Codeforces Round #310 (Div. 2) A. Case of the Zeros and Ones 水题

A. Case of the Zeros and Ones

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/556/problem/A

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

Sample Output

0

HINT

题意

10会消掉，然后问你最后剩下多少个数字

题解：

最后要么只剩下0，要么只剩下1，所以就是0的个数或者1的个数咯

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef unsigned long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[];
const int inf=0x3f3f3f3f;
inline ll read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}

//**************************************************************************************

int main()
{
    int n=read();
    string s;
    cin>>s;
    int ans=;
    for(int i=;i<n;i++)
    {
        if(s[i]=='')
            ans++;
        else
            ans--;
    }
    cout<<abs(ans)<<endl;
}