树状数组永远的概念神

#include <bits/stdc++.h>
#define int long long
#define pow_of_two(n) (n & -n == n) 
using namespace std;
typedef pair<int , int> PII;
const int N = 5e6 + 10;
int n;
int c[N];

int lowbit(int x)
{
    return x & -x;
}

void add(int x , int k)
{
    while (x <= n)
    {
        c[x] += k;
        x += lowbit(x);
    }
}

int query(int x)
{
    int res = 0;
    while (x)
    {
        res += c[x];
        x -= lowbit(x);
    }

    return res;
}

inline void solve()
{
    int t;
    cin >> t;
    int sb = 1;
    while (t -- )
    {
        cin >> n;
        string s;
        cin >> s;
        
        for (int i = 0 ; i < s.size() ; i ++  ) add(i + 1 , s[i] - '0');
        int ans = 0 , temp;
        if (n & 1) temp = n / 2 + 1;
        else temp = n / 2;

        // cout << temp << endl;

        for (int i = 1 ; i <= temp + 1 ; i ++ )
            ans = max(ans , query(i + temp - 1) - query(i - 1));
        // cout << endl;
        printf("Case #%lld: %lld\n" , sb ++ , ans);
        memset(c , 0 , sizeof c);
        // cout << ans << endl;
    }

}


signed main()
{
    ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0);
    solve();
    return 0;
}