description
stringlengths 171
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Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
x = input()
l = n // 2
r = (n + 1) // 2
fn = lambda i: int(x[:i]) + int(x[i:])
while x[l] == "0":
l -= 1
while x[r] == "0":
r += 1
if r == n:
break
ans = float("inf")
if l != 0:
ans = min(ans, fn(l))
if r != n:
ans = min(ans, fn(r))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR WHILE VAR VAR STRING VAR NUMBER WHILE VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR STRING IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def closest_nonzero(s, h, l):
index1 = 0
index = 0
for i in range(h, l):
if n[i] != "0":
index = i
break
for i in range(h, -1, -1):
if n[i] != "0":
index1 = i
break
return index, index1
l = int(input())
n = input()
half = l // 2
if l % 2 == 0:
even = True
else:
even = False
if even:
index, index1 = closest_nonzero(n, half, l)
if index == 0 and index1 == 0:
ans = min(int(n[index:]), int(n[index1:]))
elif index == 0:
ans = min(int(n[index:]), int(n[:index1]) + int(n[index1:]))
elif index1 == 0:
ans = min(int(n[:index]) + int(n[index:]), int(n[index1:]))
else:
ans = min(int(n[:index]) + int(n[index:]), int(n[:index1]) + int(n[index1:]))
print(ans)
else:
index, index1 = closest_nonzero(n, half, l)
if index == 0 and index1 == 0:
ans = min(int(n[index:]), int(n[index1:]))
elif index == 0:
ans = min(int(n[index:]), int(n[:index1]) + int(n[index1:]))
elif index1 == 0:
ans = min(int(n[:index]) + int(n[index:]), int(n[index1:]))
else:
ans = min(int(n[:index]) + int(n[index:]), int(n[:index1]) + int(n[index1:]))
index, index1 = closest_nonzero(n, half + 1, l)
if index == 0 and index1 == 0:
ans1 = min(int(n[index:]), int(n[index1:]))
elif index == 0:
ans1 = min(int(n[index:]), int(n[:index1]) + int(n[index1:]))
elif index1 == 0:
ans1 = min(int(n[:index]) + int(n[index:]), int(n[index1:]))
else:
ans1 = min(int(n[:index]) + int(n[index:]), int(n[:index1]) + int(n[index1:]))
print(min(ans, ans1))
|
FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR VAR RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
import sys
a = sys.stdin.read().split()[1].strip()
ans = int(a)
n = len(a)
k = n // 2
m = 0
i = k
while m < 3 and i < n:
if a[i] != "0":
m += 1
s = int(a[:i]) + int(a[i:])
if s < ans:
ans = s
i += 1
m = 0
i = k
while m < 3 and i > 0:
if a[i] != "0":
m += 1
s = int(a[:i]) + int(a[i:])
if s < ans:
ans = s
i -= 1
print(ans)
|
IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
b = input()
aaa = n // 2
if n & 1 == 0:
for i in range(0, aaa):
num1 = aaa + i
num = n - num1
if b[num] == "0" and b[num1] == "0":
continue
elif b[num] == "0":
ans1 = int(b[:num1]) + int(b[num1:n])
print(ans1)
break
elif b[num1] == "0":
ans2 = int(b[:num]) + int(b[num:n])
print(ans2)
break
else:
ans1 = int(b[:num]) + int(b[num:n])
ans2 = int(b[:num1]) + int(b[num1:n])
if ans1 > ans2:
print(ans2)
else:
print(ans1)
break
else:
for i in range(0, aaa + 1):
num = aaa - i + 1
num1 = aaa + i
if b[num] == "0" and b[num1] == "0":
continue
elif b[num] == "0":
ans1 = int(b[:num1]) + int(b[num1:n])
print(ans1)
break
elif b[num1] == "0":
ans2 = int(b[:num]) + int(b[num:n])
print(ans2)
break
else:
ans1 = int(b[:num]) + int(b[num:n])
ans2 = int(b[:num1]) + int(b[num1:n])
if ans1 > ans2:
print(ans2)
else:
print(ans1)
break
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR STRING VAR VAR STRING IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR STRING VAR VAR STRING IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = str(input())
ans = int(n)
mid = int(l / 2 + 1)
while mid < int(l):
if n[mid] == "0":
mid += 1
continue
l1 = n[0:mid]
l2 = n[mid:]
ans1 = int(l1) + int(l2)
ans = min(ans, ans1)
mid += 1
break
mid = int(l / 2)
while mid > 0:
if n[mid] == "0":
mid -= 1
continue
l1 = n[0:mid]
l2 = n[mid:]
ans1 = int(l1) + int(l2)
ans = min(ans, ans1)
mid -= 1
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def summ(i):
global a
b = a[:i]
c = a[i:]
if len(b) < len(c):
b = [0] * (len(c) - len(b)) + b[:]
if len(c) < len(b):
c = [0] * (len(b) - len(c)) + c[:]
ans = ["0"] * len(b)
car = 0
for i in range(len(b) - 1, -1, -1):
ans[i] = str((int(b[i]) + int(c[i]) + car) % 10)
car = (int(b[i]) + int(c[i]) + car) // 10
if car != 0:
ans = [car] + ans[:]
return "".join(ans)
def comp(a, b):
if len(a) < len(b):
return a
elif len(b) < len(a):
return b
else:
i = 0
while i < len(a) and a[i] == b[i]:
i += 1
if i == len(a):
return a
elif int(a[i]) < int(b[i]):
return a
else:
return b
def minn(i, j):
return comp(summ(i), summ(j))
n = int(input())
a = list(input())
if a.count("0") == n - 1:
print("".join(a))
elif n % 2 == 0:
t = n // 2
if a[t] != "0":
print(summ(n // 2))
else:
c = 0
j = t
while j < n and a[j] == "0":
j += 1
i = t
while i > 0 and a[i] == "0":
i -= 1
t1 = n - i
t2 = j
if t1 > t2:
print(summ(j))
elif t1 < t2:
print(summ(i))
else:
print(minn(j, i))
else:
t = n // 2
if a[t] != "0":
print(minn(n // 2, n // 2 + 1))
else:
c = 0
j = t
while j < n and a[j] == "0":
j += 1
i = t
while i > 0 and a[i] == "0":
i -= 1
t1 = n - i
t2 = j
if t1 > t2:
print(summ(j))
elif t1 < t2:
print(summ(i))
else:
print(minn(j, i))
|
FUNC_DEF ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST STRING FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP LIST VAR VAR RETURN FUNC_CALL STRING VAR FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR RETURN VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR RETURN VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF FUNC_CALL VAR STRING BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = str(input())
a1 = 0
a2 = 0
for i in range(l // 2 + 1, l):
n1 = n[i:]
n2 = n[:i]
if n1[0] == "0":
continue
else:
a1 = int(n1) + int(n2)
break
for i in range(l // 2, 0, -1):
n1 = n[i:]
n2 = n[:i]
if n1[0] == "0":
continue
else:
a2 = int(n1) + int(n2)
break
if a1 == 0:
print(a2)
elif a2 == 0:
print(a1)
else:
print(min(a1, a2))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
a = input()
i_l = n // 2
i_r = n // 2 + 1
while i_r < n and a[i_r] == "0":
i_r += 1
min1 = 2**64 - 1
if i_r != n:
min1 = int(a[0:i_r]) + int(a[i_r:])
while i_l >= 0 and a[i_l] == "0":
i_l -= 1
if i_l > 0:
min1 = min(min1, int(a[0:i_l]) + int(a[i_l:]))
print(min1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
L = int(input())
N = input()
mn = int(N)
m = (L - 1) // 2
for i in range(m, -1, -1):
if N[i + 1] == "0":
continue
mn = min(mn, int(N[0 : i + 1]) + int(N[i + 1 :]))
break
m = L // 2
for i in range(m, L):
if N[i] == "0":
continue
mn = min(mn, int(N[0:i]) + int(N[i:]))
break
print(mn)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = "0" + input()
n += 1
ind = n // 2
ans = []
i = c = 0
while ind - i > 0 and ind + i < n:
if s[ind + i] != "0":
ans.append(int(s[: ind + i]) + int(s[ind + i :]))
c += 1
if s[ind - i] != "0":
ans.append(int(s[: ind - i]) + int(s[ind - i :]))
c += 1
i += 1
if c > 4:
break
if n == 3 and s[2] != "0":
ans.append(int(s[1]) + int(s[2]))
print(min(ans))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP STRING FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR VAR NUMBER WHILE BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR IF VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
al_l = []
al_r = []
for i in range(1, n):
if int(s[i]) > 0 and i <= n // 2:
al_l.append(i)
if int(s[i]) > 0 and i > n // 2:
al_r.append(i)
if len(al_r) == 3:
break
if len(al_l) == 3:
del al_l[0]
ans = int(s)
for x in al_l:
ans = min(ans, int(s[0:x]) + int(s[x:]))
for x in al_r:
ans = min(ans, int(s[0:x]) + int(s[x:]))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR IF FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = list(input())
ans = int("".join(n))
if l == 1:
print(ans)
exit(0)
f = 0
for i in range(l // 2, 0, -1):
if n[i] != "0":
ans = min(ans, int("".join(n[:i])) + int("".join(n[i:])))
f = 1
if n[l - i] != "0":
ans = min(ans, int("".join(n[: l - i])) + int("".join(n[l - i :])))
f = 1
if f == 1:
print(ans)
exit(0)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL STRING VAR VAR FUNC_CALL VAR FUNC_CALL STRING VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL STRING VAR BIN_OP VAR VAR FUNC_CALL VAR FUNC_CALL STRING VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
num = str(input())
n = int(num)
x = len(num)
mns = n * 100
h = (x - 1) // 2
def test(i):
s = int(num[:i]) + int(num[i:])
if s < mns:
return s
return mns
ct = 0
for i in range(h, -1, -1):
