Codeforce Round #539 div2

Codeforce Round #539 div2

abstract
I 离散化三连
sort(pos.begin(), pos.end());
pos.erase(unique(pos.begin(), pos.end()), pos.end());if (pos.size() == 0) pos.push_back(0);int id = lower_bound (pos.begin(), pos.end(), q[i].time) - pos.begin();
II java输入输出带模快速幂预处理组合数逆元
III clayley's fomualtion
IV treap ???
Vprecisition fix

D 回文串

题意

输入一个回文串，你现在有一个操作：将字符串切一刀或两刀，左右平移被切下来的部分。
问最少切几刀将其变为令一个回文串？若不可能，特判掉。
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题解

猜结论：最坏的情况是2。
证明（构造）： 令len为字符串S长度
首先考虑不可能的情况：每个字符都一样；或者除了中间那个字符，其它都一样（len为奇数）。
从这一点出发，我们可以构造出一种切两刀就必定得到不同回文串的方法：
从S的头出发，while(s[i]==s[i-1])i++; 找到第一个不同字符为止，切一刀得到一个prefix,对称地在尾部切一刀得到suffix，交换prefix,suffix 即可。
以上给出最坏的情况是2的构造方法，然后我们只要考虑是否有1的解。
此时问题等价于 “字符环 ”上任意起点是否有不同的回文串。O(N*N)可解。
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代码（very tricky)

//O(n)
#include <bits/stdc++.h>
using namespace std;

int solveOdd(const string& s) {//odd                                                                                
    return 2;
}

int solveEven(const string& s) {// ???貌似利用了回文环很强的性质
    if (s.length() % 2 == 1) {
        return 2;
    }
    string lf = s.substr(0, s.length() / 2);
    string rg = s.substr(s.length() / 2, s.length() / 2);
    if (lf != rg) return 1;
    return solveEven(lf);
}


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

    string s;
    cin >> s;
    if (s.length() < 3) {
        cout << "Impossible" << endl;
        return 0;
    }
    vector<int> cnt(255);//skip init, use stl func
    for(char c : s) {
        ++cnt[c];
    }
    if (*max_element(cnt.begin(), cnt.end()) >= s.length() - 1) {//only one impossibility
        cout << "Impossible" << endl;
        return 0;
    }
    cout << (s.length() % 2 == 0 ? solveEven(s) : solveOdd(s)) << endl;

    return 0;
}

O(N^2)
#include <bits/stdc++.h>
using namespace std;

bool isPalindrome(const string& s) {//pass by reference reduce time&&space costs
    for(int i = 0; i < s.length(); ++i) {
        if (s[i] != s[s.length() - i - 1]) 
            return false;
    }
    return true;
}

bool solve1(string s) {
    string t = s;
    for(int i = 0; i < s.length(); ++i) {
        t = t.back() + t;
        t.pop_back(); //more naturally without mod,and make use of string‘s ==
        if (s != t && isPalindrome(t)) {
            return true;
        }
    }
    return false;
}

bool anyAnswer(const string& s) {
    int nt = 0;
    for(int i = 0; i < s.length(); ++i) {
        nt += s[i] != s[0];//skip an if..else
    }
    return nt > 1;
}

int32_t main() {//function programing
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);//accelerate cin

    string s;
    cin >> s;
    if (anyAnswer(s)) {
        cout << (solve1(s) ? 1 : 2) << endl;
    } else {
        cout << "Impossible" << endl;
    }

    return 0;
}

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E treap or 线段树

题意

有一个容器，里面有水
维护两个操作：
1 t s (1≤t≤10^9, −10^9≤s≤10^9),增加一个事件：从第t秒开始往里倒水的速度变为s.
2 t (1≤t≤10^9),删除第t秒的事件

然后询问
3 l r v (1≤l≤r≤10^9, 0≤v≤10^9)表示假设容器里有体积为的v水，执行l到r的所有操作，什么时候容器里没有水，如果达不到特判掉。

操作和询问共有1e5个

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题解

用treap或线段树维护所有（没被删除的）事件
结点里存(t,s) t is key
每个结点同时维护

tL,tR:子树中最小和最大的时间（左右边界）。

speedR:最大时间（右边界）的速度。

res:执行子树里所有操作后消耗（得到）的水量

mn：执行子树所有操作过程中得到的水量最小值

合并操作：LC，RC为左右儿子

tL=LC.tL, tR=RC.tR, speedR=RC.speedR

mn 和res 一同更新：

mn = 0, res = 0  
mn = min(mn, LC.mn)  
res += LC.res + LC.speedR * (time - LC.tR)  
mn = min(mn, res)  
res += speed * (RC.tL - time)  
mn = min(mn, res + RC.mn)  
res += RC.res  
mn = min(mn, res)  

