Atcoder Regular Contest 148

By xiaruize

A - mod M

很明显，答案仅有可能为1/2，因为当 M=2M=2 时，任何序列的结果均不可能超过 22

那么，可以考虑什么情况下答案为11

那么，显然我们需要找一个数，使得

{A1≡A2(mod x)A2≡A3(mod x)⋯An−1≡An(mod x)\begin{cases} A_1 \equiv A_2 (mod\ x)\\ A_2 \equiv A_3 (mod\ x)\\ \cdots \\ A_{n-1} \equiv A_n (mod\ x) \end{cases}

显然，当 x=gcd(∣A1−A2∣,∣A2−A3∣,⋯ ,∣An−1−An∣)x=gcd(|A_1-A_2|,|A_2-A_3|,\cdots , |A_{n-1}-A_n|) 时，上述条件成立，且不存在一个更大的 xx 使得条件成立

此时，若 x=1x=1 ，则答案为 22， 否则答案为 11

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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n;
int g;
int a[N];
// bool en;

int gcd(int x, int y)
{
    if (x == 0)
        return y;
    return gcd(y % x, x);
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n;
    cin >> a[1];
    for (int i = 2; i <= n; i++)
    {
        cin >> a[i];
        if (i != 2)
            g = gcd(g, abs(a[i] - a[i - 1]));
        else
            g = abs(a[i] - a[i - 1]);
    }
    if (g != 1)
    {
        cout << "1" << endl;
    }
    else
        cout << "2" << endl;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

B - dp

题目标题很诈骗qwq

暴力枚举是 O(n3)O(n^3) 的，会获得 TLE\color{orange}{TLE} 的好成绩

那么，我们考虑这三维中哪一个是可以不需要的

可以发现，被操作序列的起点是一定的，即第一个 pp . 所以，只需要　O(n2)O(n^2) 的时间即可

Code
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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n;
string s;
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n;
    cin >> s;
    string res = s;
    for (int i = 0; i < n; i++)
    {
        if (s[i] == 'p')
        {
            for (int j = i; j < n; j++)
            {
                string tmp = s;
                for (int k = i; k <= j; k++)
                {
                    tmp[k] = (s[j - k + i] == 'd' ? 'p' : 'd');
                }
                // cerr << tmp << endl;
                if (tmp < res)
                    res = tmp;
            }
            break;
        }
    }
    cout << res << endl;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

C - Lights Out on Tree

观察样例的第2个query的操作过程，我们发现，对于任意一个连通的部分，答案为 与当前连通块相连的节点个数

实现时需要注意复杂度与常数，遍历每个节点的儿子会导致超时，使用memset也有可能导致TLE

Code
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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, q;
int p[N];
int m;
int v[N];
int deg[N];
bool st[N];
int cnt[N];
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    cin >> n >> q;
    for (int i = 2; i <= n; i++)
    {
        cin >> p[i];
        deg[p[i]]++;
    }
    while (q--)
    {
        cin >> m;
        int res = 0;
        for (int i = 1; i <= m; i++)
        {
            cin >> v[i];
            st[v[i]] = true;
            cnt[p[v[i]]]++;
        }
        for (int i = 1; i <= m; i++)
        {
            res += deg[v[i]] - cnt[v[i]];
            if (!st[p[v[i]]])
                res++;
        }
        for (int i = 1; i <= m; i++)
        {
            st[v[i]] = false;
            cnt[p[v[i]]] = 0;
        }
        cout << res << endl;
    }
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

D - mod M Game

考场第一想法是，每次Alice取什么，Bob就取什么，才能保证Bob，否则Alice

这个想法如果没有 MM 是正确的，那么什么情况下会非法呢？

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2 8
1 3 5 7
1
Bob

那么，为什么会出现这种情况呢？

显然是因为原数组在模 m2\frac{m}{2} 后可以一一对应(m和n均为偶数)

所以可以特判该情况，并 AC\color{green}{AC} 本题

Code
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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 4e5 + 10;

int n, m;
int a[N];
map<int, int> mp;
vector<int> x, y;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n >> m;
    for (int i = 1; i <= n + n; i++)
    {
        cin >> a[i];
        mp[a[i]]++;
    }
    bool flag = true;
    for (auto v : mp)
    {
        if (v.sec % 2 == 1)
        {
            flag = false;
            if (m % 2 == 1)
            {
                cout << "Alice" << endl;
                return 0;
            }
            else
            {
                if (v.fir >= m / 2)
                    y.pb(v.fir - m / 2);
                else
                    x.pb(v.fir);
            }
        }
    }
    if (flag || (x == y && x.size() % 2 == 0))
        cout << "Bob" << endl;
    else
        cout << "Alice" << endl;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

待补