if num[i] != "0" and i > 0:
mns = test(i)
ct += 1
if ct >= 3:
break
ct = 0
for i in range(h + 1, x):
if num[i] != "0":
mns = test(i)
ct += 1
if ct >= 3:
break
print(mns)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR RETURN VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = str(input())
mid = int(l / 2)
m1 = int(n)
m2 = int(n)
sp = mid
for i in range(sp, 0, -1):
if n[i] != "0":
temp = int(n[:i]) + int(n[i:])
if m1 > temp:
m1 = temp
else:
break
else:
pass
sp = mid + 1
for i in range(sp, len(n)):
if n[i] != "0":
temp = int(n[:i]) + int(n[i:])
if m2 > temp:
m2 = temp
else:
break
else:
pass
print(min(m1, m2))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
mid = len(s) // 2
l = mid
r = mid + 1
while l >= 0 and s[l] == "0":
l -= 1
while r < len(s) and s[r] == "0":
r += 1
if r == len(s):
s1 = int(s[0:l])
s2 = int(s[l:])
print(s1 + s2)
elif l == 0:
s1 = int(s[0:r])
s2 = int(s[r:])
print(s1 + s2)
else:
s1 = int(s[0:l])
s2 = int(s[l:])
s3 = int(s[0:r])
s4 = int(s[r:])
print(min(s1 + s2, s3 + s4))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
l = n // 2
ans = int(s)
if s[l] != "0":
ans = int("0" + s[:l]) + int("0" + s[l:])
for i in range(1, n):
if l + i < n and s[l + i] != "0":
ans = min(ans, int("0" + s[: l + i]) + int("0" + s[l + i :]))
if l - i >= 0 and s[l - i] != "0":
ans = min(ans, int("0" + s[: l - i]) + int("0" + s[l - i :]))
if l + i >= n or l - 1 <= 0 or s[l + i] != "0" or s[l - i] != "0":
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP STRING VAR VAR FUNC_CALL VAR BIN_OP STRING VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP STRING VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP STRING VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP STRING VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP STRING VAR BIN_OP VAR VAR IF BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
k = 1
i = l // 2
j = l // 2 + 1
while i > 0 and n[i] == "0":
i -= 1
while j < l and n[j] == "0":
j += 1
ans = -1
if i >= 1:
b = int(n[:i]) + int(n[i:])
if ans == -1 or ans > b:
ans = b
if j < l:
b = int(n[:j]) + int(n[j:])
if ans == -1 or ans > b:
ans = b
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
val = input()
mid = n // 2
while mid < n and val[mid] == "0":
mid += 1
anmid = n // 2
while anmid > -1 and val[anmid] == "0":
anmid -= 1
ans = 10**n
if anmid != 0:
ans = int(val[:anmid]) + int(val[anmid:])
if mid != n:
ans = min(ans, int(val[:mid]) + int(val[mid:]))
mid = n // 2 + 1
anmid = mid
while mid < n and val[mid] == "0":
mid += 1
while anmid > -1 and val[anmid] == "0":
anmid -= 1
if anmid != 0:
ans = min(ans, int(val[:anmid]) + int(val[anmid:]))
if mid != n:
ans = min(ans, int(val[:mid]) + int(val[mid:]))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR VAR STRING VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
T = int(input())
s = input()
mid = len(s) // 2
MIN = -1
if len(s) < 3:
print(int(s[0]) + int(s[1]))
return
for i in range(mid, -1, -1):
if s[i + 1] != "0":
MIN = int(s[: i + 1]) + int(s[i + 1 :])
break
for i in range(mid, len(s)):
if s[i] != "0":
tmp = int(s[:i]) + int(s[i:])
if MIN > tmp:
MIN = tmp
break
if MIN == -1:
MIN = int(s[: len(s) - 1]) + int(s[len(s) - 1 :])
print(MIN)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER RETURN FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
v = []
for i in range(1, len(s)):
if s[i] != "0":
v.append(i)
ans = 1e1000
ch = int(s)
for i in range(0, len(v)):
if v[i] > l // 2:
ans = min(ans, int(ch // int(pow(10, l - v[i]))) + ch % int(pow(10, l - v[i])))
if i > 0:
ans = min(
ans,
int(ch // int(pow(10, l - v[i - 1]))) + ch % int(pow(10, l - v[i - 1])),
)
break
ans = min(
ans,
int(ch // int(pow(10, l - v[len(v) - 1]))) + ch % int(pow(10, l - v[len(v) - 1])),
)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
res = None
noimp = 0
for it in range(l - 1):
i = 0
if it % 2 == 0:
i = l // 2 - (it + 1) // 2
else:
i = l // 2 + (it + 1) // 2
if s[i] == "0":
continue
cur = int(s[0:i]) + int(s[i:])
if res is None:
res = cur
elif cur < res:
res = cur
else:
noimp += 1
if noimp >= 4:
break
print(res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NONE ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR NONE ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
a = input()
ans = None
i = n // 2
for q in range(n // 2):
if a[: i + q][0] != "0" and a[i + q :][0] != "0":
if ans is None:
ans = int(a[: i + q]) + int(a[i + q :])
else:
ans = min(ans, int(a[: i + q]) + int(a[i + q :]))
if a[: i - q][0] != "0" and a[i - q :][0] != "0":
if ans is None:
ans = int(a[: i - q]) + int(a[i - q :])
else:
ans = min(ans, int(a[: i - q]) + int(a[i - q :]))
if not ans is None:
break
ans1 = ans
if n & 1:
i += 1
ans = None
for q in range(n // 2):
if a[: i + q][0] != "0" and a[i + q :][0] != "0":
if ans is None:
ans = int(a[: i + q]) + int(a[i + q :])
else:
ans = min(ans, int(a[: i + q]) + int(a[i + q :]))
if a[: i - q][0] != "0" and a[i - q :][0] != "0":
if ans is None:
ans = int(a[: i - q]) + int(a[i - q :])
else:
ans = min(ans, int(a[: i - q]) + int(a[i - q :]))
if not ans is None:
break
if ans is None:
print(ans1)
elif ans1 is None:
print(ans)
else:
print(min(ans, ans1))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NONE ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER STRING VAR BIN_OP VAR VAR NUMBER STRING IF VAR NONE ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR NUMBER STRING VAR BIN_OP VAR VAR NUMBER STRING IF VAR NONE ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NONE ASSIGN VAR VAR IF BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NONE FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER STRING VAR BIN_OP VAR VAR NUMBER STRING IF VAR NONE ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR NUMBER STRING VAR BIN_OP VAR VAR NUMBER STRING IF VAR NONE ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NONE IF VAR NONE EXPR FUNC_CALL VAR VAR IF VAR NONE EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def fun(s, l):
if l == 1:
return int(s)
else:
i = l // 2
j = i
a = s[:i]
b = s[j:]
front = int(s)
while 1:
if b[0] == "0":
j += 1
i += 1
a = s[:i]
b = s[j:]
if a == "" or b == "":
break
else:
front = int(a) + int(b)
break
i = l // 2
j = i
a = s[: i + 1]
b = s[j + 1 :]
if b == "":
b = "0"
back = int(s)
while 1:
if b[0] == "0":
j -= 1
i -= 1
a = s[:i]
b = s[j:]
if a == "" or b == "":
break
else:
back = int(a) + int(b)
break
return min(front, back)
return mini
l = int(input())
s = input()
print(fun(s, l))
|
FUNC_DEF IF VAR NUMBER RETURN FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE NUMBER IF VAR NUMBER STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR STRING VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR STRING ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR VAR WHILE NUMBER IF VAR NUMBER STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR STRING VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = input()
s = input()
s = str(s)
mid = int(len(s) / 2)
num = -1
i = mid - 1
while i >= 0:
if s[i + 1] == "0":
i = i - 1
continue
s1 = s[: i + 1]
s2 = s[i + 1 :]
if num == -1:
num = int(s1) + int(s2)
else:
num = min(int(s1) + int(s2), num)
break
i = i - 1
for i in range(mid, len(s) - 1):
if s[i + 1] == "0":
continue
s1 = s[: i + 1]
s2 = s[i + 1 :]
if num == -1:
num = int(s1) + int(s2)
else:
num = min(int(s1) + int(s2), num)
break
print(num)
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
ans = 10**100001
if n % 2 == 0:
flag = 0
for i in range(n // 2):
if s[n // 2 - i] != "0":
ans = min(ans, int(s[0 : n // 2 - i]) + int(s[n // 2 - i :]))
flag += 1
if flag > 2:
break
flag = 0
for i in range(n // 2):
if s[n // 2 + i] != "0":
ans = min(ans, int(s[0 : n // 2 + i]) + int(s[n // 2 + i :]))
flag += 1
if flag > 2:
break
else:
flag = 0
for i in range(n // 2):
if s[n // 2 - i] != "0":
ans = min(ans, int(s[0 : n // 2 - i]) + int(s[n // 2 - i :]))
flag += 1
if flag > 2:
break
flag = 0
for i in range(n // 2 + 1):
if s[n // 2 + i] != "0":
ans = min(ans, int(s[0 : n // 2 + i]) + int(s[n // 2 + i :]))
flag += 1
if flag > 2:
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
ans = int(s)
cut = n // 2
if s[cut] != "0":
ans = min(int(s[:cut]) + int(s[cut:]), int(s[: cut + 1]) + int(s[cut + 1 :]))
else:
cut1 = cut
cut2 = cut
while cut1 >= 0 and s[cut1] == "0":
cut1 -= 1
while cut2 < n and s[cut2] == "0":
cut2 += 1
if cut1 > 0:
ans = min(ans, int(s[:cut1]) + int(s[cut1:]))
