题目描述：
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 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≤l≤100000) — 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.
思路：
题目是要把一个大数拆成两个小数，并且让他们加起来的和最小。
根据贪心的思想，可以想到，如果把数从中间分开，从而让两个数的位数差不多时，两者的和应该是比较小的。但是题目中有要求，拆成的两个数不能有前导零。咋办。
就把所有不是零 的元素的索引提出来加到一个vector里面，因为要取尽量中间的数，就用一下lower_bound二分找原数组长度一半的位置mid，能找到vector里接近mid的元素。
然后用这个元素所表示的索引，在原来的数组中索引附近遍历，取到不同的分割位置，选出分割后的和最小者。
但是数好像很大哎，只有用字符串来模拟了，传说中的大数加法。
代码：
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int n;//接收数字位数
string s;//接收数字
vector<int> vec;//来存值不为零的元素的索引
string rm0(string a)//删除前导零
{
int i;
for(i = ;i<a.size();i++)
{
if(a[i]!='')
{
break;
}
}
a = a.substr(i);
return a;
}
string add(string a,string b)//大数加法
{
if(a.size()<b.size()) //先把两者位数补成一样的
{
swap(a,b);
}
a = ''+a;
while(b.size()<a.size())
{
b = ''+b;
}
int c = ;
for(int i = b.size()-;i>=;i--) //从低位到高位加
{
int sum = c+(int)a[i]-''+(int)b[i]-'';
if(sum>=)
{
a[i] = (char)sum-+'';
c = ;
}
else
{
a[i] = (char)sum+'';
c = ;
}
}
a = rm0(a);
return a;
}
int larger(string a,string b)//判断a是不是大于等于b
{
if(a.size()!=b.size())
{
return a.size()>b.size();
}
else
{
for(int i = ;i<a.size();i++)
{
if(a[i]!=b[i])
{
return a[i]>b[i];
}
}
}
return ;
}
int main()
{
cin >> n >> s;
string a = "";
string b = "";
for(int i = ;i<n;i++)
{
if(s[i]!='')
{
vec.push_back(i);
}
}
int mid = lower_bound(vec.begin(),vec.end(),n/)-vec.begin();//找不为零的接近中间位置的数在原数组的索引
string ans = s;
for(int i = -;i<;i++)//在索引附近遍历一遍
{
int pos = max(,min((int)vec.size()-,mid+i));
a = s.substr(,vec[pos]);
b = s.substr(vec[pos]);
string res = add(a,b);
if(larger(ans,res))
{
ans = res;//取最小值
}
}
cout << ans << endl;
return ;
}
参考文章：
我写的可能不清楚，如果还有问题，参见推荐博客：
茄子Min，Codeforces Round #567 (Div. 2)B. Split a Number (字符串，贪心)，https://www.cnblogs.com/qieqiemin/p/11033007.html
另外本题涉及到一些实用的字符串处理函数，具体可见：
GGBeng，C++中substr函数的用法，https://www.cnblogs.com/xzxl/p/7243490.html