kick start 2018 H

Recently, I took the time to start brushing algorithm questions, and started preparing for the next test.

/**
 * 题目：https://codejam.withgoogle.com/codejam/contest/3324486/dashboard
 * 将字符串送进trie树，然后统计前缀，注意短的前缀确定的集合肯定包含了那些长的前缀的情况。
 * 除了前缀，后面的位置可以随意取。
 * O(p*n)
 */
#include <bits/stdc++.h>
using namespace std;
const int MAXSIGMA = 2;
struct Node{
    Node *p[MAXSIGMA];
    int v;
};
map<char, int> mp{
    {'B', 0},
    {'R', 1}
};

Node* newNode(){
    Node *res = new Node();
    res->v = 0;
    for(int i = 0; i < MAXSIGMA; i++){
        res->p[i] = nullptr;
    }
    return res;
};

void insert(Node *rt, string s){
    int n = s.length();
    for(int i = 0; i < n; i++){
        int c = mp[s[i]];       
        if(rt->p[c] == nullptr){
            rt->p[c] = newNode();
        }
        rt = rt->p[c];
    }
    rt->v = n;
}
map<int, int> cnt;//用来计数

int solve(Node *rt){
    for(int i = 0; i < MAXSIGMA; i++){
        if(rt->p[i]){
            if(rt->p[i]->v != 0){
                cnt[rt->p[i]->v]++;
            }else{
                solve(rt->p[i]);
            }
        }
    }
}

void mydel(Node *rt){
    for(int i = 0; i < MAXSIGMA; i++){
        if(rt->p[i]){
            mydel(rt->p[i]);
        }
    }
    delete rt;
}


int main(){
    //freopen("small.in", "r", stdin);
    //freopen("out.txt", "w", stdout);
    fstream in, out;
    in.open("large.in", ios::in);
    out.open("out.txt", ios::out);
    int t;
    in>>t;
    int n, m;
    for(int k = 1; k <= t; k++){
        in>>n>>m;
        cnt.clear();
        Node *trie = newNode();
        string s;
        for(int i = 0; i < m; i++){
            in>>s;
            auto rt = trie;
            insert(rt, s);
        }
        solve(trie);
        long long res = 0;
        for(auto c : cnt){
            res += (1LL << (n - c.first)) * c.second;
        }
        out<<"Case #"<<k<<": "<<(1LL << n) - res<<endl;
        mydel(trie);

    }
}

/**
 * 题目：https://codejam.withgoogle.com/codejam/contest/3324486/dashboard#s=p1
 * 墙只能从两边倒，而我们需要选连续的一半的墙使得value最大即可，
 * 其实我们只要选了第一块x，使得x两侧的墙中，不打算涂的数量不少于要涂的数量即可,这样我们总能保证在墙倒之前给它涂色
 * 于是转化为求最大连续子串和。
 * O(n)
 */
#include <bits/stdc++.h>
using namespace std;

int main(){
    fstream in, out;
    in.open("large.in", ios::in);
    out.open("out.txt", ios::out);
    int t;
    in>>t;
    int n;
    string s;
    for(int k = 1; k <= t; k++){
        in>>n;
        in>>s;
        int res = 0;
        int m = (n + 1) / 2;
        int j = 0;
        for(; j < m; j++){
            res += s[j] - '0';
        }
        int ans = res;
        for(; j < n; j++){
            res = res - s[j-m]  + s[j];
            ans = max(ans, res);
        }
        out<<"Case #"<<k<<": "<<ans<<endl;
    }
}

/**
 * 题目：https://codejam.withgoogle.com/codejam/contest/3324486/dashboard#s=p2
 * 容斥原理
 * res = 全部的排列 - 1对夫妻相邻 + 两对夫妻相邻 - 三对夫妻相邻...
 * 注意求组合数的时候有除法要求逆元。
 * O(n + m)
 */
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MOD = 1e9+7;
const int MAXN = 200010;
LL a[MAXN], b[MAXN]; //阶乘、2的幂

LL quickpow(LL a, LL b){
    LL res = 1, temp = a % MOD;
    while(b){
        if(b & 1) res = res * temp % MOD;
        b >>= 1;
        temp = temp * temp % MOD;
    }
    return res;
}

int inv(int x){
    return quickpow(x, MOD-2);
}

int main(){
    fstream cin, cout;
    cin.open("large.in", ios::in);
    cout.open("out.txt", ios::out);
    int t;
    a[0] = 1;
    b[0] = 1;
    for(int i = 1; i < MAXN; i++){
        a[i] = a[i - 1] * i % MOD;
        b[i] = b[i - 1] * 2 % MOD;
    }
    cin>>t;
    int n, m;
    for(int k = 1; k <= t; k++){
        cin>>n>>m;
        int nn = n * 2;
        int res = 0;
        LL c = m;
        for(int i = 1; i <= m; i++){
            if(i&1){
                res = (res + a[nn-i] * b[i] % MOD * c % MOD) % MOD;
            }else{
                res = (res - a[nn-i] * b[i] % MOD * c % MOD + MOD) % MOD;
            }
            c = c * (m - i) % MOD * inv(i + 1) % MOD;
            //c = c * (m - i) / (i + 1);
        }
        cout<<"Case #"<<k<<": "<<(a[nn] - res + MOD) % MOD<<endl;
    }
}