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 adjacentpositions 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
Input
41100
Output
0
Input
501010
Output
1
Input
811101111
Output
6
Hint
In the first sample test it is possible to change the string like the following: 
.
In the second sample test it is possible to change the string like the following: 
.
In the third sample test it is possible to change the string like the following: 
.
题意：
0 和1相遇 就相互抵消了
然后让你求 序列的最小长度
设 有n个0 m个1；
ans=str.length()-(max(n,m)-min(n,m));
1 #include<iostream>
2 using namespace std;
3 int main(){
4 int n;
5 while(cin>>n&&n){
6 cin.get();
7 int s=0;
8 int s1=0;
9 for(int i=0;i<n;i++){
10 if(cin.get()==‘0‘)s++;
11 }
12 cin.get();
13 s1=n-s;
14 cout<<n-2*(s1>s?s:s1)<<endl;
15 }
16 return 0;
17 }