DIV4(小菜鸡Codechef)

youhualiuh / 2024-02-01 / 原文

Advitiya

思路：

判断 16 17 18 的时候输出 ADVITIYA 不是就输出 WAITING FOR ADVITIYA

Code:

#include <bits/stdc++.h>

using namespace std;
    
signed main()
{
    int n; cin >> n;
    if (n >= 16 && n <= 18) {
        cout << "ADVITIYA\n";
    }
    else {
        cout << "WAITING FOR ADVITIYA\n";
    }
    return 0;
}

Judged

思路：

有5个裁判只要超过或等于4人就输出YES 否则 NO

Code：

#include <bits/stdc++.h>
using namespace std;
    
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int  t; cin >> t;
    while (t--)
    {
        int x, cnt = 0;;
        for (int i = 1; i <= 5; ++i) {
            cin >> x;
            cnt += x;
        }
        cout << (cnt >= 4 ? "YES\n" : "NO\n");
    }
    
    return 0;
}

Cookie Day

思路：

饼干要让每个人都获得至少一个，如果有一个人没获得就输出-1，并且是在一个罐子里分配，因此去遍历这个x(饼干数量)，得出公式 minn = min(minn, x / m * m); (x > m)

Code：

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

typedef long long ll;

int main() {
    ll t; cin >> t;
    while (t--)
    {
        ll n, m; cin >> n >> m;
        ll minn = 1e9 + 1;
        ll x;
        bool flag2 = 0;
        for (int i = 1; i <= n; ++i)
        {
            cin >> x;
            if (x < m) {
                continue;
            }
            else {
                minn = min(minn, x - (x / m) * m);
                flag2 = 1; 
            }
        }
        if (flag2) cout << minn << '\n';
        else cout << -1 << '\n';
    }
　　return 0;
}

Another Good String

思路：

去寻找每一次的最长连续字符先输出没有插入字符时的数量在去统计每一次插入时的最大连续字符，数据极限，我采用前缀和实现Code

Code：

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

typedef long long ll;
ll cnt;
const int N = 2e5 + 5;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    ll t; cin >> t;
    while (t--)
    {
        ll n, m; string s;
        cnt = 1;
        cin >> n >> m >> s;
        ll a[N] = {1};
        for (int i = 1; i < n; i++) 
        {
            if (s[i] == s[i - 1]) {
                cnt++;//cnt
                if (cnt > a[i - 1]) {
                    a[i] = a[i - 1] + 1;
                }
                else {
                    a[i] = a[i - 1];
                }
            }
            else {
                a[i] = a[i - 1];
                cnt = 1;
            }
        }
        cout << a[n - 1] << ' ';
        while (m--) {
            char ck; cin >> ck;
            s += ck;
            n ++;
            if (s[n - 1] == s[n - 2]) {
                cnt++;
                if (cnt > a[n - 2]) {
                    a[n - 1] = a[n - 2] + 1;
                }
                else {
                    a[n - 1] = a[n - 2];
                }
            }
            else {
                cnt = 1;
                a[n - 1] = a[n - 2];
            }
            cout << a[n - 1] << ' ';
        } 
        cout << '\n';
    }
    
    return 0;
}