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Codeforces Round 891 (Div. 3)

发布时间 2023-08-09 10:35:46作者: PHarr

A. Array Coloring

#include <bits/stdc++.h>

using namespace std;

void solve() {
    int n;
    cin >> n;
    int sum = 0;
    for( int i = 1 , x ; i <= n ; i ++ )
        cin >> x , sum += x;
    if( sum % 2 == 0 ) cout << "YES\n";
    else cout << "NO\n";
    return ;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int T;
    cin >> T;

    while (T--)
        solve();
    return 0;
}

B. Maximum Rounding

模拟一下进位就好了

#include <bits/stdc++.h>

using namespace std;

void solve() {
    string s;
    cin >> s;
    reverse(s.begin(), s.end());
    vector<int> a;
    for( auto i : s )
        a.push_back( i - '0' );
    for( int i = 1 ; i <= 5 ; i ++ ) a.push_back(0);
    int p = -1;
    for( int i = 0 ; i+1 < a.size() ; i ++ ){
        a[i+1] += a[i] / 10 , a[i] %= 10;
        if( a[i] > 4 ) a[i+1] ++ , p = i;
    }
    for( int i = 0 ; i <= p ; i ++ ) a[i] = 0;
    while( a.back() == 0 ) a.pop_back();
    reverse(a.begin(), a.end());
    for( auto i : a )
        cout << i ;
    cout << "\n";
    return ;

}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int T;
    cin >> T;

    while (T--)
        solve();
    return 0;
}

C. Assembly via Minimums

其实这里面数字的顺序是没有影响的。所以规定原始数组为递增的即可。然后比第一个数组大的数字有\(n-1\)个比，第二个数字大的有\(n-2\)个以此类推，所以比第\(i\)个数字大的有\(n-i\)个。因为读入的数字是合法的，所以我们把读入数组排序，然后每次跳\(n-i\)即可

#include <bits/stdc++.h>

using namespace std;

void solve() {
    int n , m;
    cin >> n , m = n * ( n - 1 ) / 2 ;
    vector<int> a(m);
    for( auto & i : a ) cin >> i;
    sort( a.begin(), a.end() );
    for( int i = 1 , j = 0 ; i < n ; i ++ ){
        cout << a[j] << " ";
        j += n - i;
    }
    cout << a.back() << "\n";
    return ;

}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int T;
    cin >> T;

    while (T--)
        solve();
    return 0;
}

D. Strong Vertices

一个非常经典的套路

\[a_u-a_v\ge b_u-b_v\Rightarrow a_u-b_u\ge a_v-b_v \]

所以我们把差值排序，看有多少个和最大值相同即可。

#include <bits/stdc++.h>

using namespace std;

#define int long long

void solve() {
    int n, m;
    cin >> n;
    vector<pair<int,int>> a(n);
    for( int i = 0 ; i < n ; i ++ )
        cin >> a[i].first , a[i].second = i+1;
    for( int i = 0 , x ; i < n ; i ++ )
        cin >> x , a[i].first -= x;
    sort( a.begin(), a.end());
    vector<int> res;
    for( int i = n - 1 ; i >= 0 ; i -- ){
        if( a[i].first == a.back().first ) res.push_back(a[i].second);
        else break;
    }
    sort( res.begin(), res.end() );
    cout << res.size() << "\n";
    for( auto i : res )
        cout << i << " ";
    cout << "\n";
    return;

}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int T;
    cin >> T;

    while (T--)
        solve();
    return 0;
}

E. Power of Points

这题要离线做，把区间读进来后，按照\(x_i\)排序，然后就可用前缀后缀和来做这题，具体的做法其实就是提共因式。

#include <bits/stdc++.h>

using namespace std;

#define int long long

void solve() {
    int n, m;
    cin >> n;
    vector<pair<int, int>> a(n + 1);

    int x = 0, y = 0;

    for (int i = 1; i <= n; i++)
        cin >> a[i].first, a[i].second = i, y += a[i].first;
    vector<int> res(n + 1);
    sort(a.begin(), a.end());
    for (int i = 1, cnt; i <= n; i++) {
        res[a[i].second] += y - (n - i + 1) * (a[i].first - 1);
        res[a[i].second] += (a[i].first + 1) * (i - 1) - x;
        y -= a[i].first, x += a[i].first;
    }
    for( int i = 1 ; i <= n ; i ++ )
        cout << res[i] << " ";
    cout << "\n";
    return;

