B. Split a Number
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
Dima worked all day and wrote down on a long paper strip his favorite number nn consisting of ll 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 ll (2≤l≤1000002≤l≤100000) — the length of the Dima's favorite number.
The second line contains the positive integer nn initially written on the strip: the Dima's favorite number.
The integer nn consists of exactly ll 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
Copy
7 1234567
output
Copy
1801
input
Copy
3 101
output
Copy
11
Note
In the first example Dima can split the number 12345671234567 into integers 12341234 and 567567. Their sum is 18011801.
In the second example Dima can split the number 101101 into integers 1010 and 11. Their sum is 1111. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
一下是思路：
思路很簡單，將字符串插入兩個斷點，分別從中間往左和往右尋找第一個符合題目要求的斷點位置，然後手搓一個大數加法
碼農題，思路簡單，實現複雜，寫了進160行代碼
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
char str[maxn], str1[maxn], str2[maxn], str3[maxn];
int str1_num, str2_num, str3_num, str_num;
int main()
{
int n;
scanf("%d", &n);
getchar();
for(int i = 1; i <= n; i++)
{
scanf("%c", &str[i]);
}
int cutl = n / 2, cutr = n / 2 + 1;
for(int i = cutl; i >= 0; i--)
{
if(str[i + 1] != '0')
{
cutl = i;
break;
}
}
for(int i = cutr; i <= n; i++)
{
if(str[i + 1] != '0')
{
cutr = i;
break;
}
}
for(int i = 1; i <= cutl; i++)
{
str1[str1_num++] = str[i];
}
for(int i = cutl + 1; i <= n; i++)
{
str2[str2_num++] = str[i];
}
for(int i = 1; i <= cutr; i++)
{
str3[str3_num++] = str[i];
}
for(int i = cutr + 1; i <= n; i++)
{
str[str_num++] = str[i];
str[str_num] = 0;
}
if(str1_num < str2_num)
{
int vim = str2_num - str1_num;
int num = str1_num;
for(int i = str2_num; i >= vim; i--)
{
str1[i] = str1[num--];
}
for(int i = 0; i < vim; i++)
{
str1[i] = '0';
}
str1_num = str2_num;
}
if(str_num < str3_num)
{
int vim = str3_num - str_num;
int num = str_num;
for(int i = str3_num; i >= vim; i--)
{
str[i] = str[num--];
}
for(int i = 0; i < vim; i++)
{
str[i] = '0';
}
str_num = str3_num;
}
//yun suan
int div = 0;
for(int i = str1_num - 1; i >= 0; i--)
{
int a = str1[i] - '0', b = str2[i] - '0';
str1[i] = (a + b + div) % 10 + '0';
if(a + b + div >= 10)
{
div = 1;
}
else
div = 0;
}
if(div == 1)
{
int num = str1_num - 1;
for(int i = str1_num; i >= 1; i--)
{
str1[i] = str1[num--];
}
str1[0] = '1';
str1_num++;
div = 0;
}
//printf("%s\n%s\n%s\n%s\n", str1, str2, str3, str);
for(int i = str_num - 1; i >= 0; i--)
{
int a = str[i] - '0', b = str3[i] - '0';
str[i] = (a + b + div) % 10 + '0';
if(a + b + div >= 10)
{
div = 1;
}
else
div = 0;
}
if(div == 1)
{
int num = str1_num - 1;
for(int i = str_num; i >= 1; i--)
{
str[i] = str[num--];
}
str[0] = '1';
str_num++;
div = 0;
}
if(str_num < str1_num)
{
printf("%s\n", str);
}
else if(str_num > str1_num)
{
printf("%s\n", str1);
}
else
{
int flag = 0;
for(int i = 0; i < str_num; i++)
{
if(str[i] > str1[i])
{
flag = 1;
break;
}
else if(str[i] < str1[i])
{
break;
}
}
if(flag)
{
printf("%s\n", str1);
}
else
printf("%s\n", str);
}
return 0;
}