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代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
//mt19937 gen(time(NULL));
mt19937 gen(200);

struct node {
    pair<int, int> f, s;
    node *le, *ri;
    ll mn, res;
    int tl, tr;
    node() {
        mn = 228;
    }
    node(int l, int r) {
        tl = l; tr = r;
        mn = 228;
        if (l == r) return;
        le = new node(l, (l+r)/2);
        ri = new node((l+r)/2+1, r);
    }
    node (node* a, node* b) {
        le = a; ri = b;
        tl = le->tl; tr = ri->tr;
        if (le->mn == 228) {
            res = ri->res;
            mn = ri->mn;
            f = ri->f;
            s = ri->s;
            return;
        }
        if (ri->mn == 228) {
            res = le->res;
            mn = le->mn;
            f = le->f;
            s = le->s;
            return;
        }
        f = le->f; s = ri->s;
        ll del = 1ll*le->s.second*(ri->f.first-le->s.first);
        res = le->res+del+ri->res;
        mn = min(le->mn, le->res+del);
        mn = min(mn, ri->mn+le->res+del);
    }
    void combine() {
        node tmp(le, ri);
        *this = tmp;
    }
    void update(int id, int time, int speed) {
        if (tl == tr) {
            mn = res = 0;
            f.first = s.first = time;
            f.second = s.second = speed;
            return;
        }
        if (id <= (tl+tr)/2)
            le->update(id, time, speed);
        else
            ri->update(id, time, speed);
        combine();
    }
    void del(int id) {
        if (tl == tr) {
            mn = 228;
            return;
        }
        if (id <= (tl+tr)/2)
            le->del(id);
        else
            ri->del(id);
        combine();
    }
    node* get_seg(int l, int r) {
        if (tr < l || r < tl) return new node();
        if (l <= tl && tr <= r) {
            return this;;
        }
        return new node(le->get_seg(l, r), ri->get_seg(l, r));
    }
    long double simulate(int r, ll v) {
        if (mn == 228) return -1;
        if (v+mn > 0 && v+res+1ll*s.second*(r-s.first) > 0) return -1;
        if (f == s) return s.first-(long double)v/s.second;
        if (le->mn == 228) return ri->simulate(r, v);
        long double to = le->simulate(le->s.first, v);
        if (to != -1) return to;
        v += le->res;
        ll del = 1ll*le->s.second*((ri->mn==228?r:ri->f.first)-le->s.first);
        if (v+del <= 0) return le->s.first-(long double)v/le->s.second;
        v += del;
        return ri->simulate(r, v);
    }
    long double query(int l, int r, int rr, ll v) {
        node* t = get_seg(l, r);
        return t->simulate(rr, v);
    }
};

struct query{
    int time, speed, start;
    int l, r, type;
    query() {}
};

vector <int> pos;
vector <query> q;

int main() {
    ios_base::sync_with_stdio(false);
    #ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif // LOCAL
    cout.precision(7);
    int qq; cin >> qq;
    q.resize(qq);
    for (int i = 0; i < qq; ++i) {
        cin >> q[i].type;
        if (q[i].type == 1) {
            cin >> q[i].time >> q[i].speed;
            pos.push_back(q[i].time);
        }
        if (q[i].type == 2) {
            cin >> q[i].time;
        }
        if (q[i].type == 3) {
            cin >> q[i].l >> q[i].r >> q[i].start;
        }
    }
    sort(pos.begin(), pos.end());
    pos.erase(unique(pos.begin(), pos.end()), pos.end());
    if (pos.size() == 0) pos.push_back(0);
    node* t = new node(0, pos.size()-1);
    for (int i = 0; i < qq; ++i) {
        if (q[i].type == 1) {
            int id = lower_bound(pos.begin(), pos.end(), q[i].time)-pos.begin();
            t->update(id, q[i].time, q[i].speed);
        }
        if (q[i].type == 2) {
            int id = lower_bound(pos.begin(), pos.end(), q[i].time)-pos.begin();
            t->del(id);
        }
        if (q[i].type == 3) {
            int l = lower_bound(pos.begin(), pos.end(), q[i].l)-pos.begin();
            int r = --upper_bound(pos.begin(), pos.end(), q[i].r)-pos.begin();
            if (q[i].start == 0)
                cout << fixed << q[i].l << '\n';
            else
                cout << fixed << t->query(l, r, q[i].r, q[i].start) << '\n';
        }
    }
    return 0;
}

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F 隔板法+Cayley's formula

题意

给你n,m,a,b 表示n个结点的树，结点1..n标号，最大边权为m。a,b为被选定的两个结点问有几棵a,b之间距离为m的树？

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题解

首先考虑a,b这条路径的数量，令edges为a,b之间的边数，边权的方案数可以由隔板法得出：选edges条边，使得它们和为m，方法数为C(m-1,edges-1). 结点的方案数由n-2个点选edges-1个的排列得出A(n-2,edges-1).

