Educational Codeforces Round 30 B

Balanced Substring

题意：给一个01串，找出一个连续最长的子串要求0和1的个数相同，求最长长度

思路：0当作-1， 1当作1，求前缀和，前缀和相同的2个区间，重叠的区间的和为0，即0和1的个数相同

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("
")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e5+100;x

int n,ans,a[N],sum[N];
map<int,int> f;
char s[N];

int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>s;
    for(int i=0; i<n; ++i){
        if(s[i]=='0') a[i+1]=-1;
        else a[i+1]=1;
    }
    f[0]=0;
    for(int i=1; i<=n; ++i){
        sum[i]=sum[i-1]+a[i];
        if(sum[i]==0) ans=max(ans, i);
        if(f[sum[i]]==0) f[sum[i]]=i;
        else{
            ans=max(ans, i-f[sum[i]]);
        }
    }
    cout<<ans<<endl;
    return 0;
}