A Healthy Breakfast

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
const int N = 210;
#define int long long

signed main()
{
    bool flag = false;
    string s; cin >> s;
    if (s[0] == 'R' || (s[1] == 'R' && s[2] == 'M')) flag = true;
    if (flag) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0; 
}

B - Vertical Reading

每次选择一个长度为len的区间每个区间的相同位置进行选点，而不是每隔len的长度跳一步

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
const int N = 210;
#define int long long

signed main()
{
    string s, t; cin >> s >> t;

    bool flag = false;
    for (int len = 1; len < s.size(); len ++)
    {
        for (int i = 0; i < len; i ++)
        {
            string p = "";
            for (int j = i; j < s.size(); j += len) p += s[j];
            if (p == t) 
            {
                flag = true;
                break;
            }
        }
    }

    if (flag) cout << "Yes" << endl;
    else cout << "No" << endl;

    return 0; 
}

C - Move It

每次选择同组内最小的即可

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
const int N = 1e5 + 10;
#define int long long

struct Node 
{
    int a, w;
    bool operator<(const Node& T) const
    {
        return w < T.w; 
    }
}q[N];

signed main()
{
    int n; cin >> n;
    map<int,int> mp;
    for (int i = 0; i < n; i ++) cin >> q[i].a, mp[q[i].a] ++;
    for (int i = 0; i < n; i ++) cin >> q[i].w;

    sort(q, q + n);

    int sum = 0;
    for (int i = 0; i < n; i ++)
    {
        if (mp[q[i].a] > 1) 
        {
            sum += q[i].w;
            mp[q[i].a] --;
        }
    }

    cout << sum << endl;

    return 0; 
}

D - Ghost Ants

首先考虑最简单的情况：两只蚂蚁相向而行，如果二者的距离在2t的时间内则能碰面。于是我们可以先将两个方向的蚂蚁分别进行存储，然后进行匹配。但是时间复杂度不能超过O(n^2)，所以我们需要进行优化：将两个方向的蚂蚁进行排序，然后使用二分查找得到二者距离在[0,2t]之内的蚂蚁即为答案。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
const int N = 2e5 + 10;
#define int long long
int a[N], b[N];
int n1, n2;

signed main()
{
    int n, t; cin >> n >> t;
    string s; cin >> s;
    for (int i = 0; i < n; i ++) 
        if (s[i] == '1') cin >> a[n1 ++];
        else cin >> b[n2 ++];

    sort(a, a + n1);
    sort(b, b + n2);

    int sum = 0;
    for (int i = 0; i < n1; i ++)
    {
        int l = lower_bound(b, b + n2, a[i]) - b;
        int r = upper_bound(b, b + n2, a[i] + 2 * t) - b;
        sum += r - l;
    }

    cout << sum << endl;

    return 0; 
}

E