接下来考虑剩下的结点的方案数，首先，带权边的组合数是m^(n-edges-1)，
然后对于点，我们有一个定理（证明貌似是矩阵之类的）：

设T(n，k)是n个结点上带有k个连通分量的标号（结点从1..n标号）森林的数量,
有cayley's 公式: T(n,k) = k *n^(n − k − 1).

综上所述，如果我们把a,b这条链拆开，就得到了edges+1个结点，每个结点作为一个联通分量，那么有T(n,edges+1)种森林，然后把链重新连起来，C(m-1,edges-1)*A(n-2,edges-1) 种方案，最后每个联通分量内部的边赋权值的方案数为m^(n-edges-1)，
所以答案为num[edges]=T(n,edges+1)*C(m-1,edges-1)*A(n-2,edges-1)*m^(n-edges-1).最终对num求和即得答案rep(i,1,n-1)ans+=num[edges];

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代码

import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        OutputWriter out = new OutputWriter(System.out);
        TaskC solver = new TaskC(in, out);
        solver.solve();
        out.close();
    }

    static class TaskC {

        static int n;
        static int m;

        static int[] inv;
        static int[] fact;
        static int[] invfact;
        
        static int[] powN;
        static int[] powM;

        InputReader in;
        OutputWriter out;

        TaskC(InputReader in, OutputWriter out) {
            this.in = in;
            this.out = out;
        }
        
        private void solve() {
            n = in.readInt();
            m = in.readInt();
            precalc();
            int ans = 0;
            int choosePath = 1;
            for(int k = 1; k < n; ++k) {
                if (k > m) break;
                int cur = MathUtil.mul(choosePath, cayley(n, k + 1));
                cur = MathUtil.mul(cur, binom(m - 1, k - 1));
                cur = MathUtil.mul(cur, powM[n - k - 1]);
                choosePath = MathUtil.mul(choosePath, n - k - 1);
                ans = MathUtil.sum(ans, cur);
            }
            out.print(ans);
        }

        private int cayley(int n, int k) {
            if (n - k - 1 < 0) {
                return MathUtil.mul(k, MathUtil.binPow(n, MathUtil.mod - 2));
            }
            return MathUtil.mul(k, powN[n - k - 1]);
        }

        private int binom(int n, int k) {
            return MathUtil.mul(fact[n], MathUtil.mul(invfact[k], invfact[n - k]));
        }

        private void precalc() {
            powN = new int[n + 1];
            powM = new int[n + 1];
            powN[0] = 1;
            powM[0] = 1;
            for(int i = 1; i <= n; ++i) {
                powN[i] = MathUtil.mul(powN[i - 1], n);
                powM[i] = MathUtil.mul(powM[i - 1], m);
            }
            
            inv = new int[m + 1];
            fact = new int[m + 1];
            invfact = new int[m + 1];
            inv[0] = inv[1] = 1;
            fact[0] = fact[1] = 1;
            invfact[0] = invfact[1] = 1;
            for(int i = 2; i <= m; ++i) {
                inv[i] = MathUtil.mul(inv[MathUtil.mod % i], MathUtil.mod - MathUtil.mod / i);
                fact[i] = MathUtil.mul(fact[i - 1], i);
                invfact[i] = MathUtil.mul(invfact[i - 1], inv[i]);
            }
        }
    }
    static class MathUtil {
        private static final int mod = 1_000_000_007;

        static int binPow(int a, int n) {
            int result = 1;
            while (n > 0) {
                if (n % 2 == 1) {
                    result = mul(result, a);
                }
                n /= 2;
                a = mul(a, a);
            }
            return result;
        }
        
        static int mul(int a, int b) {
            return (int)((long)a * b % mod);
        }

        static int sum(int a, int b) {
            int result = a + b;
            if (result >= mod) result -= mod;
            return result;
        }

    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void print(Object... objects) {
            for (int i = 0; i < objects.length; i++) {
                if (i != 0) {
                    writer.print(' ');
                }
                writer.print(objects[i]);
            }
            writer.println();
        }

        public void close() {
            writer.close();
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private InputReader.SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public int readInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public boolean isSpaceChar(int c) {
            if (filter != null) {
                return filter.isSpaceChar(c);
            }
            return isWhitespace(c);
        }

        public static boolean isWhitespace(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public double readDouble() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            double res = 0;
            while (!isSpaceChar(c) && c != '.') {
                if (c == 'e' || c == 'E') {
                    return res * Math.pow(10, readInt());
                }
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            }
            if (c == '.') {
                c = read();
                double m = 1;
                while (!isSpaceChar(c)) {
                    if (c == 'e' || c == 'E') {
                        return res * Math.pow(10, readInt());
                    }
                    if (c < '0' || c > '9') {
                        throw new InputMismatchException();
                    }
                    m /= 10;
                    res += (c - '0') * m;
                    c = read();
                }
            }
            return res * sgn;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);
        }

    }
}

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