if cut2 < n:
ans = min(ans, int(s[:cut2]) + int(s[cut2:]))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def ssum(s, spl):
l = s[:spl]
r = s[spl:]
p = 0
ret = ""
carry = 0
while p < len(l) or p < len(r):
dl = int(l[-p - 1]) if p < len(l) else 0
dr = int(r[-p - 1]) if p < len(r) else 0
tmp = carry + int(dl) + int(dr)
if tmp > 9:
tmp -= 10
carry = 1
else:
carry = 0
ret = str(tmp) + ret
p += 1
if carry == 1:
ret = str(carry) + ret
return ret
def smin(s1, s2):
if len(s1) < len(s2):
return s1
if len(s2) < len(s1):
return s2
for i in range(len(s1)):
if s1[i] < s2[i]:
return s1
if s2[i] < s1[i]:
return s2
return s1
def sol(s):
if len(s) % 2 == 0:
l, r = len(s) // 2, len(s) // 2
else:
l, r = len(s) // 2, len(s) // 2 + 1
while True:
if s[l] == "0" and s[r] == "0":
l -= 1
r += 1
continue
if s[l] == "0":
return ssum(s, r)
if s[r] == "0":
return ssum(s, l)
return smin(ssum(s, l), ssum(s, r))
_ = input()
s = input()
print(sol(s))
|
FUNC_DEF ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN VAR IF VAR VAR VAR VAR RETURN VAR RETURN VAR FUNC_DEF IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER WHILE NUMBER IF VAR VAR STRING VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING RETURN FUNC_CALL VAR VAR VAR IF VAR VAR STRING RETURN FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
if n % 2:
if s[n // 2] != "0":
print(
min(
int(s[: n // 2]) + int(s[n // 2 :]),
int(s[: n // 2 + 1]) + int(s[n // 2 + 1 :]),
)
)
else:
s1 = s[: n // 2]
s2 = s[n // 2 :]
mn = int(s)
mn1 = int(s)
for i in range(len(s2)):
if s2[i] != "0":
mn = int(s1) + int(s2[i:])
break
s1 += s2[i]
s1 = s[: n // 2]
s2 = s[n // 2 :][::-1]
for i in range(len(s1) - 1, -1, -1):
if s1[i] != "0":
if i == 0:
mn1 = mn
else:
s2 += s1[i]
s2 = s2[::-1]
s1 = s1[:i]
mn1 = int(s1) + int(s2)
break
else:
s2 += s1[i]
print(min(mn, mn1))
elif s[n // 2] != "0":
s1 = s[: n // 2]
s2 = s[n // 2 :]
print(int(s1) + int(s2))
else:
mn = int(s)
mn1 = int(s)
for i in range(n // 2, n):
if s[i] != "0":
s1 = s[:i]
s2 = s[i:]
mn = int(s1) + int(s2)
break
for i in range(n // 2, -1, -1):
if s[i] != "0":
if i != 0:
s1 = s[:i]
s2 = s[i:]
mn1 = int(s1) + int(s2)
else:
mn1 = mn
break
print(min(mn, mn1))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
a = int(input())
b = input()
mid = a // 2
f = mid
s = mid
while f > 0 and b[f] == "0":
f = f - 1
while s < a - 1 and b[s] == "0":
s = s + 1
ans = 1e1000
for number in [-1, 0, 1]:
keysA = f + number
keysB = s + number
fa = 0
fb = 0
sa = 0
sb = 0
if len(b[:keysA]) > 0:
fa = int(b[:keysA])
if len(b[keysA:]) > 0:
fb = int(b[keysA:])
if len(b[:keysB]) > 0:
sa = int(b[:keysB])
if len(b[keysB:]) > 0:
sb = int(b[keysB:])
if keysA >= 0 and keysA <= a - 1 and b[keysA] != "0":
ans = min(ans, fa + fb)
if keysB >= 0 and keysB <= a - 1 and b[keysB] != "0":
ans = min(ans, sa + sb)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR LIST NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
size = int(input())
number = list(input())
if size < 5:
min_num = 99999999
for start2 in range(1, size):
if number[start2] == "0":
continue
num = int("".join(number[:start2])) + int("".join(number[start2:]))
if num < min_num:
min_num = num
print(min_num)
exit(0)
start1 = size // 2
while start1 >= 0:
if number[start1] == "0":
start1 -= 1
else:
break
start2 = size // 2 + 1
while start2 < size:
if number[start2] == "0":
start2 += 1
else:
break
if start1 == 0:
print(int("".join(number[:start2])) + int("".join(number[start2:])))
exit(0)
if start2 == size:
print(int("".join(number[:start1])) + int("".join(number[start1:])))
exit(0)
sub_num1 = int("".join(number[:start1])) + int("".join(number[start1:]))
sub_num2 = int("".join(number[:start2])) + int("".join(number[start2:]))
print(min(sub_num1, sub_num2))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL STRING VAR VAR FUNC_CALL VAR FUNC_CALL STRING VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL STRING VAR VAR FUNC_CALL VAR FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL STRING VAR VAR FUNC_CALL VAR FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL STRING VAR VAR FUNC_CALL VAR FUNC_CALL STRING VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL STRING VAR VAR FUNC_CALL VAR FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
ans = -1
for i in range(int(l / 2), l):
if n[i] == "0":
continue
tmp = int(n[0:i]) + int(n[i:])
if tmp < ans or ans == -1:
ans = tmp
if i > int(l / 2):
break
for i in range(int(l / 2), 0, -1):
if n[i] == "0":
continue
tmp = int(n[0:i]) + int(n[i:])
if tmp < ans or ans == -1:
ans = tmp
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def fun():
i = m - 1
while i >= 0 and s[i + 1] == "0":
i -= 1
j = m + 1
while j < l and s[j] == "0":
j += 1
if i < 0:
a = int(s) + 1
else:
a = int(s[: i + 1]) + int(s[i + 1 :])
if j == l:
b = int(s) + 1
else:
b = int(s[:j]) + int(s[j:])
return min(a, b)
l = int(input())
s = input()
m = int(l / 2)
print(fun())
|
FUNC_DEF ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR BIN_OP VAR NUMBER STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = input()
r = input()
n = int(n)
ans = int(r)
for i in range(n // 2, 0, -1):
j = r[0:i]
k = r[i:n]
if len(j) == 0 or len(k) == 0 or j[0] == "0" or k[0] == "0":
continue
int_j = int(j)
int_k = int(k)
ans = min(ans, int_j + int_k)
break
for i in range(n // 2 + 1, n, 1):
j = r[0:i]
k = r[i:n]
if len(j) == 0 or len(k) == 0 or j[0] == "0" or k[0] == "0":
continue
int_j = int(j)
int_k = int(k)
ans = min(ans, int_j + int_k)
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def get(i):
s1 = s[0:i]
s2 = s[i:n]
j = 0
while j < len(s2) and s2[j] == "0":
j += 1
k = i - 1
while k >= 0 and s1[k] == "0":
k -= 1
if j == len(s2):
a = int(s)
else:
a = int(s1 + "0" * j) + int(s2[j : len(s2)])
if k == 0:
b = int(s1[k : len(s1)] + s2)
else:
b = int(s[0:k]) + int(s[k : len(s)])
return min(a, b)
n = int(input())
s = input()
s1 = ""
s2 = ""
if n % 2 == 0:
print(get(n // 2))
else:
print(min(get(n // 2), get(n // 2 + 1)))
|
FUNC_DEF ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR BIN_OP STRING VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = str(input())
inf = 10**10**5
minsum = inf
flag = False
for i in range(n // 2, n):
if s[i] != "0":
flag = True
minsum = min(minsum, int(s[:i]) + int(s[i:]))
if s[n - i] != "0":
flag = True
minsum = min(minsum, int(s[: n - i]) + int(s[n - i :]))
if flag:
break
print(minsum)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l, s = int(input()), input()
ans = int(s)
def check(a, b):
b = b[::-1]
b, a = list(b), list(a)
while len(b) and b[-1] == "0":
b.pop()
a.append("0")
if not len(b):
return
b = b[::-1]
b = "".join(b)
a = "".join(a)
global ans
ans = min(ans, int(a) + int(b))
def check2(a, b):
b = b[::-1]
a = list(a)
b = list(b)
while len(a) and a[-1] == "0":
a.pop()
b.append("0")
if len(a) <= 1:
return
b.append(a.pop())
a = "".join(a)
b = "".join(b)
b = b[::-1]
global ans
ans = min(ans, int(a) + int(b))
check(s[: l // 2], s[l // 2 :])
check2(s[: l // 2], s[l // 2 :])
if l >= 3:
check(s[: l // 2 + 1], s[l // 2 + 1 :])
check2(s[: l // 2 + 1], s[l // 2 + 1 :])
print(ans)
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR WHILE FUNC_CALL VAR VAR VAR NUMBER STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR RETURN ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE FUNC_CALL VAR VAR VAR NUMBER STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER RETURN EXPR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
ans = int(s)
min_len = l
pos = []
for i in range(l - 1):
if s[i + 1] == "0":
continue
ll = max(i + 1, l - 1 - i) + 1
if ll < min_len:
pos = [i]
min_len = ll
elif ll == min_len:
pos.append(i)
else:
continue
for p in pos:
ss = int(s[: p + 1]) + int(s[p + 1 :])
if ss < ans:
ans = ss
print(int(ans))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR LIST VAR ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def shTofhShiftIfZero(s, fh, sh):
curr = ""
for i in sh:
if i == "0":
curr += "0"
else:
break
fh += curr
sh = sh[len(curr) :]
if fh == "":
fh = 0
if sh == "":
sh = 0
sum1 = int(fh) + int(sh)
return sum1
def fhToshShiftIfZero(s, fh, sh):
curr = ""
fh1 = fh[::-1]
for i in fh1:
if i == "0":
curr += "0"
else:
break
newSh = fh[len(fh) - len(curr) - 1] + curr + sh
sh = newSh
fh = fh[0 : len(fh) - len(curr) - 1]
if fh == "":
fh = 0
if sh == "":
sh = 0
sum1 = int(fh) + int(sh)
return sum1
n = int(input())
s = input()
