Codeforces Round #721 (Div. 2)

Codeforces Round #721 (Div. 2)

A - And Then There Were K

打表找规律

int main() {
    IOS; VI a(1);
    while (a.back() < 1e9) {
        ll cur = a.back() * 2 + 1;
        if (cur >= 1e9) break;
        a.pb(cur);
    }
    for (cin >> _; _; --_) {
        cin >> n;
        cout << *(lower_bound(all(a), n) -1) << '\n';
    }
    return 0;
}

B - Palindrome Game (easy version)

回文串,

1 0个0, 平局
2 偶数个0, 先手必填一个位置, 后手填对称位置, 先手又变成回文(偶数) ..., 剩最后两个, 后手反转, 先手多填2个
3 1个0, 后手胜
4 奇数个0, 先手下中间, 变成情况2, 后手多下一个

不是回文,

• 串长为奇数, 中间和某个位置为0, 不论怎么下平局
• 否则, 先手填个对称位置, 后手紧接填对称位置, 串还是非回文, 且双方消耗相同, 情况无变化, 所以对称0, 对答案无贡献(就算是中间为0,贡献也是-1), 但先手可以让后手把所有非对称全下(贡献大于1), 故先手胜

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> s + 1;
        int x = 0, f = 0;
        rep (i, 1, n >> 1) f += s[i] != s[n + 1 - i];
        rep (i, 1, n) x += s[i] == '0';
        if (f == 0) {
            if (x == 0)  cout << "DRAW\n";
            else if (x % 2 == 0 || x == 1) cout << "BOB\n";
            else cout << "ALICE\n";
        }
        else if (f == 1 && x == 2) cout << "DRAW\n";
        else cout << "ALICE\n";
    }
    return 0;
}

C - Sequence Pair Weight

对于数字 1, 出现位置为

\(a_1, a_2, a_3, ..., a_k\)

贡献为 \(a_{k - 1} \times (n + 1 - a_k) + a_{k - 2} \times (n + 1 - a_k) + a_{k - 2} \times (n + 1 - a_{k - 1} ...)\)

倒着维护\((n + 1 - a_k) + (n + 1 - a_{k - 1})...\) 即可

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n; ll ans = 0; VI x;
        rep (i, 1, n) cin >> a[i], x.pb(a[i]), c[i] = 0;
        sort(all(x)); x.erase(unique(all(x)), x.end());
        per (i, n, 1) {
            a[i] = lower_bound(all(x), a[i]) - x.begin() + 1;
            ans += (ll)i * c[a[i]];
            c[a[i]] += n + 1 - i;
        }
        cout << ans << '\n';
    }
    return 0;
}