}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int T;
    cin >> T;

    while (T--)
        solve();
    return 0;
}

F. Sum and Product

这题还是很好想的，设\(a=a_i,b=a_j\)

\[a+b=x,ab=y\\ a = x-b\\ (x-b)b=y\Rightarrow b^2-xb+y=0 \]

解出一元二次方程，然后计算出\(a,b\)的值，就可以用组合数来统计答案。

#include <bits/stdc++.h>

using namespace std;

#define int long long

void solve() {
    int n, m;
    cin >> n;
    map<int, int> cnt;
    for (int i = 1, x; i <= n; i++)
        cin >> x, cnt[x]++;
    int q;
    cin >> q;
    for (int x, y, t, a, b; q; q--) {
        cin >> x >> y;
        t = x * x - 4 * y;
        if (t < 0) {
            cout << "0 ";
            continue;
        }
        a = (x + sqrt(t)) / 2, b = x - a;
        if (a * b != y) {
            cout << "0 ";
            continue;
        }
        if (a == b) cout << cnt[a] * (cnt[a] - 1ll) / 2ll << " ";
        else cout << cnt[a] * cnt[b] << " ";
    }
    cout << "\n";
    return;

}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int T;
    cin >> T;

    while (T--)
        solve();
    return 0;
}

G. Counting Graphs

这道题我赛时没有出，还是比较惭愧的。

考虑Kruskal的过程，如果对于当前加了一条边\(u,v,w\)，那么在\(u\)所在的联通快中任意一个点和\(v\)所在的联通块中任意一个点加一条权值大于\(w\)的边都是无意义的。

所以我们可以用并查集模拟 Kruskal的过程，对于当前边\((u,v,w)\)对答案产生的贡献是\((S-w+1)^{size_u\times size_v - 1}\)

把贡献累积就是答案。

#include <bits/stdc++.h>

using namespace std;

#define int long long

class dsu {
private:
    vector<int> fa;
public:
    dsu(int n = 1) {
        fa = vector<int>(n + 1, -1), fa[0] = 0;
    }

    int getfa(int x) {
        if (fa[x] < 0) return x;
        return fa[x] = getfa(fa[x]);
    }

    void merge(int x, int y) {
        x = getfa(x), y = getfa(y);
        if (x == y) return;
        if (fa[x] > fa[y]) swap(x, y);
        fa[x] += fa[y], fa[y] = x;
    }

    bool same(int x, int y) {
        x = getfa(x), y = getfa(y);
        return (x == y);
    }

    int size(int x) {
        x = getfa(x);
        return -fa[x];
    }
};

template<class T>
constexpr T power(T a, int b) {
    T res = 1;
    for (; b; b /= 2, a *= a)
        if (b % 2) res *= a;
    return res;
}

template<int P>
struct MInt {
    int x;
    static int Mod;

    constexpr MInt() : x(0) {};

    constexpr MInt(int x) : x(norm(x % getMod())) {};

    constexpr static void setMod(int Mod_) {
        Mod = Mod_;
    }

    constexpr static int getMod() {
        if (P > 0) return P;
        else return Mod;
    }

    constexpr int norm(int x) const {
        if (x < 0) x += getMod();
        if (x >= getMod()) x -= getMod();
        return x;
    }

    constexpr int val() const {
        return x;
    }

    explicit constexpr operator int() const {
        // 隐式类型转换把 MInt转换成 int
        return x;
    }

    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }

    constexpr MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }

    constexpr MInt &operator*=(MInt rhs) &{
        x = x * rhs.x % getMod();
        return *this;
    }

    constexpr MInt &operator+=(MInt rhs) &{
        x = norm(x + rhs.x);
        return *this;
    }

    constexpr MInt &operator-=(MInt rhs) &{
        x = norm(x - rhs.x);
        return *this;
    }

    constexpr MInt &operator/=(MInt rhs) &{
        return *this *= rhs.inv();
    }

    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }

    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }

    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }

    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }

    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }

    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        int v;
        is >> v;
        a = MInt(v);
        return is;
    }

    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }

    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
};

using Z = MInt<998244353>;


void solve() {
    int n , s ;
    cin >> n >> s;
    vector<tuple<int, int, int>> e(n - 1);
    for (auto &[w, u, v]: e)
        cin >> u >> v >> w;
    sort(e.begin(), e.end());
    dsu d(n);
    Z res(1);
    for( auto [w,u,v] : e ){
        int cnt = d.size(u) * d.size(v) - 1;
        res *= power( Z( s - w + 1 ) , cnt );
        d.merge( u , v );
    }
    cout << res << "\n";
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int t;
    cin >> t;
    for (; t; t--)
        solve();

    return 0;
}