目录

• A - Already 2018
• B - Kagami Mochi
• C - Otoshidama
• D - Katana Thrower

A - Already 2018

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
string s;
int main(){
    cin>>s;
    s[3] = '8';
    cout << s << endl;
    return 0;
}

B - Kagami Mochi

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n;
set<int> st;
int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        st.insert(x);
    }
    cout << st.size() << endl;
    return 0;
}

C - Otoshidama

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, y;
int main() {
    cin >> n >> y;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            int k = n - i - j;
            if (k < 0) continue;
            if(i*10000+j*5000+k*1000==y){
                cout << i << ' ' << j << ' ' << k << endl;
                return 0;
            }
        }
    }
    cout << -1 << ' ' << -1 << ' ' << -1 << endl;
    return 0;
}

D - Katana Thrower

给出n个武器的一般攻击力，以及将其扔出的攻击力，一般攻击可以重复使用无限次，但是扔出只能使用一次

问将小怪兽杀死最多需要多少次攻击

利用优先队列，首先将所有的武器按照扔出和一般攻击两种形态放到队列里，然后每次取队头，如果当前的武器是扔出形态，那么就让小怪兽的血量减去扔出形态的攻击力，否则如果是一般形态，那么就直接算可以攻击几次即可，这样相当于从后往前模拟，保证了扔出形态只可能用一次

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
struct node {
    int a, b;
    int use;
    int flag;
} a[N];
int n, h;

struct cmp {
    bool operator()(node a, node b) { return a.use < b.use; }
};

priority_queue<node, vector<node>, cmp> q;

int main() {
    cin >> n >> h;
    for (int i = 0; i < n; i++) {
        cin >> a[i].a >> a[i].b;
        q.push({a[i].a, a[i].b, a[i].a, 0});
        q.push({a[i].a, a[i].b, a[i].b, 1});
    }
    int res = 0;
    while(!q.empty()){
        if (h <= 0) break;
        node now = q.top();
        q.pop();
        //cout << now.use << endl;
        if(now.flag==1){
            h -= now.use;
            now.use = now.a;
            now.flag = 0;
            q.push(now);
            res++;
        }
        else{
            res += int(ceil(1.0 * h / now.use));
            break;
        }
    }
    cout << res << endl;
    return 0;
}