The 8th Hebei Collegiate Programming Contest

A

只需要记录是不是第一次出现并且不等于'i‘即可。

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
const int mod=1e6+3;
int st[30];
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    string a;cin>>a;
    ll res=0;
    for(int i=0;i<a.size();++i)
    {
        if(a[i]!='i')
        {
            if(!st[a[i]-'a'])res++,st[a[i]-'a']=1;
        }
    }
    cout<<res;
}

C

贪心的放，体会不用白不用这一贪心过程。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, m, a[N];
#define P pair<int, int>
vector<P> e[N];
void Totoro()
{
    cin >> n;
    priority_queue<P, vector<P>, greater<P>> q;
    vector<int> ans;
    for (int i = 1; i <= n; i++)
    {
        int l, r;
        cin >> l >> r;
        e[l].push_back({r, i});
    }
    for (int i = 1; i <= n; i++)
    {
        for (auto w : e[i - 1])
        {
            q.emplace(w);
        }
        while (!q.empty() && q.top().first < i - 1)
        {
            q.pop();
        }
        if (q.empty())
        {
            break;
        }
        ans.push_back(q.top().second);
        q.pop();
    }
    cout << ans.size() << '\n';
    for (auto x : ans)
    {
        cout << x << ' ';
    }
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    t = 1;
    while (t--)
    {
        Totoro();
    }
    return 0;
}

G

考虑段环成为链，然后进行双指针算法。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
signed main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int orig[3]{};
    for (int i = 0; i < 3; ++i)
        cin >> orig[i];
    string a;
    cin >> a;
    const int n = a.size();
    a += a;
    int res = 0;
    for (int s = 0; s < 2; ++s)
    {
        int cur[3]{orig[0], orig[1], orig[2]};
        for (int i = s, j = s; i < n; i += 2)
        {
            while (j + 2 <= n + i && cur[a[j] - '0' + a[j + 1] - '0'])
            {
                cur[a[j] - '0' + a[j + 1] - '0']--;
                j += 2;
            }
            res = max(res, j - i);
            cur[a[i] - '0' + a[i + 1] - '0']++;
        }
    }
    cout << res;
}

I

模拟题。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
#define P pair<int, int>
using ll = long long;
const int mod = 1e6 + 3;
int zi_num[5], cnt, ip_num[5], cnt2;
vector<P> e[N];
void Totoro()
{
    string zi;
    cin >> zi;
    int net_len = 0;
    for (int i = 0; i < zi.size(); ++i)
    {
        if (zi[i] == '.')
            cnt++;
        else if (zi[i] == '/')
        {
            for (int j = i + 1; j < zi.size(); ++j)
            {
                net_len = net_len * 10 + zi[j] - '0';
            }
            break;
        }
        else
            zi_num[cnt] = zi_num[cnt] * 10 + zi[i] - '0';
    }
    int n;
    cin >> n;
    int last = net_len % 8, com = net_len / 8;
    while (n--)
    {
        cnt = 0;
        cnt2 = 0;
        memset(ip_num, 0, sizeof ip_num);
        string ip;
        cin >> ip;
        for (int i = 0; i < ip.size(); ++i)
        {
            if (ip[i] == '.')
                cnt2++;
            else
                ip_num[cnt2] = ip_num[cnt2] * 10 + ip[i] - '0';
        }
        bool flag = true;
        for (int i = 0; i < com; ++i)
        {
            if (ip_num[i] != zi_num[i])
            {
                flag = false;
                break;
            }
        }
        if (flag && last)
        {
            for (int i = 8; i >= (8 - last); --i)
            {
                if ((ip_num[com] >> i) != (zi_num[com] >> i))
                {
                    flag = false;
                    break;
                }
            }
        }
        if (flag)
            cout << "YES\n";
        else
            cout << "NO\n";
    }
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    t = 1;
    while (t--)
        Totoro();
    return 0;
}

J

由于范围很大，需要对一个1e9级别的长度的字符串进行取模是不现实的，因此需要进行快速幂，考虑使用矩阵加速地推。

初始矩阵为

Res 1

0 0

变化矩阵为

10 0

w 1

这样相乘就是

Res*10+w 1

0 0

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1e9 + 7;
int cnt[20];
void muti(int c[][2], int a[][2], int b[][2])
{
    int tmp[2][2] = {0};
    for (int i = 0; i < 2; ++i)
        for (int j = 0; j < 2; ++j)
            for (int k = 0; k < 2; ++k)
                tmp[i][j] = (tmp[i][j] + a[i][k] * b[k][j]) % mod;
    memcpy(c, tmp, sizeof tmp);
}
void qmi(int c[][2], int i, int k)
{
    int res[][2] = {{1, 0},
                    {0, 1}};
    int t[][2] = {
        {10, 0},
        {i, 1}};
    while (k)
    {
        if (k & 1)
            muti(res, res, t);
        muti(t, t, t);
        k >>= 1;
    }
    memcpy(c, res, sizeof res);
}
void solve()
{
    int m;
    cin >> m;
    for (int i = 0; i < 10; ++i)
        cin >> cnt[i];
    int st = 0;
    if (m == 1 && cnt[0])
    {
        cout << 0 << '\n';
        return;
    }
    for (int i = 1; i < 10; ++i)
    {
        if (cnt[i])
        {
            st = i, cnt[i]--, m--;
            break;
        }
    }
    int o[][2] = {
        {st, 1},
        {0, 0}};
    for (int i = 0; i < 10; ++i)
    {
        if (!m)
            break;
        if (cnt[i])
        {
            int bit[][2] = {
                {1, 0},
                {0, 1}};
            int num = min(cnt[i], m);
            qmi(bit, i, num);
            muti(o, o, bit);
            m -= num;
        }
    }
    cout << o[0][0] << '\n';
}
signed main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _;
    cin >> _;
    while (_--)
        solve();
}

K

直接输出。

#include<bits/stdc++.h>
using namespace std;
#define int long long 
#define N 200010
int n,d,h,a[N],ans=-1;
map<int,int> q;
void cb(){
    cout<<"HBCPC2024";
}
signed main(){
    int t;t=1;
    while(t--)cb();
    return 0;
}