1^n+2^n+3^n+4^n will never give you summ [1..9]*(2*5)^3, and there will not be three zeros as well, thus this programm will work:



    vector <vector <int>> st(3);
        st[0].push_back(2); st[1].push_back(3); st[2].push_back(4);
        for (int i = 0; i <= 2; i++) {
            int osn = i + 2;
            for (;;) {
                int newst = (osn * *--st[i].end()) % 100;
                if (!count(st[i].begin(), st[i].end(), newst))
                    st[i].push_back(newst);
                else
                    break;
            }
        }
        int n;
        cin >> n;
        n--;
        int sum = 1 + st[0][n < 21 ? n : n  % 20 ] + st[1][n % 20] + st[2][n % 10];
        if (sum % 100 == 0)
            cout << 2;
        else if (sum % 10 == 0)
            cout << 1;
        else
            cout << 0;

Edited by author 17.10.2020 04:13