fh = s[0 : n // 2]
sh = s[n // 2 :]
miniSum = 0
sum1 = 0
sum2 = 0
sum3 = 0
if sh[0] == "0":
sum1 += shTofhShiftIfZero(s, fh, sh)
sum2 += fhToshShiftIfZero(s, fh, sh)
miniSum = min(sum1, sum2)
elif fh[len(fh) - 1] == "0":
sum3 += fhToshShiftIfZero(s, fh, sh)
miniSum = sum3
else:
sh1 = sh
sh1 = sh1[1:]
fh1 = fh + sh[0]
if fh1 == "":
fh1 = 0
if sh1 == "":
sh1 = 0
fSum = int(fh1) + int(sh1)
miniSum = min(int(fh) + int(sh), fSum)
print(miniSum)
|
FUNC_DEF ASSIGN VAR STRING FOR VAR VAR IF VAR STRING VAR STRING VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR IF VAR STRING ASSIGN VAR NUMBER IF VAR STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR STRING ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR STRING VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR STRING ASSIGN VAR NUMBER IF VAR STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER STRING VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER IF VAR STRING ASSIGN VAR NUMBER IF VAR STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
mi = int(s)
mid = l // 2
while mid < l:
if s[mid] == "0":
mid += 1
else:
break
if mid < l:
mi = min(mi, int(s[:mid]) + int(s[mid:]))
mid = l // 2 + 1
while mid < l:
if s[mid] == "0":
mid += 1
else:
break
if mid < l:
mi = min(mi, int(s[:mid]) + int(s[mid:]))
mid = l // 2 - 1
while mid >= 0:
if s[mid] == "0":
mid -= 1
else:
break
if mid > 0:
mi = min(mi, int(s[:mid]) + int(s[mid:]))
print(mi)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
S = str(input()).strip()
ans = int(S)
l = n // 2
r = n // 2
while l >= 0 and S[l] == "0":
l -= 1
while r < len(S) and S[r] == "0":
r += 1
for i in [l, r]:
for j in range(max(i - 2, 1), min(len(S), i + 2)):
if S[j] != "0":
ans = min(ans, int(S[:j]) + int(S[j:]))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER FOR VAR LIST VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
ans = 1e1000
mid = n // 2
i = mid
while i > 0:
if s[i] != "0":
a = int(s[:i])
b = int(s[i:])
ans = a + b
break
i -= 1
i = mid + 1
while i < len(s):
if s[i] != "0":
a = int(s[:i])
b = int(s[i:])
ans = min(ans, a + b)
break
i += 1
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
n = int(s)
mitad = l // 2
base = mitad + 1
while base < l and s[base] == "0":
base += 1
base = 10 ** (l - base)
a = n // base
b = n % base
sol = a + b
base = mitad
while base > 0 and s[base] == "0":
base -= 1
base = 10 ** (l - base)
a = n // base
b = n % base
sol = min(sol, a + b)
print(sol)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def calc(s, a):
return int(s[:a]) + int(s[a:])
n = int(input())
s = input()
if n % 2 == 0:
if s[n // 2] != "0":
print(calc(s, n // 2))
else:
a = n // 2 - 1
b = n // 2
while s[a] == "0" and s[b] == "0":
a -= 1
b += 1
res = -1
if s[a] != "0" and a > 0:
res = calc(s, a)
if s[b] != "0":
if res < 0:
res = calc(s, b)
else:
res = min(res, calc(s, b))
print(res)
elif s[n // 2] != "0":
res = calc(s, n // 2)
if s[n // 2 + 1] != "0":
res = min(res, calc(s, n // 2 + 1))
print(res)
else:
a = n // 2 - 1
b = n // 2 + 1
while s[a] == "0" and s[b] == "0":
a -= 1
b += 1
res = -1
if s[a] != "0" and a > 0:
res = calc(s, a)
if s[b] != "0":
if res < 0:
res = calc(s, b)
else:
res = min(res, calc(s, b))
print(res)
|
FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR STRING VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
first = [None] * (n - 1)
for i in range(n - 1):
if s[i + 1] == "0":
first[i] = -1
continue
if n % 2 == 1:
if i < n // 2:
first[i] = int(s[i + 1])
else:
first[i] = int(s[i])
elif i == n // 2:
first[i] = max(int(s[i]), int(s[i + 1]))
elif i < n // 2:
first[i] = int(s[i + 1])
else:
first[i] = int(s[i])
midl = midr = 0
if n % 2 == 1:
midl = (n - 1) // 2 - 1
midr = midl + 1
else:
midl = midr = n // 2 - 1
point1 = point2 = flag = 0
while True:
if first[midl] == -1 and first[midr] == -1:
if midl != 0:
midl -= 1
if midr < n - 1:
midr += 1
continue
if first[midl] == -1:
point1 = midr
break
elif first[midr] == -1:
point1 = midl
break
flag = 1
point1 = midl
point2 = midr
break
ans = 0
if flag == 0:
ans = int(s[: point1 + 1]) + int(s[point1 + 1 :])
else:
ans1 = int(s[: point1 + 1]) + int(s[point1 + 1 :])
ans2 = int(s[: point2 + 1]) + int(s[point2 + 1 :])
ans = min(ans1, ans2)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER WHILE NUMBER IF VAR VAR NUMBER VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
p = 1 if l & 1 else 0
while n[l + p >> 1] == "0" and n[l - p >> 1] == "0":
p += 2
if n[l + p >> 1] == "0":
print(int(n[l - p >> 1 :]) + int(n[: l - p >> 1]))
elif n[l - p >> 1] == "0":
print(int(n[l + p >> 1 :]) + int(n[: l + p >> 1]))
else:
print(
min(
int(n[l + p >> 1 :]) + int(n[: l + p >> 1]),
int(n[l - p >> 1 :]) + int(n[: l - p >> 1]),
)
)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER NUMBER WHILE VAR BIN_OP BIN_OP VAR VAR NUMBER STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
x = int(input())
li_t = list(input())
len1 = len(li_t)
len1 = len1 // 2
li_t1 = li_t[:len1]
li_t2 = li_t[len1:]
i = 0
if li_t1[-1] == "0" and li_t2[0] == "0":
while i <= len(li_t2) - 1 and li_t2[i] == "0":
i += 1
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans1 = li_t1 + li_t2
li_t1 = li_t[:len1]
li_t2 = li_t[len1:]
i = -1
while i >= -1 * len(li_t1) and li_t1[i] == "0":
i -= 1
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans2 = li_t1 + li_t2
elif li_t2[0] == "0":
i = 0
while i <= len(li_t2) - 1 and li_t2[i] == "0":
i = i + 1
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans1 = li_t1 + li_t2
li_t1 = li_t[:len1]
li_t2 = li_t[len1:]
try:
i = -1
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans2 = li_t1 + li_t2
except:
ans2 = 0
elif li_t1[-1] == "0":
i = -1
while i >= -1 * len(li_t1) and li_t1[i] == "0":
i = i - 1
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans1 = li_t1 + li_t2
li_t1 = li_t[:len1]
li_t2 = li_t[len1:]
i = 1
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans2 = li_t1 + li_t2
else:
i = 0
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans1 = li_t1 + li_t2
i = 1
li_t1 = li_t[: len1 + i]
li_t2 = li_t[len1 + i :]
if li_t1 != []:
li_t1 = int("".join(li_t1))
else:
li_t1 = 0
if li_t2 != []:
li_t2 = int("".join(li_t2))
else:
li_t2 = 0
ans2 = li_t1 + li_t2
if ans1 > ans2:
print(ans2)
else:
print(ans1)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
t = int(l / 2)
if l == 2:
print(int(n[0]) + int(n[1]))
elif l == 3:
if n[1] == "0":
print(int(n[0:2]) + int(n[2]))
else:
print(min(int(n[0:2]) + int(n[2]), int(n[0]) + int(n[1:3])))
elif l % 2 == 0:
if n[t] != "0":
print(int(n[0:t]) + int(n[t:l]))
else:
for i in range(1, t):
if n[t + i] != "0" or n[t - i] != "0":
if n[t + i] != "0" and n[t - i] != "0":
print(
min(
int(n[0 : t - i]) + int(n[t - i : l]),
int(n[0 : t + i]) + int(n[t + i : l]),
)
)
elif n[t + i] != "0" and n[t - i] == "0":
print(int(n[0 : t + i]) + int(n[t + i : l]))
elif n[t + i] == "0" and n[t - i] != "0":
print(int(n[0 : t - i]) + int(n[t - i : l]))
break
else:
for i in range(0, t):
if n[t + 1 + i] != "0" or n[t - i] != "0":
if n[t + 1 + i] != "0" and n[t - i] != "0":
print(
min(
int(n[0 : t - i]) + int(n[t - i : l]),
int(n[0 : t + 1 + i]) + int(n[t + 1 + i : l]),
)
)
elif n[t + 1 + i] != "0" and n[t - i] == "0":
print(int(n[0 : t + 1 + i]) + int(n[t + 1 + i : l]))
elif n[t + 1 + i] == "0" and n[t - i] != "0":
print(int(n[0 : t - i]) + int(n[t - i : l]))
break
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR NUMBER IF VAR NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR STRING IF VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING VAR BIN_OP VAR VAR STRING IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR IF VAR BIN_OP BIN_OP VAR NUMBER VAR STRING VAR BIN_OP VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input().strip()
k1 = n // 2 + 1
k2 = n // 2
while k1 < n - 1 and s[k1] == "0":
k1 += 1
while k2 > 0 and s[k2] == "0":
k2 -= 1
if k1 == n or s[k1:n] == "0":
print(int(s[0:k2]) + int(s[k2:n]))
elif k2 == 0 or s[0:k2] == "0":
print(int(s[0:k1]) + int(s[k1:n]))
else:
print(min(int(s[0:k1]) + int(s[k1:n]), int(s[0:k2]) + int(s[k2:n])))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER IF VAR VAR VAR VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def suu(a, b):
a = int(a)
b = int(b)
return a + b
def go():
l = int(input())
s = str(input())
mid = l // 2
pos = l - 1
while 1:
if s[pos] == "0":
pos -= 1
else:
break
if pos < mid:
print(suu(s[:pos], s[pos:]))
return
