A - Dead Pixel

The question

A resolution of a*b On your screen x,y There is a bad point at this point , Ask what is the maximum area that can be displayed

Answer key

The pixel coordinates are from 0 At the beginning , For the sake of good calculation, we directly from 1 Start counting

Just output the largest in the four areas around this point

Code

ll a, b, x, y;

void solve(){
    
    cin >> a >> b >> x >> y;
    x++, y++;
    cout << max({
    (y - 1) * a, (x - 1) * b, (a - x) * b, (b - y) * a}) << endl;
}

B - Homecoming

The question

One can make a bus or a tram to get to the stop sign quickly

The rule of riding is if you arrive j Point must be the current point to j-1 All points must be the same point , The last one is not necessarily the same

existing p money , Ask this person from 1 What station can I walk to

Answer key

We don't know where he'll start taking the bus , But we know he will arrive n This point

Arrive at the same time n The amount of money remaining at this point must also be greater than or equal to 0

So let's reverse simulate , from n set out , See what point you can't take the bus

Code

void solve(){
    
    cin >> a >> b >> p;
    cin >> s;
    int n = s.size();
    s = ' ' + s;
    if(p < a && p < b){
    
        cout << n << endl;
        re;
    }
    for (int i = n; i >= 2; i--){
    
        // cout << p << e ndl;
        if(p < a && p < b){
    
            cout << i << endl;
            re;
        }

        int j = i - 1;
        if(s[i - 1] == 'A'){
    
            while(j >= 1 && p >= a && s[j] == s[i - 1]){
    
                j--;
            }
            p -= a;
            i = j + 2;
        }
        else{
    
            while(j >= 1 && p >= b && s[j] == s[i - 1]){
    
                j--;
            }
            p -= b;
            i = j + 2;
        }
    }
    cout << 1 << endl;
}

C - Restoring Permutation

The question

Here's a number n,n It's an array b The number of elements ,b[i]=min(a[i*2-1],a[i*2]), Recover the legal array a, And let a The dictionary order of the string of elements should be as small as possible ,a The smallest element is 1, Maximum time 2*n, If a Can't recover , Output -1

Answer key

Greedy placement , If you can't play any more, you can directly output -1

Code

void solve(){
    
    cin >> n;
    vector<int> a, b(n);
    map<int, int> st;

    for (int i = 0; i < n; i++){
    
        cin >> b[i];
        st[b[i]]++;
    }
    if(count(all(b), 2 * n) != 0){
     //  Can don't 
        cout << -1 << endl;
        return;
    }

    for(auto i : b){
    
        a.pb(i);
        int j = i + 1;
        while(j < 2 * n && st[j])
            j++;

        if(j > 2 * n || st[j]){
    
            // cout << 2 << endl;
            cout << -1 << endl;
            return;
        }
        
        a.pb(j);
        st[j] = 1;
    }

    for(auto i : a)
        cout << i << ' ';
    cout << endl;
}

D - Recommendations

The question

The minimum cost of making the number of each book different

Answer key

Reference resources ：https://www.bilibili.com/video/BV127411u78f?p=4

Let's make all the books by the number of repetitions , Add one to all except the largest one

Until there is no biggest

Let's set the total cost of the current repeat quantity as sum, The biggest cost is mx, So the increased cost except the biggest one is sum-mx

Code

int n;
PII a[N];

void solve(){
    
    cin >> n;
    for (int i = 1; i <= n; i++){
    
        cin >> a[i].x;
    }
    for (int i = 1; i <= n; i++){
    
        cin >> a[i].y;
    }

    sort(a + 1, a + 1 + n);
    

    int ans = 0;
    int now = 0; //  The current number 
    int summ = 0; 
    priority_queue<int> q;

    for (int i = 1; i <= n; i++){
    
        while(q.size() && now < a[i].x){
     // now Less than a[i].x It means that we can't put a[i] Put it in 
            int mx = q.top();
            ans += summ - mx; //  contribution 
            summ -= mx; //  The total has been added up to one , new sum There is a change 
            q.pop();
            now++;
        }
        q.push(a[i].y);
        now = a[i].x;
        summ += a[i].y;
    }
    while(q.size()){
     //  Add all the last 
        int mx = q.top();
        ans += summ - mx;
        summ -= mx;
        q.pop();
    }
    cout << ans;
}