Description
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length n consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length n - 2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 2·105), the length of the string that Andreid has.
The second line contains the string of length n consisting only from zeros and ones.
Output
Output the minimum length of the string that may remain after applying the described operations several times.
Sample Input
4
1100
0
5
01010
1
8
11101111
6 题目大意：给出0和1两个数组成的序列，0和1相邻就可以配对，输出剩下不能配对的个数。 思路：配对问题，首先就可以用栈来实现，由于序列可以很大，所以用字符串来存储最好，先将第一个字符入栈， 如果它后面一个字符和它不同则将它出栈，否则就入栈；不过这个题目还可以用更简单的方法，因为0和1配对没先后 顺序，所以只要分别标记有多少个0和1，再将它们的个数相减再取绝对值，最后将这个绝对值输出即可。
#include <iostream>
#include <cstdio>
#include <cstring>
const int maxn=200000;
char s[maxn];
using namespace std;
int main()
{
int n,c,flag,kase;
while(scanf("%d",&n)==1&&n)
{
c=0;
flag=0,kase=0;
scanf("%s",s);
c=strlen(s);
for(int i=0;i<c;i++)
{
if(s[i]-'0'==1)
++flag;
if(s[i]-'0'==0)
++kase;
}
if(flag>=kase)
printf("%d\n",flag-kase);
else
printf("%d\n",kase-flag);
}
return 0;
}