if s[mid] == "0":
pos = mid
while s[pos] == "0":
pos += 1
pos2 = mid
while s[pos2] == "0":
pos2 -= 1
if pos2 == 0:
print(suu(s[:pos], s[pos:]))
elif pos >= l:
print(suu(s[:pos2], s[pos2:]))
else:
print(min(suu(s[:pos], s[pos:]), suu(s[:pos2], s[pos2:])))
return
if (
mid < l - 1
and s[mid + 1] != "0"
and suu(s[:mid], s[mid:]) > suu(s[: mid + 1], s[mid + 1 :])
):
mid += 1
print(suu(s[:mid], s[mid:]))
go()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN BIN_OP VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN IF VAR VAR STRING ASSIGN VAR VAR WHILE VAR VAR STRING VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR STRING VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER STRING FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def solve(p2, n, s):
p3 = p2
while p2 < n and s[p2] == "0":
p2 += 1
while p3 >= 0 and s[p3] == "0":
p3 -= 1
res = float("inf")
if p2 != 0 and p2 != n:
res = int(s[0:p2]) + int(s[p2:])
if p3 != 0 and p3 != n:
res = min(res, int(s[0:p3]) + int(s[p3:]))
return res
n = int(input())
s = input()
res = solve(n // 2, n, s)
if n % 2 == 1:
res = min(res, solve(n // 2 + 1, n, s))
print(res)
|
FUNC_DEF ASSIGN VAR VAR WHILE VAR VAR VAR VAR STRING VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING IF VAR NUMBER VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
min_val = n
for i in range(1, n):
if s[i] == "0":
continue
min_val = min(min_val, max(i, n - i))
ans = int(s)
for i in range(1, n):
if s[i] == "0":
continue
if max(i, n - i) < min_val - 1 or max(i, n - i) > min_val:
continue
candi = int(s[:i]) + int(s[i:])
ans = min(ans, candi)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING IF FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
num = input()
mid = int(n / 2)
if n % 2 == 1:
mid = mid + 1
ans = 0
if num[mid] != "0":
ans1 = int(num[0:mid]) + int(num[mid:])
if n % 2 == 1 and num[mid - 1] != "0":
ans2 = int(num[0 : mid - 1]) + int(num[mid - 1 :])
ans = min(ans1, ans2)
else:
ans = ans1
else:
t_mid = mid
ans1 = -1
ans2 = -1
while t_mid >= 0 and num[t_mid] == "0":
t_mid = t_mid - 1
if t_mid >= 0:
ans1 = int(num[0:t_mid]) + int(num[t_mid:])
t_mid = mid
while t_mid < n and num[t_mid] == "0":
t_mid = t_mid + 1
if t_mid < n:
ans2 = int(num[0:t_mid]) + int(num[t_mid:])
if ans1 == -1:
ans = ans2
elif ans2 == -1:
ans = ans1
else:
ans = min(ans1, ans2)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR WHILE VAR VAR VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
import sys
f = sys.stdin
out = sys.stdout
l = int(f.readline().rstrip("\r\n"))
n = f.readline().rstrip("\r\n")
ans = int(n)
p = l >> 1
while p < l and n[p] == "0":
p += 1
if p < l:
ans = min(ans, int(n[:p]) + int(n[p:]))
p += 1
while p < l and n[p] == "0":
p += 1
if p < l:
ans = min(ans, int(n[:p]) + int(n[p:]))
p = l >> 1
while p >= 0 and n[p] == "0":
p -= 1
if p > 0:
ans = min(ans, int(n[:p]) + int(n[p:]))
p -= 1
while p >= 0 and n[p] == "0":
p -= 1
if p > 0:
ans = min(ans, int(n[:p]) + int(n[p:]))
out.write(str(ans))
|
IMPORT ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = str(input())
best_len = l
best = int(n)
to_try = []
for i in range(l - 1):
if n[i + 1] != "0":
candidate = max(i + 1, l - i - 1) + 1
if candidate < best_len:
to_try.clear()
to_try.append(i + 1)
best_len = candidate
elif candidate == best_len:
to_try.append(i + 1)
for i in to_try:
best = min(best, int(n[:i]) + int(n[i:]))
print(best)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def asdf():
l = int(input())
s = input()
n = int(s)
mid = l // 2
mini = n
k = 0
for i in range(0, l // 2 + 1):
ind = mid + i
if 0 < ind < l and s[ind] != "0":
mini = min(mini, int(s[:ind]) + int(s[ind:]))
k += 1
ind = mid - i
if 0 < ind < l and s[ind] != "0":
mini = min(mini, int(s[:ind]) + int(s[ind:]))
k += 1
if k >= 5:
break
print(mini)
asdf()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR IF NUMBER VAR VAR VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF NUMBER VAR VAR VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def check(s1, s2):
if len(s1) == 0 or len(s2) == 0 or s2[0] == "0":
return -1
else:
return int(s1) + int(s2)
dummy = input()
str = input()
left = 0
right = 0
if len(str) % 2 == 0:
left = len(str) // 2 - 1
right = len(str) // 2
else:
left = len(str) // 2
right = len(str) // 2
answer = -1
while answer == -1:
it1 = check(str[0:left], str[left : len(str)])
if it1 != -1:
if answer == -1 or answer > it1:
answer = it1
it2 = check(str[0:right], str[right : len(str)])
if it2 != -1:
if answer == -1 or answer > it2:
answer = it2
left -= 1
right += 1
if left > 0:
it1 = check(str[0:left], str[left : len(str)])
if it1 != -1:
if answer == -1 or answer > it1:
answer = it1
if right < len(str):
it2 = check(str[0:right], str[right : len(str)])
if it2 != -1:
if answer == -1 or answer > it2:
answer = it2
print(answer)
|
FUNC_DEF IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING RETURN NUMBER RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = str(input())
ans = 10 ** (10**5 + 1)
i = n // 2 + 1
while i < n and s[i] == "0":
i += 1
if i < n:
ans = min(ans, int(s[:i]) + int(s[i:]))
i = n // 2
while s[i] == "0":
i -= 1
if i > 0:
ans = min(ans, int(s[:i]) + int(s[i:]))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
num = int(s)
p = n // 2 + 1
while p < n and s[p] == "0":
p += 1
p -= 1
p = 10 ** (n - p - 1)
l, r = divmod(num, p)
m = l + r
p = n // 2
while p > -1 and s[p] == "0":
p -= 1
p -= 1
p = 10 ** (n - p - 1)
l, r = divmod(num, p)
m = min(m, l + r)
print(m)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
inp = input()
inp1 = list(inp)
mid = l // 2
toLeft = 0
toRight = l
if l % 2 != 0:
if inp[mid] != "0":
opt1 = int(inp[: mid + 1])
opt2 = int(inp[mid:])
if opt1 < opt2:
ans = opt1 + int(inp[mid + 1 :])
else:
ans = opt2 + int(inp[:mid])
else:
i = mid - 1
while i >= 0:
if inp[i] != "0":
toLeft = i
break
i -= 1
i = mid + 1
while i < l:
if inp[i] != "0":
toRight = i
break
i += 1
if toLeft == 0 and toRight == l:
ans = int(inp)
elif toRight == l:
ans = int(inp[:toLeft]) + int(inp[toLeft:])
elif toLeft == 0:
ans = int(inp[:toRight]) + int(inp[toRight:])
else:
opt1 = int(inp[:toLeft]) + int(inp[toLeft:])
opt2 = int(inp[:toRight]) + int(inp[toRight:])
if opt1 < opt2:
ans = opt1
else:
ans = opt2
elif inp[mid] != "0":
ans = int(inp[:mid]) + int(inp[mid:])
else:
i = mid - 1
while i >= 0:
if inp[i] != "0":
toLeft = i
break
i -= 1
i = mid + 1
while i < l:
if inp[i] != "0":
toRight = i
break
i += 1
if toLeft == 0 and toRight == l:
ans = int(inp)
elif toRight == l:
ans = int(inp[:toLeft]) + int(inp[toLeft:])
elif toLeft == 0:
ans = int(inp[:toRight]) + int(inp[toRight:])
else:
opt1 = int(inp[:toLeft]) + int(inp[toLeft:])
opt2 = int(inp[:toRight]) + int(inp[toRight:])
if opt1 < opt2:
ans = opt1
else:
ans = opt2
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def ans(string, sep, l):
first = int(string[0:sep])
second = int(string[sep:l])
return first + second
l = int(input())
string = input()
mid = l // 2
a = 0
b = 0
for i in range(mid, 0, -1):
if string[i] != "0":
a = ans(string, i, l)
break
for i in range(mid + 1, l):
if string[i] != "0":
b = ans(string, i, l)
break
if a * b == 0:
print(a + b)
else:
print(min(a, b))
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def main():
l = int(input())
n = input()
data = []
for i in range(1, l):
if n[i] == "0":
continue
data.append((abs(l // 2 - i), i))
data.sort()
res = -1
for i in range(0, min(10, len(data))):
start = data[i][1]
x = int(n[0:start])
y = int(n[start:])
z = x + y
if res == -1:
res = z
else:
res = min(res, z)
print(res)
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
n = int(s)
mid = l // 2
ans = 10**l
for i in range(mid + 1, l):
if s[i].isdigit() and s[i] != "0":
exp = 10 ** (l - i)
fir = n // exp
sec = n % exp
ans = min(ans, fir + sec)
break
for i in range(mid, 0, -1):
if s[i].isdigit() and s[i] != "0":
exp = 10 ** (l - i)
fir = n // exp
sec = n % exp
ans = min(ans, fir + sec)
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
split = [1] * l
for i in range(l):
if s[i] == "0":
split[i] = 0
if l % 2:
c1 = l // 2
c2 = 1 + l // 2
else:
c1 = l // 2
c2 = l // 2
i = 0
while 1:
if c1 - i >= 0 and split[c1 - i]:
m1 = int(s[: c1 - i]) + int(s[c1 - i :])
else:
m1 = -1
if c2 + i < l and split[c2 + i]:
m2 = int(s[: c2 + i]) + int(s[c2 + i :])
else:
m2 = -1
if m1 == -1 and m2 == -1:
pass
elif m1 == -1:
ans = m2
break
elif m2 == -1:
ans = m1
break
else:
ans = min(m1, m2)
break
i += 1
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
a = input()
i = l // 2
ans = -1
while i > 0 and a[i] == "0":
i -= 1
r = l // 2 + 1
while r < l and a[r] == "0":
r += 1
for j in [i, i - 1, r, r + 1]:
if j > 0 and j < l:
if a[j] != "0":
if ans == -1:
ans = int(a[:j]) + int(a[j:])
else:
ans = min(ans, int(a[:j]) + int(a[j:]))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER FOR VAR LIST VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
ans = 10**100005
l1, l2 = l // 2, (l + 1) // 2
check = False
while True:
if s[l1] != "0":
check = True
m1 = int(s[:l1])
m2 = int(s[l1:])
b = m1 + m2
ans = min(ans, b)
if s[l2] != "0":
check = True
m1 = int(s[:l2])
m2 = int(s[l2:])
b = m1 + m2
ans = min(ans, b)
if check == True:
print(ans)
break
l1 = l1 - 1
l2 = l2 + 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE NUMBER IF VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
a = input()
left = []
right = []
t = ""
c = 0
idex = n // 2
if n % 2 == 0:
if a[idex] == "0":
while idex < n and a[idex] == "0":
idex += 1
if idex == n:
left = int(a)
else:
left = int(a[:idex]) + int(a[idex:])
idex = n // 2
while idex < n and a[idex] == "0":
idex -= 1
right = int(a[:idex]) + int(a[idex:])
print(min(left, right))
else:
print(int(a[:idex]) + int(a[idex:]))
elif a[idex] != "0" and a[idex + 1] != "0":
print(min(int(a[:idex]) + int(a[idex:]), int(a[: idex + 1]) + int(a[idex + 1 :])))
else:
while idex < n and a[idex] == "0":
idex += 1
if idex == n:
left = int(a)
else:
left = int(a[:idex]) + int(a[idex:])
idex = n // 2
while a[idex] == "0":
idex -= 1
if idex == 0:
right = int(a)
else:
right = int(a[:idex]) + int(a[idex:])
s1 = min(left, right)
idex = n // 2 + 1
while idex < n and a[idex] == "0":
idex += 1
if idex == n:
left = int(a)
else:
left = int(a[:idex]) + int(a[idex:])
idex = n // 2
while idex < n and a[idex] == "0":
idex -= 1
if idex == 0:
right = int(a)
else:
right = int(a[:idex]) + int(a[idex:])
s2 = min(left, right)
print(min(s1, s2))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input().strip())
n = str(input().strip())
n1 = n[: l // 2]
n2 = n[l // 2 :]
if n1[-1] != "0" and n2[0] != "0":
if l % 2 == 0:
print(int(n1) + int(n2))
else:
print(
min(
int(n[: l // 2]) + int(n[l // 2 :]),
int(n[: l // 2 + 1]) + int(n[l // 2 + 1 :]),
)
)
exit()
a1, a2 = l // 2 - 1, l // 2
while n[a1] == "0" and a1 > 0:
a1 -= 1
while n[a2] == "0" and a2 < l - 1:
a2 += 1
if a1 == 0 and a2 == l - 1:
if n[a2] == "0":
print(n)
else:
print(n[: l - 2] + n[-1])
elif a1 != 0 and a2 != l - 1:
print(min(int(n[:a1]) + int(n[a1:]), int(n[:a2]) + int(n[a2:])))
elif a1 == 0:
print(int(n[:a2]) + int(n[a2:]))
else:
print(int(n[:a1]) + int(n[a1:]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR NUMBER VAR NUMBER WHILE VAR VAR STRING VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
d = set(s)
if "0" in d:
mi = 0
m = len(s)
for i in range(n // 2 - 1, 0, -1):
if s[i] != "0":
mi = i
break
for i in range(n // 2, len(s)):
if s[i] != "0":
m = i
break
if mi == 0:
mi = int(s)
else:
mi = int(s[mi:]) + int(s[:mi])
if m == len(s):
m = int(s)
else:
m = int(s[m:]) + int(s[:m])
print(min(mi, m))
elif n % 2 != 0:
a = int(s[: n // 2]) + int(s[n // 2 :])
b = int(s[: n // 2 + 1]) + int(s[n // 2 + 1 :])
print(min(a, b))
else:
print(int(s[: n // 2]) + int(s[n // 2 :]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF STRING VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = input()
no = input()
n = int(n)
if no[n // 2] != "0":
ans = int(no[0 : n // 2]) + int(no[n // 2 : n + 1])
if n // 2 + 1 < n and no[n // 2 + 1] != "0":
b = int(no[0 : n // 2 + 1]) + int(no[n // 2 + 1 : n])
ans = min(ans, b)
if n // 2 + 2 < n and no[n // 2 + 2] != "0":
b = int(no[0 : n // 2 + 2]) + int(no[n // 2 + 2 : n])
ans = min(ans, b)
if n // 2 - 1 > 0 and no[n // 2 - 1] != "0":
c = int(no[0 : n // 2 - 1]) + int(no[n // 2 - 1 : n])
ans = min(ans, c)
if n // 2 - 2 > 0 and no[n // 2 - 2] != "0":
c = int(no[0 : n // 2 - 2]) + int(no[n // 2 - 2 : n])
ans = min(ans, c)
print(ans)
else:
x = n // 2
y = n // 2
while no[x] == "0":
x -= 1
while y < n and no[y] == "0":
y += 1
ans = -1
if y < n and no[y] != "0":
ans = int(no[0:y]) + int(no[y:n])
if x > 0:
b = int(no[0:x]) + int(no[x:n])
if ans != -1:
ans = min(ans, b)
else:
ans = b
if x - 1 > 0 and no[x - 1] != "0":
c = int(no[0 : x - 1]) + int(no[x - 1 : n])
ans = min(ans, c)
if x - 2 > 0 and no[x - 2] != "0":
c = int(no[0 : x - 2]) + int(no[x - 2 : n])
ans = min(ans, c)
if y + 1 < n and no[y + 1] != "0":
d = int(no[0 : y + 1]) + int(no[y + 1 : n])
ans = min(ans, d)
if y + 2 < n and no[y + 2] != "0":
d = int(no[0 : y + 2]) + int(no[y + 2 : n])
ans = min(ans, d)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
num = input()
ind = int(n / 2)
mini = int(num)
while num[ind] == "0":
ind += 1
if ind == n:
break
for i in range(max(1, ind - 1), min(ind + 2, n)):
if num[i] != "0":
mini = min(mini, int(num[0:i]) + int(num[i:n]))
ind = int(n / 2)
while num[ind] == "0":
ind -= 1
for i in range(max(1, ind - 1), min(ind + 2, n)):
if num[i] != "0":
mini = min(mini, int(num[0:i]) + int(num[i:n]))
print(mini)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR STRING VAR NUMBER IF VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def res(digits):
if "0" not in digits:
if len(digits) % 2 == 0:
return int(digits[: len(digits) // 2]) + int(digits[len(digits) // 2 :])
else:
mid = len(digits) // 2
return min(
int(digits[:mid]) + int(digits[mid:]),
int(digits[: mid + 1]) + int(digits[mid + 1 :]),
)
else:
mid = len(digits) // 2
i = mid
while i > 0 and digits[i] == "0":
i -= 1
closestNonZeroL = i
i = mid
while i < len(digits) and digits[i] == "0":
i += 1
closestNonZeroR = i
sum1 = (
int(digits[:closestNonZeroL]) + int(digits[closestNonZeroL:])
if closestNonZeroL > 0
else int(digits)
)
sum2 = (
int(digits[:closestNonZeroR]) + int(digits[closestNonZeroR:])
if closestNonZeroR > 0 and closestNonZeroR < len(digits)
else int(digits)
)
i = mid - 1
while i > 0 and digits[i] == "0":
i -= 1
closestNonZeroL = i
i = mid - 1
while i < len(digits) and digits[i] == "0":
i += 1
closestNonZeroR = i
sum3 = (
int(digits[:closestNonZeroL]) + int(digits[closestNonZeroL:])
if closestNonZeroL > 0
else int(digits)
)
sum4 = (
int(digits[:closestNonZeroR]) + int(digits[closestNonZeroR:])
if closestNonZeroR > 0 and closestNonZeroR < len(digits)
else int(digits)
)
return min(min(sum1, sum2), min(sum3, sum4))
l = int(input())
digs = input()
print(str(res(digs)))
|
FUNC_DEF IF STRING VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN BIN_OP FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
a = input()
ans = -1
mid = l // 2
for i in range(mid, 0, -1):
if a[i] != "0":
if ans != -1:
ans = min(ans, int(a[0:i]) + int(a[i:]))
else:
ans = int(a[0:i]) + int(a[i:])
break
for i in range(mid + 1, l):
if a[i] != "0":
if ans != -1:
ans = min(ans, int(a[0:i]) + int(a[i:]))
else:
ans = int(a[0:i]) + int(a[i:])
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
import sys
mod = 1000000007
get_arr = lambda: list(map(int, input().split()))
get_int = lambda: int(input())
get_ints = lambda: map(int, input().split())
get_str = lambda: input()
get_strs = lambda: input().split()
n = get_int()
s = get_str()
ans = int(s)
j = 0
for i in range(n // 2, n):
if j > 3:
break
if s[i] == "0":
continue
ans = min(ans, int(s[:i]) + int(s[i:]))
j += 1
j = 0
for i in range(n // 2, 0, -1):
if j > 3:
break
if s[i] == "0":
continue
ans = min(ans, int(s[:i]) + int(s[i:]))
j += 1
print(ans)
|
IMPORT ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
s = input()
vec = []
for i in range(1, l):
if s[i] == "0":
continue
vec.append(i)
vec = sorted(vec, key=lambda x: abs(x - l / 2))
ans = 10**100000
cnt = 0
for i in vec:
cnt += 1
if cnt > 4:
break
ans = min(ans, int(s[:i]) + int(s[i:]))
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
a = input()
minsum = -1
if n > 1000:
start = n // 2 - 3
end = n // 2 + 3
else:
start = 1
end = n
i = start
while start < end and start < n:
str1 = a[:start]
str2 = a[start:]
if str1[0] != "0" and str2[0] != "0":
temp = int(str1) + int(str2)
if minsum == -1 or temp < minsum:
minsum = temp
if minsum == -1 and end <= n:
end += 1
start += 1
print(minsum)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER STRING VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def sum(str1, str2):
if len(str1) > len(str2):
t = str1
str1 = str2
str2 = t
str = ""
n1 = len(str1)
