AcWing： The first 60 Weekly match

4494. having dinner - AcWing Question bank

If n And less than or equal to m and k Namely yes, The rest are no.

#include<iostream>
using namespace std;
#include<vector>
#include<algorithm>
#include<math.h>
#include<set>
#include<numeric>
#include<string>
#include<string.h>
#include<iterator>
#include<map>
#include<unordered_map>
#include<stack>
#include<list>
#include<queue>
#include<iomanip>
#include<bitset>

#define endl '\n'
#define int ll
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PII;
const int N = 10, MOD = 1e9 + 7;



signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n,m,k;
    cin>>n>>m>>k;
    if(n<=m&&n<=k)cout<<"Yes"<<endl;
    else cout<<"No"<<endl;

    return 0;
}

4495. Array operation - AcWing Question bank

Find the smallest decimal in the array , We can use small root piles , Output the number at the top of the small root heap each time .

As for each number output, if we subtract all the numbers of the group, it will obviously timeout . But we don't really have to subtract the numbers in each group , We use a variable x Record the number of outputs , For the initial 0, After outputting one number at a time, add that number to x On , Then when the subsequent number is output , All minus x Then the output （ We don't really need to reduce the number of root piles ）, In this way, you can subtract all the numbers in the array x 了 .

Note that if the number at the top of the small root heap decreases x After equal to 0, Then pop up this number , Until the heap is empty or the number at the top of the small root heap decreases x Later not 0 that will do . If it is empty, output directly 0.

#include<iostream>
using namespace std;
#include<vector>
#include<algorithm>
#include<math.h>
#include<set>
#include<numeric>
#include<string>
#include<string.h>
#include<iterator>
#include<map>
#include<unordered_map>
#include<stack>
#include<list>
#include<queue>
#include<iomanip>
#include<bitset>

#define endl '\n'
#define int ll
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PII;
const int N = 10, MOD = 1e9 + 7;



signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n,m;
    cin>>n>>m;
    priority_queue<int, vector<int>, greater<int>>que;
    for(int i=0;i<n;i++)
    {
        int x;
        cin>>x;
        que.push(x);
    }
    int x=0;
    while(m--)
    {
        while(!que.empty()&&que.top()-x==0)que.pop();
        if(que.empty())cout<<0<<endl;
        else 
        {
            cout<<que.top()-x<<endl;
            x+=que.top()-x;
        }

    }

    return 0;
}

4496. Eat fruit - AcWing Question bank

Set a two-dimensional dynamic transfer array f,f[i] [j] To i Until students , Yes j Students have different plans from the students on his left f[i] [j], initial f[1] [0]=m.

If our current classmate takes the same fruit as our last classmate , Then the number of schemes should be f[i-1] [j], And if it's different , If the last student took it 1 Fruit No , We can take 2_{m Fruit No , If you take 2 Fruit No , We can take 1 Number and 3}m Fruit No . That is, no matter which fruit the last student took , We can all take the rest m-1 Fruit , Then the number of schemes is f[i-1] [j-1]*(m-1). So the state transfer equation is ：

• The fruit of the current student is the same as that of the previous student ：f[i] [j]=f[i-1] [j].
• The current student and the previous student have different fruit numbers ：f[i] [j]=f[i-1] [j-1] *(m-1).

The final answer is f[n] [k].

#include<iostream>
using namespace std;
#include<vector>
#include<algorithm>
#include<math.h>
#include<set>
#include<numeric>
#include<string>
#include<string.h>
#include<iterator>
#include<map>
#include<unordered_map>
#include<stack>
#include<list>
#include<queue>
#include<iomanip>
#include<bitset>

#define endl '\n'
#define int ll
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PII;
const int N = 10, MOD = 1e9 + 7;



signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n,m;
    cin>>n>>m;
    priority_queue<int, vector<int>, greater<int>>que;
    for(int i=0;i<n;i++)
    {
        int x;
        cin>>x;
        que.push(x);
    }
    int x=0;
    while(m--)
    {
        while(!que.empty()&&que.top()-x==0)que.pop();
        if(que.empty())cout<<0<<endl;
        else 
        {
            cout<<que.top()-x<<endl;
            x+=que.top()-x;
        }

    }

    return 0;
}