pre[i]] != -1:
idx = l[pre[i]]
if idx < n:
dp[i] = max(dp[i], dp[idx] + 1)
else:
dp[i] = max(dp[i], 1)
l[pre[i]] = i
print(int(dp[0])) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR LIST ASSIGN VAR LIST NUMBER ASSIGN VAR BIN_OP LIST FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | a = input()
s = 0
d = [0, 0, 0]
for i in range(len(a) - 1, -1, -1):
z = int(a[i]) % 3
if z == 0:
d = [0, 0, 0]
s += 1
continue
d[z] += 1
if d[2] == 3:
s += 1
d = [0, 0, 0]
elif d[1] >= 1 and d[2] >= 1:
s += 1
d = [0, 0, 0]
elif d[1] == 3:
s += 1
d = [0, 0, 0]
print(s) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER VAR NUMBER VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER IF VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
n = len(s)
d = [0] * (n + 1)
f = [-1] * 3
r = 0
for i in range(1, n + 1):
c = int(s[i - 1])
if c % 3 == 0:
d[i] = d[i - 1] + 1
else:
r = (r + c) % 3
t = f[r % 3]
if t > -1 or r == 0:
d[i] = max(d[t] + 1, d[i - 1])
else:
d[i] = d[i - 1]
f[r % 3] = i
print(d[n]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
cursum = 0
curstr = ""
count = 0
for i in s:
cursum += int(i)
curstr += i
if (
cursum % 3 == 0
or i in ["0", "3", "6", "9"]
or len(curstr) > 1
and int(curstr[1:]) % 3 == 0
):
count += 1
cursum = 0
curstr = ""
print(count) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR LIST STRING STRING STRING STRING FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = input()
sumarr = [int(s[0])]
for i in range(1, len(s)):
sumarr.append(sumarr[i - 1] + int(s[i]))
ansarr = [int(sumarr[0] % 3 == 0) for _ in range(len(s))]
for i in range(1, len(s)):
ansarr[i] = ansarr[i - 1]
j = 1
while i >= j and (sumarr[i] - sumarr[i - j]) % 3:
j += 1
if j <= i:
ansarr[i] = max(ansarr[i], ansarr[i - j] + 1)
elif sumarr[i] % 3 == 0:
ansarr[i] = max(ansarr[i], 1)
print(ansarr[len(s) - 1]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR VAR NUMBER IF BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER |
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
-----Input-----
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
-----Output-----
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
-----Examples-----
Input
3121
Output
2
Input
6
Output
1
Input
1000000000000000000000000000000000
Output
33
Input
201920181
Output
4
-----Note-----
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | s = [int(i) for i in input()]
ans = 0
last = 0
def check(x):
Sum = 0
while x != 0:
Sum += x % 10
x //= 10
if Sum % 3 == 0:
return True
return False
for i in range(len(s)):
if s[i] % 3 == 0:
ans += 1
last = 0
else:
last = last * 10 + s[i]
if check(last):
ans += 1
last = 0
print(ans) | ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
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