n2 = len(str2)
str1 = str1[::-1]
str2 = str2[::-1]
carry = 0
for i in range(n1):
sum = ord(str1[i]) - 48 + (ord(str2[i]) - 48 + carry)
str += chr(sum % 10 + 48)
carry = int(sum / 10)
for i in range(n1, n2):
sum = ord(str2[i]) - 48 + carry
str += chr(sum % 10 + 48)
carry = int(sum / 10)
if carry:
str += chr(carry + 48)
str = str[::-1]
return str
def compare(str1, str2):
if len(str1) > len(str2):
return str2
elif len(str1) < len(str2):
return str1
for i in range(len(str1)):
if str1[i] > str2[i]:
return str2
elif str1[i] < str2[i]:
return str1
return str1
n = int(input())
str = input()
result1 = ""
result2 = ""
bl1 = False
bl2 = False
if n % 2 == 0:
u = n // 2
v = u
else:
u = n // 2
v = u + 1
while 1:
if str[u] != "0" and u != 0:
result1 = sum(str[0:u], str[u : len(str)])
bl1 = True
break
else:
if u == 0:
break
u -= 1
while 1:
if str[v] != "0":
result2 = sum(str[0:v], str[v : len(str)])
bl2 = True
break
else:
v += 1
if v == n:
break
if bl1 == True and bl2 == True:
print(compare(result1, result2))
elif bl1 == True:
print(result1)
elif bl2 == True:
print(result2)
|
FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER RETURN VAR FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN VAR IF VAR VAR VAR VAR RETURN VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER WHILE NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR NUMBER IF VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = str(input())
ans = -1
cnt = 0
le = len(s)
i = max(1, le // 2)
while i < le and cnt < 7:
while i + 1 < le and s[i] == "0":
i += 1
l = s[:i]
r = s[i:]
cnt += 1
i += 1
if l[0] == "0" or r[0] == "0":
continue
if ans == -1:
ans = int(l) + int(r)
else:
ans = min(ans, int(l) + int(r))
i = max(1, le // 2)
cnt = 0
while i < le and cnt < 7:
while i > 1 and s[i] == "0":
i -= 1
if i <= 0 or s[i] == "0":
break
l = s[:i]
r = s[i:]
cnt += 1
i -= 1
if l[0] == "0" or r[0] == "0":
continue
if ans == -1:
ans = int(l) + int(r)
else:
ans = min(ans, int(l) + int(r))
assert ans != -1
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR VAR NUMBER WHILE BIN_OP VAR NUMBER VAR VAR VAR STRING VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def solve(S):
n = len(S)
cands = []
i = n // 2
while i > 0 and S[i] == "0":
i -= 1
cands.append(i)
j = n // 2 + 1
while j < n and S[j] == "0":
j += 1
cands.append(j)
sums = []
for i in set(cands):
if i == 0:
sums.append([S, "0"])
elif i == n:
sums.append([S, "0"])
else:
sums.append([S[:i], S[i:]])
return min(int(a) + int(b) for a, b in sums)
input()
print(solve(input()))
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR LIST VAR STRING IF VAR VAR EXPR FUNC_CALL VAR LIST VAR STRING EXPR FUNC_CALL VAR LIST VAR VAR VAR VAR RETURN FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
ans = int(n)
mid = l // 2
if n[mid] != "0":
ans = int(n[:mid]) + int(n[mid:])
i = l // 2
out = False
for j in range(1, mid + 1):
if i + j < l:
if n[i + j] != "0":
temp = int(n[: i + j]) + int(n[i + j :])
ans = min(ans, temp)
out = True
if i - j > 0:
if n[i - j] != "0":
temp = int(n[: i - j]) + int(n[i - j :])
ans = min(ans, temp)
break
if out:
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR IF VAR BIN_OP VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
import sys
input = sys.stdin.readline
N = int(input())
S = input().rstrip()
L = [[] for _ in range(N)]
for n in range(N - 1):
if S[n + 1] != "0":
L[max(n + 1, N - n - 1)].append(n)
def make_sum(s1, s2):
if len(s1) < len(s2):
tmp = s2
s2 = s1
s1 = tmp
l1, l2 = len(s1), len(s2)
s2 = "0" * (l1 - l2) + s2
S = [0] * (l1 + 1)
for i in range(l1):
a = int(s2[-i - 1]) + int(s1[-i - 1]) + S[i]
S[i] = a % 10
S[i + 1] += a // 10
go = False
a = ""
for n in range(l1, -1, -1):
if S[n] == 0 and not go:
continue
else:
go = True
a += str(S[n])
return a
def smaller(s1, s2):
if len(s1) > len(s2):
return s2
elif len(s1) < len(s2):
return s1
else:
l = len(s1)
for i in range(l):
if s1[i] > s2[i]:
return s2
elif s1[i] < s2[i]:
return s1
return s1
ans = "9" * N
c = 0
for l in range(N):
update = False
for n in L[l]:
s1, s2 = S[: n + 1], S[n + 1 :]
a = make_sum(s1, s2)
ans = smaller(ans, a)
update = True
if update:
c += 1
if c > 3:
break
print(ans)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP STRING BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR NUMBER VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN VAR IF VAR VAR VAR VAR RETURN VAR RETURN VAR ASSIGN VAR BIN_OP STRING VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
import sys
l = int(input())
s = input()
n = l // 2
ans = 0
c1, c2 = int(s), int(s)
f = 0
while n > 0:
k = l - n
if s[n] != "0":
c1 = int(s[0:n:1]) + int(s[n:l:1])
f = 1
if s[k] != "0":
c2 = int(s[0:k:1]) + int(s[k:l:1])
f = 1
if f == 1:
ans = min(c1, c2)
break
n -= 1
print(ans)
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
good = l // 2
lgood = good - 1
rgood = good + 1
if n[good] != "0":
print(min(int(n[:good]) + int(n[good:]), int(n[: 1 + good]) + int(n[1 + good :])))
else:
while lgood > 0 and n[lgood] == "0":
lgood -= 1
while rgood < l and n[rgood] == "0":
rgood += 1
if lgood == 0:
a = float("inf")
else:
a = int(n[:lgood]) + int(n[lgood:])
if rgood == l:
b = float("inf")
else:
b = int(n[:rgood]) + int(n[rgood:])
print(min(a, b))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP NUMBER VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def mp():
return map(int, input().split())
n = int(input())
a = input()
b = cnt = 0
for i in range(n // 2, n):
if a[i] != "0":
t = int(int(a[:i]) + int(a[i:]))
if b == 0 or t < b:
b = t
cnt += 1
if cnt > 2:
break
cnt = 0
for i in range(n // 2, 0, -1):
if a[i] != "0":
t = int(int(a[:i]) + int(a[i:]))
if b == 0 or t < b:
b = t
cnt += 1
if cnt > 2:
break
print(b)
|
FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
ans = int(s)
ind = set()
for i in range(0, n):
if s[i] != "0":
ind.add(i)
L, R = False, False
for i in range(0, n):
if n // 2 - i > 0 and not L:
if n // 2 - i in ind:
ans = min(ans, int(s[: n // 2 - i]) + int(s[n // 2 - i :]))
L = True
if n // 2 + i < n and not R:
if n // 2 + i in ind:
ans = min(ans, int(s[: n // 2 + i]) + int(s[n // 2 + i :]))
if i:
R = True
if L and R:
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR IF BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER VAR VAR VAR IF BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR ASSIGN VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
a = input()
l = len(a)
aa = int(a)
ans = -1
pos = int(l / 2)
pos1 = pos
pos2 = pos
while pos1 < l and a[pos1] == "0":
pos1 += 1
while pos2 >= 0 and a[pos2] == "0":
pos2 -= 1
try:
n1 = a[:pos1]
n2 = a[pos1:]
ans = int(n1) + int(n2)
except:
pass
if pos1 + 1 < l and a[pos1 + 1] != "0":
n3 = a[: pos1 + 1]
n4 = a[pos1 + 1 :]
if ans != -1:
ans = min(ans, int(n3) + int(n4))
else:
ans = int(n3) + int(n4)
try:
n1 = a[:pos2]
n2 = a[pos2:]
if ans != -1:
ans = min(ans, int(n1) + int(n2))
else:
ans = int(n1) + int(n2)
except ValueError:
pass
if pos2 - 1 > 0 and a[pos2 - 1] != "0":
n3 = a[: pos2 - 1]
n4 = a[pos2 - 1 :]
ans = min(ans, int(n3) + int(n4))
if ans != -1:
ans = min(ans, int(n3) + int(n4))
else:
ans = int(n3) + int(n4)
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR VAR VAR VAR STRING VAR NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
num = int(s)
pos1, pos2 = len(s) // 2 - 1, len(s) // 2 + 1
while pos1 >= 0 and s[pos1] == "0":
pos1 -= 1
while pos2 < n and s[pos2] == "0":
pos2 += 1
for i in range(max(1, pos1), min(n, pos2 + 1)):
if s[i:][0] == "0":
continue
num = min(int(s[:i]) + int(s[i:]), num)
print(num)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER WHILE VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
length = int(input())
l = input()
min_s = None
start_pos = length // 2
end_pos = start_pos
start_count = 0
while start_pos > 0:
if l[start_pos] != "0":
n1 = l[:start_pos]
n2 = l[start_pos:]
start_count += 1
if min_s == None:
min_s = int(n1) + int(n2)
else:
min_s = min(min_s, int(n1) + int(n2))
start_pos -= 1
if start_count == 5:
break
end_count = 0
while end_pos < length:
if l[end_pos] != "0":
n1 = l[:end_pos]
n2 = l[end_pos:]
end_count += 1
if min_s == None:
min_s = int(n1) + int(n2)
else:
min_s = min(min_s, int(n1) + int(n2))
end_pos += 1
if end_count == 5:
break
print(min_s)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NONE ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER IF VAR NONE ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER IF VAR NONE ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
