AtCoder Beginner Contest 210

A - Cabbages

int main() {
    IOS;
    ll n, a, x, y, ans = 0;
    cin >> n >> a >> x >> y;
    if (n > a)
        ans = a * x + (n - a) * y;
    else
        ans = n * x;
    cout << ans;
    return 0;
}

B - Bouzu Mekuri

int main() {
    IOS;
    cin >> n;
    string s; cin >> s;
    for (int i = 0; i < n; ++i)
        if (s[i] == '1')
            return cout << (i & 1 ? "Aoki" : "Takahashi"), 0;
    return 0;
}

C - Colorful Candies

尺取, set维护

int main() {
    IOS;
    cin >> n >> k;
    VI a(n);
    map<int, int> st;
    for (int i = 0, j = 0; i < n; ++i) {
        cin >> a[i];
        ++st[a[i]];
        if (i - j + 1 > k) {
            --st[a[j]];
            if (!st[a[j]])
                st.erase(a[j]);
            ++j;
        }
        umax(m, st.size());
    }
    cout << m;
    return 0;
}

D - National Railway

枚举终点, 用multiset维护已经枚举过的点(用来当起点)

int main() {
    IOS;
    ll ans = 2e17;
    cin >> n >> m >> k;
    vector<ll> a(m, 1e15);
    for (int i = 0; i < n; ++i) {
        multiset<ll> x, y;
        for (int j = 0; j < m; ++j) {
            a[j] += k;
            y.insert(a[j] + (ll)j * k);
        }
        for (int j = 0; j < m; ++j) {
            ll cur; cin >> cur;
            if (x.empty())
                umin(ans, cur + *y.begin() - (ll)j * k);
            else
                umin(ans, min(*y.begin() - (ll)j * k, *x.begin() - (ll)(m - j) * k) + cur);
            y.erase(y.find(a[j] + (ll)j * k));
            umin(a[j], cur);
            x.insert(a[j] + (ll)(m - j) * k);
        }
    }
    cout << ans;
    return 0;
}

E - Ring MST

排序, 把一条边用到极致,

以\(0, 1, 2, 3, ...\)为根, 找模\(a_i\)的群即可, 也就是点集

\({x, (k + a_i) \ mod c_i, (x + 2 \times a_i) \ mod c_i, ..., (x + (k - 1) \times a_i) \ mod c_i}\)

其中\((x + k \times c_i) \ mod = x\)

每次将剩下的点, 用边\(i\)分割成以\(0, 1, 2, 3,..\) 为首的群, 每个群以\(1,2,3,...\) 点代表

一直循环, 直到边用完,

最后剩下的代表点只剩下一个, 即可成功

int main() {
    IOS;
    ll n, m, ans = 0;
    cin >> n >> m;
    vector<pair<ll, ll>> p(m);
    for (auto &i : p)
    cin >> i.second >> i.first;
    sort(p.begin(), p.end());
 
    for (auto &i : p) {
        ll tmp = __gcd(n, i.second);
	ans += (n - tmp) * i.first;
	n = tmp;
    }
    cout << (n ^ 1 ? -1 : ans);
    return 0;
}