容易得出分3组，去考虑，增，删，替，补，的关系

#include <bits/stdc++.h>
using namespace std;
void solve() {
    string s;
    cin >> s;
    int n = s.length();
    bool has_da = false, has_xiao = false, has_shu = false;
    for (char c : s) {
        if (isupper(c)) has_da = true;  
        if (islower(c)) has_xiao = true;
        if (isdigit(c)) has_shu = true;
    }
    int miss = 3 - (has_da + has_xiao + has_shu); 
    vector<int> re;
    if (n > 0) {
        char now = s[0];
        int cnt = 1;
        for (int i = 1; i < n; i++) {
            if (s[i] == now) cnt++;
            else {
                if (cnt >= 3) re.push_back(cnt);
                now = s[i];
                cnt = 1;
            }
        }
        if (cnt >= 3) re.push_back(cnt);
    }
    int mi = 0;
    for (auto x : re) mi += x / 3;//替换次数
    //所以用max
    if (n < 6) {
        cout << max({6 - n, mi, miss}) << '\n';
    }
    else if (n <= 20) {
        cout << max(mi, miss) << '\n';
    } 
    else {
        int del = n - 20;//删除的机会,考虑删除对替换的影响
        vector<int> m0, m1, m2;
        for (int x : re) {
            int mod = x % 3;
            if (mod == 0) m0.push_back(x);
            else if (mod == 1) m1.push_back(x);
            else m2.push_back(x);
        }
        int k1 = min(del, (int)m0.size());
        mi -= k1;
        del -= k1;
        if (del > 0) {
            int k2 = min(del / 2, (int)m1.size());
            mi -= k2;
            del -= k2 * 2;
        }
        if (del > 0) {
            mi -= del / 3;
        }
        cout << (n - 20) + max(mi, miss) << '\n';
    }
}
signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t = 1;
    // cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}