x1, x2 = l // 2, (l + 1) // 2
min_val = 10**1000000
while True:
flag = False
if n[x1] != "0":
a = int(n[:x1])
b = int(n[x1:])
min_val = min(min_val, a + b)
flag = True
if n[x2] != "0":
a = int(n[:x2])
b = int(n[x2:])
min_val = min(min_val, a + b)
flag = True
if flag == True:
print(min_val)
break
else:
x1 -= 1
x2 += 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE NUMBER ASSIGN VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
if len(s) == 2:
print(int(s[0]) + int(s[1]))
exit(0)
met = len(s) // 2
if s[n // 2] != "0" and s[n // 2 + 1] != "0" and s[n // 2 - 1 + (n == 3)] != "0":
print(
min(
int(s[: n // 2 + 1]) + int(s[n // 2 + 1 :]),
min(
int(s[: n // 2]) + int(s[n // 2 :]),
int(s[: n // 2 - 1 + (n == 3)]) + int(s[n // 2 - 1 + (n == 3) :]),
),
)
)
exit(0)
elif s[n // 2 + 1] != "0" and s[n // 2] != "0":
print(
min(
int(s[: n // 2 + 1]) + int(s[n // 2 + 1 :]),
int(s[: n // 2]) + int(s[n // 2 :]),
)
)
exit(0)
elif s[n // 2] != "0" and s[n // 2 - 1 + (n == 3)] != "0":
print(
min(
int(s[: n // 2]) + int(s[n // 2 :]),
int(s[: n // 2 - 1 + (n == 3)]) + int(s[n // 2 - 1 + (n == 3) :]),
)
)
exit(0)
elif s[n // 2 + 1] != "0" and s[n // 2 - 1 + (n == 3)] != "0":
print(
min(
int(s[: n // 2 + 1]) + int(s[n // 2 + 1 :]),
int(s[: n // 2 - 1 + (n == 3)]) + int(s[n // 2 - 1 + (n == 3) :]),
)
)
exit(0)
elif s[n // 2] != "0":
print(int(s[: n // 2]) + int(s[n // 2 :]))
exit(0)
elif s[n // 2 + 1] != "0":
print(int(s[: n // 2 + 1]) + int(s[n // 2 + 1 :]))
exit(0)
elif s[n // 2 - 1 + (n == 3)] != "0":
print(int(s[: n // 2 - 1 + (n == 3)]) + int(s[n // 2 - 1 + (n == 3) :]))
exit(0)
ult = n + 1
ult2 = -1
for i in range(n // 2, len(s)):
if s[i] != "0":
ult = min(ult, i)
for i in range(0, n // 2):
if s[i] != "0":
ult2 = i
if ult != n + 1 and ult2 != -1:
print(min(int(s[:ult]) + int(s[ult:]), int(s[:ult2]) + int(s[ult2:])))
elif ult != n + 1:
print(int(s[:ult]) + int(s[ult:]))
elif ult2 != -1:
print(int(s[:ult2]) + int(s[ult2:]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
ok = False
if n % 2:
if s[n // 2] != "0":
print(
min(
int(s[: n // 2]) + int(s[n // 2 :]),
int(s[: n // 2 + 1]) + int(s[n // 2 + 1 :]),
)
)
ok = True
elif s[n // 2] != "0":
s1 = s[: n // 2]
s2 = s[n // 2 :]
print(int(s1) + int(s2))
ok = True
if ok == False:
r = l = n // 2
while l >= 0:
if s[l] != "0":
break
l -= 1
while r < n:
if s[r] != "0":
break
r += 1
mn = int(s)
if l != 0:
mn = min(mn, int(s[:l]) + int(s[l:]))
if r != n:
mn = min(mn, int(s[:r]) + int(s[r:]))
print(mn)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = input()
mid = l // 2 + 1
ans = int(n)
for x in range(mid, 0, -1):
if n[x] == "0":
continue
ans = min(ans, int(n[:x]) + int(n[x:]))
break
mid = l // 2
for x in range(mid, l - 1):
if n[x] == "0":
continue
ans = min(ans, int(n[:x]) + int(n[x:]))
break
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
l = int(input())
n = str(input())
mn = -1
midpoint = l // 2
i = midpoint
numright = 0
while numright < 5 and i < l - 1:
if n[i + 1] != "0":
first = int(n[: i + 1])
second = int(n[i + 1 :])
if mn == -1:
mn = first + second
else:
mn = min(mn, first + second)
numright += 1
i += 1
i = midpoint
numleft = 0
while numleft < 5 and i >= 0:
if n[i + 1] != "0":
first = int(n[: i + 1])
second = int(n[i + 1 :])
if mn == -1:
mn = first + second
else:
mn = min(mn, first + second)
numleft += 1
i -= 1
print(mn)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
lim = n // 2
ans = -1
cnt = 0
for i in range(lim, 0, -1):
if cnt >= 3:
break
if s[i] == "0":
continue
res = int(s[0:i]) + int(s[i:])
if ans == -1:
ans = res
else:
ans = min(ans, res)
cnt += 1
cnt = 0
for i in range(lim, n):
if cnt >= 3:
break
if s[i] == "0":
continue
res = int(s[0:i]) + int(s[i:])
if ans == -1:
ans = res
else:
ans = min(ans, res)
cnt += 1
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = str(input())
z = n // 2
l = z
p = "9" * 113110
p = int(p)
res = p
cnt = 0
for i in range(l, n):
if s[i] == "0":
continue
cnt += 1
s1 = s[:i]
s2 = s[i:]
s1 = int(s1)
s2 = int(s2)
p = min(p, s1 + s2)
if cnt >= 3:
break
cnt = 0
for i in range(l, 0, -1):
if s[i] == "0":
continue
cnt += 1
s1 = s[:i]
s2 = s[i:]
s1 = int(s1)
s2 = int(s2)
p = min(p, s1 + s2)
if cnt >= 3:
break
print(p)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP STRING NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n = int(input())
s = input()
j = (n + 1) // 2
i = j - 1
r = int(s)
while i > 0 and s[i] == "0":
i -= 1
while j < n and s[j] == "0":
j += 1
if i > 0:
w = int(s[0:i]) + int(s[i:])
r = min(r, w)
if j < n:
k = int(s[0:j]) + int(s[j:])
r = min(r, k)
print(r)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR NUMBER VAR VAR STRING VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def cal(s, inx, len):
l = int(s[0:inx])
r = int(s[inx:len])
return l + r
n = int(input())
s = input()
res = -1
i = n // 2
flag = 1
while flag != 0:
i += 1
if i > n - 1:
break
if s[i] == "0":
continue
flag -= 1
if res == -1:
res = cal(s, i, n)
else:
res = min(res, cal(s, i, n))
flag = 1
i = n // 2 + 1
while flag != 0:
i -= 1
if i <= 0:
break
if s[i] == "0":
continue
flag -= 1
if res == -1:
res = cal(s, i, n)
else:
res = min(res, cal(s, i, n))
print(res)
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
len = int(input())
s = input()
best = 0
best2 = 0
for i in range(1, len):
if s[i] != "0":
if max(i, len - i) < max(best, len - best):
best = i
elif max(i, len - i) == max(best, len - best):
best2 = i
ans = int(s[:best]) + int(s[best:])
if best2 != 0:
anss = int(s[:best2]) + int(s[best2:])
print(min(ans, anss))
else:
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING IF FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
import sys
def main():
l = input("")
n = input("")
min = -1
n = str(n)
i = 0
j = 1
while i < int(len(n) // 2):
if len(n) % 2 == 0:
if n[int(len(n) // 2) + i] != "0":
a = n[: int(len(n) // 2) + i]
b = n[int(len(n) // 2) + i :]
sum = int(a) + int(b)
if sum < min or min == -1:
min = sum
if n[int(len(n) // 2) - i] != "0":
a = n[: int(len(n) // 2) - i]
b = n[int(len(n) // 2) - i :]
sum = int(a) + int(b)
if sum < min or min == -1:
min = sum
i += 1
if min != -1:
break
else:
if n[int(len(n) // 2) + j] != "0":
a = n[: int(len(n) // 2) + j]
b = n[int(len(n) // 2) + j :]
sum = int(a) + int(b)
if sum < min or min == -1:
min = sum
if n[int(len(n) // 2) - i] != "0":
a = n[: int(len(n) // 2) - i]
b = n[int(len(n) // 2) - i :]
sum = int(a) + int(b)
if sum < min or min == -1:
min = sum
i += 1
j += 1
if min != -1:
break
print(min)
main()
|
IMPORT FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
def main():
q = int(input())
s = input()
ans = 10**1000000
for i in range(max(1, q // 2 - 1), q):
if s[i] != "0" or s[q - i] != "0":
if s[i] != "0":
ans = min(ans, int(s[:i]) + int(s[i:]))
if s[q - i] != "0":
ans = min(ans, int(s[: q - i]) + int(s[q - i :]))
if len(str(ans)) <= max(i, q - i) and i > q // 2:
break
print(ans)
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR IF VAR VAR STRING VAR BIN_OP VAR VAR STRING IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Dima worked all day and wrote down on a long paper strip his favorite number $n$ consisting of $l$ digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
-----Input-----
The first line contains a single integer $l$ ($2 \le l \le 100\,000$) — the length of the Dima's favorite number.
The second line contains the positive integer $n$ initially written on the strip: the Dima's favorite number.
The integer $n$ consists of exactly $l$ digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
-----Output-----
Print a single integer — the smallest number Dima can obtain.
-----Examples-----
Input
7
1234567
Output
1801
Input
3
101
Output
11
-----Note-----
In the first example Dima can split the number $1234567$ into integers $1234$ and $567$. Their sum is $1801$.
In the second example Dima can split the number $101$ into integers $10$ and $1$. Their sum is $11$. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
|
n, s = int(input()), input()
for i in range((n - 1) // 2, -1, -1):
if s[i + 1] == "0":
continue
best = int(s[: i + 1]) + int(s[i + 1 :])
break
for i in range(n // 2, n):
if s[i] == "0":
continue
best = min(best, int(s[:i]) + int(s[i:]))
break
print(best)
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
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